text:
Is there an advantage to the Clemmensen or Wolff-Kishner reductions over the Mozingo reaction?
As I understand it (and please correct me if I'm wrong), ketones and aldehydes can be reduced to methylene derivatives via:

Clemmensen Reduction (acidic conditions):

Wolff-Kishner Reaction (basic conditions):

Mozingo Reaction (mild conditions)

A preference for Clemmensen or Wolff-Kishner can be established based on functional group tolerance with respect to acidic vs basic conditions.
But if the Mozingo Reaction doesn't have the drawback of requiring either acidic or basic conditions, is there a reason to ever use the Clemmensen Reduction or Wolff-Kishner Reaction instead of simply applying the Mozingo Reaction?
images for reaction formulas from their respective Wikipedia pages

The smell of the dithiane is a common issue.

A good test of your handling skills. Hydrazine has H&S issues too.

In monofunctional compounds it usually doesn’t matter. Multifunctional compounds require judicious choices.


text:
Why mixture turns brown upon heating, during esterification with H2SO4?
In the lab I was synthesising an ester with sulphuric acid as a catalyst. However the mixture turned brown and my teacher said I had burnt the acid, but what does this actually mean?

It might be worth adding some details on the process you were using to make the ester, describing it as burning the acid sounds a little unusual on his part.

I was heating the mixture under reflux but he said i was heating it too strong and as a result i 'burnt' the acid

https://chemistry.stackexchange.com/questions/60708/how-does-sulfuric-acid-react-to-heating

It isn't acid, but rest of mixture that was damaged by H2SO4 if you overheat.

what damage will be taking place then?

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text:
What is the oxidation state of oxygen in hydrogen peroxide?
In Hydrogen peroxide, the oxidation number of oxygen is "-1" instead of "-2".

But it seems to me that, the oxygen atoms have '-2' as their oxidation number as each oxygen atom here is connected to a hydrogen atom and an oxygen atom.
So, what's actually happening here?

Here is another way to think about it. Mentally break the O-O bond heterolytically (+/-). Now you have HO- and HO+. HO- has a -2 charge (normal) and HO+ has a zero charge on oxygen since in both cases hydrogen is +1. Oxygen zero, a 2-electron oxidant, will accept 2-electrons from a reducing agent to convert HO+ to HO-. This is why HOOH is an oxidant. E.g., MeSMe + HOOH --->  Me2SO  +  H2O.

Each oxygen atom is connected to a hydrogen atom(which develops -1 charge on oxygen and +1 on hydrogen) and to another $\textbf{oxygen}$ atom which contributes no charge to both(0 and 0). Similar for other oxygen atom. Hence oxygen state of each oxygen atom is -1 and for each hydrogen atom it is +1.
In general, in peroxide linkage, oxygen has -1 oxidation state

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text:
Sodium Sulfate or Sodium Peroxysulfate?
In an attempt to clean drain cleaner, I reacted it with Sodium percarbonate to clean up some hardware store variety sulphuric acid.
At first the reaction appeared successful, yielding a Nice Clear Syrupy sulfuric acid with a heavy crystallization at the bottom of the flask, but I don't think the salt that I collected is sodium sulfate.
The crystals are very long resembling fiberglass and based on images I've seen online, sodium sulfate's structure seems to be a lot like sodium chloride.
I'm not sure of the conjugate base formed during the reaction ($\ce{SO4-, SO5- or HSO4-}$) can anyone explain?
EDIT
I've since learned that when basic sodium is reacted with $\ce{H2SO4}$ in a ratio favoring $\ce{H2SO4}$, the conjugate base produced is $\ce{HSO4-}$

RE: "I thought to use Sodium percarbonate to clean up.." Do you mean that you used sodium sulfate?

@MaxW , Νο, Ι meant exactly what I said. I reacted  Sodium Percarbonate with an excess of Sulfuric Acid to yield  Sodium Sulfate and Caro's Acid (which I believe eventually becomes H2SO4), but the collected salt doesn't resemble sodium sulfate. It actually looks like if I crushed it, it'd be like monosodium glutamate.

Ok, let’s take it one step at a time:
Percarbonate is actually not a single molecule, but a double-salt of sodium carbonate and hydrogen peroxide.
The sodium carbonate reacts with sulfuric acid, resulting in carbon dioxide, sodium sulfate and water. The reaction scheme is: Na2CO3 + H2SO4 -> Na2SO4 + H2O + CO2
For safety reasons I will strongly discourage you from adding carbonate to concentrated sulfuric acid. It will most likely explode in your face!
But in theory, after having added sufficiently carbonate, Sodium sulfate will precipitate.
But getting a drying effect will be uphill, since water is produced alongside the sulfate, as you can see in the reaction scheme, on top of this is any water that comes from hydrogen peroxide, not to mention the fact that concentrated sulfuric acid is strongly hygroscopic itself.
As for the cleaning of impurities, I fail to see what neither carbonate nor hydrogen peroxide can do. HP can oxidize some organic materials, but most of these will probably already have been destroyed by the sulfuric acid. It can also oxidize some inorganic ions like sulfite (SO3), but do you suspect these to be a problem?
I don’t have detailed knowledge of the conditions needed to form peroxo(mono)sulfuric acid. Undoubtedly some will be formed when HP is added to sulfuric acid, but I am also convinced that only a fraction of the sulfuric acid/HP will react.

Since I asked that question I've learned quite a bit. It actually produce peroxymonosulfuric acid and did purify the acid to complete clarity, however there was a prohibitive amount of acid lost to Na2S04 production, not to mention the amount lost during filtration, then there was the issue of water that had to be boiled off which lead to the precipitation of the rest of the sodium sulfate. From what I understand at the ratio used, reacted the sodium percarbonate with the sulfuric acid to yield sodium bisulfate? I'm new at chemistry. I just learned to balance equations properly.

$\ce{SO3-}$?, because the way you phrased it, you look like you're talking about sulfur trioxide and not the sulfite ion. Percarbonate doesn't explode in your face unless you react it all together suddenly, like a dimwit.


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Are specific heat capacity and boiling point of a substance related or proportional to each other?
That is, if the substance has a high specific heat, will it also have a high boiling point? ( and vice versa)

No. The isobaric heat capacity of water is 4.1813 $J\cdot K^{-1}\cdot g^{-1}$. For tin it is 0.227, but tin has a much higher melting point than water.

Someone in my group thought that since water has a high specific heat, i.e., it is resistant to heat changes, it would not boil as fast as other substances. Is that because boiling is affected by vapor pressure and other factors?

According to the Ehrenfest classification, during a first-order transition (such as boiling) the heat capacity $C$ goes to infinity. This makes sense, since at the transition point the temperature does not change with input of heat and $C = dQ/dT $. This and other considerations add a little complexity to the problem, as it requires you to state which heat capacity (at which T) you are talking about.
More importantly, however, the boiling point depends on the magnitude of attractive intermolecular forces compared to kT, whereas the heat capacity depends on all the available degrees of freedom (including intramolecular). Therefore you can envision large and small molecules with similar overall solution composition (for instance hydrocarbons) - and thus similar heat capacities - but very different BPs. See for instance this
table of specific heats, where C for hexane through dodecane are very similar, whereas their BPs differ quite a bit.


text:
Pressure in Skoog Instrumental Analysis Problem 27-22b (gas chromatography)
Problem 27-22b in:
Skoog; Holler; Crouch. Principles of Instrumental Analysis. Thomson Brooks/Cole (publisher varies by country)
asks for corrected retention volumes.
A compressibility factor, $j$ is needed to compute them.
Answers that I find online all get $j = 0.915$, and a dead volume $V_M$ of $\pu{8.3 mL}$, which agrees with the solution in the back of the book.
However they seem to use a unit conversion factor of $\pu{5.17 torr psi-1}$, which is off by a factor of $10$.
In my analysis $\frac{P_i}P$ should be ($\frac{\pu{2.727 bar}  }{\pu{0.997 bar}}$), $j$ should be $0.500$, and $V_M$ should be $\pu{4.5 mL}$.
Am I wrong?

27-22
A GC column was operated under the following conditions: Column:
  1.10 m ✕ 2.0 mm, packed with Chromosorb P: weight of stationary liquid added, 1.40 g; density of liquid, 1.02 g/mL pressures: inlet, 26.1
  psi above room; room, 748 torr measured outlet flow rate: 25.3
  mL/min temperature: room, 21.2°C; column, 102.0°C retention
  times: air, 18.0 s; methyl acetate, 1.98 min; methyl propionate, 4.16
  min; methyl n-butyrate, 7.93 min peak widths of esters at base:
  0.19, 0.39, and 0.79, respectively
Calculate (a) the average flow rate in the column. (b) the
  corrected retention volumes for air and the three esters. (c) the
  specific retention volumes for the three components. (d) the
  distribution constants for each of the esters. (e) a corrected
  retention volume and retention time for methyl n-hexanoate.

I'm not at all familiar with chromotography, could you add some calculations so that those of us not famliar are more able to chime in?

Page 197,
https://www.slideshare.net/LEANNEDESOUSA/principios-de-anlise-instrumental-6ed-skoog-resolues


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Lowercase "a" in organic chemistry nomenclature?
In the IUPAC name for geosmin has a lowercase a mixed into it:
(4S,4aS,8aR)-4,8a-dimethyloctahydronaphthalen-4a(2H)-ol.
Does anyone know what it stands for or can reference me to a source that does?

'4a' and '8a' are locants which refer to a specific atom. They are fundamentally no different from the numbers '1', '2', '3', '4'... which are also locants referring to specific atoms.
The locants in geosmin come from the locants in naphthalene, which are assigned based on the following excerpt from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.


text:
Accounting for Vinylic Carbocation instability
I read that due to geometrical constraint, positive carbons in an aryl carbocation exhibit sp^2 hybridization instead of sp^1. The empty p-orbital is forced into a hybridized position via geometrical constraints. This then explains the instability of the carbocation as the positive carbon is attached to a significantly more electronegative carbon. Understanding that carbons with more "s-character" display more electronegative behavior, please clarify the following questions:
1) What is the hybridization of the positive carbon in a vinylic carbocation?
2) Are geometrical constraints present in vinylic carbocations?
3) What makes a vinylic carbocation unstable?
Origin: Why are vinylic and arylic carbocations highly unstable?

https://chemistry.stackexchange.com/questions/10509/why-are-vinylic-and-arylic-carbocations-highly-unstable

@Mithoron The link your provided is exactly where my question is derived from. I had already read everything on it before I made the post.

Then you should have linked it yourself. BTW I don't see a proper question in your post.

Better edit your question to be actual question, and don't waste time on annoying people who can tell you how to improve it.


