text:
The impossibility of 100% ionic bond
Recently, I read the definition of  oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all?

I don't think this is impossible. I can charge a capacitor and cut it in half.

@Joshua Yes, it's not really a problem to make two charged things be it anion and cation or two charge spheres, it's just that there's simply no chemical bond in there. "100% ionic bond" is not a chemical bond at all.

This is because when a force is supplied to the ionic compound it leads to dislocation of ions by merging the like charged ions together which causes great repulsion force that leads to break down of compound.

A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion.  That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation.  That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.
Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds.  It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.

Not only some attraction of cation to electrons of anion, but quantum theory has probability distribution of each orbital extending to infinity. There isn't any absolute cutoff distance from the nucleus.

Actually bonds that are $100\%$ ionic, or close to it, are possible.  But they require a little ingenuity.  Instead of combining atoms of very different electronegativity, we can build multiatomic ions in which the charge is (formally) buried or highly delocalized making covalent bonding unfavorable.  We use molecular orbital structures rather than electronegativity to drive the charge separation.
One example of this is the cyclopropenyl salt $\ce{[C3H3]^+[SbCl6]^-}$ produced by Breslow and Groves in 1970 (J. Am. Chem. Soc. 1970, 92 (4), 984–987).  To form a covalent bond between the ions would require either occupying a highly destabilized antibonding orbital in the cyclopropenyl ring, or an interaction between the ring and poorly overlapping orbitals in the bulky anion.  So we have very little covalent bonding; the molecular orbital structure does not favor it.  We could get covalent compounds by transferring a chloride ion, but this reaction is energetically unfavorable.  The salt, with its ionic bonding between the aromatic cation and complex anion, is surprisingly stable for a species with a three-membered carbon ring.  Further stabilization may be achieved and ionicity reinforced by placing substituent on the cyclopropenyl ring.
Another example is provided by compounds in which alkali metals are disproportionated into cations and anion (see Alkalide).  Because the ionization energy of alkali metals (at least from sodium on down) are so small, the cation may be stabilized by forming a complex where the (formal) positive charge is deeply buried and again, no obvious means exists for covalent bonding.  The alkalide anion must be bonded by essentially ionic attraction to the cationic complex.

Reminds me a lot about ionic liquids.

In a "100 %" covalent bond, like the $\sigma_{ss}$ bond in the dihydrogen molecule, the electron probability density is perfectly symmetrically divided between the two bonded nuclei because both have the same electronegativity.
Below: the schematised electron density $\psi^2$, for a 100 % covalent bond:

But when the two atoms have different electronegativities, the electron probability density will be higher towards the more electronegative atom, see e.g. $\ce{HCl}$. We say the bond is polarised. See below, schematised, the permanent dipole of $\ce{XY}$, with partial charges $\delta +$ and $\delta -$:

Below: the schematised electron density $\psi^2$, for a bond between atoms with differing electronegativity (highest electronegativity to the right):

If the electronegativity difference is really high, see e.g. $\ce{NaCl}$, the bond starts being so strongly polarised (electron probability density strongly skewed to towards the $\ce{Cl}$ atom) that the bond starts taking on an ionic character.
Ionic and covalent must be considered relative to each other: in a "75% ionic" bond the electron distribution is strongly skewed toward the electronegative element.
There is however no meaningful way to measure this "percentage".

There is however no meaningful way to measure this "percentage". ||  NBO analysis?

https://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist

There are "meaningful ways" and even for symmetric bonds ionic component can be assigned ^.


text:
Quantity of water in a redox problem

$\pu{76 g}$ of $\ce{FeSO4}$ are oxidized by $\pu{0.04 M}$ solution of $\ce{KMnO4}$ in presence of $80\%$ $\ce{H2SO4}$. What quantity of water must be added to the final solution of $\ce{Fe2(SO4)3}$ for the concentration to be $\pu{0.25 M}$?

I wrote the reaction:
$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$
But I have no idea how to find the amount of water. I found the quantity of $\ce{Fe2(SO4)3}$ to be $\pu{100 g}$, but I have no idea to find all the water.

Could you please double-check the amounts? I might be wrong, but something seems a bit off here. Probably the concentration of KMnO4 is 0.4 M, not 0.04 M?

This is strange; in my opinion it's a typo, because this is a way too diluted solution to start with.

To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:
$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$
where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.
From the balanced redox reaction, which it seems you have established correctly:
$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$
one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:
$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$
$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$
$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$
From (1):
$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$
Combining (2–4) into (5), and taking into account that
$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$
one can obtain the following expression for the volume of water to be added:
$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$
The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:
$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$

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text:
How to know whether to add latent heat when calculating final temperature?
From the examples in my textbook, it is usually very easy to see whether you need to account for the latent heat when calculating the final temperature. For example, if you add 2 x 13 gram ice blocks to a 250 ml cup of water, it is obvious that the ice will melt, and you will have to account for the latent heat of fusion for the ice.
However, what if it isn’t so obvious whether one of the substances will undergo a phase change? For example, what if we added 15 grams if ice to 20 ml of oil? If the final temperature is below 0, then we don’t need to account for the latent heat for ice, otherwise we will have to. How do we determine whether the ice will experience a phase change, and more generally, how do we determine whether one of the substance will experience a phase change?

It is less obvious than you seem to assume (say, if water cup in your example were already at $0^\circ$C, then the ice would not melt), but still pretty obvious.
The rule of thumb is:

Assume there is no phase change.
Calculate the final temperature.
Check whether there is a phase change in any of the substances involved as they go from starting to final temperature.
If no, we are done. If yes, assume there is a phase change and go back to step 2.

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text:
Relationship between pH and volume
I had prepared two different volume of solution: $\pu{500 mL}$ and $\pu{100 mL}$ with same concentration ($1\%$), but I obtain different $\mathrm{pH}$. Will volume affect $\mathrm{pH}$?

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The volume will not affect the pH.
However, any change in the concentration will

If correctly prepared, the pH should have been the same. Logically, consider if the pH would be different if measured in 5 separate 100 ml samples poured from the 500 ml preparation.
Why might the pH differ in your tests? Some possibilities:

One container was not completely rinsed after cleaning and had some (often alkaline) detergent clinging to it.
The preparation absorbs $\ce{CO2}$ from the air, so the layer on top is more acidic.
Error in measurement.

Unless you have a buffered system, pH may vary abruptly with a slight change in ingredients.

No, $\mathrm{pH}$ is not volume-dependent as it's an intensive property derived from another intensive property (concentration):
$$\mathrm{pH} = -\log{[\ce{H+}]}$$
The reasons for the different $\mathrm{pH}$ might be different, among them there are the following:

Contaminated beaker;
Solution absorbs carbon dioxide from the air differently due to different surface area of the liquid exposed to air;
$\mathrm{pH}$-electrode is contaminated or/and improperly calibrated;
$\mathrm{pH}$ values were measured at different temperatures.


text:
Why does Zinc not have magnetic properties despite it being a transition element?
I know Zinc has a full d-orbital which might contribute to the effect. But it is not ferromagnetic or paramagnetic like other transition element complexes. Why is that?

Well, paramagnetism and all the *ferr*magnetisms depend on unpaired electrons, so …

"has a d-orbital which might contribute to the effect" That's not the right conclusion.

Everything has magnetic properties. Zinc, in particular, is diamagnetic.

This Minute Physics video has a pretty accessible explanation of magnetism in different materials


text:
Blue litmus turning pink
I just read that carbonic acid turns blue litmus pink and not red.
Is it just because it is a weak acid. What about organic acids then? How do they respond to the litmus test?
If strength and color are related, can we determine the strength of an acid by determining if the litmus paper is more pink or more red?

Litmus paper will indeed be less or more strongly coloured depending on the pH of the solution you are measuring. Unfortunately though, there is no easy way to quantify it so you cannot use it to semi-accurately determine pH.


text:
How can atoms be influenced by light?
I was watching a SciShow video "Most dangerous Chemicals". It said that a compound called Azidozide azide explodes on being exposed to light. My question is: how does the the atom feel light. Heat can be felt (let us not go into the philosophical question why do we feel warm while standing below the sun) and can excite molecules. How can light excite molecules?

Light can excite atoms a great deal harder than heat. Related: https://chemistry.stackexchange.com/questions/86931/reactions-in-the-presence-of-sunlight/86935#86935

Light can be ‘felt’ because light is transmitted by photons. Depending on how you choose to observe light, it will either manifest itself as waves — for example in interference experiments — or as particles — for example when describing the photoelectric effect.
When described as a wave, the energy of a certain source of light will be determined by its wavelength by equation $(1)$ in which $h$ is the Planck constant and $c$ is the speed of light.
$$E = \frac{hc}\lambda\tag{1}$$
It may not immediately be obvious how this translates to the energy of a photon, but the key is the de Broglie wavelength as given in equation $(2)$.
$$\lambda = \frac hp\tag{2}$$
This equation connects all particle and wave properties of types of matter. (Indeed, due to the wave-particle dualism all particles can theoretically behave as waves and vice-versa; the wave nature of particles as large as entire molecules has been shown in interference experiments.)
When a photon with a certain energy travels through an atom or molecule, it may interact with the electrons present; often this interaction causes the photon to be absorbed and the electron to be excited to a higher energy level. As chemists, we describe this as an electronic transition from one orbital to another. The entire process can also be reversed, i.e. if an electron is excited to a higher energy level by any means, it can relax following certain quantum-mechanical selection rules back down to a lower energy level and emit a photon of a definite wavelength. The latter process can be observed in the flame colour given by sodium atoms: their 3s electron is excited to a 4p level thermally due to the heat of the flame. Upon relaxation, the electron emits a photon of $\pu{589nm}$ wavelength which we perceive as yellow. We could write that in an equation:
$$\ce{4p ->[][$\lambda = \pu{589nm}$] 3s}\tag{3}$$
If the target of a photon happens to be a molecule rather than an atom, we need to consider what type of orbitals we are exciting from and exciting into. Typically, the occupied orbitals will be bonding orbitals while unoccupied orbitals will be antibonding. Thus with the correct photon wavelength, an electron can be excited from a bonding into an antibonding orbital which we can, in layman’s terms, call ‘breaking a bond’. Whether or not the bond actually breaks depends on whether or not the atoms can dissociate apart.
A beautiful reaction to exemplify this is the reaction of chlorine and hydrogen in equimolar mixtures. Under ambient conditions, these two are metastable. If the activation energy is supplied, they will react to form hydrogen chloride.
$$\ce{H2 + Cl2 -> 2 HCl}\tag{4}$$
Activation energy can be supplied not only by a flame but also by light as long as its wavelength is short enough. A flash of blue light will work, a flash of red light will not. What happens? First, a photon excites an electron in a chlorine molecule. It will reach an antibonding, unoccupied orbital, the $4\sigma$ orbital. This causes the $\ce{Cl-Cl}$ bond to break liberating two chlorine radicals.
$$\ce{Cl2 + $h\nu$ -> [Cl2^*]^\ddagger -> 2 Cl^.}\tag{5}$$
These chlorine radicals are then energetic enough to react with hydrogen molecules to give hydrogen radicals and so on — a radical chain reaction.
$$\begin{align}\ce{Cl^. + H2 &-> HCl + H^.}\tag{6}\\
\ce{H^. + Cl2 &-> HCl + Cl^.}\tag{7}\end{align}$$
And this goes on until all the reactive species have been consumed and the radicals recombine. The reaction releases $\pu{92.3kJ/mol}$ and is therefore strongly exothermic, even explosive.
This mechanism of course also works for infrared light which we perceive as heat. However, since the energy of infrared light is very low, few reactions can be initiated by it.

