text:
Why does covalent bonding not break down if observer effect can be applied to atomic electrons?
The observer effect in quantum mechanics states that when unobserved, quantum particles such as electrons can simultaneously occupy two different states. In an atom of any element, where there are electrons around the nucleus, this effect is hinted at in some publications:

Electrons can occupy several energy levels, or orbitals, simultaneously. (Scientific American.(2012, October 9). Bringing Schrödinger's Cat to Life. Retrieved from https://www.scientificamerican.com/article/bringing-schrodingers-quantum-cat-to-life/.)

This idea is so weird. It is equivalent of saying that when unobserved, all the electrons of an atom can both be in their normal configuration and be in a totally different configuration. Only the act of observation forces the electrons into their normal configuration. Take the example of neon atom. According to the observer effect, when unobserved, its electron configuration can both be 1s4 2s3 2p3 (one of many possible alternative configurations) and 1s2 2s2 2p6. Only when observed will the electron configuration collapse into its normal version of 1s2 2s2 2p6.
I don't know if this understanding of observer effect on atomic electrons is valid. If it is, there are at least two serious problems. First, the Pauli exclusion principle will break down, because the alternative electron configuration allows more than 2 electrons in the same orbital (specified by the principal quantum number, the angular momentum quantum number, and the magnetic quantum number). Second, the foundation of covalent bonding is that when two or more atoms pool together electrons in orbitals that overlap, the total energy of the system is reduced to the minimum. If the observer effect on atomic electrons is valid, the energy required to maintain the system may get even lower than that in covalent bonding, and then covalent bonding will break down. All that's needed is for each atom to achieve an alternative electron configuration.

Thanks, everybody. I now understand this question better. Yes, the observer effect does apply to atomic electrons by allowing each of them to occupy several orbitals, subshells, or even shells at the same time. However, if we look at all the electrons and summarize their states into an electron configuration scheme, the scheme will still be the same as that obtained through the aufbau principle.

Why do you think that a superposition of different states must include invalid states? It can't.

So, does superposition suggest two simultaneous states when both states are valid? In this way, I can see that in an atom, all the electrons can be in different orbitals at the same time, but when we give the electron configuration, both states conform to the ground-energy scheme.

All of the states that are simultaneously existing describe the same overall electron density and the same total energy. The only difference is that each electron does not have defined quantum numbers.

This is a quite confusing interpretation of quantum mechanics of atoms and solids. Perhaps one should consider that covalent bonding is, in fact, the alternative electron configuration that minimizes energy of the system.

Let me add that in a molecule, not only quantum particles are involved (electrons) but also very massive "things" (nuclei), with a slow motion compared to electrons and a non-quantistic behavior. Electrons might even go crazy for some fractions of femtosecond, but their lowest-energy state is due to the environment in which they are found (around nuclei)

@The_Vinz nuclei have "non-quantistic" behavior? I don't think that is correct.

I was stating in simple terms the Born-Oppenheimer approximation :) Aside from that, as far as I remember only low-mass nuclei like H are said to be involved in quantum effects (tunneling). Note that I am mentioning only effects which affect motion, not things like interactions with magnetic fields, or nuclear excited states

I think part of your confusion stems from neglecting the fact that the electrons in a single molecule are all entangled with each other, so while an individual electron can occupy many states simultaneously, in each of those states it is entangled with all the other electrons such that the sum of states remains the same.
To start with a simple example, consider an atom of helium in its singlet ground state. Prior to measurement, neither electron has a defined spin. Both are a superposition of $+1/2$ and $-1/2$ states, but that does not mean that the helium atom can have $+1$ or $-1$ spin. The net spin is still 0. If we measure one electron's spin to be $+1/2$, the other will necessarily have $-1/2$, even though neither had defined spin prior to measurement.
We can extend this to multi-atomic molecules. Consider ground state methane. We can calculate delocalized molecular orbitals that collectively describe the charge density at any point in the molecule, with each of those orbitals being composed of a standard hydrogenic orbital of carbon (eg 2s or 2p) and athe 1s orbitals of all four hydrogen atoms. Alternatively, we can describe the same charge density distribution with four localized orbitals, each comprised of a single hydrogen 1s orbital and a hybrid sp3 orbital on carbon. Prior to making an observation, neither of these descriptions is correct, but neither is incorrect. Both correctly describe the charge density of the molecule, but both incorrectly assign electrons to specific states. Instead, the electrons are collectively in both states (and an infinite number of others that describe the same density). Because every contributing state correctly describes the charge density, there is no change in the covalent structure resulting from the superposition.


text:
Why does an electrochemical cell work?
My textbook says that an electrochemical cell works because of difference in reduction potential of two metals which cause one to lose electrons and other one to accept them. If a salt bridge is not used then the solutions of the two metals become charged and the reaction proceeds till the difference in reduction potential is cancelled by the difference in potential due to charges. All this theory seems to work well on paper, but it is never really explained why this reaction occurs.
What I mean is suppose you have a copper and a silver half cell. They are connected. The copper part does not know whether it is connected by a silver half cell or sodium have cell and I see no apparent of this information being communicated between the cells, so how does it know whether to get dissolved in the solution or precipitate out of the solution?
The book says that due to the difference in reduction potential, a potential gradient is set up which causes the charge to flow. I agree with this completely yet I am not able to grasp why this physically happens. What force causes to electrons to move from one electrode to the other? Especially when there is no salt bridge, the electrons move from positive electrode to negative electrode, which is completely opposite to anything I know.
I think that there must be some difference in local phenomena in the solution which spontaneously causes electrons to flow without any information being shared between the two electrodes. eg: At first I thought maybe copper having lesser reduction potential than silver would dissolve and precipitate out faster (basically a fast(?) equilibrium) which is completely independent of the other electrode and this difference speed would somehow cause the electrons to flow. I rejected this idea after a while but I still think difference in some sort of local phenomena in a half cell which is completely independent of the other half cell would drive the reaction forward.
Any thoughts?

Do you understand how a simple single replacement reaction works? For example, hydrochloric acid reacting with zinc?

".. I see no apparent of this information being communicated..." I fear your textbook forgot to mention the metallic wire connecting the two electrodes.

For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed at great length (way too long for an answer here) and the metal cohesive energy is the source of more than 75% of the energy in the Daniell cell (p. 1805).

@Karl the metallic wire connecting the two electrodes does not carry any information, I can very easily connect copper with a sodium half cell and it would be none the wiser. This is what leads me to believe that the process must be governed by some local phenomena.

@EdV I read the paper and your exemplary answer, but I am not able to generalize your reasoning to a copper silver cell. It is fairly easy to see that diffusion would be the driving force in a concentration cell but the driving force in this cell is not so apparent. Note that I use the word force. This is because everywhere the reaction is explained in terms of energetics but the driving force behind the reaction is not specified.

For example, according to arrhenius theory, reactions take place if two compounds collide with right amount of energy and the right orientation, diffusion takes place as the probability of gases diffusing is much higher than not diffusing, a ball falls due to gravity. It's energy is surely decreasing as it falls, but that is not the force which causes it to fall.

@Zhe Yes, a collision occurs between a zinc and Hcl and if the orientation and the energy of the collision is just right then zinc displaces hydrogen. One could argue that hydrogen could again collide with zinc chloride and displace zinc but this does not happen as the activation energy for such a reaction would be quite high as zncl2 is much more stable than Hcl . This is confirmed as the reaction is highly exothermic.

What you need to see in that example is that there is a preference for the electrons to move from zinc to proton. All you need to do is to set up a system with this kind of a preference and force the electrons to move through a wire instead.

I see it as the wire attached between the Cu and Ag couples the redox equilibria at each electrode. If the Cu and Ag electrodes are initially polished to mirror finishes and placed in solution of their respective salts, their surfaces become altered, but there is no net effect at either electrode. But adding the connecting wire couples these electrode surface equilibria: if a Cu atom is oxidized at the anode, the two electrons can either reduce a copper ion in the solution or, ultimately, reduce two silver ions in the cathode compartment. The latter is favored.

The wire carries electrons, i.e. a voltage. That's all you need. You can artificially change the voltage, and reactions reverse.