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Is the notion of orbitals different in theoretical chemistry?
I am just an average chemistry student that is currently exposed to theoretical chemistry. And I am seeing very puzzling things, such as 1p, 2f, 3g orbitals. In my attempt to reconcile this, I was thinking perhaps these are just so-called additional basis functions that are included in the calculations to better approximate the electron behaviour. Would that be correct?
Additionally, there is this idea of single-electron wavefunctions and N-electron wavefunctions, but no double-electron wavefunctions. Does this mean that there are only 1-electron orbitals in theories like DFT and HF, and no 2-electron orbitals?
This is rather weird because both these theories in theoretical chemistry and the the theory that we have learnt in high school (i.e. the theory of atomic structure that most students would know) are based on the same foundations of quantum mechanics. So how can they be so different? Could someone help to reconcile these differences?

Unfortunately, it only gets more complicated the deeper you dig. There is some explanation here: What exactly is an orbital?, but you should bear in mind that electronic structure theory is something that the average undergraduate student only barely touches. A strong background in QM is IMO mandatory to understand what some of these things mean. I'll see how far I can get with some of these queries, though, and some of it will be necessarily simplified.

The traditional "orbitals" that introductory chemistry teaches resemble what theoreticians would call spatial orbitals. These are distinct from spin orbitals, i.e. one-electron wavefunctions which describe both the spatial and spin orientations of the electron. The spatial orbitals only describe the spatial component.
For every spatial orbital, there are two possible spin orbitals, because you can have one electron with up spin and one with down spin. The simplest possible way to represent an "orbital" being filled with two electrons is by taking the product of the two spin orbitals, i.e. the product of two one-electron wavefunctions. So, the "two-electron orbitals" from introductory chemistry still have a mathematical form in theoretical chemistry, although it does depend on exactly what theory you are looking at (there are cases where the notion of a two-electron spatial orbital breaks down).
The reason why you don't see "two-electron wavefunctions" being discussed is because they have no real meaning in electronic structure theory. The N-electron wavefunction is useful because that is the way we choose to describe the entire molecule, and we can use it to extract information such as the energy of the system. The one-electron wavefunctions are useful because (1) they are used to construct the N-electron wavefunction (most simply by direct multiplication, or more properly using Slater determinants), and (2) we have developed methods to obtain the best possible forms of these. Two-electron wavefunctions are somewhere in the middle: they're not useful for describing the molecule as a whole, but at the same time there's no mathematical motivation for using them.

As for basis sets, it's a different ball game: we are no longer talking about what orbitals theoretically are ("theoretical chemistry"), but rather what is the best way to model them for a computer to get accurate results ("computational chemistry"). AOs themselves have no intrinsic physical meaning in a molecule, they merely serve as useful building blocks for MOs. Also, we don't even use the actual AOs as basis functions: we use things that resemble AOs as basis functions, to increase computational efficiency. So it is not too much of a stretch to add "unphysical" functions such as 1p, 2d, ... all for the purposes of getting better and/or faster results.
Also, orbitals in DFT have an entirely different meaning from orbitals in HF, but that's a story I'm not qualified to tell.

The problem with chemistry is that these theories cannot be taught to young kids - they are far too complicated and require far too much existing knowledge. And so we resort to teaching simple things, such as

"Orbitals are trajectories of electrons" - no.
"Orbitals are a region in which you have a 90% (or x%) chance of finding the electron" - no. Orbitals may be visually depicted by an isosurface which encompasses 90% of the electron density, but the orbital is not that region - it is a wavefunction.

For most people (cough organic chemists cough), simplified models work just fine. However, unfortunately, it can sometimes be a constant process of finding out that what you have been taught before is... wrong, or at best, heavily simplified. To understand exactly what justifies the simplification, though, you need to delve into the theory: and as I mentioned at the start, it is a very long journey. There are some good books in the Resources post, but again, a good foundation in QM is needed.

But spin orbitals are likely to be more accurate than spatial orbitals in approximating reality right? Since we know that Pauli repulsion is different for electrons with different spins. If we were to consider that each electron in an atom must be of different energy, i.e. no two electron have the same energy due to spin, then wouldn't our simplified view fall apart completely?

@TanYongBoon, that’s sort of what I was alluding to when I said the idea of a two-electron orbital breaks down. You can Google e.g. “unrestricted Hartree-Fock” for more details.

In UHF, would the 1s electrons of Li be at different energy levels, depending on their spins?

However, in Be, the 2 1s electrons would be at degenerate energy levels since the electronic shell is closed?

Actually, would Pauli repulsion (increases the energy of the system due to repulsion between electrons of the same spin) be more significant or exchange interaction (decreases the energy of the system due to exchange interaction between electrons of the same spin)?

@TanYongBoon, an easier way of looking at it is that all electrons repel each other regardless of whether they have the same spin or not (Coulomb repulsion), but electrons of like spins repel each other less (exchange interaction). That is how the maths is formulated, anyway, with $J$ and $K$ integrals/operators. And yes, in UHF, there would no longer be two "1s" electrons in Li.

First off, I think the other answer is excellent.
You seem to have two questions, so I’ll answer them separately:

1p, 2f, 3g orbitals. In my attempt to reconcile this, I was thinking perhaps these are just so-called additional basis functions [...] to better approximate the electron behaviour

Yes, exactly. Technically speaking the s, p, d, f functions are only accurate for the one-electron hydrogen atom anyway. Moreover, we usually use Gaussian functions for most basis sets, so we’re just attempting to approximate the real electron density in a multi-electron system.
As you said, we use higher angular momentum basis functions (like 1p, 2f, 3g, h, i, etc.) to better approximate the angular components (polarization) of the electron density. Sometimes we use higher principal quantum numbers to better approximate the diffuse properties of the electron density (e.g., 2s, 3s, 4s for a hydrogen atom)
It’s important to understand that most depictions of orbitals or electron density are given as surfaces. Here’s a better picture - the orbital density of a tetrathiophene molecule.

Does this mean that there are only 1-electron orbitals in theories like DFT and HF, and no 2-electron orbitals.

Mostly, yes. In typical DFT and HF, we use 1-electron orbitals. We force double occupation in theories like RHF and ROHF, but the orbitals are solutions to 1-electron equations in both HF and traditional DFT.
There are electronic structure methods that use 2-electron orbitals. These geminal theories work over 2-electron orbitals. They’re not widely used (yet) but offer some advantages for certain properties like bond-breaking.
As an example, Assessing the Accuracy of New Geminal-Based Approaches, J. Phys. Chem. A, 2014, 118 (39), pp 9058–9068.

Our study indicates that these new geminal-based approaches provide a cheap, robust, and accurate alternative for the description of bond-breaking processes in closed-shell systems requiring only mean-field-like computational cost. In particular, the spectroscopic constants obtained from OO-AP1roG are in very good agreement with reference theoretical and experimental data.


text:
Will number of moles (amount of substance) of a chemical change when I dissolve it’s solid state in a solvent so it becomes a solution
Let’s say if I have a known mass (10g) of an inorganic compound (NaCl) in its solid form and I dissolve it in water to form an NaCl solution and the total mass of the solution, let’s say 100g.
if I want to calculate the number of moles of the NaCl in the solution, does the number of moles change when i dissolve it from its solid to liquid form ?
Or is the number of moles is still = 10 / molecular weight

Number of Moles will not change if Solid doesn't react with solvent.
Here, NaCl is stable in water and solution will be 10% w/w. (10 g NaCl in 100 g solution).
If you want to calculate number of moles in 10% w/w solution, for 10 g of solution
Number of moles = (10 g solution/ NaCl molecular weight) / 10

No. Number of moles does not change until and unless NaCl reacts with water which it won't.
No of moles of any substance= amt of substance( in gms)/ molecular weight of that substance( in gms). If you dissolve a substance in anything, its molecular weight does not change and the amt you are dissolving does not change. It does not matter if its physical state changes or not. The no of moles remains constant. You may keep adding solvent but as long as you do not add some more NaCl into the solution your no of moles of NaCl will never change. But be careful while adding solvent since it will change your concentrations.
The reason why i mentioned that as long as there is no chemical reaction the no of moles will remain constant is because if there is a chemical reaction, there will be formation of products at the expense of ractants and we know that mass of reactants decreases and mass of products increases. Thus amt of NaCl will decrease in case of a reaction and change your no of moles in case of a chemical reaction involving NaCl.


text:
Why is the equilibrium constant being calculated differently here?
In my NMR textbook, it says the equilibrium constant for a dynamic process (e.g., A in equilibrium with B) is calculated by the concentration of A divided by the concentration of B.  Is this a typo?  I thought the equilibrium constant was products over reactants.

Unless they mentioned with referencen to the reverse reaction, it is a typo.

Can you add a ref to the statement?

What is the exact wording in the book?

"Up until now we have assumed that the two isomers I and II which are in dynamic equilibrium have equal energies. If this is not the case, the equilibrium mixture contains different amounts of I and II. The equilibrium constant K is then calculated from Equation (11-9), the relative concentrations of I and II being in most cases easily determined by integration of the corresponding NMR signals"  Equation 11-9: K=[I]/[II]

It's a matter of definition, really.  Equilibrium implies a reversible reaction, thus the same reaction could he rendered as $\ce{A<=>B}$ or$\ce{B<=>A}$.  So we can render the equilibrium constant as $[\ce{B}]/[\ce{A}]$ or as $[\ce{A}]/[\ce{B}]$.  We need to define which is used, of course, when we give a numerical value.


text:
Diluting topical cream (10%)
I need to dilute a face cream with 10 % active ingredient to 2.5 % and 5 % but I am confused as to how. I looked into the $$c_1V_1=c_2V_2$$ and $$\text{concentration of a solution}=(\text{amount of solute}/\text{amount of solvent})\times100$$ but not sure which would work.
Would I have to make a standard solution first and place 10 g of the cream into 100 ml of water?

To practically dilute this cream from $10\ \%$ to $5\ \%$, you would need to dilute it to a by a factor of $2$. You can see this by assuming an initial volume for your cream, in this case, $100\ \mathrm{ml}$,
and plugging into the $c_1V_1=c_2V_2$ equation. In this case that means
$$c_1:=10~\%~~~~~\\V_1:=100~\text{mL}\\c_2:=5~\%~~~~~~~$$
Solving for $V_2$ renders
$$~~~~~~~~\frac{c_1V_1}{c_2}=V_2$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=V_2~~~~~~~~~~$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=200~\text{mL}$$
Which means that the volume needed in the final solution needs to be $200~\text{mL}$. It is however, important to note that this does not mean that you need to add $200~\text{mL}$, but rather you need to add $100~\text{mL}$ of solvent to your $100~\text{mL}$ of cream for a final solution volume of $200~\text{mL}$.
Further you may calculate the dilution factor that this correlates to by using the equation
$$DF=\frac{V_\text{final}}{V_\text{initial}}$$
In this case $V_\text{final}=V_2$ and $V_\text{initial}=V_1$, so plugging into this equation renders
$$DF=\frac{200~\text{mL}}{100~\text{mL}}=2$$
You may apply these same equations to making a cream at $2.5~\%$ active ingrediant concentration and should find that if you start with the $100~\text{mL}$ of the $10~\%$ stock, you would need to dilute to a final volume of $400~\text{mL}$, which correlates to a dilution factor of $4$. If you start with $100~\text{mL}$ of the $5~\%$ solution you just made, you would need to dilute to a final volume of $200~\text{mL}$, which correlates to a dilution factor of $2$.
If you want to do this practically, then water may not be the best solvent – to continue to be able to use this as a cream, you may consider using a cream without any active ingrediant in it as your solvent – maybe something like a hand cream?