Photons in the UV or visible ranges of the EM spectrum can have sufficient energy to excite electrons. In simplified terms, this usually involves an electron entering an anti-bonding orbital, which may be sufficient to start a reaction. If that reaction yields enough energy to start the next one, you have a chain reaction. Alternatively, those electrons relax back to their ground states, photons will be emitted, and the atom or molecule will give off visible light of specific frequencies. This is called fluorescence. So in a nutshell light gives off energy and that is what will excite your molecules along with the energy from anything else that makes contact with the chemical. Ask if you need clarification.

I think this answer could be improved if it addressed how the absorption of light could result in an explosion per the OP's example, not just the emission of a photon.

Thanks for the suggestion. I'm not so sure what it is but simple warmth of light could be enough to cause it to explode perhaps because of the burning point of the chemical is so low the heat overcomes it and it ignites. And when i mentioned how the energy from contact of outside sources ignites it the friction would be that cause of explosion due to the heat generated. Feel free to correct my thinking.


text:
Difference between batch and column mode?
I have heard Ion-exchange chromatography can be used in batch or column mode, have you got any idea how are those modes?
I have looked descriptions all over books and web (Skoog for example, both versions of analytical chemistry) but found nothing clear.
Any help?

Probably batch by batch or in continuous mode.

@Alchimista but what is an ion exchange by batch?

First the sample is made to bind than eluted. https://www.slideshare.net/mobile/ParimalaGattupalli/ion-exchange-chromatography-1-4


text:
Copper pyrite representation
What is the oxidation state of iron in $\ce{CuFeS2}$? And also could you please tell it's proper representation: $\ce{Cu2S.FeS.FeS2}$ or $\ce{Cu(I)FeS2}$?

$\ce{Cu2S.FeS.FeS2}$ and $\ce{CuFeS2}$ are fine, but $\ce{Cu(I)FeS2}$ is wrong. Oxidation state in parentheses used in written names; if you want to denote monovalent copper, then $\ce{Cu^IFeS2}$

Can you please also comment on their oxidation state ie. Is  iron in +2 and copper in +1?

But you've already answered this part by typing $\ce{Cu^I}$, haven't you? :)  The structure contains  $\ce{Cu+}$ and $\ce{Fe^3+}$ cations.

In the formula Cu2S.FeS.FeS2, the 1st iron is in +2 and the second one is also in +2 by me, ie it is polysulphide ion (  S2 (2- ) ion ), so their avg will also be 2 right?

No. $\ce{Cu2S.FeS.FeS2}$ is just a convenient representation, nothing else. In reality the cubic  structure consists of array of vertex-linked $\ce{CuS4}$ and $\ce{FeS4}$ tetrahedra, that's it. There is no separate standalone metal sulfides as one would imagine.

No prob, there were debates in the past whether it's iron(II) or iron(III), but these days Mössbauer and photoelectron-emission spectroscopies alongside with magnetic measurements suggest it's iron(III).

If you are really interested in details, there is an open-access article on what's going on with copper and iron in chalcopyrites.


text:
Calculate work done by burning benzoic acid exposed to the free air
I'm self-studying and can't solve this problem in Ball's Physical Chemistry. Now, the errata shows that the answer in the book is not correct; still, I can not make sense of it.

$(2.55)$ Benzoic acid, $\ce{C_6H_5COOH}$, is a common standard used in bomb calorimeters, which maintain a constant volume. If $1.20\text{g}$ of benzoic acid gives off $31,723\text{J}$ of energy when burned in the presence of excess oxygen at a constant temperature of $24.6^\circ\text{C}$, calculate $q, w, \Delta H,$ and $\Delta U$ for the reaction.
$(2.56)$ $1.20\text{g}$ of benzoic acid, $\ce{C_6H_5COOH}$, is burned in a porcelain dish exposed to the air. If $31,723\text{J}$ of energy is given off and the system temperature is $24.6^\circ\text{C}$, calculate $q, w, \Delta H, $ and $\Delta U$.

The answers(corrected from errata) are,

$(2.55)$ $q = -31,723\text{J}, \Delta U = -31,723\text{J}, w = 0, \Delta H = -31,735\text{J}$
$(2.56)$ $q = -31,723\text{J}, \Delta H = -31,723\text{J}, w = 12, \Delta U = -31,711\text{J}$

The book never talks about heats of combustion prior. Computing using a value I found on the internet of $26.4 \,\text{kJ}/\text{g}$, I get a value for $q$, I get that $\Delta U = q + w$ then and we can solve for work. This seems to be an incorrect approach since the energy given off as heat seems to already be given.
Why is the enthalpy the same as the heat in the second case? I would assume it is because the pressure is constant, but only the external is constant, which is not what is used to determine that enthalpy is the same as heat.
Where does this work come from? The volume changes in the non-calorimeter experiment, so work on the acid is done. But how does on obtain this change?
How do I solve this problem correctly?

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text:
Theoretical limits of (NIR) spectroscopy for quantitative analysis
Edit (months later): In hindsight, I was asking about the theoretical limits of multivariate curve resolution (see Lawton & Sylvestre: Self-modeling curve resolution) in chemometric applications.

NIR spectra are commonly used to estimate analyte concentrations. The common approach is to measure the spectra of different samples with known concentration (training data) and then estimate the unknown concentrations by regression.
My question is whether we could determine analyte concentrations of any substance just from the following:
1) Infinitely precise spectra of all elements.
2) Infinitely precise spectrum of the sample.
If not, would it help if we relaxed on the NIR range assumption?
If still not, would it help if we knew the spectra of all possible molecules with infinite precision?
The first step in determining the answer is probably determining whether there are two different atoms identical in their spectra. I don't know the answer. Unfortunately my knowledge of physics and chemistry is very limited.

"Infinitely precise spectrum of the sample" ? Precise in what? BTW I find this question unclear and rather strange.

@Mithoron precise as in free from measurement errors.

Firstly, atoms do not have an IR spectrum. You mean molecules, or any kind of chemical substance. Anyway. Then, by "infinite precision", you mean noise-free. Yes of course, no two molecules have the same vibrational modes, except if it's the identical substance. But this is completely pointless, because there is no such thing as a noise-free measurement, and also because you can't just split a single absorption spectrum into the individual spectra of an arbitrary number of possibe components. This approach is 100% wilful nonsense, imo.

to continue comments by @karl, molecular spectra have transitions that have a finite spectral width and because of this spectra from one molecule will always overlap that of another to a greater of lesser extent. You only have to look at an IR spectrum to see this. This means that it can be impossible to properly measure different species as some lines will be hidden under those from different species. Having noise free signals will not help much as you will have to fit data anyway and this will in itself introduce uncertainty is assignments.

Thank you for your feedback! I will do some more reading and possibly refine my question. I'll add an analogy to mass spectroscopy, where it is feasible to generate a database of all possible peptide spectra (de novo approach) and then search for the measured spectrum in this database e.g. by the cosine distance metric. I don't hope to build such database for NIR spectra. The goal of this question is just to understand the data at hand 'from bottom up'.

The problem is the same for MS. For pure substances, you can make a database for practically every spectroscopic method, and perfectly identify everything. For mixtures, you have to put in a lot of thinking and additional constraints (limit range of possible constituents etc) before you can quantify anything.

@Karl "you can't just split a single absorption spectrum into the individual spectra of an arbitrary number of possibe components" - but isn't that what the Beer-Lambert law assumes in its $A = \sum_{i = 1}^N A_i = \sum_{i = 1}^N \varepsilon_i \int_0^\ell c_i(z),\mathrm{d}z$ (equation from Wikipedia)?

@Josef Just because you have an equation for sth. doesn't mean it has a single, unique solution. Further spectra e.g. look slightly different depending on concentration and solvent identity. Lambert-Beer has it's limits. There are far, far too many imponderables in your problem.

@Karl thank you for your reply. Of course, I'm just making sure I understand what the Lambert-Beer law is about. Regarding physical effects, this paper claims it can eliminate them, albeit under the strong assumption that the mixture is made of just two species: https://www.ncbi.nlm.nih.gov/m/pubmed/12585463/

@Josef Not "physical effects", just scattering in turbid samples. There are a dozen more effects that creep in when your sample is not perfectly dilute. The spectrum of a 1M solution of sth. does NOT necessarily look identical to the 1µM solution. Surely never perfectly identical. Even if you try to separate solvent and solute spectra and scale them accordingly.


text:
Why do bonding and anti-bonding states split in energy (mathematically)?
I am a physicist, and when I hear about the splitting of energy levels for MO antibonding and bonding states, I have a rough intuition that tells me this is reasonable to expect, but I can't think of (nor find) a good formal explanation.
My best shot at understanding this is borrowing from Avoided Crossing and saying that if I have a 2 atom system with a total Hamiltonian $ H = H_1 + H_2 $ that is the sum of each individual atom Hamiltonians, then I can take the Eigenstates for each individual Hamiltonian as a basis.
$$\begin{align}
H_1 | \psi_1 \rangle &= E_1 | \psi_1 \rangle \\
H_2 | \psi_2 \rangle &= E_2 | \psi_2 \rangle
\end{align}$$
Then the total hamiltonian in this basis becomes,
$$\begin{align}
H &= \left(\begin{matrix}
E_1 & \langle \psi_2 | H_1 | \psi_2 \rangle \\
\langle \psi_1 | H_2 | \psi_1 \rangle & E_2
\end{matrix}\right)
\end{align}$$
In the limit of the atoms being really far apart, I would assume the off diagonal terms would vanish, and then the eigenstates are just the atomic orbitals. But if the atoms come closer together, then we get off diagonal terms which causes the splitting as described in avoided crossing.
Is this correct?