@EdV it is incredible that two redox equilibria can be coupled just by connecting a wire

Maybe you would understand better with an image, or if you admit that all metals "are wanting" to loose electrons. And in order to do that, they must find another atom which is weak enough or "humble" enough to accept to work "against its will". In other words, some atoms are more "willing" than others. It is like a fight, with a "strong" and a "weak" atom. In your copper-silver cell, copper is more assertive : it imposes its electrons to the "weak" silver atom. Silver cannot emit its own electrons. It is even obliged to accept electrons from copper and have these electrons react with its proper ions. producing more silver atoms. Some atoms are stronger than copper. Zinc for example, opposed to copper, imposed its electrons to copper. This order of "strength" is experimental, and cannot be derived by calculations.
Now to speak more scientifically, you may replace the present notion of strength by the reduction potential. The more the reduction potential is negative, the more the metal has a tendency to loose its electrons.
Anyway, a salt bridge is absolutely necessary for the cell to work. If there is no salt bridge, it is not a cell, and it produces no chemical reaction, and no electric currant.

Complement, required by Manit.
If you want a local phenomena to explain the cell, you may consider the following development. You may admit that copper "wants" to create more $\ce{Cu^{2+}}$ ions in solution. These ions are positively charged. So the copper compartment must attract negative ions to compensate the new positive charges appearing around the copper plate. As a consequence, the new $\ce{Cu^{2+}}$ ions attract negative ions from where they are available, namely in the other compartment. And as $\ce{Ag}$ is not able to create $\ce{Ag+}$  ions with the same force, Silver plate must admit that it looses its negative nitrate ions to send them to the copper compartment. And as a consequence, it must admit that the positive ions $\ce{Ag+}$ should also "disappear" from the compartment. And the only way to make this charge disappear from the compartment is to accept electrons, so that the $\ce{Ag+}$ ion is transformed into metallic silver.
OK ?

Your explanation makes a lot of sense if , say we put a copper spoon in a silver nitrate solution, the nitrate ions will cause the copper to come into the solution and now there would be a competition between silver and copper ions and then copper would win, forcing the silver ion to get deposited on the copper spoon. But in this case the copper ion is not in direct contact with silver and that is why I believe the reaction must be governed by some local phenomena completely independent of the metal in the other half cell.

Ok I get it. So basically you are saying that both copper and silver are always applying force on the electrons in the wire, but when the switch is open there is no source or sink of electrons hence nothing happens but when the switch is closed the force due to copper overcomes the force due to sliver and hence a net force acts on the electrons.

I it is not like fight or something it depends solution pressure and osmotic pressure

An important thing is not to limit yourself to the idea of electrochemical cell with metals, being dissolved or deposited.
More general image is the general redox half-reactions:
$\ce{<oxidized form> + n e- <=> <reduced form>}$
The oxidized form may, or may not be a metallic ion like $\ce{Cu^2+}$ or $\ce{Ag+}$.
The reduced  form may, or may not be a metal of the electrode, like $\ce{Cu(s)}$ or $\ce{Ag(s)}$.
E.g. one of industrial energy storage types uses cells with this aquaeous vanadium system with inert electrodes:
$$\begin{align}
\ce{V^2+(aq) &<=>[discharging(anode)][charging(cathode)] V^3+(aq) + e-}\\
\\
\ce{VO2+(aq) + e- + 2 H+(aq) &<=>[discharging(cathode)][charging(anode)] VO^2+(aq)  + H2O}
\end{align}$$
There are ongoing reactions in both directions at electrodes, even if not connected to any circuit. If an electrode happens to have the equilibrium potential, both reactions have the same rate with the zero net electron production.
If an electrode has a lower than the equilibrium potential, the reduction reaction consuming electrons is faster and the electrode potential raises towards the equilibrium one. Unless the potential is externally forced, e.g. at alectrolysis. Similarly, if an electrode has a higher than the equilibrium potential, the oxidation reaction producing electrons is faster and the electrode potential falls down towards the equilibrium one.
Electronic chips are known to be using electronic charge pumps to internally generate needed voltage levels. Electrode systems can be analogically considered as chemically powered charge pumps.
If 2 electrodes are connected by a galvanic circuit, the current disbalances their potentials and "the chemical charge pumps" start feeding and collecting the electrons to/from the electrodes. If the chemical processes are fast, the cell is a hard voltage source with low internal resistance ( like car acid lead batteries ). If they are slow, the cell is a soft source with high internal resistance.

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Electrophilic Substitution with Ceric Ammonium Nitrate and Lithium Bromide as reagents
The 2020 IChO Preparatory Problem List (source here) had the following reaction as part of the solution to the first question:

(CAN = Ceric Ammonium Nitrate).
What is the mechanism for this 'Electrophilic Substitution' of the ring? I have not encountered such a reaction before. As far as I'm aware, CAN is an oxidizing agent, whereas Lithium Bromide would dissociate to give $\ce{Br-}$, which is not an electrophile. Aromatic Nucleophilic Substitution (ArSN) is out of the question here, as the ring is highly activated by the methoxy group.
The only probable reaction I can think of is that CAN successively oxidizes $\ce{Br-}$ to $\ce{Br2}$, then to $\ce{Br+}$, but this seems unlikely, and I could not find such a reaction mentioned anywhere in literature. A mechanism for this reaction, preferably along with a reference, would be much appreciated.

Is CAN and KBr a good enough similarity?

https://shodhganga.inflibnet.ac.in/bitstream/10603/119795/10/10_chapter%204.pdf The reaction is present with a yield of 90% Page 24

I'm inclined to think that it is more likely that CAN is oxidising the aromatic to a radical cation intermediate that is then captured by Br-. If you have access (I do not) this review may cast some light https://pubs.acs.org/doi/10.1021/cr068408n

@Waylander: I think OP's suggestion is correct. This is similar to iodination of salicylamide in the presence of $\ce{NaI/NaOCl}$ (https://doi.org/10.1021/ed085p1426).

You are correct as the answer below demonstrates

@Waylander I wrote the answer btw :) Also, thanks for the reference.

First of all, the linked paper clearly points out the following:

Hint: In second step (A $\rightarrow$ B), combination of lithium bromide and cerium(IV) ammonium nitrate (CAN) is used as a brominating reagent

From a problem solving perspective, this step is trivial for someone who has read the problem well (not me, unfortunately :)

Ceric Ammonium Nitrate is a very strong oxidizing agent and has the ability of oxidizing benzene compounds first to Nitrobenzene, then to quinone or quinoid derivatives1. Assuming this line of reasoning works, the quinoid derivative could then undergo nucleophilic addition with $\ce{LiBr}$, followed by workup with acid to give the p-subsituted benzene derivative. This could be a probable reaction mechanism, if the substitutent was $\ce{-OH}$ instead of $\ce{-OMe}$ and if an acid workup step was involved. Also, this mechanism can give the unwanted product 4-bromo-3-isopropylphenol, as shown below:

However, this reaction mechanism is very speculative, and it also won't hold good here, because the substituent is $\ce{-OMe}$. Some further research on the links provided in the comments2 yielded the following results:

Alkyl aromatics were brominated in the side chain using sodium bromide and CAN. Asakura et al. reported bromination of C-5 of uracil nucleosides using CAN and lithium bromide in acetonitrile.

This is also the reaction present in (3), which seems to have been lifted without a citation to Asakura et al.

Finally, Asakura et al. 4 provides a suitable mechanism for this reaction, which I correctly assumed arose from the oxidation of $\ce{Br-}$ to $\ce{Br2}$, and then to $\ce{Br+}$

References:

Nikishin, G. I., et al. “Reaction of Aromatic Hydrocarbons with Cerium Ammonium Nitrate.” Bulletin of the Academy of Sciences of the USSR Division of Chemical Science, vol. 24, no. 6, June 1975, pp. 1328–29. doi:10.1007/BF00922077.
Nair, Vijay, and Ani Deepthi. “Cerium(IV) Ammonium NitrateA Versatile Single-Electron Oxidant.” Chemical Reviews, vol. 107, no. 5, May 2007, pp. 1862–91. doi:10.1021/cr068408n.
Cerium (IV) Ammonium Nitrate mediated oxidative transformations of diaryl methylenecyclopropanes. Shodhganga link
Asakura, Junichi, and Morris J. Robins. “Cerium(IV)-Mediated Halogenation at C-5 of Uracil Derivatives.” The Journal of Organic Chemistry, vol. 55, no. 16, Aug. 1990, pp. 4928–33. doi:10.1021/jo00303a033.

