text:
Why is the auto ignition temperature of petrol higher than that of diesel?
I can't find any reasonable explanation please help.

Auto ignition temperature is defined like this by Wikipedia:

The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark. This temperature is required to supply the activation energy needed for combustion.

Gasoline (or petrol for non-americans) has an auto-ignition temperature about 50 °C higher than diesel.
The answer boils down to explaining why the typical chemicals making up the two fuels have different activation energy barriers to combustion in air. This reaction basically involves the constituent molecules getting enough energy to react with oxygen.
We can get a crude (and probably oversimplistic) understanding by looking at the typical components of the two fuels. Diesel tends to be made up from linear saturated hydrocarbons; petrol is typically a mixture of branched and unsaturated hydrocarbons. So, a little simplistically, the typical component if diesel will have a larger surface area than a typical component of petrol (branched chain hydrocarbons are more "bunched up" than their linear counterparts with the same number of carbons). To put it another way, in linear hydrocarbons each carbon is has, at least, two hydrogens between it and any surrounding oxygen molecules from the air but in unsaturated or branched hydrocarbons the component carbon atoms are often buried inside because they are surrounded by other carbons (consider the molecule shapes of linear hexane and 2,2 dimethyl butane which have the same number of carbons).
This is a very simplistic way of looking at the question but give a little insight. In typical diesel compound there are simply more opportunities for oxygen to attack the molecule and kick off a reaction because the amount of the molecule exposed to the outside is higher. By being more compact (lots of carbons buried deep inside the molecule) there is less opportunity for this in the components of petrol.
For the reaction to kick off some of the molecules in the mix have to have enough thermal energy to react when they meet. And oxygen has to hit the hydrocarbon hard enough or there is no reaction. Once an oxidation reaction starts, it releases energy increasing the likelihood of another molecular reaction. At a certain point when there is enough energy from new reactions and the mix doesn't spread this out too quickly, this becomes a runaway process and the mixture goes bang. This process is–at the molecular level–easier for diesel than petrol because the chances of an oxygen causing a chemical changes when hitting a hydrocarbon is higher.
Of course a lot of other factors are involved but this crude picture gives at least some useful insight. This was discovered experimentally in the early days of engine design as the designers observed that different extracts from oil had different combustion properties in engines and refined the way oil was distilled to give them the behaviours they wanted in their engines.

at self ignition temp the compound suddenly reacts explosively ie rate suddenly increases by a lot. how can the opportunistic view of collision which u suggested explain it's not a gradual increase in rate but a sudden one?

another way of putting it would be, how is this chance of collision related to The self ignition temperature?

@nevertookcoursesbutwhy The molecular picture makes clear the underlying process that oxygen has to bang into a hydrocarbon with enough force to cause a reaction and more surface area increases the probability of that happening. The reaction itself releases energy increasing the thermal energy of other molecules in the mix. Once enough molecules have reacted, this becomes a runaway process and the mix goes bang.


text:
Why does coconut oil spoil much faster than groundnut oil?
I have observed that various foods cooked using coconut oil spoil much faster than foods cooked using groundnut oil. One good example is Asian coconut milk curry. Even with proper refrigeration, this food item cannot last more than 2 or 3 days typically. This is in very sharp contrast to food items cooked with groundnut oils which can last for much longer, possibly a few weeks. I have been looking up various Internet sources and some have cited the presence of natural enzymes, such as lipoxygenases and peroxidases, which are present in these coconut oils, while oils like groundnut oil and palm oil, do not contain these natural enzymes.
Looking at the chemical composition of the fats was also something I did. As shown here, most of the fatty acids found in coconut oil are saturated fatty acids, with lauric acid making up a large part of them. In fact, as shown here, groundnut oils seem to contain more unsaturated fatty acids than coconut oil, which we would expect to be more reactive and liable to degradation. Thus, the difference in the composition of the fats in the oils may not be what is responsible for the observed quicker rate of spoilage of coconut oil.
I would like to ask if anyone knows more to share regarding this issue. For now, it seems that the presence of enzymes may be the likely underlying reason for the faster rate of spoilage of coconut oils.

Do you have more than your personal experience as evidence of this difference in stability? Have you tested different brands etc? My first instinct was also to think about degrees of saturation and I would have followed the same reasoning as you. Also, the enzyme hypothesis sounds weak. I am not sure enzymes would remain in significant concentration after the sort of treatments used to purify oils, even less tolerate the treatment and remain active in the oil.

I think this thing about coconut oil spoiling is reasonably common to experience among people in my country. It is a common understanding that coconut milk-based foods spoil very fast. It does not just apply to curry, even certain traditional peranakan cakes and desserts made with coconut milk also spoils incredibly fast.

I'd just add that coconut milk and coconut oil are probably worlds apart in treatment. The coconut milk is pasteurized and stabilized in some way, but retains a menagerie of compounds not found in simple oil. They may have in common: there may be free organic acids that accelerate the "saponification" of the milk and oil.

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text:
Quantitative MO theory- how is the overlap integral and the weighting coefficients of the molecular orbital actually evaluated?
Title. In my inorganic chemistry course, we learn about SALCs and qualitative MO treatment, with only a fleeting reference to the S integral and actual molecular wave functions (only hydrogen, $\ce{H2}$, was discussed).
I'm curious about how the S integral is evaluated for MOs in slightly more complex molecules i.e. $\ce{BF3}$, $\ce{CO}$, $\ce{H2O}$, and how the weighting coefficients are found, as they aren't discussed at all at my level.
Can someone explain this? Thanks!

In essence, you're asking how we find the MOs of a chemical species, I suppose. That's quite a broad topic; there are several ways of doing it, each with different levels of rigour, you may want to check out e.g. Chpt 15 of Levine's Quantum Chemistry (7th ed.).

In essence, one picks a computational level, chooses a system to run the calculation on (typically a molecule) and lets a computer and an appropriate program do the heavy numeric and algebraic lifting. No really, that's it.
In more detail: One would want to solve the multi-electron, multi-nucleus Schrödinger equation (that's already an approximation, because we ignore relativistic effects here). That's really difficult. So we make more approximations, such as fixed nuclei, a product ansatz for the wave-function (made up of one-electron wave-functions) and often a finite-size basis set. A basis set are functions that can be linearly combined into molecular orbitals; a large one usually gives better results, but make the computation longer. This choice is part of the computational level. Other, more mathematical and technical choices make up the other part (and this is a science in itself).
For a molecule like $\ce{C2H6}$, the computational time can vary between several days to weeks and less than a second, depending on sophistication of the computational level and its accuracy. In the end, we obtain an approximate wave-function (containing molecular orbitals), from which approximate properties such as energy or NMR shifts can be obtained. The computer is useful in this by calculating integrals (partially analytic, mostly numeric) and keeping track of many very large matrices.
At times, the problems can be worked out using pen and paper, making use of high symmetry and tiny basis sets. This leads to easily interpreted, but not very accurate wave-functions.

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text:
Existence of single ion activities
I have used Davies equation to find what I believed was single ion activities and multiplied them on concentrations to define the thermodynamic dissociation constants for any kind of buffer to find equilibrium conditions. Thereby I could reproduce major findings in recent publications which apparently used the activity coefficients obtained from Davies equation in a similar way.
Now I hear from a reviewer:

Unfortunately a misunderstanding about calculated "single-ion activity coefficients" is being promoted in the manuscript.
From a mathematical expression for the mean activity coefficients an expression for single ion activity coefficients can be derived. But these single ion activity coefficients have no physical or thermodynamic meaning.

I know a heated debate on this issue has been going on. Wonder if any proposals for armistice could be suggested.
Perhaps to clarify my predicament, what I understand as single-ion activity are such as the "fs" in $$K_{12}=f_1[H^+]f_2[HA^{2-}]/f_1[H_2A^-]$$ to define the thermodynamic dissociation constant of a buffer.


text:
Equation for overall heat transfer coefficient

The heat flow (in $[\pu{J/s}]$) through the wall of a tube of length $L$ and inner radius $R_0$ is
$$Q = (2πR_0L)U(T_{1\infty} - T_{2\infty}),$$
where $T_{1\infty}$ is the bulk temperature in the inside and $T_{2\infty}$ is the bulk temperature on the outside. The overall heat transfer coefficient reads as:
$$\frac{1}{UR_0} = \frac{1}{h_1R_0} + \frac{\ln{(R_a/R_0)}}{k} + \frac{1}{h_2R_a} \tag{1}$$
where $R_0$ and $R_a$ are the inner and outer radius of the tube, respectively, $h_1$ and $h_2$ are the inner and outer heat transfer coefficients, and $k$ is the thermal conductivity of the wall.
How does Eq. (1) simplify for the $T$-profile shown on the left?

$U = k/R_0 \cdot 1/\ln(R_a/R_0)$
$U = h_1$
$1/U = 1/h_1 + 1/h_2 \cdot (R_0/R_a)$
$1/UR_o = (\ln(R_a/R_o)/ k ) + (1/h_2R_a)$

I had seen a similar $T$-profile in one of our exercises and there they used Eq. (1), however when selecting option 1, I get "incorrect answer". How can you tell from looking at the $T$-profle what the equation for the overall heat transfer coefficient will be?
All help and hints are appreciated!

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Note that a screenshot or picture of an exercise is not searchable.

This is the heat equation analogue to Ohm's law, in which resistances are additive (in series), and hence the inverse conductances are additive.
The way to reason is that the full equation (1) for the heat conductivity (1) contains three terms: one for the interior of the tube, one for the tube wall, and a third for the exterior. When you look at the profile, it becomes evident that, since temperatures immediately adjacent inside and outside the tube are equal, the tube has infinite heat conductance ($k=\infty$).
The right answer is therefore 3.

Why is ($k=\infty$) because the temperatures adjacent to the inside and outside of the tube are equal? And also, how does this lead to 3 being the correct answer?

Please check choice 3. Does it match equation 1 if you remove the middle term?

If there was resistance to heat flow across the tube there would be a gradient across it. The absence of that gradient suggests that $k=\infty$


text:
Alkoxide as a leaving group in hydrolysis of ester
In the base catalysed hydrolysis the hydroxide ion adds to the ester to form a tetrahedral intermediate, after which the alkoxide ions leave and forms carboxylate ion and alcohol.
Also we find the rate of reaction to be quite high.
So my questions are:

Most importantly, alkoxide is a very strong base. So why is it acting as a leaving group here (apparently, a very good leaving group due to the high rate of reaction)?
Why is the tertrahedral intermediate not as stable as the corresponding ones of carbonyl compounds?
Why is this reaction essentially irreversible, considering the fact that alkoxide is a stronger base the hydroxide ion? (Therefore it should be able to displace the hydroxide in the hypothetical backward reaction.)