You are (essentially) correct and the off-diagonal terms are related to so-called "overlap" between atomic orbitals (AOs). It is common to talk of the AOs "overlapping" to form molecular orbitals (MOs), and as you rightly said, the off-diagonal terms (or "overlap") goes to zero in the limit of infinite bond length. The term "avoided crossing" is not commonly seen in introductory MO theory, but I think it is fair to describe it as such.
Of course there are some slight subtleties. Let's stick to a one-electron system such as $\ce{H2+}$ for now. $\psi_i$ refers to the 1s orbital on hydrogen atom $i$ and $E$ is the energy of a 1s orbital on hydrogen. The minor issue is that $\hat{H} \neq \hat{H}_1 + \hat{H}_2$ (following your notation where $\hat{H}$ is the molecular Hamiltonian and $\hat{H}_i$ is the Hamiltonian for hydrogen atom $i$), because:
$$\begin{align}
\hat{H} &= -\frac{\nabla^2}{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{R} \\
\hat{H}_1 &= -\frac{\nabla^2}{2} - \frac{1}{r_1} \\
\hat{H}_2 &= -\frac{\nabla^2}{2} - \frac{1}{r_2} \\
\end{align}$$
(I used atomic units. $r_i$ is the distance between the electron and nucleus $i$, and $R$ is the distance between the two nuclei). So actually, it is more like
$$\hat{H} = \hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}$$
and therefore the first term in the matrix
$$\begin{align}
\mathbf{H}_{11} &= \left<\psi_1\middle|\hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}\middle|\psi_1\right> \\
&= E + \left<\psi_1\middle|-\frac{1}{r_2}\middle|\psi_1\right> + \frac{1}{R} \neq E
\end{align}$$
(since $\langle\psi_1|\hat{H}_1|\psi_1\rangle = E$, and also $\langle\psi_1|(1/R)|\psi_1\rangle = (1/R)\langle\psi_1|\psi_1\rangle = 1/R$).
For more information I refer you to Atkins' Molecular Quantum Mechanics 5th ed., pp 262–266. The terminology can be a bit confusing and also Atkins does not use atomic units (in his text he uses $j_0 = e^2/4\pi\varepsilon_0$ which is simply equal to $1$ in atomic units). Adjusting for this, the final quoted formulae for the matrix elements are:
$$\alpha = \mathbf{H}_{11} = \mathbf{H}_{22} = E - j' + \frac{1}{R}; \qquad j' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_1\right>$$
(as found earlier), and
$$\beta = \mathbf{H}_{12} = \mathbf{H}_{12} = \left(E + \frac{1}{R}\right)S - k'; \qquad S = \langle\psi_1|\psi_2\rangle;\quad k' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_2\right>$$
If you now solve the generalised eigenvalue equation $\mathbf{Hc} = E\mathbf{Sc}$ (the simple $\mathbf{Hc} = E\mathbf{c}$ only works when your basis set is orthonormal, but in this case it is not) you obtain the two eigenvalues
$$E_\pm = E + \frac{1}{R} - \frac{j' \pm k'}{1 \pm S}$$
Note that when the atoms are infinitely far apart both $S$ and $k'$ vanish, so (1) the off-diagonal element $\beta$ vanishes and (2) the "bonding" and "antibonding" orbitals are of the same energy, i.e. $E_+ = E_-$.


text:
Electrophilic addition to alkenes
Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.

The first step is the attack of $\ce{H^+}$
On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:

Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.

These two would lead to different products but I'm interested in the major one.
The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.

Yeah, there should be bromonium here, but it's not what is called non-classical carbocation.

But sir I have been told that the inductive effect is so strong that it lead the the 1st pathway. But I feel it should be the one with bromonium ion. As in anchimeric effects during substitution reactions. Is there a strong justification for that?

Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom

Why would it go with the classical carbocation pathway? You have not given the explanation for that, please elaborate a bit.

And it's not a cyclic carbocation btw. It's bromine that's carrying the positive charge (if you are referring to this).


text:
Bromonium ion or Mesomeric effect ( intermediate during addition to alkenes)
It's mentioned in Peter Sykes , A Guidebook To Mechanism In Organic Chemistry , 6th edition, 13th impression by Pearson, Pg. No. 185 , that on addition of $\ce{HBr}$ to bromoethene the only product obtained is 1,1-dibromoethane. It's due to the positve mesomeric effect of bromine to stabilize the carbocation in this case, however the rate is remarkably slower as compared to addition to ethene.
1
Another reaction corresponding to addition of $\ce{Br2}$ to ethene proceeds with the formation of a bromonium ion intermediate.
2
My question is, why can't the first reaction just proceed with a bromonium ion intermediate, like (28) above can form a bromonium ion and give 1,2-dibromoethane as a product (there may be a possibility that the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion)? And if bromonium ion is unstable (that is if it can't be formed in 1st reaction) why does it form at all in reaction 2? Why can't the Bromonium ion in the 2nd reaction just undergo a 1,2- hydride shift and convert into the "more stable" ion like 29b. It should also lead to formation  of geminal dibromide .It's a circular argument.
It's clearly mentioned that bromonium ion and carbocation are canonical forms of each other , which means you expect them to convert into bromonium ion whenever there's a positively charged carbon near a bromine atom.
3
Here's a similar example of phenonium ion.
[4]: https://i.stack.imgur.com/0j5Ww.jpg .
The carbocation doesn't stay the way it is in structure 4. It rather forms 3. Notice the bond between phenyl ring and the carbon atom on the right. It's not a completely solid bond.

Surely, you can also formulate bromonium formation for the first reaction. But clearly, the carbocation in alpha position to the brom atom will be better stabilized due to the mesomeric stabilization.

If you then consider that formation of the carbocation is likely the rate determining step, it makes sense to assume that essentially only the pathway via the alpha-carbocation is populated.

If bromonium ion can't be formed in first reaction why does the second reaction form it? It should also favour carbocation on the alpha position.

But still it's encouraging to hear that Bromonium ion can be formulated formulated for the first reaction.

Anything can be formulated only question if it is reasonable. In the second reaction though no alpha cation can be formed.

Can you please elaborate why can't bromonium ion break and undergo 1,2 hydride shift to form alpha carbocation?

Now we are getting somewhere. So you propose a 1,2 hydride shift after the bromonium ion has been formed. Thats a good question. I dont know for sure why this doesnt happen. Maybe the bromide is just the better nucleophile.

In other words the hydride shift is likely slower than the bromide attack.

That may be the possibility. But I'm glad you understood what I was trying to say. Thank you.

After some consideration I would say that the bromonium ion is likely even more stable than the alpha-bromonium ion you proposed after a possible hydride shift, which would explain all observations.

You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:

the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:

The bromonium would only be able to form after a rotation of the $\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).

First of all, structure (9)-bromonium ion intermediate and structure (10)-the open intermediate aren't canonical forms of each other simply because "In most resonance, $\sigma$ bonds are not involved(except in hyper-conjugation but what we have here is not hyper-conjugation), and only the $\pi$ or unshared electrons are put in different ways and canonical forms don't really exist(its our imagination) but reaction intermediates like bromonium ion and structure (10)-type intermediates have been isolated under special conditions."
Had structure(9) and (10) been canonical forms of the same molecule, we would've never been able to isolate structure(9) and (10). The positive charge in structure(10) might be stabilized by an attraction for Br but that does not involve a full bond.(see the image below)

The possibility that "the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion" sounds unreasonable to me because the carbocation will always form on the bromine side as the +M effect of bromine that stabilizes (29a) & (29b) is way more than its destabilizing -I effect.
In no way is bromonium ion intermediate less stable than structure (10).
Ab initio molecular orbital studies show that bromonium ion intermediate is more stable than the intermediate in reaction 1(structure-(10)).
In the addition of HBr to bromoethene, it follows the convention: "the electrophile attacks the substrate before the nucleophile does". Electrophile need not actually be a positive ion but can be the positive end of a dipole or an induced dipole, with the negative part breaking off either during the ﬁrst step or shortly after. So, the formation of the carbocation after the attack of the proton is inevitable.
But due to the fact that in reaction 1, the electrophile is hydrogen ion, the reaction takes place through the mechanism involving structure (10) and not the one in reaction 2 involving the bromonium ion intermediate(9).
I hope you realize that (2) never forms (4). Your statement saying that (4) forms (3) interconvert among themselves is not correct rather (3) and (4) are separate species formed through different set of mechanisms. None of the structure is a canonical form of the other.

The above reaction clearly shows that (12) and (9a) are not really canonical forms of each other. I really don't get the sense in which that statement on pg. 180 has been written but it definitely doesn't specifically mean that (9) and (10) are canonical forms of each other.
Don't get confused by the reaction that I mentioned above, Peter Sykes says This neighbouring group participation by bromine does not ofcourse prove that addition to alkenes proceeds via cyclic bromonium ions, in the past,the reaction was just used to show that such species(cyclic bromonium ions) are not just mere assumptions. That's all!

Structure 10 or 28 and bromonium ion intermediate are not canonical forms of each other...didn't i say that earlier??...because two canonical structures of the same compound have the same number of sigma bonds...that's a basic thing dude...

Can you please tell me the problem that you are still facing

@gowreeshmago I've edited the answer, hope its understandable now.

Its mentioned in Peter sykes that they are canonical forms. It's not me saying that.

Can you tell the page number.

Pg 180 its written bromonium ion is an alternative canonical form of carbocation 10

give me a min..

edited the answer again.

You yourself showed bromonium ion can be formed if there's a carbocation at beta position.

Look at the comments of logical×2 .

didn't you notice the reaction conditions, its a special reaction that has occurred under specific conditions so that the desired cyclic bromonium ion can be showed that it is actually formed in some reactions(I'll specify about it in my answer), but the point is in reaction 1 of your question the cyclic bromonium ion won't be formed because the $\beta$ carbocation is formed in a very less concentration, as compared to (29a). There are many proposed ways for a reaction to occur but most of them form the minor products (like the cyclic bromonium ion way does for reaction 1)

And you can rely on my statements, most of them have been taken from Jerry March which is kind of the most trustworthy book for organic chemistry. Believe me

I saw you asking "If bromonium ion can't be formed in first reaction why does the second reaction form it? It should also favour carbocation on the alpha position." I guess you didn't read my answer properly, I've mentioned why that is not feasible(or better if I say "won't count for the major product").