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On the spin-adaptation of 2nd order Møller-Plesset perturbation theory
$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2] { {\left<#1} \left|#2 \right>}$
In spin adaptation (common in electronic structure theory) we approximate that the spin-up and spin-down spin orbitals (herein assumed to be real and represented by $\chi$ with some placeholder spin coordinate $\vec{s}$ and spatial coordinates $\vec{x}$) have the same spatial orbitals (represented by $\phi$ with spatial coordinates $\vec{x}$). This allows us to eliminate the spin components of the molecular orbitals (MOs) in exchange for an 8-fold speed-up of two-electron integral calculation and storage:
$\chi(\vec{x_1})=\phi(\vec{r_1})*\omega(\vec{s_1}) \rightarrow \phi(\vec{r_1})$
I am trying to spin-adapt the spin-orbital, second-order MP2 energy correction (using canonical MOs):
$E^{(2)}=\frac{1}{4}\sum_{ijab}^{SO} \frac{\Bra{ij}\Ket{ab}^{2}}{\Delta _{ab}^{ij}}$
(where $\Delta _{ab}^{ij}=\epsilon{_i}+\epsilon{_j}-\epsilon{_a}-\epsilon{_b}$, where i and j represent occupied MOs in the reference solution, where a and b represent unoccupied MOs in the reference solution, and where $\epsilon_m$ denotes the orbital energy of the $m^{th}$ molecular orbital) to arrive at the closed-shell (spin-adapted) expression:
$E^{(2)}=\sum_{ijab}^{SF} \frac{2\Bra{ij}{ab}\left.\right>^{2}-\Bra{ij}{ba}\left.\right>\Bra{ij}{ab}\left.\right>}{\Delta _{ab}^{ij}}$
Below is my attempted derivation (I use SO and SF to denote whether the indices i,j,a,b denote spin-orbital or spin-free MOs, respectively):
$E^{(2)}=\frac{1}{4}\sum_{ijab}^{SO} \frac{\Bra{ij}\Ket{ab}^{2}}{\Delta _{ab}^{ij}}$
$=\frac{1}{4}\sum_{ijab}^{SF}
\frac {\Bra{i_\alpha j_\alpha}\Ket{a_\alpha b_\alpha}^{2}
+\Bra{i_\alpha j_\beta}\Ket{a_\alpha b_\beta}^{2}
+\Bra{i_\alpha j_\beta}\Ket{a_\beta b_\alpha}^{2}
+\Bra{i_\beta j_\alpha}\Ket{a_\alpha b_\beta}^{2}
+\Bra{i_\beta j_\alpha}\Ket{a_\beta b_\alpha}^{2}
+\Bra{i_\beta j_\beta}\Ket{a_\beta b_\beta}^{2}
} {\Delta _{ab}^{ij}}$
(note: there are other possible terms, but they equate to zero due to orthonormality of spin component. Also note that I have assumed spin-up and spin-down MO energies are equivalent, allowing me to write only one denominator)
We now expand the anti-symmetrized two electron integrals using $\Bra{ij}\Ket{ab}=\Bra{ij}\left.ab\right> - \Bra{ij}\left.ba\right>$ and the orthonormality of the spin component, $\Bra{i_{\omega_{1}}}\left.i_{\omega_{2}}\right>=\delta_{\omega_{1},\omega_{2}}$, keeping only non-zero terms:
$E^{(2)}=\frac{1}{4}\sum_{ijab}^{SF}
\frac {{\Bra{ij}\Ket{ab}}^{2}
+\Bra{ij}\left.ab\right>^{2}
+\Bra{ij}\left.ba\right>^{2}
+\Bra{ij}\left.ba\right>^{2}
+\Bra{ij}\left.ab\right>^{2}
+{\Bra{ij}\Ket{ab}}^{2} } {\Delta _{ab}^{ij}}$
Where I have kept the ordering of terms to facilitate term-by-term comparison to the previous energy expression. We can, of course, combine these terms:
$E^{(2)}=\frac{1}{4}\sum_{ijab}^{SF}
\frac {2{\Bra{ij}\Ket{ab}}^{2}
+2\Bra{ij}\left.ab\right>^{2}
+2\Bra{ij}\left.ba\right>^{2}} {\Delta _{ab}^{ij}} = \frac{1}{2}\sum_{ijab}^{SF}
\frac {{\Bra{ij}\Ket{ab}}^{2}
+\Bra{ij}\left.ab\right>^{2}
+\Bra{ij}\left.ba\right>^{2}} {\Delta _{ab}^{ij}}$
Almost there! Now we expand the anti-symmetrized two electron integrals to get:
$E^{(2)}= \frac{1}{2}\sum_{ijab}^{SF}
\frac {\Bra{ij}\left.ab\right>^{2}
-2\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right>
+\Bra{ij}\left.ba\right>^{2}
+\Bra{ij}\left.ab\right>^{2}
+\Bra{ij}\left.ba\right>^{2}} {\Delta _{ab}^{ij}}$
Which, upon combining terms and factoring out a 2, yields:
$E^{(2)}= \sum_{ijab}^{SF}
\frac {\Bra{ij}\left.ab\right>^{2}
-\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right>
+\Bra{ij}\left.ba\right>^{2} } {\Delta _{ab}^{ij}}$
This is SO close, but not equal, to the spin-adapted energy expression for MP2 (reproduced below):
$E^{(2)}=\sum_{ijab}^{SF} \frac{2\Bra{ij}{ab}\left.\right>^{2}-\Bra{ij}{ba}\left.\right>\Bra{ij}{ab}\left.\right>}{\Delta _{ab}^{ij}}$
But we're in trouble, as $\Bra{ij}{ab}\left.\right>^{2} \neq \Bra{ij}{ba}\left.\right>^{2}$, so the above expressions apparently do not equate. Additionally, two-electron integral symmetries do not help us out. At the time of writing, I had not found the solution. But in the nick of time, I found the workaround, and have posted it below.

$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2] { {\left<#1} \left|#2 \right>}$
It turns out the trick is not a matter of exploiting symmetry or alternate derivations, but rather a simple bookkeeping trick!
$E^{(2)}= \sum_{ijab}^{SF}
\frac {\Bra{ij}\left.ab\right>^{2}
-\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right>
+\Bra{ij}\left.ba\right>^{2} } {\Delta _{ab}^{ij}}$ can be broken into three sums:
$E^{(2)}= \sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right> } {\Delta _{ab}^{ij}}
+\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ba\right>^{2} } {\Delta _{ab}^{ij}}$
The beauty is that the indices of each summation are independent of the indices of the other summations. As such, I am free to rename a as b and b as a in the third summation:
$\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ba\right>^{2} } {\Delta _{ab}^{ij}}= \sum_{ijba}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ba}^{ij}}$
Which, because a and b behave the same in the summation (both over unoccupied MOs) and in the denominator (both represent the energy of an unoccupied MO), can be rearranged as:
$\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ab}^{ij}}$
(In pseudo math, $\sum_{ijba}^{SF} = \sum_{ijab}^{SF}$ and ${\Delta _{ab}^{ij}}= {\Delta _{ba}^{ij}}$). Putting it all together, we get
$E^{(2)}= \sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right> } {\Delta _{ab}^{ij}}
+\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ab}^{ij}} = 2\sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\left.ba\right> } {\Delta _{ab}^{ij}} = \sum_{ijab}^{SF} \frac{2\Bra{ij}{ab}\left.\right>^{2}-\Bra{ij}{ba}\left.\right>\Bra{ij}{ab}\left.\right>}{\Delta _{ab}^{ij}}$
Which proves the equality.