I once learned that "%" denotes mass-percentage. volume-percentages were indicated as "Vol%". Is that not true anymore?


text:
Explain the structure of HNC (Hydrogen Isocyanide)

What are the different types of bonds present in the structure?
Does Nitrogen donates one electron from its lone pair to carbon and acquires a positive charge?
How do I determine the Oxidation state of Nitrogen and Carbon?

In such type of questions, we can't directly determine Oxidation state by the normal algebra method.
First you have to draw the structure.
1.In a covalent bond, the more electronegative element gets a contribution of -1and the other element gets a contribution of +1 charge(valid for both sigma and pi bonds. Treats double double bond as two single bonds and triple bonds as three singles bonds).

Bonds between same element have contribution 0 to each element.
In a coordinate bond, if the bond is directed from less electronegative element to more electrononegative element, the more elecronegative element gets a contribution of -2 and the less electrononegative element gets a contribution of +2.

4.in case the coordinate bond is directed from more electrononegative to less elecronegative elements, the contribution is 0 to each element.
5.To these 4 steps for each and every bond. The value of Oxidation state of an element is sum of all contributions due to all other elements/bonds.
In your question, due to C-H bond, carbon gets -1 and hydrogen +1. Due to the triple bond between C and N, N gets -3 and C gets +3.
Summing all,
Oxidation state of hydrogen is +1
Oxidation state of carbon is -1+3=+2
And oxidation state of nitrogen is -3

Is one of the bonds between N and C, an ionic bond?

No. Only in solvents like water it disocciates into H+ and CN-.

In the structure we see that there are four bonds around nitrogen.Could you explain which type of bond each of them is?

I thin your statements three and four are wrong .They are supposed to be opposite.That is,if coordinate bond is directed from more electronegative element to less electronegative element then we should neglect the contribution of cordinate bond.                                                                                Reference: http://www.sciencehq.com/chemistry/oxidation-number.html

Oh sorry it was a typo. I have edited it now.

Does carbon have a lone to account for the negative charge?


text:
What is the configuration of dimethyl[tris(trifluoromethyl)]-λ⁵-phosphane?
I know that $\ce{(CH3)2P(CF3)3}$ is geometry of trigonal bipyramid and according to Bent's rule $\ce{CH3}$  goes to axial positions and $\ce{CF3}$s remains on equatorial positions. However, as per Bent's rule more electronegative ligand goes to axial position.
Now my question is how to determine which is more electronegative between $\ce{CH3}$ and $\ce{CF3}$?

You can note that $\ce{-CH3}$ shows electron releasing (+I) effect while $\ce{-CF3}$ shows strong electron withdrawing effect. Now, you can treat inductive effect as the organic analogue of electronegativity, giving axial positions to $\ce{-CF3}$  and equatorial ones to $-\ce{CH3}$ in accordance with Bent's rule.


text:
Does fluoride anion ever act as a strong field ligand?
In the case of $\ce{[Fe(H2O)6]^3+}$ there are $5$ unpaired electrons as water acts as a weak field ligand. But instead of fluoride acting as a weak field ligand in the $\ce{[MnF6]^4-}$ , why does it act as a strong field ligand? In what conditions does it act as a strong field ligand and a weak field ligand?

Can anybody share their thoughts on this question? Many also consider $\ce{F-}$ to be SFL with $\ce{[NiF6]^2-}$ but not $\ce{[NiF6]^4-}$

Even with [PtF4]2- behaves as a strong field ligand I think (low spin)

You're asking the wrong question. Whether F- "behaves" as a strong- or weak-field ligand in these complexes has nothing to do with fluoride itself, because fluoride is a constant factor in all of these examples. It's not some magical property of fluoride that's being turned on or off depending on what complex it's in. You should really be asking about the metal and its oxidation state.

@orthocresol Yes that's why I added another example, wherein both the central metal atom and the ligands are same, the only difference being the oxidation state of the metal atom. So my question is rather which "magical property" of the metal/ its oxidation state is being turned "on/off" in such cases?

May be it related to ligand EN and metal oxidation state?this what CFT tell us.

@SirArthur7 - $\Delta_o$ always increases with metal oxidation state in complexes with the same metal and same ligands. There are lots of conceptual ways to think about this, but all involve closer interaction between ligand and metal when oxidation state is higher.

I actually do not see a source actually calling $\ce{MnF6^{4-}$ is low spin.  Is the Emperor actually wearing any clothes?


text:
Why does a balloon filled with helium float if there is so much more air on top of it than below it?
Also what about Newton's third law? Wouldn't buoyancy be flawed because a balloon also displaces the air that is on top of it thus a downward force?

Does air pressure increase with altitude or decrease with altitude?  It is the external pressure variation acting normal to all the surfaces of the balloon that gives rise to a net upward force on the balloon.

I'm voting to close this question as off-topic because it belongs on Physics.SE (assuming it isn't duplicate)

The pressure caused by air is a function of altitude.
Think about it: If there were less pressure below the balloon, then the air around it would rush in and fill the lower pressure until it's equalized.
In other words, the air is not an isolated column above and below the balloon. There is also the air adjacent to this column which exerts pressure; effectively, there is "more air" below the balloon than above it, in terms of air pressure.
Since there is more air weighing down on the lower altitude, the pressure below the balloon is greater than the pressure above the balloon.
Speaking of buoyancy, it's the same principle to explain why objects float in water. The greater pressure below the balloon is the result of a force exerted in all directions—including upwards. The pressure pushing downward, opposing the upward force, is the weight of the air above the balloon plus the weight of the balloon itself. Helium is about 15% the density of air (0.179 vs 1.225 g/L). That volume of air displaced by the balloon is contributing less force than normal air would. Therefore, the downward force is less than the upward force, so the balloon experiences a net force upward and floats up into the sky.
If we had an object denser than air, say a block of iron, then the lower pressure above the object, plus the density of the iron, would exceed the air pressure below, and so it would fall.

You're not looking at just the column of air that the balloon is in. You also need to look at an adjacent column of air.
Now compare the weight of the two columns of air at the same height as the bottom of the balloon.
Which is greater? Obviously, the one without a balloon shaped hole in it.
If this were a regular balloon, that hole would be filled with air and the weights would be essentially the same, but it's helium so the weight due to the gas inside the balloon is a lot smaller.
So the adjacent column of air is pushing up harder that the column of air above the balloon. It therefore rises.


text:
Why do we use pV term independent of U in the equation H = U + pV?
I have read the following definition of enthalpy in my textbook:

A substance has to occupy some space in its surroundings depending upon its volume ($V$). It  does against the compressing influence of the atmospheric pressure ($p$). Due to this,the substance possesses an additional energy called $pV$ energy which is given by the product of pressure ($p$) and volume ($V$) of the system.
The sum of internal energy and $pV$ energy of any system, under given set of conditions, is called enthalpy. It is denoted by $H$ and is also called heat content of the system. Mathematically, it may be put as $$H = U + pV$$

Let's suppose we have some water in a test tube at room temperature open to atmosphere. It will have internal energy which consist of kinetic energy of  molecules, chemical energy, rotational energy and some other forms of energy. In order to occupy some space in its surroundings, the molecules of the water should vibrate in such a way that they cancel out the compressing influence of the air molecules present in the atmosphere.
So if we try to measure the internal energy of the system, isn't the energy required to make space for the surroundings already included in the internal energy of the system. Why do we add an extra term $pV$ along with internal energy $U$ to measure the heat content of the system?

This is an excellent question, which makes total sense.  In my judgment, the author is desperately trying to provide a physical interpretation to the enthalpy property.  Also, in my judgment, all it serves to do is confuse the student (for no reason).  Enthalpy is just a convenient property to use in analyzing many kinds of thermodynamic systems, rather than always writing U+PV.

Most of the books and articles on the internet use the same definition for enthalpy which is confusing me.

could you provide a ref (which textbook are you quoting?)

Like I said, there is good reason why you are confused.  I guarantee your confusion will vanish once you disregard your book's futile attempt to provide a physical interpretation of enthalpy, and, instead, merely regard enthalpy as a convenient thermodynamic function to work with in analyzing many different types of thermodynamic problems.

@TryHard this is from a 11th grade chemistry textbook from a local Indian author,but I see this defintion in all those other textbooks.

I don't know if I've managed to clarify this for you. I can ellaborate a little but I would keep in mind that a lot of thermo relations are a form of book keeping. The labels help you keep track of things. With some practice you get used to the relationships between the variables. Just be careful using a relation such as H=-TS. It is rarely used in this form, more often you'll use $\Delta U=\Delta H - \Delta (PV)$ or the differential form. Also, rem that $\Delta H=q_P$ (heat for a process at constant pressure and no nonPV/chem work).

@ Chester Miller "merely regard enthalpy as a convenient thermodynamic function to work with in analyzing many different types of thermodynamic problems. " I am trying to know why enthalpy is a convenient state function in solving thermodynamic problem.How can you define enthalpy in your own words?

Typo in my previous comment: " Just be careful using a relation such as H=TS."

Further error in my previous comment: "$H=TS+\mu N$"

I read the supposed original provided in the link. This is not a duplicate, and even if the original attempts to answer a similar question, it is about as useful as reading a wikipedia article describing the Legendre transform. The original poster is searching for physical intuition.

Try Hard is right,the question posted in the link is entirely different from what I am asking so it is unable to clear my doubt,marking it as a duplicate will make other people less trustfull towards this site so I suggest that this question should be reopened.

You are asking me how I would define enthalpy.  I would define it by the equation H=U+PV.  That's it.  Nothing more, nothing less.

@Chester Miller same question ,why do we add an extra term $PV$ along with $U$ to measure the heat content of the system when we already count that in $U$?(discussion of topics such as enthalpy might seem insignificant but it helps in further learning when we know the basic terms)

If you did work, this requires more heat.

@Chester Miller Still one confusion,when we are talking about Heat Content(enthalpy) I think we are considering  the total energy the system  has at a state ,then we should not consider the $PV$ term as this energy is spent in doing work.Or am I mistaken about my definition of enthalpy?

You are mistaken about the definition of enthalpy.  The total energy of a close system is its internal energy.  And I would very strongly encourage you not to use the term Heat Content to describe enthalpy.  Heat depends on process path, and enthalpy is independent of any process; it is a physical property of the material.

@Chester Miller So what is the physical significance of the term enthalpy?

It doesn’t need to have a particular physical significance.  It is just a convenient parameter to work with in many kinds of problems.  Why are you wasting your valuable time obsessing over this rather than doing something much more worthwhile such as getting practice solving problems.