And the bromonium ion won't really undergo a 1,2-hydride shift because bromonium is itself more stable than the species it will form after the shift.


text:
Why the DMM (multimeter) register negative resistance when measure a drop of hydrochloric acid
I noticed that when I put the multimeter electrodes (in the ohmmeter mode) in a drop of hydrochloric acid with mild concentration (15%) the multimeter displays negative values. Can someone explain this behavior?

Because you have made an electrochemical cell.

The multimeter probe tips are dissolving and setting up a galvanic cell that augments the current flow thru the ohmmeter.
The probe tips may be of the same metal, but the ohmmeter voltage will favor the dissolution of one over the other, and the cell thus formed will assist the applied voltage.
Quickly switching to voltmeter (or milliammeter) mode will show a cell voltage (current).
Using the ohmmeter to measure resistance of a non-corrosive solution (e.g., NaHCO3) will show some (positive) resistance.
Inert electrodes (Pt, Au) will show a positive resistance.


text:
If ethyl bromide is treated with alcoholic silver nitrate, which product will be formed?
Silver nitrate promotes ionisation , so the ethyl bromide should ionise, the alcoholic species should attack (by loan pair of oxygen) so finally an ether should be formed , but the answer given in my book is nitroethane and ethyl nitrite.

Do you mean ethyl nitrile ($\ce{CH3CH2CN}$) or ethyl nitrite ($\ce{CH3CH2ONO}$)? I find it hard to see how ethyl nitrile could be formed at all.

The book probably assumes that the 'alcohol' is not a very nucleophilic one, so the nitrate ion attacks; and as the nitrate ion is bidentate... Everything adds up, except for the alcohol not participating at all; I'd also have thought it would do something, but there you go.

@user6376297 how nitrate can add to form nitroethane, this is my confusion, I can see formation of nitroethane when nitrite is used, but when nitrate is used , how nitroethane form ?

related https://chemistry.stackexchange.com/questions/88515/reaction-of-tert-butyl-bromide-with-silver-nitrate

@Mithoron, the dupe question is alkyl halide + silver "nitrite" but this question specifically asks about silver "nitrate".

@NilayGhosh It's OP's typo, pretty obvious from context.


text:
How to choose alcohol solvent for organic extraction?
Assuming I have an organic substance that is known to be "soluble in alcohol" (e.g. capsaicin) I want to extract, which alcohol should/can I use? I know that both ethanol and methanol are commonly used, but what about:

isopropanol
glycerol
n-Butanol
1,4-Butanediol
tert-Butyl alcohol
1-Pentanol
tert-Amyl alcohol

In general, which alcohols are suitable/unsuitable for organic compound extraction (e.g. which should be the best solvent in terms of amount of solute dissolved in given volume of solvent), and why? Is there any a priori way to determine e.g. if isopropanol extraction will be OK for a given "soluble in alcohol" organic substance?

Note: I am aware of this discussion, but it doesn't cover the specific nature of alcohol extraction.

Need to consider how you're going to isolate the extract once you've done the extraction. Some of your possible solvents have high boiling points and will not be easy to remove.

@Waylander going up to 300C is OK with me; most of them are boiling @ <150C, one way or the other.

'Soluble in alcohol' probably just means soluble in ethanol. Of course it may well be soluble in other alcohols as well.

What are the foreseen uses of the extraxted? How many solvents are available to you? How do you suppose to recover the extracted?  Once one know details as these,  then one might can easily answer. Cost and toxicity can matrer. And so on.  You mention that you can heat up to 300 °C , but capsaicin does not care of that as it melts at 62°. And so on.

The optimal solvent will extract the greatest amount of desired product and the least amount of difficult-to-remove substances.
The extraction is only the first step; separation from solvent and other extractives follows. Try one, then another solvent to see if there is a difference. Maybe number of carbons or hydroxyls is important; you can also mix alcohols, and/or add water.
Capsaicin has a phenolic group; perhaps it would be extracted from its source with alkali (NaHCO3 or Na2CO3).


text:
Atropisomerism of naphthyl alcohol

In March's Advanced Organic Chemistry (6th ed.), it's stated that the above two are atropisomers. I don't get why is it so. How is the topmost $\ce{C}$ locked in its place? Surely, there's steric hindrance due to presence of two methyls but how does that prevent rotation? It can certainly rotate with the methyls being "rigid" in their positions. Also, are they chiral? I believe they aren't because they are superimposable mirror images.

Think of the di methyl hydroxy group as the fins of a fan and planar hydrogen of naphthalene jutting out. Rotation will be restricted.

The hindered rotation is due to the hydrogens at the naphthyl moiety. The following shows the rotation calculated at the DF-B97D3/def2-SVP level of theory. To better visualise this, I have chosen a mode with large atoms (not actually using the van-der-Waals radii, because that looked very strange; click here for a ball and stick version):

We can observe three minima and three transition states corresponding to that rotation, the depiction you have posted is slightly inaccurate as it shows one local minimum and a transition state.
From the three states that should be observable, the global minimum is achiral (1), while the other two are enantiomers (2, 3). The latter are connected via a low lying transition state (TS-2-3); probably because of the hydrogen bonding interaction guiding it. The global minimum on the other hand is separated via a very high energy barrier. These transition states (TS-1-2, TS-3-1) are again enantiomeric to each other.
I have calculated the associated energy barriers in $\pu{kJ/mol}$ at the DF-B97D3/def2-TZVPP level of theory.
\begin{array}{c*{6}{r}}\hline
\text{State} &
\Delta E_\mathrm{o} &
\Delta G^{\pu{50 K}}&
\Delta G^{\pu{75 K}}&
\Delta G^{\pu{100 K}}&
\Delta G^{\pu{298.15 K}}&
\Delta G^{\pu{500 K}}\\\hline
%
\mathbf{(1)}     &
    0.0 &  0.0 &  0.0 &  0.0   &  0.0  &  0.0\\
\mathbf{\text{(TS-1-2)}}     &
   32.9 & 32.9 & 32.9 & 33.1   & 35.7  & 39.7\\
\mathbf{(2)}     &
    0.7 &  0.7 &  0.7 &  0.8   &  1.4  &  2.3\\
\mathbf{\text{(TS-2-3)}}     &
   12.0 & 12.1 & 12.1 & 12.3   & 14.9  & 18.9\\
\mathbf{(3)}     &
    0.8 &  0.8 &  0.8 &  0.8   &  1.5  &  2.3\\
\mathbf{\text{(TS-3-1)}}     &
   32.9 & 32.9 & 33.0 & 33.1   & 35.7  & 39.8\\\hline
\end{array}

From these values we can estimate rate constants for the energy barrier with the Eyring equation, which employs the transition state theory:
$$
k = \kappa \frac{k_\mathrm{B}T}{h}\exp\left\{\frac{-\Delta G}{RT}\right\}$$
I use $\kappa = 1$, i.e. I am ignoring it, to get a guesstimate at which temperature we will be able to distinguish between all, or some of the states.
\begin{array}{c*{5}{l}}\hline
k/\pu{s-1};\ T= &
\pu{50 K}&
\pu{75 K}&
\pu{100 K}&
\pu{298.15 K}&
\pu{500 K}\\\hline
%
\mathbf{(1) \to (2)}     &
   \pu{4.9E-23} & \pu{1.8E-11} & \pu{1.1E-5} & \pu{3.5E6}  & \pu{7.5E8}\\
\mathbf{(2) \to (3)}     &
   \pu{2.6E-1}  & \pu{5.4E3}   & \pu{7.8E5}  & \pu{1.5E10} & \pu{1.1E11}\\
\mathbf{(3) \to (1)}     &
   \pu{2.8E-22} & \pu{5.9E-11} & \pu{2.8E-5} & \pu{6.1E6}  & \pu{1.3E9}\\
\hline
\end{array}
From this we can see, that at room temperature the rotation is pretty much free with a couple million transitions per second. At around $\pu{100 K}$ the is hindered; from the global minimum (1) to the adjacent minima the methyl groups simply cannot pass. Without haven calculated the values, I'd guess that this rotation starts to become hindered at around $\pu{150 K}$.
The other barrier still is easily manageable. That essentially means, that the enantiomers still readily interconvert. You will spectroscopically observe a superposition, which gives rise to the schematic structures that were used in the book. Even at $\pu{75 K}$ this barrier is taken a couple of thousand times per second
Only around $\pu{50 K}$, the interconversion between the enantiomers will also be hindered, and you should be able to distinguish all three states.
TL;DR:
You will start seeing atropisomerism at about $\pu{150 K}$, but at that temperature you'll only see the superposition of 2 and 3 (probably what is depicted in the book, i.e. left in your question) and the achiral, global minimum (on the right).

Side note:
It should be fairly obvious, that non covalent interactions also play an important role in this molecule. The barrier between the enantiomers 2 and 3 is probably significantly lower because of the hydrogen bond. However, one shall not forget, that this interaction is also present in the local minima to a not insignificant amount. The following pictures show the non covalent interactions analysed with NCIPLOT-3.0 and plotted with VMD; green areas are weakly attractive, blue areas are stronly attractive (hydrogen bonds), red areas are repulsive.


text:
Why is the oxidation state of Carbon (-4)?
Based on my understanding, atoms try to achieve the electronic configuration of the nearest noble gas, by losing or gaining electrons. For example, Sodium is closest to Neon, so loses an electron so the outer shell is full (2nd shell is full). This doesn't include the transition metals or the lanthanides or the actinides, which I realize follow something different because of their orbitals.
In the case of Carbon, it is equally distant from Neon as it is from Helium. There's the octet rule, but why exactly does it prefer Neon's configuration over Helium?

Valency don't have a sign. Do you mean oxidation state?

And -4 Is far from the only possible oxidation state for carbon.  We can get any whole number from -4 to +4 inclusive, with some fractions thrown in on the side (e.g. cyclopropenyl ion =+(1/3)).