Spin adaptation is certainly a valuable thing in correlation calculations. Unfortunately, I only know of simple schemes for 2 different cases: (1) closed-shell singlets ($S = 0$, $m_S = 0$) and (2) closed shell triplets ($S = 1$, $m_S = 0$). The formula you have cited is closed-shell singlet adaptation. Since the above answers prove the result algebraically, I thought I could show the same result using diagrams which are a very useful tool for, among other things, deriving MPn energy and wavefunction corrections. The Hamiltonian used in post-Hartree Fock correlation calculations is the normally-ordered form
$$H_N = F_N + V_N = \sum_{p,q = 1}^{\#\,\mathrm{spinorbitals}} \langle p | f | q \rangle N[a^p a_q] + \dfrac{1}{2} \sum_{p,q,r,s = 1}^{\#\,\mathrm{spinorbitals}} \langle pq | rs \rangle N[a^p a^q a_s a_r] $$
where $a^p_{(p)}$ are the fermionic creation (annhiliation) operators, $H_N = H - \langle \Phi_0 | H | \Phi_0 \rangle$, $|\Phi_0\rangle$ is the Hartree-Fock reference Slater determinant, $N[\ldots]$ denotes normal ordering with respect to the Fermi vacuum $|\Phi_0\rangle$, and $\langle p | f | q \rangle$ and $\langle pq | rs \rangle$ denote the usual Fock and 2-electron integral matrix elements in the MO basis. The expectation value $\langle \Phi_0 | H | \Phi_0 \rangle$ is simply the Hartree-Fock electronic energy. In the usual MP partitioning of the Hamiltonian, where $F_N$ is the unperturbed part and $V_N$ is the perturbation, the MP2 energy correction is given by
$$\Delta E_{MP2} = \langle \Phi_0 | V_N R^{(0)} V_N | \Phi_0 \rangle$$
where $R^{(0)} = \sum_{n\neq 0} \dfrac{|\Phi_n\rangle \langle \Phi_n |}{-\Delta_n}$ is the reduced resolvent operator for the unperturbed Hamiltonian and $\Delta_n \equiv \Delta_{a_1\ldots a_n}^{i_1\ldots i_n} = \sum_{\gamma = 1}^n (\epsilon_{a_\gamma} - \epsilon_{i_\gamma})$ is the usual MP energy denominator. Using the diagrammatic rules of many-body perturbation theory, we have the following Goldstone diagrams which I've drawn and evaluated on paper using the usual notation that $i,j,k,\ldots$ denote occupied MOs while $a,b,c,\ldots$ denote unoccupied MOs (the rules can be found in books like Szabo & Ostlund):

Spin-adaptation diagrammatically is done by associated a weight factor of $2^l$ where $l$ is the number of closed loops in the Goldstone orbital diagrams. Hence this kind of spin-adaptation can only be done with Goldstones. In the following, we will denote $h$ as the number of internally-contracted hole (occupied) lines, $w$ as the topological weight of the diagram, and $s$ as the sign of the diagram. I'll also show explicitly how this $2^l$ factor can be seen by breaking up of spinorbital states $|p\rangle = |P\rangle \otimes |\sigma_P\rangle$ where $P$ is a spatial orbital index and $\sigma_P = \pm \frac{1}{2}$ is the spin projection.
Diagram I: $w = \frac{1}{2}$, $h = 2$, $l = 2$, $s = (-)^{l+h} = 1$
$2^l$ Rule:
$$\dfrac{1}{2}2^2\sum_{IJAB} \dfrac{\langle IJ|AB\rangle \langle AB|IJ\rangle}{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B} = 2\sum_{IJAB} \dfrac{\langle IJ|AB\rangle \langle AB|IJ\rangle}{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B}$$
Explicitly integrating out spins:
$$\sum_{ijab} \langle ij|ab\rangle \langle ab|ij\rangle  = \sum_{IJAB} \langle IJ|AB\rangle \langle AB|IJ\rangle
\sum_{\sigma_A \sigma_B \sigma_I \sigma_J}\langle \sigma_I | \sigma_A \rangle \langle \sigma_J | \sigma_B \rangle \langle \sigma_A | \sigma_I \rangle \langle \sigma_B | \sigma_J \rangle $$
Switching the middle two inner products (since they're scalars)
$$ = \sum_{IJAB} \langle IJ|AB\rangle \langle AB|IJ\rangle
\sum_{\sigma_A \sigma_B \sigma_I \sigma_J}\langle \sigma_I | \sigma_A \rangle \langle \sigma_A | \sigma_I \rangle \langle \sigma_J | \sigma_B \rangle  \langle \sigma_B | \sigma_J \rangle $$
and using resolution of identity in the spin-$\frac{1}{2}$ space $\sum_{\sigma_A} |\sigma_A \rangle \langle \sigma_A | = \sum_{\sigma_B} |\sigma_B \rangle \langle \sigma_B| = 1$
$$ = \sum_{IJAB} \langle IJ|AB\rangle \langle AB|IJ\rangle
\sum_{\sigma_I \sigma_J}\langle \sigma_I  | \sigma_I \rangle \langle \sigma_J | \sigma_J \rangle  = (2)^2 \sum_{IJAB} \langle IJ|AB\rangle \langle AB|IJ\rangle$$
Diagram II: $w = \frac{1}{2}$, $h = 2$, $l = 1$, $s = (-)^{l+h} = -1$
$2^l$ Rule:
$$-\dfrac{1}{2}2\sum_{IJAB} \dfrac{\langle IJ|BA\rangle \langle AB|IJ\rangle}{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B} = -\sum_{IJAB} \dfrac{\langle IJ|BA\rangle \langle AB|IJ\rangle}{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B}$$
Explicitly integrating out spins:
$$\sum_{ijab} \langle ij|ba\rangle \langle ab|ij\rangle =
\sum_{IJAB} \langle IJ|BA\rangle \langle AB|IJ\rangle
\sum_{\sigma_A \sigma_B \sigma_I \sigma_J} \langle \sigma_I  | \sigma_B \rangle \langle \sigma_J | \sigma_A \rangle \langle \sigma_A | \sigma_I \rangle \langle \sigma_B | \sigma_J \rangle $$
and using resolution of identity again to set $\sum_{\sigma_A} |\sigma_A \rangle \langle \sigma_A | = 1$
$$= \sum_{IJAB} \langle IJ|BA\rangle \langle AB|IJ\rangle
\sum_{\sigma_B \sigma_I \sigma_J} \langle \sigma_I  | \sigma_B \rangle \langle \sigma_J | \sigma_I \rangle \langle \sigma_B | \sigma_J \rangle $$
using $\langle \sigma_I | \sigma_J \rangle = \delta_{IJ}$
$$= \sum_{IJAB} \langle IJ|BA\rangle \langle AB|IJ\rangle
\sum_{\sigma_B \sigma_I} \langle \sigma_I  | \sigma_B \rangle \langle \sigma_B | \sigma_I \rangle $$
using $\sum_{\sigma_B} |\sigma_B \rangle \langle \sigma_B | = 1$
$$= \sum_{IJAB} \langle IJ|BA\rangle \langle AB|IJ\rangle
\sum_{\sigma_I} \langle \sigma_I  | \sigma_I \rangle = 2\sum_{IJAB} \langle IJ|BA\rangle$$
So we can see that the $2^l$ rule really works (and of course, there is a more general proof that covers all cases)
Thus, the total MP2 correction in closed-shell singlet spin-adapted form is:
$$\Delta E_{MP2} =  \sum_{IJAB} \dfrac{2\langle IJ| AB\rangle \langle AB | IJ\rangle - \langle IJ | BA \rangle \langle AB | IJ \rangle }{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B}$$
Only in the case of real orbitals (usual case), we can write this as the more compact
$$\Delta E_{MP2} =  \sum_{IJAB} \dfrac{2\langle IJ| AB\rangle ^2 - \langle IJ | BA \rangle \langle AB | IJ \rangle }{\epsilon_I - \epsilon_A + \epsilon_J - \epsilon_B}$$


text:
Stereo-isomerism in 7-methylbicyclo[2.2.1]hept-2-ene

The above are two compounds.

The objective is to find if the compounds are identical/enantiomers/diastereomers

I can see a plane of symmetry if we keep the $\ce{H-C-CH3}$ in the plane. So, concluding it as identical looks great.
but, if so, how can one transform the first compound to the second?

Applying, CIP rules, we might invalidate the P.O.S. The solution given for this question is "identical".
Update:(attempt)
I have somehow manipulated the bonds such that I can prove the identity. Even though it is extremely-sterically hindered as indicated in the comments section (Now moved to chat). The forceful manipulations are done keeping in mind, not to break any bonds and only .

However this again proves that the molecules are not identical.
Now, since it is proposed that these molecules are different, can I get a lab or computer synthesized version of both molecules? Upon many search attempts, I could get this document which discusses about the same cyclohexene ring but Deuterium instead of the $\ce{CH3}$ at the bridge head and substituted oxygen ions.