The energy U of a system consisting of a single component and phase is
$ U = -PV + TS + \mu N$
As you rightly point out, U includes the energy required to create the volume by expanding against the surrounding (constant) pressure, namely the -PV term. Removing this contribution gives H.
Therefore
$ H = U + PV = TS + \mu N$
Equating TS with the "heat content" of the system leads to the statement in the textbook (this is my interpretation of this textbook's author's meaning). This kind of algebraic manipulation can lead to all sorts of confusion.
EDIT
Some people recoil (and downvote) when they see an expression such as
$ U = -PV + TS + \mu N$
as if this were somehow unphysical. Well, it does suggest that there is such a thing as an absolute value of U rather than only relative values, and it is hammered into our heads in school that there are only relative, not absolute, energies. But the point here is that there is an absolute value, relative to an empty system, and you can use N (the number of particles) as your simple scaling constant. The expression above captures the extensive property of U. You could equally write
$ \Delta U = -P \Delta V + T \Delta S + \mu \Delta N$
and no one will blink. Well, if you define
$\Delta U = \Delta N \times U_m$
$\Delta S = \Delta N \times S_m$
$\Delta V = \Delta N \times V_m$
with
$\Delta N = N-0$
you obtain an expression equal in spirit to the sacriligeous one above.
Usage of terminology such as "heat content" is potentially confusing, but we can use similar arguments to those used in explaining the PV term's meaning as work to create the system from scratch by increasing the particle number, while keeping intensive properties such as T and P constant, to suggest that the term TS is the heat that must be exchanged with the surroundings during creation of the system (from 0 to N particles). This is more of theoretical than experimental interest, in the same way we can hypothetically charge an object from 0 to a finite charge to compute its energy (relative to no charge).
Note that saying that H represents the heat content is confusing, since the application of the equation is not usually in the spirit I have just outlined, and if you look at
$ H = U + PV = TS + \mu N$
there is an additional term $\mu N$ term which cannot be easily associated with the concept of "heat content".
So, being generous, I would think the author of the book meant by the statement was, very loosely, that changes in H, that is $\Delta H$, are usually associated with the heat exchanged by a system at constant P (subject also to other restraints but more on that in other questions and answers referring specifically to the meaning of H).

Comments are not for extended discussion; this conversation has been moved to chat.


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Aldol condensation in acidic medium

Attempt:
I did aldol condensation in acidic medium followed by $\ce{E1cB}$ to get 1,3-diphenylbut-2-enone but the answer given is:

I am unable to devise a mechanism to reach the above product from the reactant. Could someone please suggest the mechanism?

Do another aldol onto the ketone of your 1,3-diphenylbut-2-enone

I have drawn a three step mechanism as opposed to the concerted step reaction proposed in the other answer.

This mechanism has the added benefit of not leaving any room for formation of any side products as there is only one source carbon of acidic hydrogens in each compound.

The only mechanism I see is, you write the three molecules in enol form (acidic medium) and try to form a cyclic compound in a concerted step as you did in forming benzene from acetylene and then as it is acidic medium, you gotta eliminate water molecules from the new cyclohexane to end up at the product.


text:
Optimum pH range for formation of Schiff bases
What is the optimum range of pH for the formation of Schiff Bases by the addition of hydroxylamine to an aldehyde or a ketone?
According to Peter Sykes' A Guidebook to Mechanisms in Organic Chemistry,
"The initial addition step or the dehydration step can be made rate limiting at will, depending on the pH of the solution. At neutral and alkaline pHs it is generally dehydration, that is the slow and rate limiting step while at acidic pHs it is the initial addition which is slow and rate limiting. This clearly has significance in preparatory terms, and formation of such derivatives of carbonyl compounds tends to exhibit pH optima."
My doubts are :

How does the pH change the rate limiting step? That is, why is dehydration the rate limiting step at neutral and alkaline pHs and why is addition the rate limiting step at acidic pHs?
What is the significance in preparative terms that Sykes is taking about?
How will this significance determine the pH and what is that pH (or the range of pHs)?

Dehydration requires protonation of the hydroxy intermediate, this clearly is slower at neutral or alkaline pH. Hydroxylamine is basic so at acidic pH it protonates making it non-nucleophilic hence lower pH slows the rate.
Time taken for the reaction to go to completion. If the rate is slowed then other competing reactions (e.g. self-condensation) can and do reduce your yield.
The pH will vary for each substrate and needs to be experimentally determined. This is where a literature search saves you a lot of grief.

Can you tell me how to find the relevant literature? I am new to this and can't seem to find any papers or any book on the web that carries the information I need. I specifically require the pH when the substrate is acetaldehyde.

This depends on the resources available to you. Do you have access to online databases such as Scifinder or Reaxys? If not a Google search on "acetaldehyde oxime synthesis" turns up multiple references to papers, patents and procedures such as this http://www.prepchem.com/synthesis-of-acetaldoxime/  which looks perfectly viable and contains further references

Thanks for the link. While it doesn't mention the pH, it does have further references. How do I access the referenced papers? I could not find the paper by searching its name on Google.

You need some form of subscription through your college library/employer to access American Chemical Soc/Royal Society of Chemistry papers.


text:
What happened to this stainless steel tool when exposed to vinegar?
Based on an article I found online about how to disinfect hygiene tools I've been soaking nail clippers and similar items in a mixture of white vinegar, dish soap, and warm water for a few hours in a plastic sandwich container.
So far it's been fine, but recently did it with a stainless steel cuticle nipper and after taking it out and rinsing it and drying it off, the metal had turned a dark gray color it's giving off a foul odor.
What happened to the metal, and is the tool still safe to use?
For reference, here is a photo of the tool (on top) and a different one showing what it used to look like (bottom).

From what material was the container holding the solution made?

Updated to reflect that it was a plastic sandwich container. Side note: I only use the container for this and will never use it for food storage.

A plastic container rules out a galvanic reaction with the container.   It looks like the acid in the solution dissolved the stainless steel coating.  Not all coatings or stainless steel is the same so it is hard to say exactly what happened.  If the tool smells bad then get rid of it.  Bad smells are one way our body warns us that something is not safe   Homebrew solutions involving mixing cleaners is risky.  Scalding hot water (or steam) is a much safer way to disinfect tools because it will not leave a harmful residue.

What grade of stainless steel was the clipper made of? Was it labeled stainless steel, or is this an assumption because it looked shiny or similar to some other corrosion-resistant object (which may itself be stainless steel or not)?
Test with a magnet: if they are non-magnetic, the clippers could be stainless steel. Without knowing more about the source of the clippers, I could guarantee you that they are not stainless steel - after, they were not stain-less even in vinegar.
It is difficult to imagine any kind of heat treatment or surface treatment (short of plating with a more active metal) that could result in this kind of generalized corrosion.

(+1) Good answer. I would bet the clippers are an aluminum or zinc alloy, maybe plated with nickel. If they are really stainless steel, they are awful.

This answer covered it very well . I could add that martensitic and ferritic stainless steels are magnetic , but these stainlesses would not be darkened or otherwise affected by your cleaning. Possibly there is some deposit from the cleaning solution. Another guess is German Silver, actually a brass made silver color by addition of several per-cent nickel , But I don't even like that guess.


text:
Mirror planes of symmetry in bicyclo[2.2.2]octane?
One of the questions in my book has answer given as one mirror plane of symmetry perpendicular to the 3-fold rotation axis, but I think it has 3 more, each parallel to the 3-fold rotation axis as I have attempted to show below.

This picture is really bad... and not useful.

Gah, I just deleted a comment about it, because @A.K. already told it in his edit, and now question isn't unclear, but is telling something you hadn't known...

I would speculate that the ring has a twist to it which would mean it has no mirror symmetry. Either way you are either right and the answer is wrong or you are both wrong and the molecule has a twist in it. If it were Barrelene, then I would agree you have additional mirror planes, but more information is needed.

Even if the bridges aren’t flat, they are conformationally flexible enough such that on a sufficiently long timescale they behave as if they are flat. Whether that counts as being a mirror plane is up to interpretation... see eg https://chemistry.stackexchange.com/q/61107/16683 Here the barrier to “flipping” is definitely much lower than in the cyclohexane case.

@orthocresol I guess that depends on what you want to do with the mirror planes.


text:
Is there a thermodynamic driving force for racemisation?
Why does racemisation occur?
My doubts are:

Is it due to entropy?
If it is due to entropy, how is the reduction of optical activity by racemisation a consequence of increased thermodynamic stability?
Will the entropy increase or decrease? I have learnt in thermodynamics that it is increase in entropy that favours thermodynamic stability.

No the racemisation occurs because of thermodynamic instability. In other words there is enough thermal energy to allow racemisation.

Yes, there is, and it’s called entropy. @MaxW I’m not sure what you’re trying to say. If something occurs because of instability then it follows that the products are more stable than the reactants.

@orthocresol My only hesitation in saying yes to the question is just to make clear that it is not the reduction in optical activity causing thermodynamic stability, but that the reduction is a consequence of it being more thermodynamically stable (due to entropy) to have a mixture, which happens to be optically inactive.

@orthocresol  I have edited my question after reading your comments to include what remains as my doubt

Having thought about it a little (unfortunately, I don't always have time to write answers on the spot), I'm finding it surprisingly difficult to formalise why a racemate has a larger entropy than an enantiomerically pure compound. It is intuitive, to me at least - the best parallel I can draw is to the entropy of mixing of two ideal gases - but I'm not sure how to do the maths. Your second question is basically "why do thermodynamically favoured processes occur?", to which I cannot give any real answer apart from the Second Law of Thermodynamics. And your third question answers itself.

(@orthocresol, here's my attempt at formalization, if you're interested.)

Thermodynamics is not usually helpful in understanding racemisation: think mechanisms and kinetics instead
The thing about enantiomers is that, from a thermodynamic point of view, they are the same so any process is not being driven by differences in the energy between the molecules.
What matters, if racemisation is to occur, is that there is some accessible pathway that allows interconversion. For example, in acidic or basic ethanol solution, (3R)-3-phenyl-2-butanone will racemise via the (achiral) enol form of the molecule. There is a small amount of the enol present which interconverts to the (chiral) molecule but without remembering which chiral molecule the enol was formed from. This will, ultimately, give a racemic mixture. If no such mechanism existed the molecule would not interconvert. What matters is that some such pathway exists.
The notorious medicine thalidomide is an interesting example. One enantiomer is a useful therapeutic, the other a dangerous teratogen. But making a pure enantiomer doesn't help as the two forms interconvert in the body (by a complicated mechanism).
These conversions are not driven by the different stabilities of the end products but by the inability of non-chiral intermediate molecules in an equilibrium with the chiral starting materials to remember the chirality of their progenitor. This, in effect, increases entropy. But that entropic force is only possible if there is an available reaction mechanism that passes through a non-chiral form allowing the increased disorder that results from a chiral starting material turning into a racemic mixture.