Basically, Carbon forms covalent bonds in majority of compounds. It doesn't mean that it has lost its electrons to the atom to which it is bonded to - rather it shares its electrons so as to get noble gas configuration.
Oxidation state is a hypothetical number assigned to each atom in a molecular structure (refer this) which assumes all bonds to be purely ionic. So it doesn't mean that that the carbon atoms in $\ce{C_2H_2}$ having an oxidation state of $-1$ have gained $1$ electron each to result in 5 electrons in the valence shell.
If you meant the $\ce{C^{4-}}$ (carbide) ion, carbon actually prefers this because removing 4 electrons is difficult as it is quite endoergic as it requires nearly 14000 kJ/mol (refer this) to form a $\ce{C^{4+}}$ basically due to its small size, which means that electrons are closer to the nucleus and we can remove it only at the expense of high energy. Thus, $\ce{C^{4-}}$ is comparatively easy to form as it is easier to add electrons into the valence shell.

$\ce{C^{4-}}$  is impossible to form.  No atomic ion with 2 or more negative charges is bound.  Only when imagining a compound as purely ionic does  $\ce{C^{4-}}$  come to mind.

Yes. I was just comparing the ability to remove or add four electrons to a carbon atom


text:
NMR Spectra Oddities
I've noticed that as the chain length of n-alkanes increase, their number of carbon environments tend to 5 peaks, and no more, no less. Why is this? I've looked on multiple NMR (C13 NMR, I might add) databases and they show that there are 5 carbon environments. The outer carbons are all distinct environments like you'd normally expect, but the inner part of it (e.g. the innermost carbons in tridecane, C13H28) is all one environment. I hope somebody could shed some light on it. It's stumped both me and my chemistry teachers.

The environments are the same only to within the limits of the spectrometer's resolution and other experimental details; remember that every measurement comes with uncertainties, some of which are baked into the very fabric of the universe.
If you had an "infinitely precise" NMR spectrometer, all the peaks would be distinct (except when symmetry demands they be the same). In fact, every single possible compound would be uniquely identifiable by its atoms' specific shifts, shared with no others. In principle, you'd be able to assign a carbon peak at 123.6207426[...]964109454164[...] ppm to riboflavin, whereas a peak at 123.6207426[...]964109454165[...] ppm would indicate monobrominated hexabenzocoronene intercalated into a strand of DNA, and so on. Of course this is just a thought experiment and would never be possible in practice, even though it would make characterisation of complex molecules and mixtures a dream...

Yes, I thought that may be possible. However, is that the only reason, do you think? Would there be another reason that it works too, apart from simply the spectrometer only able to be so precise? And why 5 peaks? Why not four, or six? Do you know why the number of peaks doesn't keep on increasing even after, say, dozens and dozens in the carbon chain? Thank you for the comment; it really helped. I'm just wondering if that's the only explanation for it.

I'm afraid that trying to quantify at what point the signals will start to overlap is probably quite hard from theoretical considerations alone. It makes sense for it to happen though; the further apart things are, the weaker they interact with each other. You might find it interesting to check the 13C NMR of a sequence of 1-functionalized alkanes, e.g. 1-hexanol, 1-heptanol, 1-octanol, 1-nonanol, 1-decanol, .... See, for example, here.

Specifically, to get such precise peak positions, you would need to have a lifetime of the excited state of the spin system in the order of days. I.e. a T2 of many hours. That's not a limitation of the spectometer, but of the very sample itself.

There is a very simple anwer: these methylene groups are practically chemically equivalent. It is impossible to make a specific reaction with one of them.
Chemically equivalent hydrogens have the same chemical shift, and that's it.
Additionally, when two protons with similar CS couple, they create a higher order spectrum instead of two doubletts. The closer the CS are, the more the outer peaks get suppressed and the inner coalesce. Classic transition from AB to AA', the roof effect becomes so pronounced that you only see one peak, perhaps with a funny shape.
With 13C+decoupling, you seem to be slightly better off because there are no interactions (line splitting) visible. Still each line has an intrinsic bandwidth, because the relaxation times are finite, and the magnetic field has a finite homogenity. And the temperature distribution over the sample has a finite homogenity. Etc.
And of course you have in fact still interaction with surrounding protons, only they cancel out in the spectrum. Doesn't mean they're not there. Because otherwise you'd get relaxation times in the order of days with decoupling, which is not the case.

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


text:
What's the reaction mechanism for the oxidation of primary alcohols (e.g. ethanol) to carboxylic acids using acidified KMnO4?
I've seen mechanisms for $\ce{MnO4-}$ in base reacting with primary and secondary alcohols to oxidise them, but what about $\ce{MnO4-}$ in acid?
All the ones I've found have either been referring to $\ce{MnO4-}$ in base or just showed the redox reaction occurring. I've been trying to look for one for a while now, but I still haven't found a satisfying one (i.e. one that goes beyond showing the reaction in redox equation form).
I hope anyone who knows anything about it can answer me on that, and as thoroughly as you can. Thanks.

related https://chemistry.stackexchange.com/questions/4143/reaction-mechanism-for-oxidation-of-primary-alcohol-to-carboxylic-acid https://chemistry.stackexchange.com/questions/49542/mechanism-of-the-oxidation-of-alcohols-with-kmno4

also https://chemistry.stackexchange.com/questions/9001/difference-in-the-functioning-of-alkaline-and-acidified-kmno4 https://chemistry.stackexchange.com/questions/43283/mechanism-for-oxidation-of-primary-alcohols-to-carboxylic-acids but I don't see proper duplicate.

I was looking for a mechanism specifically about acidified KMnO4. I can't seem to find one very easily, not even here on Stack Exchange. I hope someone has an answer for it.

Well, it looks like doesn't it? And it's not particularly strange, redox mechanisms are notoriously obscure and often debatable. Even if you find it there's high probability it's more or less wrong.


text:
Ionic Character, on which factors do it depend
I was watching a lecture of chemistry teacher. He gave the following question:
Q Arrange in terms of increasing ionic character: LiF,NaF,KF,RbF,CsF?
My try: I tried two approaches:
Approach 1: As F is common to all compounds, therefore its metal that affects the ionic character. As we move down the group as in this case, the shielding effect increases and hence it becomes more easier to abstract an electron from the outermost shell or in other words, Ionization energy decreases as we move down the group, therefore CsF>RbF>KF>NaF>LiF should be the order.
Approach 2: Li, Na, K, Rb, Cs have same electrovalency. As Cs has more larger size, the positive charge intensity on the atom decreases and thus by formula of electrostatic force F=-kq1q2/r^2, the electrostatic force of attraction decreases and thus lattice energy decreases and therefore order is:
LiF>NaF>KF>RbF>CsF.
According to me lattice energy should determine ionic character as lattice energy determines ionic bond strength I think and therefore approach 2 should be more legitimate but when I checked the answer, the answer by Approach 1 was correct. Why is it so? Which of the approach is right and which is wrong and why? Is there any other approach?

When you are considering lattice energy as the parameter for judging the ionic character of compounds, you are assuming two things.

The constituting elements of this alkali metal halides are purely ionic,and thus the Lattice energy (U) is the released energy when those two isolated gaseous ions forms the crystal lattice. But, in practice the constituting ions are never purely ionic (especially smaller alkali metal halide such as $\ce{LiF} $ or $\ce{NaF}$.)
The ions constituting the crystal lattice are hard non-elastic spheres. They hardly (in general) obey this postulate. This postulate is again not totally valid for smaller alkali metal halides due to distortion of their electron cloud due to large polarization.

So, it is seen that if you just theoretically say that the electrostatic force as simply Coulomb's law and thus lattice energy decreases, it doesn't conclude anything as it is seen that the Lattice energy is not a proper parameter for judging ionic character.
The correct approach in judging the covalent character present in an ionic compound is to follow Fajans' rules. According to this rule, $\phi _{cation}$ is inversely proportional to $r_{cation}$ and directly proportional to $Z_{cation}$ Here, $Z_+$ for all the cation is same. But radius of $\ce{Li^+}$ is smallest and ionic radius gradually increases from $\ce{Li^+}%$ to $\ce{Cs^+}$. So, $$\ce{$\phi$_{Li^+} > $\phi$_{Na^+}  > $\phi$_{K^+}  > $\phi$_{Rb^+}  > $\phi$_{Cs^+} }$$ Here  $\phi$ is the measure of polarising power of cation. As, the value of $\phi$ increases, the polarising power increase and thus the anion's electron cloud gets attracted towards cation nucleus which tends to form a kind of overlap between the orbitals of cation and anion, which increase covalency and decreases ionic character.
So, according to Fajans' rules, increasing ionic character (i.e. more covalency to less covalency) is, $$\ce{LiF < NaF < KF < RbF < CsF}$$

thank you. I got it. I have to work on understanding Fajan's rule.


text:
Langmuir method
I have done a biosorption study using a protein as a biosorbent and six different concentrations of $\ce{Pb}$ ions. I need to show the efficiency of the biosorbent, I was told to use the Langmuir isotherm to represent these results. Thus far, I
have plotted a graph of $q$ vs. $C_\mathrm f$ using the formula
$$q = (C_\mathrm i-C_\mathrm f)\times\frac{V}{m}$$
I have also derived my $R$ value from the straight line obtained. How do I get to the Langmuir isotherm from here? What do I need to calculate and what I should plot to represent the data in Langmuir isotherm? Also, I am a biologist so I would appreciate step by step help as I am very confused with all the different equations for Langmuir.

Could you label your variables, so it's clear what $\mathrm{q}$, $\mathrm{C_i}$, $\mathrm{C_f}$, $\mathrm{V}$, $\mathrm{m}$ and $\mathrm{R}$ are?

Sure, Ci -- initial concentration of Pb ions (ug/L) Cf- residual concentration of Pb ions (ug/L), V- volume of the reaction (L), and m - mass of biosorbent (g).

So $\mathrm{q}$ is the amount of absorbed material per mass of biosorbent, is that right?

Introduction
The Langmuir isotherm is a theoretical equation that describes the reversible adsorption of a non-interacting adsorbate on a surfaced that contains a fixed number of identical sites. Therefore, the main assumptions of the model are:

There is a maximum amount of adsorbate that can be fixed to the surface, $\mathrm{S_{max}}$. In other words, the adsorption can reach a saturation point.
All sites are equivalent, and there are no other factors (for instance roughness) that make sites more available or more binding than others.
Sites are not affected by the status of neighbouring sites; all sites are independent. So, neither the local nor the global degree of saturation impacts how likely it is for a single adsorbate particle to adsorb or desorb from a site.