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

7-Methylbicyclo[2.2.1]hex-2-ene exists as two diastereomers, anti-1 and syn-1, which differ by the relationship of the methyl group and the double bond. Both compounds are meso with the bridgehead carbons stereogenic and chirotopic (R/S) while C7 is stereogenic but achirotopic (r/s). The plane of symmetry  in each one does not produce an enantiomer of itself. Stereoisomers have the same molecular formula and the same atom connectivity. You probably know that the stereoisomers (R,R)- and (S,S)-tartaric acid are enantiomers and bear a diastereomeric relationship with (R,S)-tartaric acid, which is a meso-compound. A pair of meso-compounds with the same molecular formula and the same atom connectivity may also be diastereomers. Just like structures 1, the central carbons (C3) of ribitol and xylitol are stereogenic and achirotopic. The properties of these two pentaols are different as are the properties of anti-1 and syn-1.


text:
What does an electron's spin of 0.5 and minus 0.5 signify?
While teaching me magnetism, my teacher told me about the spin of an electron. He told me that the spin of .5 means that if we rotate the electron twice counter-clockwise on its axis, we would have the same face(picture) of the electron. And if we rotate an electron of -0.5 spin twice in the opposite direction, we would have the same face. But if rotate it only once(360⁰), we would not get the same face/picture of an electron.
How can this be possible? Or am I just making a mistake somewhere? A simple explanation would be greatly appreciated for I am only a tenth grader.

In my understanding, the wavefunction of fermion particles like electrons is not identical wrt 360 degree rotation, but has the opposite value. Superposition of these 2 values has the zero norm, leading to zero probability. It then leads to zero probability 2 fermions can share the identical quantum state. But I am very far from quantum mechanic expert, so  so I cannot provide detailed reasoning.

A general hint: Do not try to force classical mechanics pictures on quantum mechanics. Sometime this may work, but often not and only lead to wrong ideas. Quantum mechanics is very different and unintuitive at first.

To complement Feodoran, for the given value of the electron  mass and maximal possible electron radius limit, there is no classical way how an electron could have such a high value of the spin angular momentum, so the classical idea of a fast spinning ball cannot be applied.

The particular value $\pm 0.5$ means the multiplier of the Planck constant $\hbar$, providing the value of the preferred axis projection of the electron spin angular momentum $\pm 0.5 \cdot \hbar$. ( Note that $\hbar = \frac{h}{2 \pi} $)

See https://www.youtube.com/watch?v=cd2Ua9dKEl8 and https://www.youtube.com/watch?v=z_6B2M12H9w .  A potentially relevant example of how 2 pi rotation is not enough but 4 pi can be : https://www.youtube.com/watch?v=Nat-EsReXtQ .

It is very tempting (and often also very useful!) to picture electron spin as an angular momentum vector, similar to a spinning top. Using this analogy, there are two properties (or numbers) of this angular momentum vector that we need in order to describe the electron spin. The first one is the spin itself and this is often designated the symbol $s$. The second number is the projection of the spin on a particular axis (often the $z$ axis) and this is assigned the symbol $m_s$.
You probably learned already that not all values of $s$ and $m_s$ are allowed in quantum mechanics. From measurements we know that an electron has a spin of $s=1/2$ (always positive) and its projections can be $m_s=+1/2$ or $m_s=-1/2$. The quantum numbers $s$ and $m_s$ are just numbers we use to label a particular state of the electron, but when we do measurements we also measure units. The units of angular momentum are J s (Joule second) and from comparing the classical definition of angular momentum with the rules of quantum mechanics, we can derive that one unit of angular momentum in quantum mechanics has a value of $\hbar$. This means that the angular momentum of the electron has a value which is basically half of what you would expect based on classical arguments and sometimes people like to emphasize the weirdness of this by saying things like "an electron has to rotate twice around its axis to come back to its original position". Of course this not true because the angular momentum of the electron is not a consequence of a rotation at all (the electron is a point particle). In addition, the sign of the projection of $s$ has nothing to do with this hypothetical rotation either. The sign of $m_s$ does influence the interaction of the electron with an inhomogenous magnetic field. If you have a magnetic field gradient along the $z$ direction, electrons in one state move up while electrons in the other state move down and you will see two separated spots on a position sensitive detector behind the magnet. This is the basis of the famous Stern-Gerlach experiment. Note that you have to perform these kind of experiments with neutral systems that behave like a single electron because a charged electron would experience a much stronger interaction between its charge and the magnetic field (Lorentz force) than between its spin and the magnetic field (Zeeman interaction).
What I find very interesting about half-integer spin is that there is no orientation possible for which the electron spin does not interact with the magnetic field (which would correspond to $m_s=0$ or a perpendicular orientation of the electron spin with respect to the magnetic field).

rather spin number $s=1/2$ versus spin itself $ \hbar \sqrt{s(s+1)} = \hbar \left(\sqrt{sqrt(3)/2}\right) \simeq 0.866 \hbar$, where those $\pm 0.5 \hbar$ are the projection of $0.866 \hbar$ along the preferred axis.

In a way the electron does have to "rotate" through 720° to "get back to original", because a "rotation" through 360° converts the spin wavefunction to the negative of itself ($\exp(\mathrm{i}2\pi\hat{S}_z)|\alpha\rangle = -|\alpha\rangle$, $\exp(\mathrm{i}2\pi\hat{S}_z)|\beta\rangle = -|\beta\rangle$, something to do with spinors...? I know very little...). But I am not really convinced that that is what OP's teacher means...

@orthocresol You are correct of course that the $\hat{S}_z$ is the generator of rotation, although it does not really corresponds to a physical rotation of the electron itself. There is certainly a lot more to the interpretation of spin that I did not address here.

To stress that spin is not really a rotation, maybe a note on where this spin = rotation analogy is coming from might be helpful. AFAIK this ONLY because the mathematical description of spin (the operators and their commutators) is very similar to angular momentum.

the electron is a point particle  Wellll..... not in an orbital it isn't.

@J... not as an orbital ;)

@J... no, one may lead to ideas like electrons moving on trajectories through some confined space. The other one not. Also, you did a very similar thing.

@J... I didn't mean you would imagine it. But people just learning about QM might get some not wrong, but distorted, ideas. I am well aware that this terminology is frequently used like this, and I do to. But just because is it used, does not automatically make it a good idea. And since you started pointing out inaccuracies (which depending on context might just be fine as is), I just pointed out you are doing something similar.

@J... no, you are misinterpreting me again. My point is not really on topic about this question.

@J... Well, you have to start somewhere. But why do you assume that is all I am doing? (Okay, you assumed right. But still, the best strategy would be to try and start a discussion on different levels, including here.)

The strange properties of half-integer spin are one of those mysterious facts that make physics interesting. It's part of the geometry of a mathematical object called a 'spinor'. The name comes from making an analogy with a 'vector', but related to spin.
Spinors are fairly advanced mathematics, but there is an intuitive way to think about them that works a lot of the time. A spinor can be thought of as an ordered pair of oriented reflecting planes passing through a point. 'Ordered' means that we distinguish the first and second plane. 'Oriented' means they have a front and a back face. If you do two reflections one after the other, you get a rotation, so any spinor has a rotation associated with it.
Now if the two planes are identical and facing the same way, the first and second reflection reverse one another, and you get the identity. As you twist one of the planes with respect to the other, the corresponding rotation is through twice the angle between the intersecting planes, about an axis along the line where the two planes meet. When the planes get to 45 degrees apart, they produce a 90 degree rotation. When they get to 90 degrees apart they produce a 180 degree rotation. When they are 135 degrees apart the result is a 270 degree rotation. And when the planes are 180 degrees apart, so they are lined up back-to-back, you get a 360 degree rotation. Note that because the planes have a well-defined front and back, two planes back-to-back are the opposite of two planes facing the same way.
The distinction is invisible if all you can see is the rotation that results, but the physics of electrons can tell the difference between planes back-to-back or facing the same way.
If you keep on rotating one plane with respect to the other, when the first plane has turned 360 degrees and is back to where it started, the rotation produced by the pair of reflections has turned a full 720 degrees. This is what is meant by saying you have to rotate a spinor 720 degrees to get back to where you started. It's the rotation represented by the pair of planes that is being spun.
I should say, the above picture is not quite right. The spinor is really representing some kind of oriented angle in space. It's the intersection axis and the angle between the planes that matters, not the particular planes themselves. So if you rotate both of the planes together about their common axis (rather than just rotating one with respect to the other) you get the same spinor. Similarly, any pair of perfectly aligned planes is the identity and always the same, and any pair of perfectly opposed planes is its negative. I realise that's probably somewhat unclear - it's hard to give an intuitive geometric picture of a very unfamiliar concept.
Now I'm going to get more technical, for anyone who wants to look into it more deeply. There is an extension to the geometry of vectors called 'geometric algebra', which was originally discovered by the English mathematician William Kingdon Clifford back in the days when they were inventing vector geometry. Clifford's geometric algebra was actually invented before vectors were fully developed and accepted, but vectors took over and Clifford's ideas were abandoned and forgotten in physics for a long time.
Geometric algebra has several different sorts of object in it, and unifies them all into a single structure. Scalars, which have no direction or extent; vectors, which represent a one-dimensional extent, a line, a directed length, or a reflection; bivectors, which represent a two-dimensional extent, a plane, a directed angle, or a rotation; and trivectors which represent a three-dimensional extent, or a volume. Scalars, vectors, bivectors, and trivectors can all be added and multiplied together freely, so you can add a vector to a scalar and then multiply the result by a bivector and so on.
The 'spinors' in any space are 'the even sub-algebra' of the geometric algebra: the elements made up entirely of even-dimensional components. So in 3D space, as described above, this consists of just the scalars (0-dimensional) and bivectors (2-dimensional) and linear combinations of those. And in 3D this happens to be the space of products produced by multiplying two vectors together, each vector representing a reflection. When you multiply vectors in geometric algebra, you get the sum of a scalar (the dot product) and a bivector (closely related to the cross product) that describe how the vectors are oriented with respect to each other.
In 3D, the spinors (scalars + bivectors) also happen to be the same as the 'quaternions', which you may also have heard of. In 4D and higher, things get more complicated.
Geometric algebra gives a much more intuitive, 'geometric' picture of spinors, which otherwise are just mysterious abstract entities that you have to take on trust without ever really understanding them, which is what most physicists do.