It is irreversible because the alkoxide deprotonates the carboxylic acid wihich is the initial product of the hydrolysis.

@Waylander So,why can't the carboxylate anion once again take up a proton either from the aqueous phase, or from the alcohol, because we have seen earlier in the mechanism scheme of this reaction only the alkoxide can be removed as a leaving group?

For (2), tetrahedral intermediates with a leaving group are not stable. Not sure what corresponding ones you're talking about, but those are probably not stable.

@YUSUFHASAN It can, but the proton on it is the most acidic (its an acid after all) so it is the most likely to be removed in a system where protons are being exchanged between all the species present.

@Waylander But,what I really want to know is related to (1), why is the alkoxide even leaving in the first place? Because,if such a bad leaving group is being removed,then it seems to me that after a few unsuccessful collisions, a protonated carboxylic acid can be formed, and the alkoxide can initiate a reverse reaction which may soon become favourable under thermodynamic control.

@Waylander In fact, I have had this doubt since we started studying esters and their reactions in our class: why does the alkoxide leave with such a great rate(we were told it is fast) in nucleophilic acyl substitution? It occurs in various places: Claisen condensation(attack of the deprotonated ester on the other) and this hydrolysis reaction,to name a few. I am not at all convinced by the these mechanisms where we remove the alkoxide after addition to the carbonyl, can you please explain this anomaly(apparently) to me?

@Zhe What the OP is referring to are the intermediates formed by the addition of,say, a Grignard reagent to carbonyl compound. Here, we can give answer as a Td complex of the form O-MgX+ before the aqueous workup to create an alcohol,but in case of esters, we have got to remove the alkoxide when the Grignard reagent is added even before the aq. workup, and we cannot give the answer as an O-MgX+ intermediate(at least in the questions we do;I know the OP personally,so I know what he is referring to), why is this so? You say Td intermediate with leaving grp aren't stable,is OR- a leaving grp?

@YUSUFHASAN here's the thing; primary and secondary alkoxides are not substantially worse leaving groups than OH -. What makes the reaction proceed is that it is irreversible. OH- adds and leaves again many times but that is a dead end because it regenerates SM.

@Waylander But in my O.chem book, they have given a statement as Cl- reacting with R-OR'(an ether) will not give R'-Cl and OR- because "OR- is a strong base and so, a poor leaving group", what about this?

@Waylander Please address this issue of leaving group capacity(or lack of it) of alkoxide in a detailed answer if it is possible, because both me and the OP are highly confused about the very fundamentals of esters reactions,in particular, nucleophilic acyl substitution(also,if it possible, could you talk about the attack of an enolate attacking on a second ester and eliminating alkoxide in the Claisen condensation?)

related https://chemistry.stackexchange.com/questions/72456/ro-alkoxide-as-a-leaving-group https://chemistry.stackexchange.com/questions/41331/why-does-a-carbonyl-reform-in-ester-reactions

@ YUSUF HASAN: A good way to appreciate how tetrahedral alkoxide intermediates collapse is to look at the pKa of the possible leaving groups. In saponification, alkoxide eliminates RO- whose conjugate acid is ROH (pKa ~16) as opposed to eliminating RCH2- (conjugate acid RCH3 (pKa 50). The tetrahedral intermediate is not stable. Adding RMgBr to R'(C=O)R" forms RR'R"CO-. It stops here, it is stable, with no elimination (RH, R'H, RH" pKa ~50) until protonation.

@user55119 But what about the example I cited in comment 9 above? Why does alkoxide not leave then? Or,to put it in other words, why does the Td complex have to collapse, when the potential leaving groups are alkoxide and R-?I normally don't request an answer on the SE network as I am aware of the policies,but I would like to make an exception here,and probably go out of my way to request you to provide an answer if it is possible,depicting conclusively where the alkoxide can or cannot act as a leaving group, and why do esters have an uncanny tendency to hydrolyse in basic medium. Thanks!


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Applications of magnesium sulfate monohydrate MgSO4 · H2O
How is magnesium sulfate monohydrate $\ce{MgSO4 * H2O}$ used for in chemistry and how does it work as a valuable laboratory drying agent?
I could not find too much information in my book or the internet besides the basics (chemical formula, physical or chemical characteristics, etc.), but I did find what it can be used in. For example, fertilizer, feed additives for livestock, nutrient flavor enhancer, pharmaceuticals, leather and fur production, coloring fabrics, fermentation in beer, paper production, sewage treatment, cosmetics, but nothing about how or why chemistry labs would use it. What is its purpose?
Each time I tried to look it up other products that sounded similar but not the same appeared such as $\ce{MgSO4 * 7H2O}$ , manganese sulfate, manganous sulfate, etc.

I hope the additional information was sufficient enough to prove I put effort into solving this problem on my own.

manganese sulfate and manganous sulfate are not hydrates of magnesium sulfate. There is big different between $\ce{Mg}$ and $\ce{Mn}$.

Briefly, anhydrous magnesium sulfate ($\ce{MgSO4}$) is commonly used as a desiccant in organic synthesis due to its affinity for water (Wikipedia). Anhydrous $\ce{Na2SO4}$ and $\ce{CaCl2}$ are also used in similar manner.

Is magnesium sulfate monohydrate MgSO4⋅H2O the same as MgSO4? What is the reason for MgSO4⋅H2O creation? What is its purpose?

If it were the same, it would be written the same. As to the purpose: like most chemical compounds, it has none.

Update. Here is what I have gathered about what drying agents are generally used for in chemistry labs. - In "organic labs" it is difficult to remove water from compounds. Organic compounds, in particular, are very reactive with water. That is why it is important to remove the water. It is to make sure water is not reacting with the created compounds.

Magnesium sulfate is frequently used in the laboratory, especially after aqueous work-up.
Aqueous work-up is a common technique in the lab to get rid of residual impurities after completion of a reaction.
For this the organic solution of the reaction is cooled to room temperature and mixed with water. Specific impurities will then diffuse into the aqueous phase and can be separated with a separatory funnel.
Unfortunately some residual water will stay in the organic phase and this can have negative impact for following characterization of the compound.
Therefore magnesium sulfate is added, which is able to catch the residual water in its crystal lattice. After filtration of the solid magnesium sulfate, the water is reduced to a non-significant amount and the compound ready for further investigation.

The question is specifically about  magnesium sulfate monohydrate. Your answer is about the use of anhydrous magnesium sulfate.

basically it can be used similarly.
The crystal lattice bears space for up to 7 molecules of H2O/ MgSO4

First, MgSO4 is not some zeolite with cavities for storing water, it's hydrates have totally different structures based on the number of crystal water! Saying there is "space to bear" is just wrong.  Second, it's moderately soluble in water and sparingly in ether, so by adding MgSO4 to water–organic system you will just dissolve it. That's not how MgSO4 is used for drying at all. Third, heptahydrate is not the crystallohydrate with the highest water content, dodecahydrate arguably is, but this is not the point.

Fourth, if you want to dry organic solvents for synthesis or GC analysis, you must use anhydrous MgSO4.


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How to evaluate the expectation value of the Hamiltonian for a three electron Slater determinant?
I'm having trouble with a few concepts dealing with slater determinants and many electron systems. In particular, when dealing with 3 electron systems, I know that
$$
\psi_0 = \frac{1}{\sqrt{6}} \det
\begin{bmatrix}
    1s(1)\alpha(1) & 1s(2)\alpha(2) & 1s(3)\alpha(3) \\
    1s(1)\beta(1)  & 1s(2)\beta(2)  & 1s(3)\beta(3) \\
    2s(1)\alpha(1) & 2s(2)\alpha(2) & 2s(3)\alpha(3)
\end{bmatrix},
$$
where the $1s$ and $2s$ are spatial functions and $\alpha$, $\beta$ are spin functions.
Performing the determinant expansion yields 6 terms, and it turns out that $\psi_0$ is an eigenfunction for the unperturbed Hamiltonian (in atomic units)
$$
\hat{H}_0 =
-\frac{1}{2} \nabla^{2}_{1} -\frac{3}{r_1} -\frac{1}{2}\nabla^{2}_{2}
-\frac{3}{r_2} -\frac{1}{2}\nabla^{2}_{3} -\frac{3}{r_3}
$$
 and that the resulting eigenenergy is
$$E^{0}=E_{1s}+E_{1s}+E_{2s}.$$
How does one show this?
My confusion mainly stems from the difficulty in acting the Hamiltonian on the determinant expression above. For example, how does one compute (in bra-ket notation)
$$
\frac{1}{\sqrt{6}} \hat{H}_0
| 1s(1)\alpha(1) 1s(2)\beta(2) 2s(3)\alpha(3)\rangle?
$$
This is only the first term in the expansion, and once I have some intuition on how to do this properly, then the general pattern should emerge quite clearly. Any help and guidance would be greatly appreciated!

2 hints: First notice that the Hamiltonian has same terms for all 3 electrons, so we can just stick with the first electron ($\hat h_i$), and multiply its result by 3 in the end.
Second, an intermediate result of the Hamiltonian acting on the ket vector is not needed. Directly consider the expectation value $\langle \hat h_i\rangle$. Notice which electronic coordinates the operator is acting on, and which not. Also keep in mind that the orbitals are form an orthonormal basis.

So breaking the Hamiltonian into 3 components and using the first term in the expansion as an example, only $\hat{h}_{1}$ acts on the first term since it leads with $1s(1)$? Sorry, I'm having trouble in interpreting and understanding your second hint.

Of course the whole Hamiltonian acts on all terms. But we can break it into the smaller parts  ($\hat h_1+\hat h_2+\hat h_3$) and apply the sum rule of integration, i.e. consider the terms separately. Next the different terms of the Slater determinant can be considered separate as well. So the first term is $\langle 1s(1)\alpha(1)1s(2)\beta(2)2s(3)\alpha(3)|\hat h_1|1s(1)\alpha(1)1s(2)\beta(2)2s(3)\alpha(3)\rangle$, which can be rearanged to $\langle 1s(1)\alpha(1)|\hat h_1|1s(1)\alpha(1)\rangle\langle 1s(2)\beta(2)|1s(2)\beta(2)\rangle\langle 2s(3)\alpha(3)|2s(3)\alpha(3)\rangle$.

By the orthonormality of spin orbitals, that term evaluates to $1s(1)$, correct? So going about this way, I'm going to end up dealing with 36 terms (many of which go to $0$).