With these assumptions, an expression can be found (it can be derived in terms of chemical equilibrium, chemical kinetics or statistical mechanics) that expresses the relationship between the adsorbed amount of material, $\mathrm{S}$, and the concentration of the adsorbate in the adjacent phase, $\mathrm{C}$:
$\mathrm{S = \frac{K_a C}{1 + K_a C}S_{max}}$
where $\mathrm{S}$ is the amount of adsorbed material, $\mathrm{S_{max}}$ is the maximum possible amount of adsorbed material (in the saturation point), $\mathrm{K_a}$ is the equilibrium constant (written in the direction of adsorption) that regulates the adsorption/desorption of an adsorbate particle on a single site, and $\mathrm{C}$ is the concentration of the adsorbate on the adjacent phase.
It is very common to make the equation intensive by expressing the amount of adsorbed material as a fraction of the maximum possible amount, called coverage ($\mathrm{\theta}$):
$\mathrm{\theta=\frac{S}{S_{max}}=\frac{K_a C}{1+K_a C}}$
Linearisation
The main disadvantage of that formulation of the Langmuir isotherm is that it's not linear, and fitting data to that expression is not straightforward. Consequently, there are various linearisation schemes that have been employed in the literature:
Hanes-Woolf: $\mathrm{\frac{C}{S}=\frac{1}{K_a S_{max}}+\frac{C}{S_{max}}}$
Lineweaver-Burke: $\mathrm{ \frac{1}{S} = \frac{1}{K_a S_{max}} \frac{1}{C}  +\frac{1}{S_{max}} }$
Eadie-Hofstee: $\mathrm{S=S_{max}-\frac{1}{K_a}\frac{S}{C}}$
Scatchard: $\mathrm{\frac{S}{C}=K_a S_{max}-K_a S}$
All of them allow for a linear relationship involving $\mathrm{S}$ and $\mathrm{C}$ to be plotted. Although in principle all equations are equivalent to the formulation above and should lead to the same values, in practice they affect the fitting of real data differently.
In my experience, Lineweaver-Burke is the most commonly used as it introduces less bias in the estimation of the values as other linearisations, but it has the drawback of being less precise at low values of $\mathrm{S}$.
Lineweaver-Burke
So, to fit your data to a Lineweaver-Burke equation, you need to plot $\mathrm{1/S}$ (y axis) vs. $\mathrm{1/C}$ (x axis). Your points will then fit a linear equation of the type
$\mathrm{y=a+bx}$
where
$\mathrm{y=\frac{1}{S}}$
$\mathrm{x=\frac{1}{C}}$
$\mathrm{a=\frac{1}{S_{max}}}$
$\mathrm{b=\frac{1}{K_a S_{max}}}$
Your example
As far as I can tell, you express the adsorbed amount of material as a load $\mathrm{q}$ with units of micrograms of adsorbate per gram of biosorbent:
$\mathrm{q=\frac{n_{ads}}{m}=\frac{\Delta C·V}{m}=(C_f-C_i)\frac{V}{m}}$
so, in the formulation above, $\mathrm{q}$ corresponds to $\mathrm{S}$ and $\mathrm{C_f}$ corresponds to $\mathrm{C}$, and your Lineweaver-Burke expression is
$\mathrm{\frac{1}{q}=\frac{1}{K_a q_{max}}\frac{1}{C_f}+\frac{1}{q_{max}}}$
Plot $\mathrm{1/q}$ vs. $\mathrm{1/C_f}$ for your points and fit them to a linear equation. Your y-intercept corresponds to $\mathrm{1/q_{max}}$ and your slope corresponds to $\mathrm{1/K_a ·1/q_{max}}$. From there you can obtain the values for your equilibrium constant $\mathrm{K_a}$ and your saturation load $\mathrm{q_{max}}$.

"$K$ is the equilibrium constant that regulates the adsorption/desorption of an adsorbate particle on a single site" Correct me if I'm wrong, but isn't the $K$ for the adsorption and desorption processes different?

$k$ (the kinetic coefficient) for adsorption ($k_a$) and desorption ($k_d$) will in general be different, but $K$ (the kinetic pseudo-equilibrium constant)  is precisely the quotient of these two kinetic coefficients. As with a true thermodynamic equilibrium constant, we can write it in both directions, but these are mathematically equivalent ($K_d = K_a^{-1}$ by definition).  By convention, in adsorption chemistry, we always use the adsorption expression ($K_a=k_a/k_d$), so it often gets simplified to $K$. I've edited the answer to unambiguously identify that it's $K_a$.


text:
Which of - carboxylic group or carbaldehyde group - has more negative inductive effect?

Which has more -I effect $\ce{-COOH}$ or $\ce{-CHO}$, and why?

I tried to analyze it on my own but I was confused on what to do. My teacher today taught it by shifting of electrons from $\ce{C=O}$ to the formation of a charge separated structure. Still, I was confused about it.
Mostly, I'm confused on why we are even allowed to create a charge separated structure. Even with it, it seems $\ce{-COOH}$ should have more -I effect. But I am not sure. I need help here.

Are you sure your teacher explained the -I effect by "the formation of a charge separated structure" and the -M effect instead?

Yes, he made the charged structure and I don't know why it was feasible. I'm not sure under what conditions we are allowed to do so. Sometimes these things seem to be bit mechanical and goes over my head.


text:
How can I attempt to inhibit the hydrolysis/degeneration of aspartame?
I've surfed the internet, looking for some possible inhibitors for the hydrolysis process of aspartame, but couldn't find any. I'm wondering if such a thing is possible and if yes, how should I proceed?
I'm writing an essay about how the rate of degradation of aspartame could be slowed down and wanted to attempt using some compounds to inhibit the reaction.

Hydrolysis under what conditions? In vivo? In aqueous solution?

aqueous solution

Aspartame is hydrolysed in both acid and alkaline solution with the methyl ester being the first site hydrolysed, so control of the pH of the aqueous solution is important. Addition of buffer solutions would enable control of the pH, keeping it close to neutrality.
https://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html


text:
Vibration motion at the imaginary frequency for TS of the Claisen rearrangment
I understand that the vibration along the reaction coordinate is different for the transition state , as the motion of the atoms takes TS towards products/reactants.
However, what would be the vibrational motion corresponding to the imaginary frequency of the Claisen rearrangement?

I started at PM7 niveau, where it looks like this:

and at the DFT niveau (DF-B97-D3/def2-SV(P)) it is not that different:

To summarize: Id say, it looks "as expected".
The transition state needs to connect both local minima. When you follow the animation of the normal modes above, you can see bond forming and bond breaking between either the two carbons on the left top or the carbon and the oxygen on the bottom right.

#p pm7 opt=(ts,noeigen,calcall)
8        0.416763000     -1.336451000     -0.208978000
6        1.497193000     -0.113742000      0.149606000
6       -0.772891000     -1.071748000      0.210613000
6       -1.509741000     -0.011936000     -0.327122000
6        0.829456000      1.060608000     -0.253868000
1        2.339023000     -0.475266000     -0.444938000
1        1.609073000     -0.346478000      1.210997000
6       -0.368126000      1.414823000      0.371361000
1       -0.923592000      2.282901000      0.026388000
1       -0.520920000      1.227340000      1.430860000
1        1.106186000      1.529032000     -1.191942000
1       -2.506791000      0.192973000      0.036038000
1       -1.388722000      0.264355000     -1.370246000
1       -1.103706000     -1.651275000      1.071122000

#p b97d3/def2SV/W06 opt=(ts,noeigen,calcall)
8       -0.556373000     -1.341873000     -0.246607000
6        1.267703000     -0.946693000      0.176479000
6       -1.313091000     -0.435583000      0.272443000
6       -1.504992000      0.814802000     -0.295637000
6        1.395538000      0.372483000     -0.314454000
1        1.717678000     -1.782296000     -0.375359000
1        1.166045000     -1.105900000      1.259763000
6        0.768640000      1.425447000      0.328791000
1        0.744341000      2.432184000     -0.112679000
1        0.518449000      1.360808000      1.397182000
1        1.705654000      0.503456000     -1.362984000
1       -2.163740000      1.551184000      0.190272000
1       -1.227963000      0.987609000     -1.343598000
1       -1.692271000     -0.594810000      1.314532000

That's true, it's exactly what you'd expect.
However, because of the bonds breaking and forming at the TS, would you account for the BSSE when computing the activation barrier for the reaction?

I never really saw someone using BSSE correction, if it is not directly about it. From what I could see, the common way is to use bigger basis sets.


text:
Why are acids and bases are defined as both donors and acceptors of species in the acid-base theories?
So, a Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. On the other hand, a Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor.
Why is it that in these acid-base theories, acids and bases are defined as both donors and acceptors of species? Or is there a part of acid-base theory I don't quite understand?

I honestly don't understand your question. How else do you want acids or bases to be defined as? Some ions will always be donated into the solution, and some will always be accepted from it. What other way is there for acids or bases to function?

These are primarily old theories that were kind of evidence or observation based. These are not completely correct theories as there are may loop holes in them. For eg. many types of compounds with significant acidic and basic characters cannot be predicted by these theories. Also, to a certain extent the Bronsted and Lewis concepts are complementary.Wikipedia says - ""

A Lewis base is often a Brønsted–Lowry base as it can donate a pair of electrons to H+; the proton is a Lewis acid as it can accept a pair of electrons. The conjugate base of a Brønsted–Lowry acid is also a Lewis base as loss of H+ from the acid leaves those electrons which were used for the A—H bond as a lone pair on the conjugate base.

and

In 1923, Lewis wrote An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The Brønsted–Lowry acid–base theory was published in the same year. The two theories are distinct but complementary.

Basically, proton is a $H^+$ ion and we all know from our simple understanding that compounds that give $H^+$ or proton are acids and those that accept $H^+$ or proton are bases.

They're distinct alright, but complementary as mentioned. I prefer the Brønsted-Lowry theory though... It's easier to understand for me.

@KianStevens You know, its not absolutely fine I  would see. For eg. if you are reading Friedel Crafts Alkylation you would come across the need of a Lewis acid like AlCl3 or BF3. If you are acquainted with the Lewis theory, you would face there as AlCl3 is not predicted by the Bronsted Lowry concept.


text:
Is the Hartree-Fock procedure the same thing as the self-consistent-field (SCF) theory?
Is it right to think that the Hartree procedure is called the SCF theory, as the notion of self-consistent field comes from approximating multi-electron wavefunction as multiple one-electron wavefunctions. Since we don't know the final charge density, we must iterate until it is self-consistent?