text:
Why is borole not considered aromatic
I came across a problem which stated that borole isn't aromatic. I thought that borole is similar to pyrrole so it should be aromatic, but that isn't the case. Can I know why isn't it aromatic?
I was also wondering if we replace the nitrogen in pyridine with boron then will the compound remain aromatic?

Draw the pi orbital system, then fill in the available electrons. You will hopefully then be able to answer your questions.

I see that boron has 2 electron available. So won't they go into resonance?

@Safdar Boron has empty p orbitals, but can resonance of borole not involve delocalisation of electrons in the vacant p orbital from the neighbouring pi bonds. Won't this cause charge seperation and negative charge formation on boron. This idea is different from pyrrole where the ring has negative charge but can it not occur?

A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane.
For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole):

Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the obitals going from left to right are $0.13$, $1.9$ and $1.9$ respectively.

[emphasis mine]
The point to be taken here is the sentence in bold, the occupany in the empty p-orbital of boron [left most diagram and first value given] is small compared to the other two π-bond orbitals. This means that there is no complete delocalisation in borole. Only the double bonds are in conjugation.
A reason for this is that the orbital of Boron is an empty p-orbital and so the total number of electrons that can potentially delocalise is $4$ and so not of the form $4n+2$. If anything, the compound would become "anti-aromatic" if conjugation were to happen and so that delocalisation doesn't take place.
Now, on your second question, the lone pair of nitrogen in pyridine doesn't contribute to the aromaticity since the lone pair is out of plane of the ring. So adding placing boron shouldn't affect the aromaticity.

Simply for the beginners in organic chemistry, this explanation of aromaticity is good enough:

An aromatic (or aryl) compound contains a set of covalently bound atoms with specific characteristics:

A delocalized conjugated $\pi$ system, most commonly an arrangement of alternating single and double bonds.
Coplanar structure, with all the contributing atoms in the same plane.
Contributing atoms arranged in one or more rings.
A number of $\pi$ delocalized electrons that is even, but not a multiple of $4$. That is, $4n + 2$ $\pi$-electrons, where $n = 0, 1, 2, 3,$ and so on. This is known as Hückel's rule.

Hence, borole ($\ce{C4H5B}$), which has mimicked pyrrole ring follows first three rules but the forth. It is planner with contributing atoms arranged in a single ring in same plane. It has a conjugated $\pi$-system with consecutive $\mathrm{p}$-orbitals along the ring, but one of them is empty (still valid to count for aromaticity; e.g., tropylium ion, $\ce{C7H7+}$). However, borole has missed the most important rule: Hückel's rule. It does not contain $4n + 2$ $\pi$-electrons, therefore not aromatic.


text:
Why is aniline more basic than pyrrole?
In a question asking to compare the basicity of pyrrole and aniline, the explanation given for aniline being more basic than pyrrole was as follows:

Basicity depends on the availability of lone pair of $e^-$ on Nitrogen atom.
In pyrrole, lone pair of $e^–$ present on $\ce{N}$ are involved in resonance with the ring which provide aromatic character to the ring.
On protonation of pyrrole, the electron will not be above to participate in resonance, While anilines aromatic even though the involvement of line pair of nitrogen of $\ce{—NH2}$ group is not there and lone pair of aniline is delocalized in benzene ring but due to deocalization benzene loss its aromatic character so, the delocalization is not that strong as in pyrole molecule.
Lone pair is more available of protonation. Hence, aniline is more basic than pyrrole.

But, why should the aromatic character get destroyed? It is still a planar structure and should be aromatic. What is an intuitive explanation?

There would be no aromaticity in pyrrole after protonation since there is no lone pair to allow for Hückel's rule for aromaticity (the $4n+2$ rule). So, upon protonation, you are destroying aromaticity in pyrrole. Also the lone pair is completely delocalised in pyrrole and it cannot easily participate as a base.
However, in aniline, there is a partial loss of the lone pair due to resonance but the lone pair of nitrogen is still available for protonation. Aromaticity is not lost in the resonance structures of aniline.
Therefore aniline is more basic than pyrrole but is less basic compared to pyridine (where the lone pair is out of plane of resonance).


text:
Are there methods for iron content determination in spinach that do not require thiocyanate?
I am looking for a way to measure the iron content of spinach for a chemistry experiment. Unfortunately I am not sure that thiocyanate will be available and I'm therefore trying to find an alternate method. The most common one seem to be a titration with potassium permanganate. The issue with this method is that other much more abundant metals in spinach would also react with the permanganate.
I had the idea to ash the spinach, dissolve the ashes in water. Then use a displacement reaction since iron is the least reactive metal in the chemical makeup of spinach. I am not sure exactly how but I thought it might be possible to use a compound to precipitate the more reactive ions out of solutions.
Do any of you have some ideas regarding this experiment?

I suppose you need to dissolve ash in acid, otherwise iron could end as insoluble oxides ( or metal if highly heated with carbonized organics ). The iron oxidation state after dissolution can be uncertain, as oxides tend to be non-stoichiometric and reducting environment during burning undefined.

Through spectrometer: https://chemistry.stackexchange.com/questions/18435/how-to-determine-the-iron-content-in-spinach-with-a-spectrometer?r=SearchResults

Or through titration: https://chemistry.stackexchange.com/questions/42866/how-to-determine-the-iron-content-in-spinach-from-a-titration-with-permanganate?r=SearchResults

@NilayGhosh yes I am looking for a method other than the spectrometer since I may not have access to KSCN. The problem with that titration is that many other metals will react with the permanganate.

@Poutnik How could I proceed from them to determine the amount of iron in the solution?

Hm, with avoiding instrumental techniques ,including spectrometers, it would be difficult, as manganometry may not be specific enough. But anyway, only about 2% of spinach iron ( 2-2.7 mg/100g) would be available for a body, as it is mostly blocked by oxalate, compared to 15-35% by meat sources, where iron is mostly bond in the heme.

@Poutnik I thought oxalate could be "broken down" by cooking/ashing? Yeah manganometry seems like it would be imprecise since so many other metals would react with the manganate. Thats why I was thinking of using another compound before the manganate to precipitate all the iron out of the ash solution, given iron is one of the least reactive metals in spinach. Do you think there would be a way to do this?

I have read oxalate content drops by spinach cooking from 1000mg/100g to one half.

Spinach is not specially rich in iron. $100$ g spinach contains $3.1$ mg Fe. $100$ g beans contains $6.1$ mg Fe. Lentils  $8.6$ mg/$100$ g.. Soja$ 8.4$ mg/$100$ g. Nutmegs $4.5$ mg/$100$ g.   Ref.: K. Diem. C. Lentner, Tables scientifiques, CIBA-Geigy SA, Basel, Switzerland, $1972$, p. $511 - 515$.