Yes, there should be 6 surviving terms, and 30 terms being 0 (I think).


text:
Chemical shifts of OH protons
The chemical shift of a proton is an indication of the amount of shielding it receives from the electron density on it or other magnetic anisotropy effects present in the molecule. For the protons of the carboxylic acid functionality, the chemical shift typically exceeds $\pu{10 ppm}$. Baseline estimates given for $\ce {CH, CH2, CH3}$ protons are $ 1.7, 1.3$ and $\pu{0.9 ppm}$ respectively. However, we are often able to observe the chemical shifts of alcohol protons in the range of $\pu{1-2 ppm}$. This is rather counterintuitive since the electronegativity of oxygen is markedly higher than that of carbon and hence we would observe a mucher greater deshielding of the $\ce{OH}$ proton. For alcoholic protons, there is likely to be the presence of hydrogen bonding effects. Is the presence of these $\ce {H}$ bonding effects responsible for the less than expected deshielding?

Hydrogen bonding leads to deshielding in alcohols. Isolated ethanol in the gas phase has a shift of $\pu{0.55 ppm}$, while the shift is $\pu{0.8 ppm}$ in neat ethanol, so there is a shielding effect due to H-bonding. However, the OH shift is extremely sensitive to self association and/or solvent, resulting typically in greater deshielding. See for instance the table of shifts in reference [1].
According to Hore (in his Oxford NMR book), the charge distribution around the proton is polarized (mainly because the proton is attracted by the H-bond acceptor), effectively deshielding it. In any event, the hydrogen bond acceptor $\ce{O}$ transfers charge to the donor $\ce{O}$. The intervening H apparently gets the hard end of the bargain.
[1]: Fulmer et al, Organometallics 2010, 29, 2176–2179

Thanks for the gas phase NMR data. That is highly appreciated. However, the effect of hydrogen bonding seems quite counter-intuitive to me as electron density, in the form of a lone pair, is donated to the hydrogen atom during H bonding. Hence, it is puzzling that the effect is overall deshielding.

Hore's book is not especially illuminating about the source of the deshielding effect in terms of the specific electronic distribution about the proton. You have a valid point that the hydrogen bond acceptor donates charge to form the bond, which would partially reside near the H atom, so this does seem counterintuitive. It might be that the distribution is stretched, ie becomes thinner in the radial direction about the bond. Therefore the average shift is reduced. But I speculate. I only addressed one misconception behind your question.


text:
What is the molecular formula of 2-ethynyloctahydro-1H-indene?
I thought the formula for this $\ce{C11H15}$, but it's not. I need assistance in finding the proper formula:

You forgot the $\ce{H}$ on the triple bond! :-)

I suspect you forgot one hydrogen at the terminal carbon in alkyne group $\ce{R-C#C-H}$. With this in mind the formula would be $\ce{C11H16}$:


text:
How to convert but-2-ene to but-1-ene?
How can we convert but-2-ene to but-1-ene? Is it even possible, since but-2-ene is the more stable one.

You are right that a disubstituted alkene will be more stable than a monosubstituted. This tendency towards forming highly substituted alkenes is known as Zaitsev's rule, but it is not without exception. In simple elimination reactions, more sterically hindered bases cannot access the secondary hydrogen as well as the primary hydrogen, and the monosubstituted product will dominate. Hofmann elimination is another reaction that favors the formation of less substituted alkenes, also for steric reasons.

Though but-2-ene is more stable than but-1-ene it's still possible to convert them into each other easily as follows:

Add $\ce{HBr}$ into but-2-ene solution it will result in formation of 2-bromobutane.

Now add $\ce{C(CH3)OK}$ to the obtained 2-bromobutane, as it is a bulky base, due to steric factors it forms the Hoffman product via a Hoffman elimination reaction

Sorry that the second picture is not the perfect representation, but it still should aid understanding.


text:
Why BiCl5 isn't stable?
I read this in a textbook:

$\ce{Bi(V)}$ is very unstable and is a good oxidizing agent.

Why does it happen that way? Is it because in $\pu{+5}$ oxidation state $\ce{Bi}$ pulls in more electrons and hence gets reduced fast or there's a different concept?
Also then $\ce{BiCl5}$ should be more stable than $\ce{BiCl3}$ because it's getting the electrons that it needs, which is not true.

In one word, it's due to inert pair effect. If you want to go deeper, you can pull in Relativistic effect to ultimately conclude the cause of inert pair effect and explain the reason of stability of $+3$ oxidation state rather than $+5$ oxidation state.

I require a proper explanation for this. I know inert pair effect is affecting this, but how? That's the main question.

As far as I can tell BiCl5 doesn't exist. So "not very stable" was a serious understatement. It would be significantly stronger oxidant then elemental chlorine which marks some serious problem with it's hypothetical synthesis.

@Mithoron why would it be a strong oxidant? Is the reasoning in my question right?

Well, your inert pair effect makes the pair harder to use - one needs stronger oxidant to oxidate Bi (III) to Bi (V) and therefore Bi (V) compounds are stronger oxidants themselves.

The reason behind this is the mainly inert pair effect .
In $\ce{BiCl3}$, due to the much electronegativity difference between $\ce{Bi}$ and $\ce{Cl}$, Chlorine atom forms bonds with almost pure $p$ orbitals of $\ce{Bi}$, and the lone pairs on $\ce{Bi}$ are of almost pure $s$ character. Thus, $\ce{Bi}$ atom doesn't actually utilise much of its $s$ orbital electrons in forming the bonds which is energetically much more preferable. ( If it seems non-obvious, recall Bent's rule which says that more electronegative atoms prefer to form bonds with orbitals with more $p$ - character.)
On the other hand, if $\ce{Bi}$ is present in $\ce{BiCl5}$, it has no other option other than utilising its $s$ orbitals for bonding. The equatorial bonds will mainly consist $s$ character and $p_x, p_y$ orbitals will also participate in forming the equatorial bonds. So, $s$ character in the equatorial bonds will be slightly lesser than $33$% (again due to Bent's rule). And the axial bonds will consist of mainly $p_z $ and $d_{z^2}$ orbitals and little bit of $s$ mixing may also be there as it is spherically symmetric. So, overall central atom $\ce{Bi}$ has to use its $s$ electrons in bonding which is difficult energetically.
Now, the participation of $s$ electrons is difficult in bonding is due to the relativistic contraction of $s$ orbitals. It becomes closer to the nucleus and more stable in case of heavy elements like $\ce{Au, Hg, Tl, Pb, Bi}$ etc. Hence, the central atom needs to pay more energy to involve those stable electrons in bonding. That's why, the $s$ electrons become kind of inert and thus these heavy elements don't prefer higher oxidation states to avoid participation of these inner $s$ orbitals in bonding.

This contraction is relative to the 2nd period and 3rd period elements,I presume?

@YUSUFHASAN Yes !


text:
Why sp3 hybridisation in NH3 but bonding with pure p orbitals in case of other hydrides of group 15
According to my book, the following line was given relating to the bond angle in Group 15 elements

The bond angle in PH3, AsH3 and SbH3 is close to 92° which suggests that the orbitals used for bonding are close to pure p-orbitals.

I wanted to ask if there is any reason for having sp3 hybridised orbitals for N, but not so for other elements of that group in their hydrides. Or is it completely based on experimental observations?

The answer to this question is actually much more complicated than most textbooks make it seem. There's a great explanation in Abright, Burdett and Whango "Orbital Interactions in Chemistry" Ch. 9, which I'll try to summarize here.
As explained in this answer: Why is the inversion barrier larger in PH3 than it is in NH3?, the lone pair orbital in a planar XH3 molecule would be pure $p$. Thus the bonding orbitals average to $sp^2$. As the bond angle decreases from 180, the loss of symmetry makes possible mixing between that lone pair orbital (originally a pure central atom $p$ orbital) and an anti-bonding orbital involving the central atom $s$ orbital. This "s-p mixing" stabilizes the lone pair orbital at the expense of the (empty) antibonding orbital. This mixing necessarily also involves the bonding orbital that involves the central atom $s$ atomic orbital.
In nitrogen, the low energy of the atomic $s$ orbital and the strong overlap interaction of that $s$ orbital with the hydrogen $s$ orbitals means that the bonding orbital made from the atomic $s$ orbital is quite low energy and the antibonding orbital is quite high energy. As a result, mixing with the $p$ lone pair orbital is less favorable, so less mixing occurs and the angle does not deviate as far from 180.
With the larger central atoms, the interaction of the $s$ orbital with the hydrogen $s$ orbitals is less, so the antibonding orbital is not as high energy. Likewise, the $s$ bonding orbital is not as low energy. So mixing with the lone pair $p$ orbital is more favorable, and the bond angle deviates farther from 180. The figure below (from Albright, figure 9.9) plots the energy of the occupied orbitals. The lowest line represents the $s$ bonding orbital and the top line the lone pair.

The key point here is that more $s$ character is mixed into the lone pair orbital as you move down from NH3.

I would recommend you read Linus Pauling’s Nature of the Chemical Bond, an excellent book for VBT.
Basically, the reason for hybridisation is to increase bond strength, which is maximised at 2 when one has 25% s and 75% p character.
However, having a lone pair in an s orbital is more stable than having it in a hybridised orbital (due to promotion energy), so in cases with lone pairs we end up with the lone pairs having more s character and the bonding pairs having more p character. This is the same for any compound with lone pairs.
Case in point- the lone pairs in water have 44% s character, while the bonding pairs have 6% each.
To answer your question, nitrogen is much more electronegative than hydrogen, so the interaction is somewhat ionic. Hence, the bond angle increases just like in water- the predicted bond angle for water would be about 90 degrees, but that is obviously not the case.
At the end of the day, VBT is much more complex than one might think if one learns it in organic chemistry.

I would like to add that a pure s orbital has bond strength of 1, while a pure p orbital has bond strength of sqrt 3.

Also, hybridisation is a mathematical concept, so the notion of it costing energy is simply Pauling’s way of explaining why lone pairs always have more s character.

While Pauling's ideas are very important in historical context and are useful for introductory qualitative understanding of bonding, the quantitative aspects are largely out of date. For example, your number of 44% s character in H2O lone pairs is MUCH higher than is generally accepted now based on MO theory and computational approaches.

I guess Pauling still advocated for bunny ears on water, at least that is what this number reflects, that has been disproved though on various computational and experimental levels.

That’s definitely true, but considering the OP asked for a valence bond treatment that’s the best hybridisation” answer I could give.


text:
Solubility of solids in liquids
How does crystallisation happen and why?
I have been told that when a solid is dissolved in a liquid solvent, an equilibrium is set up between processes of dissolution and crystallisation. But if this is so then why does all solute dissolves (I am talking about an unsaturated solution), isn't it supposed to recrystallise simultaneously too?
I am certainly missing a lot and most importantly what crystallisation is and how and why it happens?
I don't want to know anything about the separation technique using the principle of  crystallisation. Just want to know the principle itself.