Thanks a lot. While I understood most of your answer, I think I don't understand how inability to remember chirality of progenitor molecule the reason for increasing entropy. Please say more on that

@AnubhabDas Imagine the non-chiral compound A interacting with B to regenerate the original chiral molecule C. Which enantiomer you get depends on which orientation (or which "side") of A the second molecule B approaches from. But that will be random in most situations meaning that a random mixture of the enantiomers of C will result.

Racemization can indeed be understood from an entropic perspective. While an activation energy barrier exists for spontaneous racemization (hence the importance of kinetics), it is ultimately the increase in entropy that drives racemization.
A simple model: consider a series of $N$ coins, and label heads $H$ and tails $T$. A solution of enantiomers corresponds to a sequence $$\underbrace{HHH\cdots HHH}_{N\;\text{occurrences}},$$ with $N \sim 10^{24}$. Racemization corresponds to picking up a random coin in your sequence and flipping it over. Kinetics determines how quickly you can pick up and flip coins. Eventually, if the kinetics are fast relative to laboratory timescales, we expect some equilibrated sequence like $$\underbrace{HTT\cdots HTH}_{N\;\text{occurrences}},$$ which corresponds to a racemate. The original enantiomeric solution corresponds to just one sequence of coin flips, and the equilibrated racemate to $2^N$ possible sequences. The increase in entropy is then $$\Delta S = k_\text{B}\ln\frac{2^N}{1} = Nk_\text{B}\ln 2.$$ (We could instead have modeled this scenario as doubling the volume of an ideal gas, with the same results. I find coin flips more intuitive.)
The optical activity of your solution corresponds to the excess number of heads over tails in the average sequence. In the enantiomeric solution all our coins are heads, so we have $N$ excess heads. In the racemate we have $0$ excess heads on average.

There are also a few statements from matt_black's answer that I'd like to clarify.

While both an enantiomeric solution and a racemate have the same internal energy $U$, they do not have the same free energy $G$ (or $F$), because the free energy depends on entropy.
The entropic force for racemization is always present regardless of its rate. Kinetics controls this rate, but cannot affect the thermodynamics. I dislike calling entropy a force, because it's really statistical in origin, but that's just me.

This answer surely completes my question. I really wish I could accept more than one answer. Well, if I understood correctly, you said that the racemate has equal number of heads and tails. Will that not mean than any N/2 flips will give heads and rest tails? That will lead to the number of possible sequences as N choose N/2 because all we have to do is choose which flips are head and rest will be tails.

@AnubhabDas, the racemate will have equal numbers of heads and tails on average. It is not a constraint: each coin can independently be heads or tails.

In addition to @matt_black 's answer, I think it's prudent to add some mechanisms that have partial racemisation instead of complete racemisation, which would have happened in case entropy completely determined the reaction products.
In an $S_N1$ reaction, we have a possibility for the nucleophile to attack the carbocation from either side. However, there are multiple cases where the carbocation doesn't form before the attack of the nucleophile. In these cases, there's a marked difference between the two chiral forms.


text:
Does Cyclooctatetraene decolourise Br2 water and why?

I am thinking yes because it is not aromatic according to Huckle's rule. However, why in terms of orbitals etc?
I am a first year and so I am new to these concepts.


text:
AFM of Smooth Amorphous Carbon
I have been having trouble getting decent resolution images and clear data of smooth amorphous carbon thin films prepared by RF bias assisted vacuum deposition. I was using a DLC tip in tapping mode and was able to get some data but is not exactly repeatable, with many more films to analyse acquiring results has been difficult and time consuming. With rms values ~0.125 and variation of hybridisation from CDE contour plots this may be an awkward task with the old machine we have and the point to point variation of conductivity.
Are there are suggestions for tips, scan parameters, scan modes etc. I am no expert with AFM by any stretch so any advice would be helpful
Much appreciated!


text:
Constructing two body Hamiltonian for Helium atom in hydrogen slater orbitals single particle basis
I am trying to construct a two body states to calculate the matrix elements of electronic Hamiltonian of Helium atom. $$H=-\frac{1}{2}\nabla^{2}_{1}-\frac{1}{2}\nabla^{2}_{2}-\frac{2}{\vec{r}_{1}}-\frac{2}{\vec{r}_{2}}-\frac{1}{|\vec{r}_{1}-\vec{r}_{2}|}$$
Constructed the following two body (slater determinant) states (spinless fermion so spin states are excluded), $$|\psi_{1s}\psi_{2s}\rangle, |\psi_{1s}\psi_{2p_{x}}\rangle,|\psi_{1s}\psi_{2p_{y}}\rangle,|\psi_{1s}\psi_{2p_{z}}\rangle,|\psi_{2s}\psi_{2p_{x}}\rangle,|\psi_{2s}\psi_{2p_{y}}\rangle,|\psi_{2s}\psi_{2p_{z}}\rangle,|\psi_{2p_{x}}\psi_{2p_{y}}\rangle,|\psi_{2p_{x}}\psi_{2p_{z}}\rangle,|\psi_{2p_{y}}\psi_{2p_{z}}\rangle$$
where each ket is actually slater determinant for example,$$|\psi_{1s}\psi_{2s}\rangle:=\frac{1}{\sqrt{2}}[\psi_{1s}(\vec{r}_{1})\psi_{2s}(\vec{r}_{2})-\psi_{1s}(\vec{r}_{2})\psi_{2s}(\vec{r}_{1})]$$
I am trying to construct the matrix elements for the above hamiltonian with 10 two body slater determinant states (i have used only $1s$,$2s$ and $2p$ single particle states).While constructing the matrix ($10\times10$ matrix), I have a problem in evaluating the coulomb repulsion term. $$\int\frac{1}{\sqrt{2}}[\psi_{1s}(\vec{r}_{1})\psi_{2p_{x}}(\vec{r}_{2})-\psi_{1s}(\vec{r}_{2})\psi_{2p_{x}}(\vec{r}_{1})]\frac{1}{|\vec{r}_{1}-\vec{r}_{2}|}\frac{1}{\sqrt{2}}[\psi_{1s}(\vec{r}_{1})\psi_{2p_{x}}(\vec{r}_{2})-\psi_{1s}(\vec{r}_{2})\psi_{2p_{x}}(\vec{r}_{1})]d^{3}\vec{r}_{1}d^{3}\vec{r}_{2}$$
How do I evaluate those integrals if not analytically, How do i do numerically this six dimensional integration. Can someone suggest a numerical technique to evaluate such an integral or a python code or C code ?

Is there any reason why not analytically? And what about the 1-electron integrals? The basic principle to solve 1- and 2-electron integrals are the same. So maybe start there.

I really don't have any idea about solving it numerically. I am completely new to numerical integration . IF you suggest some already existing codes in python or C it would be nice and be very helpful. I thought there would be no analytical solution.In this case one electron integral is very simple.

Packages based on analytic expression for GTOs are already linked in one of your previous questions: https://chemistry.stackexchange.com/questions/103409/solving-two-electron-integral-numerically
If you want to learn about numerical integration, I would start with simpler things, the harmonic oscillator in 1D to 3D for example.

Can you please suggest some book or some reference for that ?

Again, please refer to the references in your previous question. As far as I understand you, the two questions are identical. Which makes this question either a duplicate or "unclear what you're asking".

@Tyberius your "possible duplicate" comment is a duplicate ;)

@Feodoran if that were a comment flag, I would flag myself.

@Tyberius what is a "comment flag"?

@Feodoran you can flag a comment (usually located on the left under the upvote icon) as spam, abusive, obsolete, or for some other moderator intervention. I was just joking that I wouldn't be able to mark my comment as duplicate of yours.


text:
What does FeCl2 mean?
I'm doing a lab for a class , and in the manual, they right the compound as Fe(Cl)$_2$. I looked up that compound, and it has a very high boiling point, but the compound during the lab was already in a liquid state, so I'm pretty confused. Specifically what was created was made from dissolving iron in HCl in a boiling water bath. I don't know much about chemistry, so if you can try to dumb it down, that would be much appreciated!

Why does a high boiling point disagree with it being a liquid or dissolved in solution? If the boiling point is high, that means it will likely be a liquid if the temperature is lower than that (or a solid if the temperature is even lower). @user69873

It should be labeled "FeCl2, water solution".  You have a water solution.  Like most metal chlorides, ferrous chloride dissolves readily into water.
Keep it in an airtight container.  Ferrous ion in solution gets oxidized easily.


text:
Celsius to Kelvin
A temperature conversion question asks the following:
"The temperature of a sample of a substance changes from 10°C to 20°C.  How many Kelvin degrees does the temperature change?"
The correct answer is 10 K, but my question is, why wouldn't you just convert the difference (10°C) to 283 K?
Thanks in advance.

Let's put it this way. The temperature of a substance was 10°C; how many K was that? Then it changed to 20°C; how many K is that? Now what is the difference?

Degrees K are the same size/value as degrees C. The question is asking about the change not the absolute value.

The point is, we are measuring $\textbf{change}$ in temperature. Not the temperature itself.
Celsius scale has 100 divisions from 0 °C to 100 °C. Kelvin scale also has 100 divisions from 273 K to 373 K. The length of each scale is same. So 100 divisions of Celsius scale = 100 divisions of Kelvin scale or 1 division in each scale is equal.
So ∆°C = (100 − 0)/100 = 1. And ∆K = (373 − 273)/100 = 1 are their change in temperature per division which is same for both.
So ∆°C = ∆K(change)
And K = °C + 273.


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Exceptions to the Nitrogen Rule in mass spectrometry
The Nitrogen Rule in mass spectrometry states that (organic) molecules containing no or an even number of nitrogen atoms will have even masses, and molecules containing an odd number of nitrogen atoms will have odd masses.
Examples (all monoisotopic masses):
No nitrogen, even mass:
H2, 2 Da
HBr, 80 Da
C2H6, 30 Da
One nitrogen, odd mass:
NH3, 17 Da
C5H5N, 79 Da
 etc.
Often there is a little disclaimer saying that the nitrogen rule is only valid for molecules that do not contain "exotic atoms." What are some examples for molecules do not follow the nitrogen rule?

Just look at periodic table. P and Cl have odd masses.

@MaxW - Yes, but P tends to be tri or pentavalent, so that cancels out. Consider phosphoramide $\ce{H6N3OP}$, mass 95 Likewise with Cl - its monovalent. Chloramine $\ce{H2NCl}$, mass 51 (and 53). What you need is an even valent atom with an odd mass or an odd valent atom with an even mass (like N).

Vanadocene, Manganocene and Cobaltocene all seem good bets, taking advantage the fact that all the first series transition metals have a +2 valence state, and then picking the odd massed ones. Europium II is another place to look.
In main group chemistry it's harder.