$100$ g fresh spinach contains $662$ mg $\ce{K}$, $106$ mg $\ce{Ca}$,  $65$ mg $\ce{Cl}$, $62$ mg $\ce{Na}$ , $62$ mg $\ce{Mg}$, $51$ mg $\ce{P}$, $27$ mg $\ce{S}$, $3.1$ mg $\ce{Fe}$, $0.82$ mg $\ce{Mn}$, $0.2$ mg $\ce{Cu}$. From the same reference as in my previous message.

Don't go reinventing the wheel, there are literature values for the iron contents of vegetables, and that literature comes with procedures how they were determined.

@Karl I understand, but my means are more limited

@ Hugo. Iron(III) can be titrated by EDTA $0.02$ M at pH$ 2.5$ and $45$°C. The indicator is thiosalicylic acid, which goes from red to colorless. If you can obtain this indicator, I will give you the detailed process. 1 mL EDTA $0.02$ M corresponds to $1.2$ mg Fe. This fits your needs. Ca and Mg do not disturb. Titration of Fe(II) can be done by KMnO4, but the iron must first be reduced by a Walden column (metallic silver). I also may give you the process, if you have got such a column.

@Maurice, What reference book do you have? It seems to have a lot of classical analyses.

@ Farooq. You are right. My reference book contains more than one hundred inorganic titrations with EDTA. It is a wonderful booklet called "Méthodes d'analyses complexométriques avec l'EDTA", 116 p., edited by Merck, Darmstadt, 1973. Unfortunately it is written in French.

@Maurice Any change you could send me a picture of that specific titration? My first language is French so that won't be a problem.

@ Hugo. I don't know how to add a picture to my message. So I copy the titration here. The sample can contain up to $35$ mg Fe. Dilute it up to $100$ mL. Adjust the pH to about $2.5$ with $\ce{NaOH}$ or $\ce{HCl}$ $ 0.1$ M. Check with indicator paper pH $0-6$. Add $1$ mL indicator ($5$ g sulfo-$5$-salicylic acid in $95$ g water). Heat to $45$°C and titrate the red solution with $\ce{Na2EDTA}$ $0.02$ M up to discoloration. This method gives only Fe(III). If the total iron content $\ce{Fe(II) + Fe(III)}$ is to be determined, the sample must be heated to boiling with $1$ mL conc. nitric acid.

@Maurice Merci beaucoup

I've used the ferrozine method before.  This review gives three different methods, one of which is the ferrozine method, and at a first glance, none of them require thiocyanate.  https://aslopubs.onlinelibrary.wiley.com/doi/pdf/10.4319/lom.2013.11.113

There are iron testing kits in water which are available at an affordable price. I have not used them but read more about them here Testing iron in well water. You can test your spinach extracts for a yes no type answer if iron is present in significant quantities or not. Just keep in mind that PopEye's spinach is really not that rich in iron. Too much iron is a liver toxin.
I am afraid to share the bad news for you that with the resources you have, you cannot reliably determine iron content in spinach. Micronutrients or minerals cannot be determined by classical methods because they are present in a complex matrix and in very small quantities (parts per million range). They need instrumental analysis (absorption by molecular complexes, atomic emission or atomic absorption). I saw a couple of papers from ResearchGate which talk about the titration of spinach extract with permanganate. Nothing can be more wrong than this procedure in analytical chemistry. They are basically titrating all the organic stuff with it. Those methods are horribly wrong. Someone on SE Chemistry posted a query on how to determine iron in spinach by titration Iron in spinach...that is all incorrect. Yes, you will generate some numbers with those burette readings but those numbers have no significance.
Regarding your ashing method, that will not work either, because you need at several kgs of spinach to get a small amount (milligram) of iron. What should be the ashing temperature? How would you isolate iron from the ashes? Selective precipitation to get rid of other minerals? Nope, that is a bigger headache. For that you would need more reagents.
If this is a school project, make sure you have the resources after searching the literature. If this is an home experiment- classical methods will not work.

As you are looking for inventive ways to possibly measure Fe presence for a school presentation, may I suggest a crude (but likely interesting) visual aid in possibly determining relative iron concentration.
One idea is that Fe ions in the presence of acidic H2O2 will engage in radical production. The latter reaction in the presence of suitable light over time could result in a cyclic reaction system (for example, photo-Fenton). The latter is a conceivable path to magnify even small transition metals presence.
The latter radicals (especially, the powerful hydroxyl radical) can breakdown organic matter, including, for example, a strongly colored dye over time, resulting in eventual de-colorization. The relative periodic time evolution for the latter could be a manner to simply display/document apparent differences in even small relative iron (also copper, manganese,..) concentrations.
Here is a citation from the literature as an example, Effect of iron salt on the color removal of water containing the azo-dye reactive blue 69 using photo-assisted Fe(II)/H2O2 and Fe(III)/H2O2 systems.
So, spinach grown in iron-rich soil, reduced to ash added to acidic H2O2 (plus perhaps more acid added 1st to the spinach,  further lowering pH) in the presence of a photo-sensitive dye subject to photolysis, should display more rapid de-colorization (as captured by periodic snapshots) than one grown in iron-poor soil.
You may experiment with various ways of numerically quantifying the tentative iron content, as deemed viable given limited resources (including funds which alternatively, could purchase an iron testing kit, available here, for example, at $99, or less expensive water testing strips here).
Note: I am assuming one does not have the time to actually grow the same variety of spinach. Otherwise, I would simply speculate that iron-richer spinach may possess different coloration (and perhaps other discernable quantifiable attributes) for the purpose of a simple relative classification as to iron content.
This would conceptually be based, relatedly to above, on naturally occurring photosynthesis. To quote from a supporting source:

Iron, one of the structural elements of organic components that play an essential role in photosynthesis and nitrogen assimilation of plants, is available at extremely low concentrations in large parts of the Southern Ocean's surface waters. We tested the hypothesis that photosynthesis is the primary target of iron stress in phytoplankton living in this specific environment, resulting in a reduced carbohydrate production.

On the suggested coloration argument relating to iron content, also a confirming reference, to quote:

Leafy vegetables featuring a deep, dark green color, such as spinach and collard greens, are usually high in iron.


text:
Why is sodium bicarbonate added to lower the pH?
I use salicylaldehyde, chloroacetic acid and sodium hydroxide for the synthesis of o-formylphenoxyacetic acid. In the procedure it is written that after the reaction I need to add sodium bicarbonate to lower the pH to $9-10$.
I don't understand why I need to add $\ce{NaHCO3}$, why it is able to lower the $\mathrm{pH}$ and what is the reaction that occur when I add it. Can someone help me?

Poutnik has given an excellent answer to your question:

Why $\ce{NaHCO3}$ is able to lower the $\mathrm{pH}$ and what is the reaction that occur when I add it?

It is also important that your curiosity shown in the title:

Why is sodium bicarbonate added to lower the $\mathrm{pH}$?

Since I did not know of your synthetic procedure, I guess this is for your workup procedure partial purification. I also guess chloroacetic acid is your limiting reagent. Thus, looking at your product and excess reagent, you have your intended product (2-formylphenoxyacetic acid as its sodium salt) and starting compound, salicylaldehyde (as sodium phenolate) as minor impurity (it is also possible to have hydroxyacetic acid as its soldium salt in minute amount).
Therefore, it is important to maintain $\mathrm{pH}$ of the solution so that you may able to remove some impurity by solvent extraction. If you go lower on $\mathrm{pH}$ (suppose $\mathrm{pH} \lt 2$), then all organic materials would go to the organic layer ($\ce{2-CHO-PhO-CH2CO2H + 2-CHO-PhOH + HO-CH2CO2H}$). However, If you maintain $\mathrm{pH} \approx 7-8$ (suppose you add enough $\ce{NaHCO3}$), then most of 2-CHO-PhOH would go to the organic layer ($\ce{2-CHO-PhO-CH2CO2^-Na+ + HO-CH2CO2^-Na+}$ stay in water layer). To do so:
$$\mathrm{pH}= 8.50 = 10.32 + \log {\left(\frac {[\ce{CO3^2-}]}{[\ce{HCO3-}]} \right)} \ \Rightarrow \  \left(\frac {[\ce{CO3^2-}]}{[\ce{HCO3-}]} \right) \approx 0.015$$
Keep in mind that above calculation is approximation (actual value might deviate from this but not much). Considering you may use just enough $\ce{OH-}$ in the initial reaction and use saturated $\ce{NaHCO3}$ solution to extract, this molar ration would even go lower and  would reach $\mathrm{pH} = 8$ (when molar ratio is about $0.0048$).
$$\left(\frac {[\ce{2-CHO-PhO-}]}{[\ce{2-CHO-PhOH}]} \right) = 10^{8.00-8.34} \approx 0.46$$
Thus, more than $55\%$ of $\ce{2-CHO-PhOH}$ would go to organic phase upon extraction. Hence you may able to remove most of $\ce{2-CHO-PhOH}$ from the mixture with multiple extraction with saturated $\ce{NaHCO3}$ solutions.
Note: All of these calculations are approximations, just enough to show the effectiveness of $\mathrm{pH}$ adjustment.