I edited your question because it's a very good one, but I think your opening line blurred the emphasis.

@TryHard Thanks. It makes much more sense now.

I have been told that when a solid is dissolved in a liquid solvent, an equilibrium is set up between processes of dissolution and crystallisation.

Once a solid is mixed with a solvent, dissolution and crystallization reactions happen simultaneously. That part is correct. In some cases, an equilibrium is reached. However, in other cases the dissolution goes to completion, so this statement is a bit too general. It might be better to say "a competition between dissolution and crystallisation" ensues which sometimes leads to equilibrium.
At the beginning, when the concentration of solute is still low, the rate of crystallization is low. As more of the solid dissolves, the crystallization rate increases because the concentration of solute increases.
Under saturating conditions, at one point the rate of crystallization matches that of dissolution, and you arrive at an equilibrium. Under sub-saturating conditions (if the concentration does not reach the solubility limit), the crystallization rates never reaches that of the dissolution rate, and the amount of solid decreases until it is gone. The process never reaches equilibrium (it goes to completion, i.e. the reactant is depleted).

isn't it supposed to recrystallise simultaneously too?

Both in saturated solutions and in solutions that are not saturated, solute particles will collide (with the solid or with other solute particles). However, it is much easier for a solute particle to attach to a solid than for two solute particles to start a nucleus of crystallization. Thus, while solute particles do collide, they will not form a solid when concentrations are sub-saturated. In fact, you can make super-saturated solutions that lack solid because making crystallization nuclei requires higher solute concentration than adding solute particles to an existing solid.

The principle of solubilization — in the ideal case — is closely related to that of melting. Think of solubilization as a two step process:

melt the solute;
mix the solute and the solvent.

Of course, step 1 involves raising the temperature to the melting point, or at least adding energy (in the form of heat) at the melting temperature in order to break attractive bonds between solute molecules. We will consider only the case where we are already at the melting point (an isothermal process), so you only need to add heat in order to turn the solid into liquid.
Step 2, on the other hand, is spontaneous if the solute and solvent mix ideally. You don't have to add heat (energy) for this step to proceed, assuming both substances are liquids and mix ideally.
Now what happens when you solubilize something is that you need to add less heat than you would have needed in step 1 to melt all of the solute, since some of the required (free) energy comes from the process of mixing.
By a similar line of argument, based on the properties of ideal solutions, you can relate (approximately) the solubility to the melting point and the heat of fusion as follows:
$$\log(x_2) = \frac{\Delta H^\circ_\mathrm{fus}}{R}\left(\frac{1}{T_\mathrm{fus}}-\frac{1}{T}\right)$$
Since $\Delta H^\circ_\mathrm{fus}$ is generally positive (heat is required to melt something) then increasing $T$ tends to increase the solubility of solids.
Crystallization is just the reverse of solubilization! And yes, there is always an equilibrium between soluble/insoluble solute, it's just that sometimes the soluble or insoluble part is negligible, depending on the values of $\Delta H^\circ _\mathrm{fus}$, $T_\mathrm{fus}$, and $T$.

Basically the process of recrystalisation is to dissolve all of your sample in a solvent which mainly dissolves the impurities and just fewer of your desired compound.
If the sample is now dissolved in the refluxing solvent a clear solution results.
If this solution is cooled down slowly, the formation of crystals starts and these crystals strongly prefer to form a crystal lattice (composed of one component only) in order to gain as much lattice enthalpy as possible. Therefore all impurities stay in solution and can be filtered off afterwards.

You missed a several of points. (1) Recrystallization works best when there is a large difference between the solubility of the substance in hot and cold solvent. (2) The increase in purity depends on the substance, impurities and solvent. (3) Since the substance will be at least somewhat soluble in the cold solvent, some substance will remain in solution.


text:
Reagents in Krapcho's decarboxylation
Krapcho's decarboxylation is a reaction involving esters with a electron-withdrawing group $\beta$ to the carbonyl group and halide ions. It is typically conducted with $\ce {LiCl}$ and $\ce {DMSO}$ at high temperatures. However, in one of Krapcho's original papers, $\ce {NaCN}$ was also used alongside $\ce {DMSO}$.
What interests me is the first step of the reaction which involves the nucleophilic subsitution, likely by an $\ce {S_N2}$ pathway, of the alkyl group of the ester functionality. This is seemingly weird as the $\ce {C-O}$ bond usually does not break unless in acid catalysis. This is the case for the cleavage of ethers using $\ce {HX}$. I suppose the use of $\ce {LiCl}$ was to allow the $\ce {Li^+}$ to coordinate to the oxygen atom, allowing the $\ce {C-O}$ bond to cleave more easily. Why is it that $\ce {NaCN}$ can also do the job?

CN- is much stronger nucleophile.

So as mentioned the Krapcho-Decarboxylation follows a SN2-mechanism and is therefore working best with a methylester.
The solvent choice for DMSO (polar aprotic) is due to low stabilisation of the chloride anion rendering it highly reactive for the nucleophilic attack. The resulting methyl chloride is gaseous at room temperature and leaves the reaction mixture (le Chatelier -> formation of product strongly favoured)
The leaving group of this substitution reaction decays into CO2 and the decarboxylated product. (Again: formation of a gas)
I think that the destabilisation of the cyanide in DMSO and the formation of stable acetonitrile can be considered as the driving force in the second reaction.


text:
What happens during Glycolysis?
Can I get brief explanation of the reactions that occur under Glycolysis? What do the curved lines represent?

Glycolysis is a process of which converts Glucose(Described as C$_{6}$ in your diagram), to two Pyruvate (2C$_{3}$). Then in downstream of glycolysis, Pyruvates are then converted to AcetylCoA to be used for many things in body. Glycolysis is part of a metabolic pathway, and in total it uses 1 Glucose, 2 ATP, 2 NAD+ to produce 2 Pyruvate, 4 ATP and 2 NADH. So those curve lines represents the Total input and output of energy. Each step is facilitated by different enzymes, so you could search for enzymes in each step for more information about each reaction. Wikipedia has table of each reactions.

What do the curved lines represent?

This is a schematic diagram of glycolysis where the metabolites are shown on the left side and the cofactors on the right. Intead of writing a reaction $\ce{A + B -> C + D}$, you can write it as $\ce{A -> D}$ with a curved arrow indicating that B is also a reactant and C is also a product.
In the case of formation of ATP, there are two reactions that are coupled. In the case of formation of NADH, there is a single redox reaction with two half reactions, a reduction half reaction yielding NADH, and an oxidation half reaction oxidizing the intermediate in glycolysis.

Can I get brief explanation of the reactions that occur under Glycolysis?

There is not enough information in the diagram to discuss the reactions that occur. What you do see is that glucose is broken down into two smaller molecules, and that this goes along with the production of two molecules of ATP (from ADP) and two molecules of NADH (from NAD+). This describes the net reaction, skipping many of the details (such that two molecules of ATP are hydrolized in early steps, and four molecules of ATP are formed in later steps).


text:
What takes place in this transformation of the 1,3-dicarbonyl?
I was just looking at a particular synthesis method on the Organic Chemistry Portal when I saw a particularly puzzling step, involving a 1,3-dicarbonyl compound. The step which I am puzzled by is shown here:

The second step, acid-catalysed esterification is unambiguous. It is the first step which opens the ring which concerns me. The fact that esterification occurs in the second step seems to suggest that the product of the acid-catalysed ring opening in the first step is a carboxylic acid and I simply cannot see how it can possibly form. Would someone kindly enlighten me on step 1?

Step 1 is probably in error and should be aq. hydroxide. Usually this cleavage is done on non-enolic beta-diketones but it can occur with enolic beta-diketones.

This is a reverse Claisen.

@Zhe But how do you explain the increase in the number of oxygen atoms? There is some sort of oxidation taking place right?

@user55119 Is it possible if you elaborate further as to what is the mechanism exactly?

I assume step 1 is hydrolyze the ester. Step 2 is reverse Claisen + reform ester. The reverse Claisen converts a ketone into an ester using the alcohol. Are you calling that the oxidation?

There is no net oxidation. Yes, an oxygen has been added but a ring has been lost.  Hydroxide saponifies the ester. The reverse (retro) Claisen condensation forms a carboxylate and a ketone. Step 2, esterifies both carboxylic acids.


text:
Valence bond theory and molecular orbital theory — how do they complement one another?
Based on my readings, valence bond theory (VBT) and molecular orbital theory (MOT) tend to complement one another in explaining a molecule, but I don’t understand how VBT helps us explain the properties of a compound (other than bond angles). How does VBT complement MOT in aiding our understanding of molecules?

VBT is for kids who ain't ready to digest MOT.

@Ivan Neretin I’m not talking about the nonsense one learns in organic chemistry, but the actual mathematical VBT model Pauling proposed that computational methods still use (somewhat) today.

I would be cautious on calling nonsense chemical reasoning that allowed to sketch down structural formulae such as those of benzene or porphyrin well before of VB and MO theories and techniques such as AFM.

@AlchImista what I was referring to was not Pauling’s VBT model, but the oversimplified “sp/sp2/sp3” stuff you learn in introductory College Chemistry.

You are correct that Valence Bond Theory (VB) is still very active in the chemical bonding community along with MOs. I recommend this article/conversation by Roald Hoffmann about this topic (Acc. Chem. Res. 2003, 36 (10), 750–756). Succinctly, they both are capable of providing a description of bonding in molecules; however, VB theory is better at demonstrating the existence of localized bonds while MO theory is more generalizable to delocalized systems (and ultimately the foundation of predictive computational chemistry). If you have an organic molecule, for example, the VB picture is more helpful for understanding reactivity because these molecules (for some reason) usually behave according to the intuition accessible within such a local bonding framework. MO theory could give you a numerically exact answer, but the wavefunction is in general hard to interpret since it's delocalized across the entire molecule.
One topic that might be of interest to you in this field is MO Localization. This is the field of study which aims to take the MOs and the high accuracy wavefunction and transform them into orbitals that reflect chemistry as intuitively as VB theory. (One can take arbitrary rotations of MOs without changing the observables of the wavefunction). This problem is technically ill-defined, however, there are many approaches that have done this to great success. For example: Natural Bond Orbitals (NBOs), Adaptive Natural Density Partitioning (AdNDP), Intrinsic Bond Orbitals (IAOs), and many others. As the titles suggest, there is a belief that localized (VB-like) bonds are real things hidden in the true wavefunction.