The reason for the rule is that, outside of transition metal complexes, stable free radicals are relatively rare. If all electrons are paired and we are talking about molecules, not ions, there must also be an even number of protons. Thus an odd number of neutrons must be present to arrive at an odd molecular weight, and the only elements with the most abundant isotope having an odd number of neutrons are beryllium and nitrogen, although dysprosium is kinda close.
In an undergraduate lab you wouldn't be given an unknown with beryllium in it because it's so dangerously toxic, so that leaves you with nitrogen unless you have a deuterated sample.
But there are stable free radicals like nitric oxide (neurotransmitter) or chlorine dioxide (fumigant) that violate the rule. Molecular oxygen doesn't count because it's a diradical and so has an even number of electrons.
EDIT: I spoke too soon about abundances of nuclides with an odd number of neutrons. Here is a table of such nuclides with abundance $> 20\%$ taken from http://atom.kaeri.re.kr/nuchart/ :
$$\begin{array}{rl} \text{Nuclide} & \text{Abundance} \\
\hline
\sideset{^{9}}\ {Be}& 100\%\\
\sideset{^{14}}\ {N}& 99.636\%\\
\sideset{^{105}}\ {Pd}& 22.33\%\\
\sideset{^{129}}\ {Xe}& 26.4006\%\\
\sideset{^{131}}\ {Xe}& 21.232\%\\
\sideset{^{163}}\ {Dy}& 24.896\%\\
\sideset{^{167}}\ {Er}& 22.869\%\\
\sideset{^{195}}\ {Pt}& 33.78\%\\
\sideset{^{207}}\ {Pb}& 22.1\%
\end{array}
$$
Of these $\sideset{^{195}}\ {Pt}$ is the most abundant isotope of $Pt$ in addition to $\sideset{^{9}}\ {Be}$ and $\sideset{^{14}}\ N$ previously mentioned.


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Please explain the reason why E2 mechanism occurs
I known as that when the weak bases are reacting with 2° alcohol, E1 mechanism occurs. But, in the picture, E2 mechanism is occurred. Why E2 mechanism occurred?
In step [3]
Weak base : pyridine

Did I get it wrong? It is smith organic chemistry 4e, 309 page


text:
Nomenclature of Coordination compounds containing nitrogen
As Per rules , amine when present as a ligand is written as "ammine" with double m's , so a compound having a complex ligand such as ethane-1,2-diamine , will be written with 1 'm' amine or double 'm' ammine?

As per rules, double m is used when the ligand is just $\ce{NH3}$, ammonia:
Cite from IUPAC Nomenclature of Inorganic Chemistry, Recommendations 2005:

IR-7.1.3 Representing ligands in additive names
Additive names are constructed by placing (sometimes modified) ligand names as prefixes
  to the name of the central atom. For anionic ligands, the anion endings ‘ide’, ‘ate’ and ‘ite’
  (see Section IR-5.3.3) are changed to ‘ido’, ‘ato’ and ‘ito’, respectively,
  when generating
  these prefixes. Names of neutral and cationic ligands are used unchanged, except in a few
  special cases, most notably water (prefix ‘aqua’), ammonia (prefix ‘ammine’), carbon
  monoxide bound through carbon (prefix ‘carbonyl’), and nitrogen monoxide bound through
  nitrogen (prefix ‘nitrosyl’) (cf. Section IR-9.2.4.1).

E.g. in $\ce{[CoCl(NH3)5]^{2+}}$ pentaamminechloridocobalt(2+).
For amines, the ligand name is unchanged, e.g.
$\ce{[Co(en)3]^{3+}}$ tris(ethane-1,2-diamine)cobalt(3+).


text:
Acetal Formation Mechanism
In this example, why is there is a ring formation with the attack of the OH from the intermediate itself rather than a continuation of addition? I know my proposed product is wrong, however I just want an explanation as to why this occurs.

Careful! The connectivity in the diol changed half-way through drawing that mechanism (I see an oxygen connected to a carbon that is all of a sudden connected to 3 carbons). In addition, in the second line, the step between structures 2 and 3 is not a step; those are resonance structures, not separate species.

In addition, it is likely that the primary hydroxyl group adds to the protonated aldehyde prior to the secondary hydroxyl.

Intramolecular reactions are always preferred over intermolecular.
Remember that these reactions are reversible so even if the intermolecular attack proceeded, the product would be unstable with respect to intramolecular attack by the free OH that is within the molecule.

Why would the product of the intermolecular reaction be unstable?

It is unstable with respect to intramolecular reaction, otherwise it is stable. Under the reaction conditions the oxygens of the acetal will continue to protonate, since the free OH within the molecule is adjacent it is more likely to react than the OH of another molecule.

I support Waylander's answer above--the reaction proceeds to form a ring because an intramolecular reaction is favored over an intermolecular reaction. However, I would like to clarify that this is mainly due to kinetic reasons and not due to thermodynamic favorability as suggested before ("Intramolecular reactions are always preferred over intermolecular"). The oxygen in the hydroxyl group will attack the positively charged carbon atom at a higher rate than an oxygen atom in a hydroxyl group on another diol molecule because of the former's proximity (intramolecular vs intermolecular proximity).

That's not correct; thermodynamic favourability is definitely at work here. Entropy favours a reaction converting two molecules (carbonyl + diol) to one molecule (acetal), rather than one with three molecules (carbonyl + 2 diol) to one molecule (acetal). As was already pointed out, the reactions are reversible, so the outcome is likely controlled by this thermodynamic consideration rather than the kinetic consideration (which you are correct in mentioning; it's just not the crux of the issue here).


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Mechanism of Dow's process
Does Dow's process follow Benzyne mechanism or ArSn2 mechanism?In some books I have seen that Dow's process to convert halobenzene to phenol follows ArSn2 mechanism.But nucleophilic substitution in benzene ring mostly follows benzyne mechanism.

Benzyne Mechanism (more specifically, elimination-addition mechanism) are followed only in the cases where we use nucleophiles, which can also act as strong bases. Whereas, Addition-Elimination mechanism is followed in all the other cases. $\ce{OH-}$ is not  a strong enough base to perform Substitution reaction by Benzyne mechanism. It generally undergoes $S_N Ar$ (Addition-Elimination) Mechanism.

Thanks Soumik.Your answer is quite satisfactory.

But isn't OH- a strong base? In such a case it would form a benzyne intermediate, right?


text:
Simple question about mixing two electrolytes
Suppose we have two 0.1M solutions in water. What is their conductivity and how will it change when we mix them, and why?
A) NaCl and KCl:
Okay, so both are strong electrolytes and conduct well but what happens when they get mixed and why?
B) CH3COOH and NH3*H2O:
Both are weak electrolytes and will conduct electricity. Again, I don't know what happens after mixing and why?
C) HCl and NaOH
Both strong electrolytes, they neutralize each other in a reaction. So, the result is no conductance (or very weak?)
Thank you for your answers.

The first case is pretty straight forward. As you mentioned, both $\ce{NaCl}$ and $\ce{KCl}$ are good electrolytes.
Assuming the $\ce{NaCl}$ solution is there in a beaker, and you are adding $\ce{KCl}$ solution from a burette, the conductivity of the solution will not change significantly. This is because although two strong electrolytes are added, the number of ions per unit volume remains same as both solutions have the same concentration. However, none of $\ce{NaCl}$ and $\ce{KCl}$ will dissociate completely because of the common ion effect. This effect will, however, not be significant.
The second and third cases are best analysed using conductometric titration curves.
Let's consider the case of $\ce{HCl}$ and $\ce{NaOH}$ first. I will assume $\ce{HCl}$ is present in the beaker and $\ce{NaOH}$ is being added from the burette. The $\ce{OH-}$ ions being added will neutralise the $\ce{H+}$ ions from the solution immediately and will form the weakly ionising electrolyte $\ce{H2O}$. However the $\ce{Na+}$ ions and $\ce{Cl-}$ ions will remain in solution.
So, effectively, $\ce{H+}$ is being replaced by $\ce{Na+}$. Since the ionic mobility of $\ce{Na+}$ is lower than the ionic mobility of $\ce{H+}$, the conductivity of the mixture decreases. However, the phenomenon reverses on reaching the equivalence point. After the equivalence point, there is no more neutralisation of $\ce{H+}$ and $\ce{OH-}$ and they remain dissociated in solution on further addition of $\ce{NaOH}$. So, the conductivity increases from this point until $\ce{NaOH}$ is added.
Here is the conductometric titration curve

The final case is that of $\ce{CH3COOH}$ and $\ce{NH4OH}$. Again, I will assume that $\ce{CH3COOH}$ is present in the beaker and the $\ce{NH4OH}$ is added from the burette.
This case is a little different from the previous case. The neutralisation of $\ce{H+}$ and $\ce{OH-}$ will occur in this case too. There will be an initial decrease in conductivity as there is decreased in dissociation $\ce{CH3COOH}$ due to common ion effect. However, after that, there will be a rapid increase in the conductivity till the equivalence point is reached. This is because $\ce{NH4OH}$ converts the weak electrolyte $\ce{CH3COOH}$ to $\ce{CH3COONH4}$ which is a very strong electrolyte and ionised almost completely. After the equivalence point is reached, the conductivity of the mixture becomes indifferent to further addition of $\ce{NH4OH}$ as $\ce{NH4OH}$ is a weak electrolyte and hence, does not contribute to conductivity on further addition. Here is the conductometric titration curve -

I think:
A) When you mix both you have better conductivity by increment of ions in solutions.
B) none conductivity When you mix both you have a amine, then When you mix both the conductivity decrease.
C) When you mix both the conductivity increase (little) because you have complete disossiated ions (Na+ and Cl-) in the solution.

I object. In (A), when you mix two 0.1M solutions, you don't obtain a solution with higher concentration of ions, hence no increase. (B) is wrong because you won't be getting an amine in ambient conditions. (C) is wrong for more subtle reasons.


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What are solvents that can dissolve sebum that is safe for topical use?
I do not know much about these properties and I need some feedback regarding these.
Sebum is a semi-fluid that in humans contains a mixture of triglyerides, cholesterol, ester waxes, and squalene. 57.5% of sebum is triglycerides and fatty acids. The next most abundant molecules are ester waxes, which comprise 26% of sebum. Squalene makes up 12%, and cholesterol is the least abundant lipid in sebum.

Depend, for what. I think DMSO can use it for dissolve. PD. Sorry my poor english.

Dissolving sebum (the oily substance that protects and waterproofs our skin) sounds like a neat chemical way of removing it - a way to remove skin oiliness. But the solvents that dissolve sebum in a test tube are likely to be too aggressive for extensive use, or to have an oily nature in themselves.
Of the safe and common solvents that might be useful, isopropanol (91% if available, or 70%) would be somewhat invigorating - if not used in excess. It will almost certainly dry the skin, perhaps too much - so that you resort to applying a skin lotion to provide some oiliness.
An alternate way of removing sebum from skin would be to emulsify it with a detergent. Sodium stearate is a common, safe detergent (actually, a soap) which is somewhat soluble in water and available commercially (Ivory Soap is 99.44% pure!). Applied to a washcloth, it will remove much of surface sebum. Rinse, lather, repeat.
Some chemical and cosmetic improvements have been made to other products: added colors, fragrances and acids to reduce the pH from neutral to slightly acidic, for a gentler action on skin; added water for semi-liquid soap (Soft Soap).

thank you this is really helpful for me. A bit far fetches but, do you have any ideas on what to use on the hair that will dissolve the sebum without clogging one's hair follicles?