These are two different questions, and really should be divided. (It's site policy to only ask one question at a time, in a given Question.) Please re-ask your second paragraph in a new question. Check out the [help] for more information on site policies.  Thanks for asking here -- welcome to the site!

HF and SCF are used interchangeably quite often (whether you agree with it or not).

To those who think this question is too broad: it has a concrete answer which is given below and it is definitely not too broad. It is a fundamental part of the theory that is not discussed in modern literature.

As an extension of HF and SCF being used interchangeably, I often see (and use myself) SCF to cover both HF and DFT, as the Kohn-Sham equations are solved using an SCF procedure that is identical to the one used for HF.

No.
The Hartree-Fock method encompasses a set of assumptions used to formulate and solve a series of equations embodying a mean-field, time-independent Schrödinger equation.
The self-consistent field method is a computational technique used in myriad applications where the equations to be solved are implicit—that is, where it is impossible to isolate an analytically solvable expression for a variable and its derivatives.  To the best of my knowledge, any solution method involving a "guess-and-check" approach can be considered a 'self-consistent solution method'.
The Hartree-Fock method is an example of a use-case for a self-consistent field procedure, but they are not synonymous.

Excellent answer. I would note that the author's confusion likely stems from old literature using the phrase 'SCF' when describing HF, as back then it was the first widely used SCF method. Nowadays, as you've pointed out, there are numerous SCF methods that are not HF.


text:
SCF procedure performed for some quantum-chemical methods, but not others. Why?
The SCF procedure is used to find solutions to the Schrödinger equation, and obtain energy of the molecular system. It uses a "guess-and-check" approach, where an initial guess is needed to solve the equation, giving the charge density, which can be used again and again until the results become self-consistent.
So, why don't some quantum-chemical methods perform SCF, while others like HF do?

Most computation methods write the wavefunction($\Psi$) in terms of Slater determinants ($\Phi$)
$\Psi = \sum_i^L c_i\Phi_i$
which in turn are made from MOs ($\phi$), which are linear combinations of basis functions ($\chi$)
$\phi_i = \sum_\mu^K C_{\mu i}\chi_\mu$
The molecular energy depends on the MO coefficients ($C_{\mu i}$) in a non-linear fashion, so if they are variationally optimized then you need to iterate (i.e. SCF). This is the case for HF (where $L$ = 1 and $c_1$ = 1) and MCSCF (where both $\{c_i \}$ and $\{C_{\mu i} \}$ are optimized).
For methods like CI, perturbation theory, and coupled cluster, the MO coefficients are not changed, so you don't need an SCF.

It can also be said more generally that there are two main methods for solving the Schroedinger equation: variation and perturbation. When you use variation, you alter some coefficients of basis "things" (basis functions in SCF, Slater determinants in CI or both in MC-SCF) to make the initial guess of WAVE FUNCTION for the solution more and more consistent with the Hamiltonian of the system. When you do perturbation it is the HAMILTONIAN itself which is gradually improved to arrive at more precise wave function (which should be already made consistent with the initial Hamiltonian, though). Perturbation method is very general and hence versatile, used not only for energy calculations; historically, it was the first expansion beyond the HF - DFT came in much later and does not compute the actual wave function, only the electron density (and also the energy therefore).
Answering @Umkay comment on whether DFT uses SCF:
DFT (more precisely, Kohn-Sham version of it) DOES perform SCF. The crucial point is that DFT uses a different approach to the description of the system than the usual SCF/perturbation methods. These methods try to find the most reasonable approximation to the wave function of the system; DFT instead takes notice of many properties being a direct function of election density instead of the wave function itself; and, as the density has only three arguments (x, y and z) instead of 3*N for N-electron wave function, it tries to go round and compute the density not from the "best-approximated" wavefunction.
It appears, though, that quite efficient way for this is to construct a fictional system which is defined to have the same density as the "true" one but has a wave function which is less undertaking to compute. Namely, this system have elections without direct interaction between them (i.e., without corresponding terms in the Hamiltonian) but, like in HF, having some self-consistent term describing their interactions with less computational burden. The difference with the HF is that KS-DFT uses this term to describe interactions between electrons which arguably can be described by such a term: the correlated motion of [distinct pairs of] electrons. In HF it is only the electrostatic field created by all the electrons which is described in this way (KS-DFT uses also that but not just that). The probable impossibility to have such CORRELATIONAL term is supported by the still ongoing search for the "best" density functional. Hence also the debate on whether DFT counts as an ab initio method. More importantly, it means that the orbital-related properties computed with DFT are quite nominally corresponding to those of the real system - DFT was simply not designed to get them, after all. On the other hand, one may say that to get density "good enough", you probably also need the "fictional system" close enough to the real one. For example, it was once shown by Stowasser and Hoffman in 1999 that for ionization properties KS-DFT orbitals look very similar to the "actual" orbitals for this case - the Dyson orbitals - computed with some high-level method.
But KS-DFT still makes use of SCF for computing the wave function of its "fictional" system. And there are the double-hybrid functionals which also employ perturbation approach. But these are not quite consistent with the initial formulation of DFT, because, as A. Becke noted in 2014 review, they use unoccupied orbitals which are merely by-products of employing SCF for DFT as they have no direct connection with the electron density at all.

Welcome to Chem.SE, Igors! Nice to see you made the hop over from CCL.

Since DFT does not compute the actual wave function, it doesn't perform SCF, correct? Because I read somewhere on this site that SCF covers both HF and DFT, as the Kohn-Sham equations are solved using the SCF procedure identical to HF.

@hBy2Py: I just got to read the CC Daily mail and saw that I am capable of answering :-)

@Umkay: I have edited the answer itself, I hope it does rely to your comment.


text:
What are the consequences of having an element with an atomic number above 137?
In other words, is an element such as Feynmanium the last chemical element that can physically exist on a planet such as Earth?

137 is in no way special.

I think he is referring to a predicted limit for Bohr atoms. When you look at the shell model of atoms 137 is the limit, because 138 is supposed to have electrons that are moving faster than the speed of light. I think when you add modern quantum mechanics you can show that the number should be higher but I can't find that reference anymore.

Almost duplicate: https://chemistry.stackexchange.com/questions/8073/the-last-elements-atomic-number

The critical value "137" occurs in calculations based upon relativistic theories that treat the nucleus as a point, such and Dirac theory or Sommerfeld theory.
However, when the nucleus is correctly treated as having a non-zero size, a significantly higher value is calculated.
The most recent calculations I see are in New method for solving the $ \bf Z > 137$ problem and determining hydrogen-like energy levels (2014) Phys.-Usp. 57, 189.  (alternative link to pre-print):

Following the development of the Dirac theory in 1928, Dirac [reference 2], Darwin [reference 3] and Gordon [reference 4] obtained expression (1) as a result of exact solution of the Dirac equation in the Coulomb field of a point charge (-Ze). ...  Formula (1) [results in a complex number] if Z [is greater than 137]...the complexity of energy levels in (1) is often called the “Z>137 catastrophe”.
In 1945, Pomeranchuk and Smorodinsky [6] considered
an atomic system with [the nucleus being a sphere of radius $0.8 \times 10^{-12}$cm and having uniform charge within the nucleus and found 175 to be the critical value].

The article goes on to explain that since the 1945 calculation, various calculations, with various nuclear charge distribution models, have calculated critical values in the 170-175 range.  The article's own calculations find 178.

I believe you are overthinking it, and that by a wide margin. The "critical" value 137 occurs in calculations based upon non-relativistic (and not even completely quantum) Bohr theory, when you equate the electron velocity and the speed of light. Refuting it does not require anything nearly as sophisticated. Its reappearance on a different level of theory is hardly relevant, though of course not accidental.

@IvanNeretin I think I'm biased because of a previous question that asked whether the 137 still occurs in relativistic theories.  https://chemistry.stackexchange.com/questions/41645/relativistic-effects-in-element-137-feinmanium-and-above

@IvanNeretin https://chemistry.stackexchange.com/questions/41645/relativistic-effects-in-element-137-feinmanium-and-above?noredirect=1&lq=1#comment67038_41645


text:
Truncated CI Wavefunction
In a calculation of for example a Li-Atom in a DZ basis we have 4 MO.
For the alpha string (containing two electrons), that means we have for a FCI calculation the determinants:
|12>, |13>, |14>, |23>, |24>, |34> (electrons in MO 1 and 2, 1 and 3 and so on).
If I wanted to do a truncated CI calculation (CIS for example) only singly excited wavefunctions would contribute.
My (maybe extremely dumb) question now is: Are the determinants |23> and |24> singly excited or doubly excited? Is it the excitation
$$a^\dagger_3a_1|12> = |32> = -|23>$$
 or is it
$$a^\dagger_2a_1 a^\dagger_3a_2|12> = |23>$$
 Or is it both? And the coupling (1,-1) so to say determines whether this in used as a single or doubly excited determinant?
EDIT: From the "Molecular Electronic Structure Theory" book by Helgaker, Jorgensen and Olsen, from the chapter 11.8.5 (Direct CI algorithms for RAS Calculations) I read "In CISD theory, there are three such classes (string classes): class 0, with no electrons in virtual space; class 1, with one electron in virtual space; and class 2, with two electrons in virtual space".
This for me reads as if |23> would count as a class 1 string and not a class 2.

Both. If you think about it from the context of CC theory, the first determinant will come from $\hat{T}{1}$, and the second will come from $\hat{T}{2}$ and $\hat{T}_{1}^{2}$. The prefactors handle overcounting.

As for the book statement, without any more context, I would view this as a partitioning that avoids the overcounting by simplifying the number of strings before generating the determinants.

Also, to be more exact, I know $\hat{T}{2}$ is not the same as $\hat{T}{1}^{2}$, but I don't think it's important here.

@pentavalentcarbon yes you're right, my error was in thinking it's both |23>, but one is |32> and the other |23>. Writing the question actually made myself realise my error 5 minutes later :D

But isn't that a separate issue? I have to admit that because I don't do CI/CC I can't immediately see what you're talking about. I also don't see there being a problem with with question, it's a good one.

I also don't know about the coupling, maybe if I tried to work it out...

@pentavalentcarbon Well one problem would be that it kind of answers itself.