Is MO localisation related to PNO methods (for example, DLPNO-CCSD(T), et al.)?


text:
Which reactions occur during the oxidation and dissolution of bromide/bromine?
I'm trying to understand the fate of bromide anions during and after they undergo anodic oxidation. I have a solution of 94% DMF/6% water (by volume) with 0.1 M tetrabutylammonium  bromide (TBAB) as the supporting electrolyte. I've used this electrolyte to reduce a cathode at -2.5 V or less v. the Ag/AgCl reference electrode. During the reaction, the solution near the platinum counter electrode (the anode) turned yellow. I would assume this is due to the reaction:
$$\ce{2Br–(aq)-> Br2(l) + 2e– }$$
The next day, I noticed that the yellow liquid in the solution had completely disappeared. I'm assuming that the bromine had dissolved into the solution into $\ce{Br-}$ ions again. However, I'm not 100% sure of the chemical reaction associated with this. Also, would there be any side products associated with the dissolution? Or, do we end up with the species we initially started with in terms of bromine?
I ask because I would like to be able to reuse the electrolyte multiple times and so want to figure out exactly how the electrolyte changes after the reaction. At the cathode, there will be reactions such as the hydrogen evolution reaction which will make the solution more basic. However, I'm not sure what happens with the bromine.

A possibilty is that elementary bromine formed -- causing the the yellow color -- then solubilized in water yielding bromine water.

(source)
It does not dissolve well (2.8% ref -- but do not forget the high density (about 3.10 g/mL) of liquid bromine here). It then may disproportionate into then colorless hypobromit and bromide
$\ce{Br2 + H2O -> Br- + BrO- + 2H+}$ (ref.)
especially under light, forming an acidic solution.  The hypobromit may act as an oxidizer.
To draw an analogy, chlorine gas does dissolve into water, too (ref, ref).
Proceed with caution since bromine is corrosive, though.

Thanks for this answer. I think you're probably right. I do wonder if there is some reaction with the DMF, though. I recently learned that bromine and DMF violently exothermically react (in situations without water, at least). I couldn't figure out what reaction this is exactly (would be good to know). However, I'm thinking that this is not a good system if there is that safety hazard.

[Link to info about bromine and DMF] (https://webwiser.nlm.nih.gov/getSubstanceData.do?substanceId=24&displaySubstanceName=&STCCID=&UNNAID=&selectedDataMenuItemID=7&catId=64)

@Electrolyzer Interesting OSHA-related data base!

@Electrolyzer I speculate a brief reduction of a tiny amount of elementary bromine at a very low concentration, e.g. in a sweep of cyclovoltammetry (e.g. DOI: 10.1021/acs.jchemed.7b00361, OA) is lesser dangerous than extended presence of concentrated (neat) bromine in DMF in an excited state (like heat, sun light).  Otherwise, NBS-DMF would not be listed as a mild monobromination reagent at room temperature in Leo Paquette's encyclopedia EROS (cf. DOI: 10.1021/jo00393a066, or scheme 1 of DOI: 10.1021/ja00040a063 in the preview). CV indeed uses NMe4BF4 instead to limit redox on your sample.

Good point. I imagine in very small amounts, it would be safe as suggested by the references. In my case, I will be running at a high current for possibly hours, so can't have amides+bromine building up! It appears that acetonitrile also works for my reaction and is compatible with bromine, so I'll go with that. I think the solution to reusing the electrolyte is just to use an anion that oxidizes to something that can be removed easily (like a solid or gas). I wonder if chlorine gas from Cl- ions would dissolve into acetonitrile...


text:
Why is guanidine considered a strong base?
I've studied $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15,$ and anything with a $\mathrm{p}K_\mathrm{a}$ lower than should be an acid. Guanidine has a $\mathrm{p}K_\mathrm{a}$ of $13.8$ and is being called by Wikipedia a strong base. Why is it so?
According to Wikipedia, ammonia is a weak base. Then why is its $\mathrm{p}K_\mathrm{a}$ much higher than the $\mathrm{p}K_\mathrm{a}$ of guanidine?

Equivalent resonance in amidine bases

It's pKa  of conjugated acid ($\ce{pK_{bH^+}}$) - it's protonated form is about as acidic as water.

due to the stability of carbocation formed after protanation

Wikipedia notes that if a base (in water solution) has a conjugate acid whose $\mathrm{p}K_\mathrm{a}$ exceeds $13,$ it's a strong base.  The criterion is (apparently) based on complete or at least predominant dissociation up to $0.1$ molar.  Guanidine makes it, ergo a strong base.

We know that ammonia is a weak base than why is its pka much higher than Guanidine
https://en.m.wikipedia.org/wiki/Ammonia

Conjugate acid pKa < 13.

But Wikipedia states pka of Guanidine<pka of ammonia

No.  The number for guanidine is for its conjugate acid while the number for ammonia is for NH3 itself, the conjugate acid for ammonia would be about pKa=5.  Conjugate acid pKa is what counts for identifying a strong base.  It's inconsistent nomenclature, yes that sucks.


text:
Why do aromatic compounds have an upfield shift upon coordination to metals?
Why do aromatic compounds have an upfield shift upon coordination to metals e.g. ferrocene? The 1H NMR shows a single environment at ~4.1 ppm. This is in contrast to the usual aromatic shift of between 6-9 ppm. At the moment, I have the following rationale, although I am not convinced about 2/3:
1) Coordination of the aromatic group to the Fe essentially localises the double bonds, as Fe will backbond into the aromatic $\pi$*-orbitals.
2) Bonding of the aromatic group to the Fe will disrupt the ring current which would also cause an upfield shift. However, there seem to be several papers online which suggest the presence of ring currents even in sandwich complexes?
3) There are 6 $\pi$-electrons for only 5 carbons in Cp, therefore there are essentially 1.2 $\pi$-electrons per Cp carbon whereas there is 1 $\pi$-electron for each carbon in Cp. Therefore Cp is more shielded than benzene naturally and will have a lower shift.

Your "3)" goes in more or less right direction, but what you wrote is rather dire simplification of what needs MO theory to be explained properly, I think. Definitely not 1) or 2)

Hi, thanks for the reply. How could you rationalise the fact that aromatic groups such as benzene also go to a lower shift upon coordination to metal ions then?

There are bonds between HOMO and LUMO of the aromatic ligand and metal. Metal and ligand donate electron density to each other and ligand may wind up with less ten it started.

You should compare apples and apples. The $\delta$ ($\ce{CDCl3}$) for benzene is 7.36 ppm and that for bis(benzene)chromium is 7.16 ppm. reference

Chemical shift data dug up for a number of iron compounds in the website of Hans Reich reveals that the shift is not unique to aromatic compounds, but rather appears due to Fe(2+):

The base shift for a vinyl proton (C=C-H) is 5.25 ppm.
The following is a series of spectra from magritek, illustrating the effect of Fe(2+) complexation on the shift of the cyclopentadienyl ligand.

Sandwich compounds based on magnesium do not show such large upfield shifts:

This seems to eliminate your conjecture 3 as an explanation, since cyclopentadienyl is aromatic and addition of Fe(2+) causes an upfield shift, whereas Mg(2+) does not cause such a large one.
Now Fe(2+) complexes may be paramagnetic or diamagnetic depending on the the ligand (low/high field). Paramagnetic metals have unpaired electrons that can induce strong dipolar magnetic fields. In porphin, for instance, paramagnetic Fe(3+) results in strong shielding in the aromatic ligand, with one proton shifted upfield ~20 ppm. A diamagnetic Zn(porphin) complex does not show such strong shielding. Ferrocene is diamagnetic.
The quick and sloppy answer therefore appears to be that the shift observed in ferrocene is mainly due to Fe(2+), even though the compound as a whole is diamagnetic.
Regarding the bonding arrangement, the wikipedia has this to say:

In terms of bonding, the iron center in ferrocene is usually assigned
  to the +2 oxidation state, consistent with measurements using
  Mössbauer spectroscopy. Each cyclopentadienyl (Cp) ring is then
  allocated a single negative charge, bringing the number of π-electrons
  on each ring to six, and thus making them aromatic. These twelve
  electrons (six from each ring) are then shared with the metal via
  covalent bonding. When combined with the six d-electrons on Fe2+, the
  complex attains an 18-electron configuration.

To add a little more detail, Mulay and Fox hypothesize on the nature of metal-ligand bonding in ferrocene based on the magnetic anisotropy:

The anisotropy between K3 and the average of K1 and K2 was found to be
  $\pu{49.5×10—6 cgs units}$, and is evidence in favor of a single dπ—pπ bond between each of the rings and the iron atom. Pauling's "resonance
  structures" also seem plausible; the studies do not support Fischer's
  "complete π‐electron donation" structure. Theoretical calculations
  based on Langevin's formula for diamagnetism show a close agreement
  with the experimental value.

A more detailed account of bonding in ferrocene can be found in Miessler and Tarr's textbook.
[6]: Magnetic Anisotropy of Ferrocene and Chemical Bonding, J. Chem. Phys. 38, 760 (1963); https://doi.org/10.1063/1.1733734, L. N. Mulay and S.N.D. Mary Eleanor Fox


text:
Why this redox equation occurs on basic medium?
I have the following redox equation that I need to balance:
$$\ce{MnO4-(aq) + SO3^2-(aq) → MnO2(s) + SO4^2-(aq)}$$
The problem says that it happens on basic medium, but I don't see the $\ce{OH-}$ which I usually use to identify the basic medium.
How do I identify that this reaction happens on basic medium? Is it something that the problem should clarify?

You are given a net ionic equation, which is the reduced form of a complete ionic equation: that's why you see neither $\ce{H2O}$ nor $\ce{OH-}$.
You can deduce that the medium should be slightly alkaline a priori only if you:

either know that manganate $\ce{MnO4^2-}$ formed in alkaline medium hydrolyses when there is a lack of reducing agent (here, sulfite $\ce{SO3^2-}$) to manganese(IV) oxide $\ce{MnO2}$ and that $\ce{MnO2}$ is also formed near neutral $\mathrm{pH}$ values;
or you compose the redox reaction as you were asked to and notice that two $\ce{OH-}$
are produced:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} & |\cdot 2 \tag{red}\\
\ce{\overset{+4}{S}O3- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} & |\cdot 3 \tag{ox}\\
\hline
\ce{2\overset{+7}{Mn}O4- + 3\overset{+4}{S}O3- + H2O &→ 2\overset{+4}{Mn}O2 + 3\overset{+6}{S}O4^2- + 2 OH-} \tag{redox}
\end{align}
$$
As for why your problem states the redox reaction takes place in basic medium, it's just an extra hint assisting you in balancing the oxidation and reduction half-reactions (number of oxygens in oxides and oxoanions, for instance), but you could've balance the equation all right even if there were no information about the alkalinity of the medium.