Not sure how technical you want to get, but if the commercial shampoos do not appeal to you, how about sodium stearate (in solution may be easier to handle)? The key is lather, rinse, repeat. Each time you do this, you may remove 80-90% of the oils from hair. Finally, your hair will be squeeky clean - you can feel it. Probably any of the commercial soft soap handwashing liquids will do as well.


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Why is CO2 not in the C∞v point group?
It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group.  However, in all the character tables I can find, it's listed as a D∞h.  What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?

In short, the two O atoms are equivalent, but $C_{\infty h}$ fails to acknowledge this, so you add more symmetry elements that switch these two atoms, and so you end up with $D_{\infty h}$ .

Okay, I almost got that.  What symmetry operations would switch the O's which wouldn't apply to a C∞v (assuming that's what you meant?) group?

Why, many. Think of all these perpendicular $C_2$ axes, to begin with.

Oh, of course.  Duh, my stupid.  Sorry, and thanks!  If you have time to post a complete answer, I'll accept.

Crazy how it is; now it's clicked I can't believe I didn't get it before.  :)

For concrete examples, consider the two linear species: $\ce{OCO}$ versus $\ce{OCN-}$.

Gotcha.  So linear H3+ would be $D∞h$, since there is a horizontal plane of symmetry through the centre of the middle hydrogen?

Both are linear with a $C_{\infty}$ axis, but $D_{\infty h}$ has a center of inversion and $C_{\infty v}$ does not.

And that all-important $sigma_h$ mirror plane perpendicular to the rotation axis, giving the name of the point group.

@GeoffHutchison - Yes but for a linear molecule you don't need to identify a center of inversion, $i$, and a mirror plane perpendicular to the rotation axis ($\sigma_h$). For a linear molecule, if the center of inversion exists so must the perpendicular  mirror plane, and if the perpendicular  mirror plane exists so must the center of inversion.

Of course - I just wanted to point out the mirror plane because it’s in the name of the point group. Exactly as you say, you can deduce one by the other.

It will be easier to understand by giving an example of molecules:
$D_{\infty h} \to \ce{CO_2}$,
$C_{\infty v} \to \ce{HCN}$.
Both are linear molecules, however $\ce{CO_2}$ has an inversion, $\ce{HCN}$ does not.


text:
Alcohol protection with ethers
Various ethers are used for hydroxyl protection. My question is about their removal. I know that there are various methods but I want to specify on the use of acids.
Certain groups are removed easily like Trityl, by using aqueous acetic acid. If I'm not mistaken, the mechanism is Sn1, as the trityl carbocation is stabilised through delocalisation on the phenyl groups.
However, other similar groups like benzene or PMB are not removed that easily. Can't they follow the same logic, as the postive charge could again be delocalised in the aromatic ring? Is it because the delocalisation is not good enough, as in the case of trityl?
Also, I've noticed that in order to remove smaller groups like methyl, haloacids are mainly used. Why can't a more weak acid do the job?

PMB can be removed under acid conditions, but it needs strong acid as the benzyl cation is not as stable as the trityl - 20% TFA in DCM is the standard in our labs, though hydrogenation is a better and milder way. Phenyl ethers are not protecting groups,  PhO- is a decent leaving group. To remove OMe you first need to protonate it and for that you require strong acid and you also need a good nucleophile as the counter ion.

@Waylander My queastion about PMB or Bn is why it can't be removed by a weaker acid since the positive charge that would be formed (Sn1) can be delocalised just like in Tr removal. I didn't mention phenyl but benzene group as protecting. As for the OMe I asked why can't a weaker acid do the job. I'm already familiar with what is being used in general. The why is what

A weaker acid will not protonate PMB or Bn

What would obstruct it? Sterically they are more open than triyl.

Protons are not really affected by steric factors being quite small

I had in mind some kind of bulky acid. Why would a weaker acid not suffice then?

Let us continue this discussion in chat.

You are right.In case of benzene and PMB, the delocalization is not good enough. So a strong acid is required.
As for the case of OMe, a strong acid as well as strong nucleophile will be needed. A weaker acid can't do the job. Let me explain why.
Note that carbon oxygen bond is quite strong.
So what we need is a strong nucleophile. But in case of OMe as a protecting group, even a stronger nucleophile will not work if the oxygen is not heavily protonated due to two reasons:-
1. Strength of C-O bond.
2. SN1 can't occur as methyl carbocation not stable.
Here acid is a stoichiometric reagent, not simply a catalyst. If a weaker acid is used, it will just establish a protonation-deprotonation equilibrium. However, a stronger acid will shift the equilibrium towards protonation.
And haloacids are best suited.
Conclusion being that a weaker acid will render nucleophile incapable to cleave ether in case of OMe.

authors: Αντώνιος Κελεσίδης


text:
Why do you drop the charge in the crossover method?
When I do the crossover method of crossing ions, my teacher always tells us to drop the charge, but won't tell us why. I was wondering if we add it back in later, and if not, why does it disappear?

Let's figure out the molecular formula for calcium chloride.
Calcium has a +2 charge. Chloride has a -1 charge.
So with the crossover rule we get
$$\ce{Ca_{-1}Cl_{+2}}$$
But for molecular formulas a -1 is impossible. The coefficients must be positive integers (usually...). Also have a 1 is superfluous, so it is just $\ce{Ca}$ and not $\ce{Ca_1}$.
Now for the chlorine the + is superfluous, so it is just $\ce{Cl_2}$ and not $\ce{Cl_{+2}}$.
I noted that the coefficients must be positive integers. A case where that is not true is when there is a double salt. Say a mixture of NaCl and KCl. The cations and the anions must balance so the formula can be written as
$$\ce{Na}_x \ce{K}_{1-x}\ce{Cl}$$
thus $(x) + (1-x) = 1$ and the charges balance.
Frankly I find this whole rule thing much to do about nothing. The net charge must be 0. Calcium has a +2 charge and Chlorine has a -1 charge. So you need two -1 charges to balance the +2 charge. Thus $\ce{CaCl2}$. Note that the coefficients must be the lowest common multiple, so it is $\ce{CaCl2}$ and not $\ce{Ca2Cl4}$ which also has a net 0 charge.
So to me the only difficult part is figuring out the charges on the ions. Once you know that, the rest is easy. Making a special rule just seems to be overkill.


text:
Interpretation of H as T goes to absolute zero
Euler's identity for the enthalpy states that (for a single component and phase)
$ H = TS + \mu N$
What are the implications of this expression in the context of the third law of thermodynamics (consider a "a perfect crystalline substance")? My textbooks describe the behaviour of $\Delta S$, $S$ and $\Delta G - \Delta H$ but not of $\Delta H$ or $H$.
Is it safe to assume that as $C_p$ and $S$ go to zero, so too does $\Delta H$ for any process at 0 K?

I should add that this problem follows from a "heated" dispute with @Mithoron during my attempt to answer this problem: https://chemistry.stackexchange.com/questions/103885/why-do-we-use-pv-term-independent-of-u-in-the-equation-h-u-pv/103948#103948

The third law is not my area of expertise and I'm genuinely interested to know the answer.

Any decent thermodynamics text has a section on the general thermodynamic relations. The fundamental thermodynamic relation is $dQ = T dS = dE + p dV$ which defines $dE = T dS - p dV$ taking $S$ and $V$ as independent variables. Thus you have $E = E(S,V)$. If instead you take $S$ and $p$ as independent, you can rewrite the fundamental relation as $dH = T dS + V dp$ by defining $H = E + pV$, with $H = H(S,p)$. Notice that this is all defined through differentials, not absolute values. There is no need, nor no reason, to assume that either $S$ or $H$ must go to zero as $T$ approaches zero.

@JonCuster Well, that is sort of the Planck statement of the 3rd law of thermo, but ok. I've been wrong enough today with my misstatement $H=TS$ that I deserve to be doubted...

The entropy of a system at absolute zero is a constant, but the constant does not have to be zero. Many systems have a degenerate ground state and thus a non-zero entropy at T=0.

@JonCuster Ok, consider the question for "a perfect crystalline substance"

https://chemistry.stackexchange.com/questions/19541/is-not-possible-to-find-absolute-value-of-internal-energy

Short answer: No. There's no reason to expect that this should be the case.

My standard reference for classical thermodynamics is Callen's Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Ch. 11 is relevant here.
Planck's statement of the third law is, according to Callen, "The $T=0$ isotherm coincides with the $S=0$ adiabat." In addition to the theoretical ramifications of having an isotherm and an adiabat coincide, the Planck statement also assigns to this adiabat $S=0$. This is physically grounded in statistical mechanics.
Let $\Delta$ represent the change in a quantity of a reversible isothermal process. Then, at zero temperature:

$S = 0$, by the third law.
$\Delta S = 0,$ again by the third law.
$\Delta G-\Delta H = T\Delta S = 0,$ because $T \to 0$.
$$\frac{\text{d}\Delta H}{\text{d}T}-\frac{\text{d}\Delta G}{\text{d}T} = \lim_{T\to0}\frac{\Delta H-\Delta G}{T} = \lim_{T\to0}\Delta S = 0,$$ once again by the third law. (This shows why the Gibbs free energy and the enthalpy are essentially equivalent thermodynamic potentials at low temperatures!)

But there is no reason to expect $\Delta H = \Delta G = 0$. Regardless of temperature, some energy is still required to effect a change in the system.

Fair enough, I was about to answer my own question. The "left over" term in H or G is $\mu N$. N obviously can't go to zero. And $\mu$? Presumably not...


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Why Isn't Huckles rule applicable to more than 3 fused rings

In the following compound (Pyrene) why isn't Huckle's (4n+2) rule applicable? I do know it is applicable to only mono-cyclic ring systems but what changes when there are more than 3 fused rings?
Edit:
A similar question has been posted previously but that doesn't mention as to why only the peripheral bonds are to be considered.

@thewitness please do see the edit

That is because the double bond in the centre doesn't take part in resonance. It is quite similar to the lone pair in pyridine and the "second" lone pair in furan where we don't consider them as a part of π electrons while applying Haeckel's Rule. We require a complete conjunction to classify a compound into aromatic or non aromatic. In case there is no complete conjugation, it is non aromatic.

So in any of such compounds the double bonds not along the periphery do not involve in resonance?

It depends on whether or not it can participate in resonance.

So the peripheral rule isn't applicable everywhere?

@thewitness " double bond in the centre doesn't take part in resonance" -it's impossible, also its taking part in resonance isn't a problem at all.