In configuration interaction (and in coupled cluster) we typically think of excited determinant being caused by single particle excitations of the form $\hat{t}^a_i = \hat{a}^\dagger_a\hat{a}_i$ where $i$ labels a virtual orbital and $a$ labels an occupied orbital, causing an excitation of the Hartree Fock reference determinant $\Psi_0$.
The resulting CI expansion would go:
$$
\Phi_\text{CI} = \Psi_0 + \sum_{i\in occ\\ a \in vir} \hat{t}^a_i \Psi_0 + \sum_{i,j\in occ\\ a,b \in vir} \hat{t}^{ab}_{ij} \Psi_0 + \sum_{i,j,k\in occ\\ a,b,c \in vir} \hat{t}^{abc}_{ijk} \Psi_0 + \: ...
$$
Hence, in the calculation on Li mentioned above, we would not consider the excitation $\hat{t}^2_1 = \hat{a}^\dagger_2\hat{a}_1$ in the first place as it would annihilate the Hartree Fock determinant $|12>$ containing a particle in orbital 2.
By this logic the determinant $|23>$ is a singly excited determinant caused by $\hat{t}^3_1= \hat{a}^\dagger_3\hat{a}_1$, or class 1 as described by Helgaker.
Note:
As determinants $|23>$ and $|32>$ only differ by a multiplicative constant (-1) only one needs to be considered and the resulting coefficient/cluster amplitude will take care of the relative sign.


text:
Difference between 'Preferred IUPAC Name' and 'Systematic IUPAC Name'
When looking at different chemicals, I come across the 'preferred IUPAC name' and the 'systematic IUPAC name'. I've noticed that when I name chemicals, I name them by their 'systematic IUPAC name', as I've been taught to name them using this system.
When is the 'preferred IUPAC name' used? I sometimes come across it when looking at aldehydes, ketones and carboxylic acids. However, the fact that it's 'preferred' makes me wonder whether chemists use this method more often than the other system.

The problem with systematic IUPAC names is that many compounds can have more than one systematic IUPAC name. A compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC. Thus, asking for the systematic IUPAC name does not make sense for such compounds.
Therefore, current IUPAC recommendations (currently only for organic chemistry) include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including also the many retained names.
Whether the preferred IUPAC name is also actually preferred by chemists cannot be generalized and depends strongly on the context.

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


text:
The reactions of grignard reagent
Question

Find y in this question :

The $\ce{ -CH2-Cl}$ below according to me will not take part the reaction (or please correct me) thus my answer is y=2 but actually it's y=3

This is surely from an older version of MS Chauhan. I remember this being an error in the book which was corrected in the recent edition.

The third Grignard reagent molecule could extract a proton from the methylene group which is in an electron withdrawing environment.  This type of enhanced acididity on carbon is referred to by the term "carbon acid" .
While this reference deals with carbonyl groups, conjugation with a phenyl ring can do the same thing to the methylene group.
We do not see any change of the methylene group in the final product because it's just deprotonated, and the acid workup brings the proton back.  We would, however, see the third Grignard reagent being protonated to its parent hydrocarbon, in this case methane which bubbles off.

What about the third equivalent of MeMgBr coupling with the benzyl chloride to produce an ethyl group. Alternatively, ethoxide may cause SN2 displacement of chloride or, less likely, the intermediate magnesium alkoxide of the product may displace chloride and effect polymerization.

Possible, but apparently not considered by the text.  Grignard reagents are known to preferably act as Bronsted-Lowry bases instead of nucleophiles at the drop of a hat, and conjugation of the methylene group with an aromatic ring drops the hat.  The resulting deprotonation would kill the coupling.  Here It is noted that benzyl chloride +Mg gives 98% Grignard reagent and only 2% coupling.

If allyl MgX is prepared under standard conditions, the yield is modest owing to coupling. Must be run dilute.


text:
How to make a pH 6.8 Tris buffer?
To preface this, I am a biologist.
I need to make a $\mathrm{pH}=6.8$, $\pu{1 M}$ Tris buffer. We have access to Tris base and Tris Hydrochloride. (Unfortunately, none of the fancy Tris compounds will work in my situation.)
I tried making a straight Tris base buffer, but before I could lower the $\mathrm{pH}$ to 6.8 (using concentrated $\ce{HCl}$), it crashed drastically. I assume that I got past the buffering capabilities.
Are there any recommendations for this? Thank you.

You need $\mathrm{pH}$ of your Tris buffer @ 37 degrees, I believe.

From Wikipedia...

The conjugate acid of tris has a pKa of 8.07 at 25 °C, which implies that the buffer has an effective pH range between 7.5 and 9.0.

So you're outside the range where TRIS is an effective buffer.
I assume by "crashing" you mean that a precipitate formed. A buffer is also rated based on its "buffer capacity." Effectively this depends on the the concentration of the buffer. Thus solubility of the buffers comes into consideration.

Tris is short for tris(hydroxymethyl)aminomethane. Tris has a $\mathrm {pK_a}$ of $\mathrm {8.1}$ and a $\mathrm {pH}$  level between $\mathrm {7}$  and $\mathrm {9}$. Although that range may change at physological temperature, I doubt it go as low as $\mathrm {pH} = 6.8$ without sacrificing  $\mathrm {[B]/[A]}$, which may go as low as $\mathrm {\approx0.05}$.
Thus, I think you should use $\mathrm {BIS-TRIS-propane}$ instead. $\mathrm {BIS-TRIS-propane}$ (1,3-Bis[tris(hydroxymethyl)methylamino]propane) is two Tris molecules connected by 1,3-diaminopropane. It has $\mathrm {pK_a}$s of $\mathrm {6.8}$ and $\mathrm {9.0}$, and effective $\mathrm {pH}$ range of $\mathrm {6.3-9.5}$.
I'm not sure how soluble it in deionized water, but worth trying ($\mathrm {$1/g}$). Probably a good idea to keep your stock solution in $\mathrm {0.5 M}$ range.

Another choice: Ypu may use N,N-Bis(2-hydroxyethyl)-2-aminoethanesulfonic acid (BES) (CAS Number: 10191-18-1), instead. It has p$K_\mathrm{a}$ of $7.193$ at $\pu{25 ^{\circ}C}$ (Ref.1) and useful pH range of $6.4-7.8$. BES is therefore better suited for pH control in the buffer region of biochemical interest (Ref.1,2). Sigma-Aldrich sells it in 99% pure form (~$1.0/g).
BES is soluble to the extent of $\pu{3.2 M}$ at $\pu{0 ^{\circ}C}$ (Ref.2) and can be easily purified from aqueous alcohol. Two equimolal buffer solutions consist of BES and sodium besate (sodium salt of BES), each at a molality of $\pu{0.02 mol kg^{-1}}$ and $\pu{0.10 mol kg^{-1}}$, respectively have shown p$K_\mathrm{a}$ value of each solution as $6.97$ at $\pu{37 ^{\circ}C}$ (Ref.1). On the other hand, a solution prepared with $\pu{0.03 mol kg^{-1}}$ of BES and $\pu{0.02 mol kg^{-1}}$ of sodium besate has shown p$K_\mathrm{a}$ value of $6.76$ at $\pu{37 ^{\circ}C}$ (Ref.1).
References:

R. N. Roy, E. E. Swensson, G. LaCross, Jr., C. W. Krueger, “Standard buffer of N,N-bis(2-hydroxyethyl)-2-aminoethanesulfonic acid (Bes) for use in the physiological pH range 6.6 to 7.4,” Anal. Chem. 1975, 47(8), 1407–1410 (DOI: 10.1021/ac60358a076).
N. E. Good, G. D. Winget, W. Winter, T. N. Connolly, S. Izawa, R. M. M. Singh, “Hydrogen Ion Buffers for Biological Research,” Biochemistry 1966, 5(2), 467–477 (DOI: 10.1021/bi00866a011).


text:
When to disconnect tubing in vacuum filtration?
The UCLA site says in step 6 to disconnect the vacuum tubing before turning off the vacuum and the second link from Leeds says in step 11 to turn off the vacuum source first and then disconnect the tubing.
I've been taught both ways and am confused about the difference in methods.
When should one actually disconnect the tubing and what is the rationale for each method?

First disconnect the tubing else the vacuum in the erlenmayer flask can suck in water (if you use a water pump). This can be tedious even if your filtered through phase has to be discharged (e.g. solvent wastes separation). If your target is the liquid, than is detrimental.

What about a regular vacuum pump?

I would do the same but with a normal setup nothing shall happen. It depends on pump,  what was used before.... With a line or a pump there should be a trap in between vacuum and erlenmayer. ...

If you were to turn off the pump while leaving the object attached then if you have an oil filled pump or a water pump then you will have a suck back which will either damage the pump or contaminate the thing you are filtering, or if you are very unlucky both.
If you have a house vacuum system then you should close the valve / tap between your filtration and the vacuum manifold before you remove the hose to your flask. If you do it the other way around then you will interfere with the use of the house vacuum by others.
I think you need to know if you have your own vacuum pump or if you are using a house vacuum system where a centralised pump serves many people.


text:
Fried potatoes absorbing oil
Suppose we bring a pan and put in it some oil and turn on the heat, before the oil could get very hot and nearly boiling, we put on the potatoes that start frying.
Someone told me that if the potatoes were put in oil before it gets really hot, then they will absorb more oil than if they were put in oil after it gets really hot.
Is this true?? If yes then WHY?
Note:(given that in both cases they stay the same period of time immersed in oil).

How did you try to solve the problem? It will help the answerers to answer your question!

It is an example of the Leidenfrost effect. Ideally the water being expelled from the food would actually keep the oil from contacting the food itself by forming a barrier of steam (vaporizing water) around the food.

Try it first by weighing out similarly shaped potatoes before and after frying using the two methods described in order to calculate the amount on oil absorbed from the difference (do this multiple times and obtain a standard deviation so you can make sure that the difference is not due to experimental errors) and after the real answer is obtained one can start thinking why (or actually have your friend do it so he can prove you that he is correct).
My prediction is that is will make no difference on the basis that oil and water don't mix and considering that potatoes contain significant amount of water they wont let much oil in before the temperature increases (which will evaporate the water and allow the oil to move in the potato). In both cases you ll end up with potatoes saturated in fried oil.

Thank you, but potatoes don't contain only water, but other components as well, so maybe oil would be absorbed by those other components. And maybe as well to water but for a short time, because I realize that after serving the fried potatoes, after some time, the potatoes tend to leak some oil..

Potatoes are 80%  wt water. Try the experiment above and i think you ll see no difference.