RE: "but you could've balance the equation all right even if there were no information about the alkalinity of the medium." -- There is a point here which has not been clearly expressed. In basic solution $\ce{MnO4^- -> MnO2}$, but in acidic solution $\ce{MnO4^- -> Mn^{+2}}$.

@MaxW What do you mean? OP is already given the product and it's $\ce{MnO2}$, not $\ce{Mn^2+}$. The rest can be deduced from this very fact.

That is my point. Since the reduction product is $\ce{MnO2}$ the solution must be basic. If the solution were acidic then the reduction product would be $\ce{Mn^{+2}}$.

@MaxW I see, and if the medium were strongly alkaline, then permanganate were reduced to manganate. But what's the point of listing all possibilities unrelated to the terms of OP's question?

The OP asked how the solution being basic was relevant. "The problem says that it happens on basic medium, but I don't see the $\ce{OH−}$ which I usually use to identify the basic medium."

@MaxW Yes, that's true, and I think I addressed that fully. So? I'm sorry I really don't get what you are trying to imply. I must have a short circuit or something:)

LOL, I think we're having a violent agreement. The OP obviously didn't understand that the product $\ce{MnO2}$ meant that the solution must be basic. So to an experienced chemist the additional information that solution was basic is superfluous.

@MaxW Yeah, that's why I suggested that mentioning alkalinity was only a hint and not a must-have info required for problem solving. Maybe the way I wrote it was cumbersome (English is not my first language — I bet you can tell:) ), so if you see how it can be better formulated, feel free to edit my answer.

Your English is very very good. I studied French as a second language but forgot most since I never really used it outside of class. So like most Americans I know English and Greek. If it isn't English, it is Greek to me. ;-)


text:
Reaction mechanism for the conversion of acyl hydrazide to acyl azide
How is an an acyl hydrazide converted into an acyl azide in the presence of $\ce{HNO2}$? According to me, the $\ce{-NH2}$ group of hydrazide will get converted into $\ce{-N2+}$ through $\ce{HNO2}$. However, I am unable to proceed after that.

In ChemSE you can Accept a answer by giving a TICK beside the answer please read This

You are correct acyl hydrazide can be converted into acyl azide by adding $\ce{HNO2}$ also you wrote that the $\ce{NH2}$ group of hydrazide will be diazotisated making a $\ce{N2+}$ group hence I am assuming that you know the mechanism, after diazotisation the adjacent $\ce{N}$ donate the lone pair to the $\ce{N2+}$ group making the final product as acyl azide.
The mechanism scheme is in the following picture :


text:
Calculate Gibbs energy of dinitrogen pentoxide using its vapor pressure and sublimation
I am having trouble with the following problem:

Dinitrogen pentoxide, $\ce{N2O5}$, is a solid with a high vapor pressure. Its vapor pressure at $\pu{7.5 °C}$ is $\pu{100 mmHg}$, and the solid sublimes at a pressure of $\pu{1.00 atm}$ at $\pu{32.4 °C}$. What is the standard Gibbs energy change for the process $\ce{N2O5(s) → N2O5(g)}$ at $\pu{25 °C}$?

Okay so my attempt is to first use the Clapeyron equation to find $\Delta H$ for usage in the reversible process entropy formula. The Clapeyron equation is $$\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ Plugging everything in I got $\pu{50807.79 J}$.
Next I found the change in entropy in a reversible process:
$$\Delta S = \frac{\Delta H_\mathrm{vap}}{T_\mathrm{vap}}.$$
Now here is where I have a problem. I calculated the enthalpy of vapor pressure. So I would expect to use the temperature that corresponds to the enthalpy of vaporization. Problem is, the book lists this as the wrong answer. I got within a few significant figures when I used $T_\mathrm{sublimation}$.
I calculated the final answer by using
$$\Delta G = \Delta H - T\Delta S$$
The corresponding values are plugged in. I am not sure if I even solved the problem correctly, but if I am, why is that temperature used?

First, here is a way to solve it that avoids calculating entropy:

Calculate the enthalpy from the given data.
Calculate the vapor pressure at 25 $^\circ$C. This is equal to the equilibrium constant for the process.
Calculate $\Delta G^\circ$ as -RT ln(K).

This way is efficient because you already set up the Clapeyron equation in the first step and you just use it again in the second. You could even solve 1. and 2. graphically, and then go from there.

I got within a few significant figures when I used Tsublimation.

You need to calculate $\Delta S^\circ$, i.e. the change in entropy when all species are at standard state. At the normal temperature of sublimation, $\ce{N2O5(g)}$ is at the standard state, so that is the temperature you need to use. The enthalpy is not concentration dependent, but the entropy and the Gibbs free energy are.

I am not sure if I even solved the problem correctly[...]

You solved it correctly in that you got the correct answer. You did not solve it as the problem author intended, I suspect, because you had to look up the normal sublimation point (unless you used the Clapeyron equation to estimate it).


text:
Choosing the right wavelength for calibration curve and for analysis of actual results
Introduction:
I am currently conducting a series of experiments in which I am studying the transport of polystyrene nanoparticles (PNPs) through sand and soil. This is done in sand/soil columns, into which I am injecting the PNPs solution using a peristaltic pump.
The PNPs are fluorescent, so I am using a fluorimeter for detection. During the experiment, I am collecting the fluid that is leaving the column in a fraction collector, and as a result I am constructing a breakthrough curve that is supposed to show me the behavior of the PNPs compared to an inert tracer (bromide) that is also found in the inlet solution.
In order to construct the breakthrough curve, I need to prepare a calibration curve that will allow me to convert the intensity of the collected fractions into concentrations. For the calibration curve I am preparing several samples with known concentrations.
Question:
For preparation of the calibration curve, I am choosing the wavelength that gives me the max intensity in each sample's emission spectrum, and then for the breakthrough curve I am taking the intensities in that same wavelength from each fraction for the breakthrough curve. In the example below it is 635 nm.

Then, when I came to take the appropriate intensities of my actual results, I obtained this graph:

The X-axis range is the same: 575-700 nm. I need to choose a certain wavelength and take the data pertaining to it.
Can I actually use these intensities, or would it be wrong because at 635 nm I don't have a max point?

It looks  like you have some material that is suspended in solution and scattering and hence giving a sloping background that is obscuring your data.  Is it possible that your sample has decomposed?

Adding to @porphyrin, applying to standards for your calibration and a typical sample containing your NPs eluted: if it takes 2 min to record your spectrum once, does a next spectrum looks reasonably the same (regarding position of the maximum recorded, and counts/s on the detector) if recorded a twice / multiple times just after the initial recording / 10 min later? Different to potential decomposition of your NPs: perhaps your soil samples equally contain a quencher (e.g., oxygen, chloride)?  Does tonic water (as a source of quinine) pass the soil samples still a little bit fluorescent?

Your peak is buried by scattering as @porphyrin suggested. You can see two inflections around the expected maximum. Consider that fluorescence quantitative analysis is tricky.  Dealing with np makes it potentially difficult as there can be a concentration depending scattering that shift your baseline continuously though I do not see this in your calibration. It could simply be extremely fine sand. Actually you should make a calibration using water filtered to just sand and see what you get. I suspect something like the current and actual measurements.

Thanks everyone. I tried a few more things in the meantime: I added bromide (same concentration as in inlet solution) to my standards to see if it was interacting with the NPs, but the intensities seem to not be affected (i.e. the graph looks similar to the first graph above). Then I tried adding some sand (clean sand, i.e. sand that has been treated with HCl and lots of water) to the standard samples WITH the bromide. The results of this step were a bit different, since high concentrations gave the bell-shape, while lower concentrations gave the descending, more-or-less linear shape.

You should really use water previously passed through the sand while doing the calibration.  You should run into similar problems if sand is responsible.  Another  point I did not want to bother before. Wouldn't better work with absorption? Or absorption of your np stay buried in water cutoff?

@Alchimista, I AM using water that already passed through the sand, this is my background water. Why would it be better to work with absorption? I was under the impression that with fluorescent particles one should work with emission spectrum.

It is generally tricky to make quantitative determination by fluorescence as it is an absolute measure and has a complicated intensity-concentration, even prone to vary due to concentration depending factors.  Indeed working with np is a hybrid and not straightforward situation. We have suspensions rather than solutions.  But if you have a double beam abs spectrophotometer and a band out of water cut off, then I would give it a try.

Fluorescence would be perfect for a yes there are np/no np response. Also, it is true that is more sensitive. But it requires that everything goes smoothly

Also I noticed that basically you have not a useful calibration curve. The concentration of the sample is well below that used to make the "calibration". That can be why you measure similarities percolating water through sand at just low concentration.


text:
Entropy change of isothermal irreversible expansion of ideal gas

Two mole of an ideal gas is subjected to isothermal expansion from $\pu{2 atm}$ to $\pu{1 atm}$ at $\pu{300 K}$. Calculate entropy change of the system, surrounding and total, if the process is irreversible.

Since
$$ΔS_\mathrm{system} = nR\ln\left(\frac{P_1}{P_2}\right),$$
I got the $ΔS$ for the system as $1.4R$. I know that for reversible process the entropy should not change, hence the entropy of surroundings decreases by $1.4R$. But how do we calculate the same for an irreversible one?

there are lots of other questions that address this on the site, maybe you can take a look at one of those. See eg https://chemistry.stackexchange.com/questions/84590/calculation-of-entropy-for-an-isothermal-irreversible-expansion?rq=1

Here is a cookbook recipe for determining the change in entropy for a system that has suffered an irreversible process:
THE RECIPE

Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system
Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states.  This is the most important step.
Devise a reversible (alternative) path between the same two thermodynamic equilibrium states (end points).  This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path.   For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic.  You can even separate various parts of the system from one another, and subject each of them to a different reversible path, as long as they all end up in their correct final states. Plus, there are an infinite number of reversible process paths that can take you from the initial state to the final state, and they will all give exactly the same value for the change in entropy.  So try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4).
For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path.  This will be your change of entropy S.  That is, $\Delta S=\int{\frac{dq_{rev}}{T}}$ , where the subscript rev refers to the reversible path.

For further details and worked examples, see:
Reference https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

I am being required to qualitatively tell is $\Delta S_{surr}$ =0 or greater than 0 too. Do you have any insight on how to do this too?