Huckle's rule is  applicable to specific rings not whole compounds. If compound has at least 1 aromatic ring, then it's aromatic.


text:
Mole fraction concept
In a certain titration experiment,
1) when $\ce{KIO3}$ solution is added to $\ce{KI}$, the $\ce{I-}$ ion is oxidised to the triiodide complex, $\ce{I3-} $
$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1} $
2) Ascorbic acid immediately reduces the triiodide complex to the colourless $\ce{I}^- $ ion.
$\ce{C6H8O6 + I3^- + H_2O -> C6H8O7 + 3I^- + 2H+}\tag{2} $
3) When all the ascorbic acid is reacted with  $\ce{I3}^- $ , the excess $\ce{I3}^- $  will react with the starch solution to form a deep blue $\ce{I3}^- $  starch complex.
$\ce{I3^- \text{ + Starch} -> I3^-}\text{- starch complex}\tag{3}$
At the experiment I have the moles of $\ce{KIO3}$ let’s say $0.1$ moles.
If I want to find the moles of $\ce{C6H8O6}$ can I say that the mole fraction is -> $ \ce{IO3^-} : \ce{I3-} : \ce{C6H8O7} $ is $1:3:1$ ? Meaning the moles of $\ce{C6H8O7} $ is $0.1$ moles

Try using equivalents concept. It is more effective

Combining equation (1) and three times equation (2) gives an overall reaction
$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1} $
$\ce{3C6H8O6 + 3I3^- + 3H_2O -> 3C6H8O7 + 9I^- + 6H+}\tag{2a} $
$\ce{IO3- + 3C6H8O6 -> 3C6H8O7 + I^- + 6H+}\tag{4}$
So $\ce{IO3- \text{ to} C6H8O6}$ is 1::3.


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How can there be an enthalpy of formation for gaseous water?
Under standard conditions( 1 bar and 298.15 K), water will be in the liquid state. So how can you find the enthalpy of formation for gaseous water at standard conditions(as found in the back of all chemistry textbooks)?

Vaporising it in calorimeter.

It's a hypothetical state that can be used in actual heat balance calculations so you don't need to include the heat of vaporization in the actual calculation separately.  This is discussed in most thermodynamics textbooks.

The only explanation that comes into my mind is that,
$$\ce{H2 (g) + \frac{1}{2} O2 (g) -> H2O (g) <-> H2O(l)}$$
 And
$$\Delta H_f^\circ(\ce{H2O (g)}) = \Delta H_f^\circ (\ce{H2O (l)}) + \Delta H_\mathrm{vap} (\pu{25^\circ C}) = -285.8 + 44.99 = \pu{-241.8 kJ/mol}$$ almost as tabulated depending the sources, here the calculation is perform with tabulated values found in the 87th Edition of the Handbook of Chemistry and Physics.
This is a hypothetical state but actually, vapour water is in equilibrium with liquid water even under the standard state. It's just that this equilibrium is very unlikely to go in direction of formation of the gas water. But if you find an ingenious way to vaporize water under these conditions then the heat of formation of gaseous water will be the same as I just calculated.

Also in the 87th Edition of the Handbook of Chemistry and Physics, it is said (5-4):

The standard state of a pure gaseous substance is that of
  the substance as a (hypothetical) ideal gas at the standard
  state pressure.

Although I've never figured that out and I can't really tell you about the need for this value, it exists. Probably you'd like to have this kind of values when performing a calculation considering a particular state but as far as I used to do, I've never used this one.


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Is citric acid is a anti foaming agent?
I added citric acid in my dish washing liquid. It stopped foaming.  Iwant my liquid again to foam. because I prepared 4 litres of liquid. Please let me know how can i get back my liquid foam. please answer immediately.

Any acid is an anti-foaming agent (if you have an anionic surfactant, that is).

Citric acid is not a defoaming agent in itself. It is routinely used for pH adjustment in many commercial hand-dishwashing agents.
But it depends on the surfactant in the product. As Ivan tells, anionics will at some point be protonated by acids, and thereby loose their surface activity –and foaming capability.
Commercial anionic surfactants like ethersulfates will only do so at rather low pH, outside what is relevant for dish-washing products. But if your product is based on saponified oils it’s another story. Many fatty acids are weak acids (pK in the rage of 4-5) and will be significantly protonated at pH even as high as 5-6.
By the way: making good foam is as much an art as a science :-). If you want to dig deeper in the subject, I can recommend “Prof Steven Abbot –Surfactant Science, Principles and practice”. It is freely available on the web.

If your dishwashing liquid is an anionic surfactant which is protonated by the citric acid, you may have noticed a precipitate of some kind, or perhaps an unusual oiliness. If you can, check the pH.
A possible cure would be to raise the pH to 8 or 9 and stir or shake. Borax should do the trick. Sodium carbonate (washing soda) could work, but would probably give some foam from evolution of CO2, so do it little by little. More alkaline materials like sodium or potassium hydroxide would be less safe for home use because of possible overdosing.
Grease removers are generally not acidic, because most greases are more dispersible/removable with alkaline aqeous surfactants.


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Why don't ions form crystal lattices in water and other polar solvents?
Ionic compounds are in the form of crystal lattices, and when they are dissolved in a polar solvent, they are bonded with the corresponding cations/anions of the polar solvent. For example, $Na^+$ bonds with $OH^-$ and $Cl^-$ bonds with $H^+$ when NaOH is dissolved in $H_2O$.
The reason why $NaCl_{(aq)}$ conducts electricity is because $Na^+$ and $Cl^-$ do not form crystal lattices with $OH^-$  and $H^+$  respectively. But why don't they form crystal lattices?

They actually form: ions can be dissolved up till a certain concentration, and beyond that, they form ionic lattices just as you predict.
The easiest way to describe the scenario as an equilibrium between ions-in-water vs ions-in-lattice:

If the ion-water interaction is very strong, the ions can be very stable in water, even more than in an ionic lattice, so ions will be in the solution.
If the ion-water interaction is weak compared to the solid, then the solid phase is much more stable, and only a very low concentration of ions can be in the solution.

Looking at only interaction energies is a very simplistic view as it does not consider entropic factors, yet can describe at least part of the story. Entropy is generally a huge driving force for any kind of dissolution and allows some ions to be in solution even if the solid state energetically more stable.

Your statement 'For example, Na+ bonds with OH− and Cl− bonds with H+ when NaOH is dissolved in H2O' is not correct if the ions are in solution.
In an ionic solid each ion is surrounded by other ions of the same charge and those of opposite charge, in NaCl each Na$^+$ ion is surrounded by 6 Cl$^-$ and vice versa. The interaction between the ions is essentially Coulomb in nature with energy that varies with distance $r$ as $E \sim 1/r$. This is a long range interaction and so choosing a 'victim' ion this can interact with all its many neighbours, not just nearest neighbours,  thus many interactions have to be summed up. The sum is called a Madeling Sum and the cohesive energy produced is written in the simple form $E\sim se^2/r_0$ where $s$ is the Madelung sum (-1.748 for NaCl), $r_0$ the closest separation of any two ions and $e$ the electronic charge. However the ions could just approach one another and the ionic solid could just collapse. Why does this not happen? What prevents this is the Coulomb repulsion between electrons (and between nuclei) as any two ions (or atoms) approach one another and the Pauli Exclusion principle. This ensures that only two electrons can fill an energy level and so energy is needed to promote an electron from one energy level into an unfilled one on the other atom as they approach. Both these effects are repulsive in nature and vary as $\sim 1/r^{12}$ so are very short range.
Having established that there is a binding energy holding a solid together when an ionic solid is added to a polar solvent we expect it to dissolve and free ions to be produced. Overall if this is to happen the free energy change $\Delta G$ must be negative. This is composed of two parts the enthalpy change (potential energy)  $\Delta H$ and the entropy change $\Delta S$ as $\Delta G = \Delta H -T\Delta S$, where $T$ is the temperature. If the free energy is to be negative then there are four possibilities depending on the sign of the change $\Delta H$ and $\Delta S$. The one situation that cannot occur if the solid is to dissolve is $\Delta H \gt 0, \; \Delta S \lt 0$ as this always makes $\Delta G$ positive. It will always dissolve if $\Delta H \lt 0, \; \Delta S \gt 0$ as $\Delta G $ is negative. The other cases depend on particular values.
We expect that $\Delta H$ will be positive on dissolution as the crystalline solid exists and it will take energy to make it dissolve. (This will cool the solution if heat is not supplied to keep the temperature constant). As there is a larger volume for the ions to occupy in solution than in the solid the entropy change is expected to be positive. Simply having more places to occupy will increase the entropy. However, the solid does not have zero entropy, the ions can move albeit by a small amount about their equilibrium positions in the crystal and waves of vibrational motion called phonons exist and so there is already certain amount of entropy. This is lost on dissolution and has to be exceeded if the entropy  change is to be positive. Also it is necessary to consider the entropy change in the solvent. As the ions are charged, in a polar solvent the dipoles are attracted to the charge, and partially order around the it. This leads to a decrease in entropy of the solvent as there are ions plus an 'atmosphere' of a few solvent molecules around each one. All these terms need to be considered, but as, say, NaCl does dissolve then the entropy change must be positive.
When the ion is dissolved, why do pairs of ions not form?  This is because the solvent is polar and when the dielectric constant of the solvent is considered the energy varies as $E\sim 1/(\epsilon r)$ where $\epsilon$ is the dielectric constant, for water this 78 and 37 for acetonitrile,  but for an ionic crystal  it is only about 2, similar to that for non-polar hexane. This large dielectric constant means that the interaction of one ion to another is  $\epsilon$ times smaller compared to that at the same distance in the crystal or in a non-polar solvent. In other words, in a polar solvent the ions are effectively independent. The interaction is small and comparable to the thermal motion jostling the molecules and so the number of times they form pairs is vanishingly small. This means that the solution will conduct electricity.

"What prevents this is the Coulomb repulsion between electrons (and between nuclei) as any two ions (or atoms) approach one another..." Are you talking about the coloumbic repulsion between two like ions (Cl- - Cl- or Na+ - Na+) as they try approaching the victim ion? "...and the Pauli Exclusion principle" More energy is needed to elevate electrons to the 4s shell in Cl- and 3p shell in Na+: Is that what you're saying?

"We expect that ΔH will be positive on dissolution as the crystalline solid exists and it will take energy to make it dissolve." So energy is absorbed to carry out the ionisation. "This will cool the solution if heat is not supplied to keep the temperature constant." Does this mean that the absorbed heat energy will get used up during ionisation?

And from your explanation of entropy, do you mean to say that for NaCl, the entropy change due to spacing out significantly exceeds the (absolute value of) the entropy change due to other factors such as restricting the little freedom of ions in crystal structure, electrostatic forces restricting freedom of ions, etc.? And does the relative permittivity of free space of the polar solvent also contribute to a positive change in entropy as electrostatic forces that constrain freedom are reduced?

(a) re Pauli, yes more energy would be needed, (b) the atoms are always ionised in crystal and solution so no ionisation is needed. (c) yes if dimers are not formed as dielectric is large then dielectric  indirectly contributes to entropy