@RishiNandhaVanchi What are your thoughts on this?

One way to proceed is as follows. Start from two things you know from the problem statement:

You have an ideal gas
The process is isothermal

Those two conditions mean the energy of the system does not change ($\Delta U=0$) during the process. It follows, from the first law, that
$$
q_\mathrm{sys} = \Delta U -w_\mathrm{sys} = -w_\mathrm{sys}
$$
You are only told the process is spontaneous. But you can assume that the expansion was against a constant pressure $P_\mathrm{ext}=P_\mathrm{fin}=\pu{1atm}$, so that
$$ w_\mathrm{sys}=-P_\mathrm{ext}\Delta V = -nRT(1-\frac{P_\mathrm{fin}}{P_\mathrm{ini}})$$
Finally, the entropy change of the surroundings is
$$ \Delta S_\mathrm{surr} = \frac{-q_\mathrm{sys}}{T_\mathrm{surr}}= \frac{w_\mathrm{sys}}{T_\mathrm{surr}}$$
Since the process is isothermal $T_\mathrm{surr}=T_\mathrm{sys}=T$ and
$$ \Delta S_\mathrm{surr} = -nR(1-\frac{P_\mathrm{fin}}{P_\mathrm{ini}})$$
Plugging in values into this equation should give the desired answer ($-R$).

The problem statement seems to  imply that $P_{ext}=P_{fin}$

@ChesterMiller Thanks, I know, I didn't write that explicitly, but now made it a little tidier.


text:
What is the significance of molar mass
What exactly is the significance of molar mass in chemistry? Why not just mass?

Say, a coin weighs 2 g. Then a thousand coins weighs 2 kg. How come these values (2 and 2) are similar, and why are they in different units?

Thank you that was surprisingly the most useful answer I have gotten throughout but are you sure it is correct

An answer? I never gave you any. I merely asked a rhetorical question; if it made you think in the right direction, good for you. Just what part of my comment could be mistaken for something that may or may not be correct? That a certain real-world coin actually weighs 2 g? I never said that. Or that 1000 times 2 g makes 2 kg? Well, yes, I guess I can say so with a good deal of confidence.

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass

molar mass of a substance is the mass of one mole of a substance

This is not quite correct. If you said the speed of a car is the distance it travels in one hour, cars would go at speeds of 120 miles (race car) or 30 miles (in town). However, speed is defined as distance traveled per time. As a result, speed has the dimension of distance divided by time, and a typical unit (at least where I live) is miles per hour.
Molar mass is defined as the mass per amount of substance. The dimensions are mass divided by amount of substance, and typical units are g/mol.

Why does molar mass have the same value as molecular mass?

If the molar mass is given in g/mol and the molecular mass is given in u (unified atomic mass unit) or Dalton, the number is the same. The reason is that both the mole (at least until May 2019) and the Dalton are based on the mass of the carbon-12 isotope. A mole of a substance is defined as the amount that has the same number of particles as present in 12 grams of carbon-12. The Dalton is define as 1/12th of the mass of a carbon-12 atom.
From these definitions, you will find that the molar mass of the carbon-12 isotope is 12 g/mol, exactly, and the molecular mass of carbon-12 is 12 Dalton, exactly.

What will change in May 2019?

The mole will be defined as "One mole contains exactly 6.02214076×10^23 elementary entities." The molar mass of carbon-12 will be an experimentally determined quantity.

I'd change the sentence "If the molar mass is given in g/mol and the molecular mass is given in u..." to If the molar mass is given in g/mol and the molecular mass is given in u/atom ..."

I would argue that molar mass by definition is the mass of 1 mole of substance. Otherwise you can call it something else. The term "molar" speaks for itself, just as in other molar quantities.

@Try_Hard : As long as it is clear that the dimensions are mass divided by amount of substance, either definition is fine. I'm not a native English speaker, so to me the mass of 1 mole of substance sounds like the dimensions would be mass, while the mass per amount of substance (or the mass divided by the amount of substance) sounds like the dimensions are mass divided by amount of substance. Some definitions just add "expressed in g/mol" to make that clear, which is fine too.

@MaxW : I think it is fine to say in the text that this refers to the mass of one atom (or call the quantity $m_\mathrm{atom}$), but I'm not so sure about putting it into the units. What is the dimension of "atom"? What is the value of "atom". If it is 1, the statements are still true. The definition of molecular mass already includes that it refers to a single molecule (or atom).

Kirsten basically the Avogadro number will be taken as exactly that. So we shall expect in principle little change in the atomic mass for atoms different than C 12. It is a shame but I cannot quickly figure it out


text:
Finding the color of crystal violet theoretically
I am trying to find the color of an organic molecule, crystal violet, theoretically.
I suspect that the color is due to the large conjugated system. My question is, how would I go about finding the energy levels of such a system? I know how to solve the Schrodinger equation for a particle in a ring or a box (i.e. benzene, or a poly-diene) to solve for energy levels and find the lowest-wavelength transition. But how would I do it for this sort of molecule?

You need something more sophisticated than particle in a box, probably (at least) Huckel theory.

Using some decent software.

You would have to make ridiculous simplifications. For starters, this molecule is not perfectly planar.

The photophysics of this and similar molecules is very interesting. The excited state decays non-exponentially and strongly depends on solvent viscosity and both effects are due to twisting of phenyl groups, so the excited state is definitely not planar.

@Oscar Lanzi.  It is of course planar around the center. The fact that conjugation will depend or will force planarity is right one should determine as per the question. The comment you refer to is unnecessary, it can be made for every polyene, even.

For a reliable and experimentally comparable answer, you really need to use a numerical method rather than a pencil and paper estimation. For an organic system this size you'll probably be safe using time-dependent density functional theory (TDDFT). Most computational chemistry packages have this built in. Off the top of my head, I know that Gaussian (commercial–standard for organic molecules like this), and GAMESS (free) both have TDDFT built in.
If you're intent on using a toy method you could use Hückel theory as orthocresol suggested. You could in principle tune the parameters of such a model until the excitation energy matches the experimental result, but I don't see what additional physical insight this would give you.

Here's about limitations of pinging https://chemistry.meta.stackexchange.com/questions/3889/pinging-in-comment-to-suggest-answering-question

"... don't see what additional physical insight ..." Word.


text:
Amber paramfit "monomer (or ligand)" problem
I am trying to parameterize a novel molecule which has atoms not included gaff or parm force field. I finished parameterizing vdw-bond and angle force field. Also I did PES Scan for dihedrals and I wrote them in a frcmod file as truncated fourrier integral. Yes, actually most probably I will not need paramfit parameterization but I wondered and I tested paramfit.
Here is tutorial:
http://ambermd.org/tutorials/advanced/tutorial23/index.html#Introduction
I succesfully finished this tutorial and after this I tried to apply paramfit to my "monomer" and "not included any force field" molecule. But xleap gives some error message and I locked completely.
I changed gen_conformers.sh (Don't click if you don't want to download!) for my molecule and I couldn't obtain different conformations.
Then I tested where was the mistake. I created leapin file like this:
source oldff/leaprc.ff99
addAtomTypes { { "B" "B" "sp3" } }
addAtomTypes { { "O" "O" "sp3" } }
addAtomTypes { { "C" "C" "sp3" } }
a=loadmol2 bor.mol2
loadamberparams bor.dat
impose a {1 1}, { {"H1" "O2" "B3" "O4" 72.000} }
impose a {1 1}, { {"O12" "B3" "O4" "C5" 19.000} }
impose a {1 1}, { {"B3" "O4" "C5" "C6" 108.000} }
impose a {1 1}, { {"B3" "O12" "C13" "H18" 11.000} }
saveamberparm a a_1.prmtop a_1.mdcrd
check a
quit

then I run this file with this command:
xleap -f leapin

but after that leap.log file:
  Checking Unit.
Building topology.
Building atom parameters.
Building bond parameters.
Building angle parameters.
Building proper torsion parameters.
Building improper torsion parameters.
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C2 H5 C1 H6
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C1 H2 C2 H3
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C1 H2 C2 H4
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C1 H3 C2 H4
 ** Warning: No sp2 improper torsion term for  H-H-C-H
        atoms are: H2 H3 C2 H4
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C4 H10 C3 H11
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C3 H7 C4 H8
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C3 H7 C4 H9
 ** Warning: No sp2 improper torsion term for  C-H-C-H
        atoms are: C3 H8 C4 H9
 ** Warning: No sp2 improper torsion term for  H-H-C-H
        atoms are: H7 H8 C4 H9
 total 7 improper torsions applied

and after that, even if I changed angles many many times in "impose" command
 impose a {1 1}, { {"B3" "O4" "C5" "C6" 108.000} }

I obtained the same *.inpcrd files and I couldn't generate different conformation gaussian job file.
So how can I apply paramfit for a ligand? (or monomer)

amber has a mailing list, you may have better luck on that.
http://lists.ambermd.org/mailman/listinfo/amber

What is "monomer or ligand" ? We're not a helpdesk here, questions should make sense to other people, too.

I'm voting to close this question as off-topic because it's a question about how to use AMBER.

I think opposite. The question is about applying genetic algorithm (G.A) to dihedrals. And paramfit is the only software can do this.( Why I am saying this? here: https://doi.org/10.1002/jcc.23775 ) So even if user don't want to talk about paramfit he/she has to talk about paramfit. Another problem that the user faces is not to be able to save the related coordinate files by turning the dihedral angles.(xleap)
Charmm, tinder, gromacs softwares are already using "amber ff " so in this respect, all molecular simulations have to do with "amber" thus problem is  more complex than it looks.

Erdem, I think I found answer, ( Iwas facing with the same problem for about 1 month) I tested for my molecule and I got true results: to obtain rand conformers; Instead of xleap use another program and skip this step of the paramfit tutorial with these conformers (you will need to change *.dat files according to your dihedral angels, this is fatal important). You can use this program:
http://en.freedownloadmanager.org/Windows-PC/ArgusLab-FREE.html
Good Luck!


text:
Gabriel Phthalamide Reaction
Can we use acid chloride in place of alkyl chloride in Gabriel Phthalamide Reaction?
Maybe it can be used as process to synthesize amides.

In practice, there's no reason to; you can just use ammonia.

@orthocresol I mean theoretically if it's a possibility.

NO, this is not a viable way to make amides.
Phthamide anion will react with acyl halides, but the problem arises with the second step, the removal of the phthalimido group. The usual conditions described here of hydrolysis with hydroxide or reaction with hydrazine will result in removal of the acyl group to reform phthalimide plus the carboxylic acid, in the case of OH- hydrolysis, or the acyl hydrazide if using hydrazine. This is because the acyl group is the more electrophilic site for the nucleophile.


text:
Why doesn't the particle in a ring model have a potential energy term?
I was thinking of the "particle on a ring" problem as a subdivision of a bigger problem, i.e. the hydrogen atom. While in the hydrogen atom problem, the Hamiltonian contains a potential energy term, the Hamiltonian in the "particle on a ring" problem does not.
Why is it so? What are its implications in the mathematical solution to the problems?

The short answer is that the particle in a ring is defined to have no potential energy term (at least, not beyond what Karsten Theis has pointed out). Once you introduce something else into the system, then it is no longer really the particle on a ring problem any more.
However, if you want to relate it to the hydrogen atom, then note that in the hydrogen atom, the Coulomb potential is a function of only $r$, i.e.
$$V(r) = -\frac{1}{r}$$
in atomic units. In the particle on a ring model system, the particle is already constrained to be at a particular value of $r$, so even if we were to hypothetically introduce a Coulomb-type potential $\propto 1/r$, this would just be a constant. The effect of having a constant potential energy is just to shift every eigenstate up in energy by the same amount; it has no real physical effect.
What are the differences in the mathematical solutions? Well, you already know it, surely; the eigenstates of the hydrogen atom are atomic orbitals, whereas the eigenstates of the particle on a ring are simply $\exp(\mathrm im\phi)$ with $m \in \mathbb{Z}$ to satisfy the boundary condition.
Of course, this $\exp(\mathrm im\phi)$ term is indeed part of the mathematical form of the atomic orbitals. This is because when you solve the Schrödinger equation for the hydrogen atom, you can successively separate out different degrees of freedom. Firstly you separate the bit which depends on $r$, which goes on to become the radial wavefunction $R(r)$. Then you need to solve for the angular wavefunction $Y(\theta,\phi)$, and the way of doing this is to again use separation of variables, i.e. assume $Y(\theta,\phi) = f(\theta)g(\phi)$. That gets you to a differential equation which looks like $(-1/g)(\mathrm d^2g/\mathrm d \phi^2) = m^2$, which has $g = \exp(\mathrm im\phi)$ as its solutions. I'm glossing over some details here, but that's the general idea; consult e.g. Griffiths' Introduction to Quantum Mechanics for the nitty-gritty bits.

There is a potential energy of some sorts: The potential energy is constant within the ring, and infinitely high outside of it. This is an idealized situation similar to the particle in a box (one-dimensional, two-dimensional, three-dimensional).
Even though these models are artificial and ideal, they can be used to estimate properties of certain sets of molecules. For example, you can discuss the spectra of molecules with linear conjugated double bonds using the model of the one-dimensional particle in the box.
When setting up the Hamiltonian, you implicitly introduced a infinitely high potential by boundary conditions and choice of coordinate systems. The free electron would move in three dimensions. For the one-dimensional particle in the box, the choice of just one coordinate constrains the electron to one dimension, and the boundary conditions of an electron density of zero at the edges of the box constrains the electron to a certain length of box. For the particle in a ring, you can choose a cylindrical coordinate system, keep the radius constant, the height at - say - zero and keep the angle as the only coordinate of the Hamiltonian and the wave function.


text:
Why would decreasing ionic strength of buffer disrupt hydrogen bonds
I'm reading about cDNA synthesis in my textbook and it mentions that:

The captured mRNAs are subsequently released and collected, for example, by decreasing the ionic strength of the buffer so that the hydrogen bonds will be disrupted

So my question is why would this happen? What does the ionic strength of the buffer have to do with the hydrogen bonds the mRNA has formed?

At high ionic strength, the poly-A tails of mRNA bind oligo-dT that is covalently attached to beads in a column (source). The binding interaction is based on base stacking and hydrogen bonding. When you decrease the ionic strength, the repulsion between the (negatively charged) phosphate groups becomes stronger (is shielded less by counter ions), lowering the binding strength to the point that the mRNA elutes.
The length of the oligo-dT on the beads is chosen such that the mRNA binds sufficiently strong at high ionic strength, but is released at low ionic strength. The appropriate length of oligo-dT is probably on the order of 8-10 nucleotides.

decreasing the ionic strength of the buffer so that the hydrogen bonds will be disrupted

This statement does not capture the situation accurately. What is really being disrupted is the overall binding of mRNA to the beads. The hydrogen bonds, as short range interactions, should not be directly affected by ionic strength. The stacking interactions might weaken with decreasing ionic strength. Long range ionic interactions, however, are significantly weakened by increasing ionic strength because counter-ions bound to the phosphate backbone will strongly shield the ionic interactions (water also shields it compared to vacuum or non-polar solvents). In this case, the ionic interactions are repulsive, so binding will get weaker at low ionic strength because that is when the repulsive interactions are strongest.

so the ions in the solution have the opposite of a dielectric effect, where instead of reducing attraction between opposite charges, they reduce repulsion between charges?


text:
Heating Dry Benzenediazonium Chloride
If we heat BDC with water then phenol is formed. I was wondering what would happen if we heated dry benzenediazonium  chloride. I know that on heating benzenediazonium fluoroborate we obtain fluorobenzene but then it (benzenediazonium  fluoroborate) is insoluble in H2O so there is no confusion over there. On the other hand BDC can be obtained in both forms.

If I had to guess, I'd say chloride ion would displace nitrogen giving chlorobenzene.  If this is wrong, inquiring minds want to know.

I think it just explodes on heating

please read (this)[http://reag.paperplane.io/00000121.htm] under the heading precautions

@AdvilSell Thank You. You are right that it will explode. Read it in a few other places as well. However I am surprised that they explode. If you come to know more about , perhaps with more details then please post it up.

Strongly related: https://chemistry.stackexchange.com/questions/76469/why-should-the-temperature-be-maintained-at-0-5-c-in-a-diazotisation/76550?r=SearchResults#76550


text:
Conflicting definitions of Lewis acids
I have struggled in attaining a solid grasp of the Lewis acid/base concept largely because so many sources seem to offer conflicting definitions.
For instance, Klein's Organic Chemistry mentions that Brønsted–Lowry acids and bases are strictly a subclass of Lewis acids and bases. Other sources say this is true of bases but that a Brønsted acid need not be a Lewis acid and vice versa.
A common example of a non-Lewis acid provided by Wikipedia is $\ce{HCl(g)}$. The rationale being that it has no available empty atomic and/or hybrid orbitals to accept an electron pair (it does have an empty antibonding orbital though — see below).
Another source says hydronium is only a Lewis acid if it is "informally" treated as a hydrogen ion $(\ce{H+})$.
There appears to be further disagreement on whether the adduct formed in the Lewis acid/base reaction is necessarily a product or whether it can effectively appear as a transition-state before splitting apart to give the products. For example, say something like
$$\ce{NH3 + H2O <=> NH4OH <=> NH4+ + OH-}$$
Other users on this site suggests the approach should be to consider molecular orbitals. I will further note that $\ce{HCl}$ has an empty antibonding orbital so this definition, although arguably the best I can think of, still necessarily conflicts with multiple other sources labeling $\ce{HCl}$ as a non-Lewis acid.
Any help on resolving some of these apparent conflicts would be appreciated.

There is a very nice article in Helvetica Chimica Acta titled "Confusing Quantitative Descriptions of Brønsted-Lowry Acid-Base Equilibria in Chemistry Textbooks – A Critical Review and Clarifications for Chemical Educators", Helvetica Chimica Acta – Vol. 97 (2014). It is really comprehensive 31 paged article.

@M.Farooq
A common example of a non-Lewis acid provided by Wikipedia is $\ce{HCl(g)}$,How?
Can we consider $\ce{HCl(g)}$ in the following reaction as :Bronsted acid or Lewis acid ?
$$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$$

@AdnanAL-Amleh, Can you share the link? I want to see the context.

@M.Farooq : thanks. Its the 6th line of this post.

@AdnanAL-Amleh, Please see this https://sites.google.com/site/funwithchemi/the-team

When Lewis proposed his definitions, orbitals were in the domain of physicists...I still would like to see the entire context.

@AdnanAL-Amleh See here: https://www.vedantu.com/question-answer/is-hcl-a-lewis-acid-class-11-chemistry-cbse-610a056f42eeb369b74e2a48
"Hence we can say that H+ is a lewis acid, not HCL"
Also here: https://www.quora.com/Why-is-HCl-a-Lewis-acid
This was in general the point of my question...regarding inconsistent definitions of lewis acids. You have some sources say HCL is a lewis acid, and others say it is not.

Moeller's Inorganic Chemistry textbook from 1952 has a fairly long discussion of different theories of acids, citing a lot of primary literature. For Lewis acids, it cites some reviews by W. F. Luder from the 1940's and some of Lewis' original work. Specifically, it says that Lewis in a paper in 1938 (J Franklin Inst. 226:293) described "phenomonological criteria" for Lewis acids and bases, which include the ability to displace weaker Lewis acids and bases.
My impression from Moeller's description of Lewis' work is thus that he defines an acid as a molecule in which one atom has an incomplete valence shell of electrons. He further views a Bronsted acid as a complex between a Lewis acid ($\ce{H+}$) and a Lewis base (the conjugate base). In a reaction like $\ce{NH3 + AcOH -> NH4+ + AcO-}$, he would say that the stronger Lewis base $\ce{NH3}$ is displacing the weaker Lewis base acetate from the Lewis acid $\ce{H+}$.
If I have time, I'll try to dig up some of those primary sources. Here's a quick list if anyone else wants to go after them:

G. N. Lewis; Valence and the Structure of Atoms and Molecules. The chemical catalog co. (1923)
G. N. Lewis; J Franklin Inst 226:293 (1938)
W. F. Luder; Chem Rev 27:547 (1940)
W. F. Luder and S. Zuffanti; The Electronic Theory of Acids and Bases, Wiley and Sons (1946)
W. F. Luder, W. S. McGuire and S. Zuffanti; J. Chem. Ed. 20:344 (1943)
W. F. Luder; J. Chem. Ed. 22:301 (1945)

UPDATE: The original 1923 Lewis text is available for free from Google Books here:https://books.google.com/books?id=2hxDAAAAIAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false
The most relevant passage is from p.142:

Yeah, that's pretty much this. BTW you can use mathjax to improve post https://chemistry.meta.stackexchange.com/a/3046/9961

@Andrew
A common example of a non-Lewis acid provided by Wikipedia is $\ce{HCl(g)}$,How?
Can we consider $\ce{HCl(g)}$ in the following reaction as :Bronsted acid or Lewis acid ?
$$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$$