Reference:

Arthur Harry Gerber, “Refractory compositions,”  European Patent 1992, EP0530943A1 (Date of filing: 30/03/1992; Application number: 92302800.5)(PDF).

Thanks for your complementary answer. The title question and questions in the body have been worded in unclear way, if the real question was why NaHCO3 or why to lower pH.

I apologize for not being clear with my wording, I have just started using the forums, next time I will be more precise.
In any case, both answers were very useful and clarified all my doubts.

This answer addresses the original interpretation of the question "Why $\ce{NaHCO3}$ ?“ rather than "Why to lower pH ?“. For the latter, see the Mathew's answer.
The reason of using $\ce{NaHCO3}$ is the reaction:
$$\ce{HCO3- + OH-  <=> CO3^2- + H2O}$$
First, near all hydroxide is converted to carbonate, and then an excess of bicarbonate shifts $\mathrm{pH}$ below 10.
The final mixture forms a carbonate/bicarbonate $\ce{pH}$ buffer with the excess of bicarbonate, because the 2nd dissociation constant of carbonic acid $\mathrm{p}K_\mathrm{a2}=10.32$ and
$$\mathrm{pH}=10.32 + \log {\left(\frac {[\ce{CO3^2-}]}{[\ce{HCO3-}]} \right)} $$
Bicarbonates are ideal to shift strongly alkaline $\mathrm{pH}$ to just medium or mildly alkaline $\mathrm{pH}$, without overrunning to neutral or acidic region.


text:
Why is the zero of standard enthalpy of formation a convention?
The standard enthalpy of formation $\Delta H_f^°$ of pure elements is zero by definition. Why is that a convention? It is true that enthalpy is defined unless a constant (like energy and entropy), but enthalpy of formation is actually a variation of enthalpy, so we don't really care.
Also, how could we possibly choose something else than zero if there's no heat transfer?
Note: I'm using General Chemistry (Ralph Petrucci)

Entropy is absolute. No standards required. Energy, as you note, does require a reference. I am hard pressed to think of a reference besides elements in their standard state that would be so simple, universal, and reasonably self-consistent.

Energy definitely needs a reference but energy change does not.

Well, the enthalpy of formation of pure elements from pure elements is indeed zero.

Note that by the Chemistry SE site policy ( in contrary to e.g. Physics SE), using LaTeX or MathJax formatting for question titles is highly discouraged, as it affects information indexing/searching.

Enthalpy is similar to gravific energy. It needs a reference. For gravific energy, the zero is arbitrary, and it is at the sea level. For enthalpy considerations, it is also arbitrary, The zero has been once decided to be defined by the pure elements. Other choices could have been possible, like the situation where all atoms are separated from one another. But the scientists all agreed to accept that the zero enthalpy is obtained when the elements are in their pure state at ordinary temperature and pressure.

Zero enthalpy is not the same as zero enthalpy of formation, as isobarically, dH=n.Cp.dT.

Maybe the problem here is how I'm reading your question. The enthalpy of formation of elements is by definition zero because the reaction is trivial (e.g., $\ce{H2 -> H2}$) and since the reactant is, by definition, pegged to zero, the delta is also zero.

The problem is that enthalpy is defined unless an arbitrary constant, but that constant elides when working with entropy change, so I'm saying you can't just define some arbitrary value for it. It is naturally zero because no energy is exchanged (like in the trivial reaction you wrote),  we are not setting it to be zero (this is what it looks like to me).

The definition is something like 'formation of pure elements in their normal physical state at 1 atm and at 298.15 K'  (and may now be 1 bar or a different temperature). But the important point is normal physical state so some are solids, some liquids and some gasses. So making all these zero saves a lot of extra fiddling about compared to that  if we chose, say,  free atoms at some huge temperature as the reference. For example consider then the calculation forming  solid carbon or gaseous Cl2.

It starts with this:  The standard enthalpy of formation of any substance, $\Delta H^\circ_\mathrm{f}$, is (by convention) defined as the enthalpy change for the reaction at 1 bar and a specified temperature (usually $\pu{298.15 K}$), in which the product is 1 mole of that substance, and the reactants are its component elements in their respective standard states.
For instance, $\Delta H^\circ_\mathrm{f}$ for $\ce{H2O_{(l)}}$ is equal to $\Delta H^\circ$ for the following reaction at standard state:
$$\ce{H2_{(g)} + 1/2O2_{(g)}->H2O_{(l)}},$$
since $\ce{H2_{(g)}}$ and $\ce{O2_{(g)}}$ are the respective standard states for hydrogen and oxygen.
Once you've accepted this definition as the convention for determining $\Delta H^\circ_\mathrm{f}$ for any substance, it directly follows that the value of $\Delta H^\circ_\mathrm{f}$ for any element in its standard state must be zero.  For instance,  $\Delta H^\circ_\mathrm{f}$ for $\ce{H2_{(g)}}$ is equal to $\Delta H^\circ$ for the following reaction:
$$\ce{H2_{(g)} -> H2_{(g)}},$$
which is necessarily zero.
To give an analogy: Suppose you define the "altitude of formation", $\Delta z_f$, of any location on earth as the change in altitude necessary to reach that location from sea level.
Hence $\Delta z_f$ for the summit of Mt. Everest is $\Delta z$ for the altitutude change associated with:
$$\ce{sea level -> summit of Everest},$$
which is 29,029 feet.
It necessarily follows, from this convention, that $\Delta z_f$ for any location at sea level is zero, since that would be equal to $\Delta z$ for:
$$\ce{sea level -> sea level}$$.

Ooooh! Thank you so much, this is what I've been looking for so far. Maybe I was misinterpreting the other answers, though. I'll accept your answer because I consider it to be the clearest one. The key word was "once you have accepted this convention, it directly follows..." that is coherent with how I expected it to be. Some books, after defining the enthalpy of formation, say we choose (not that it directly follows) the zero for pure elements as a convention. I suppose that was bad phrasing or maybe mistranslations in the italian edition.


text:
Why is Cl2-P-O-Cl an unfavourable structure for POCl3?
Why is the structure $\ce{(Cl)2P - O - Cl}$ unfavourable compared to the two structures in the picture?
The latter in the picture and the one I stated both have zero formal charge, so I don't really see why the structure I made isn't favourable.

Both O and Cl favour making bonds to P than to each other.

Is there a quick way to explain this? If not no problem, thank you for answering.

Compare their electronegativities ( P 2.1 Cl 3.0 O 3.5, Al approx.) and eventually enthalpies of bond formation, if available. Compounds with Cl-O bonds are energy rich.

Additionally, there would be reducing P(+III) next to oxidizing O-Cl. Water and fire in the same place.


text:
Turning sun into water
I don't know if this question belongs here or on physics stackexchange, but I believe it leans more towards this site.
A friend of mine had a strange but funny shower thought. He said:

"What would happen if we bring one atom of oxygen for every two atoms of hydrogen to the sun, would that turn it into water?

Now, there are some hurdles, I know, but let's bend the rules a bit and suppose we could overcome the immense heat and radiation on sun's corona, and we brought the oxygen and hydrogen together, would that actually happen? What if the sun collided with a huge concentration of oxygen without it decomposing the oxygen molecules?
Again, it's crazy, I know, but would the sun turn into water?

Even if water forms it will get decomposed due to the high temperature of the sun.

Not enough oxygen on Earth to make a dent ... the sun is too big.

It does not lean towards chemistry, but astronomical speculation. Most likely outcome would probably be a really weird supernova.

Why not to ask about water stability at temperatures and P such as those at the various sun layers? As it is not a chemistry question.

I’m voting to close this question because it's a speculation question whose response would use astronomical principles, if any.

in the sun hydrogen is in the plasma state, which makes it impossible to form bonds (because in the plasma state electrons are separated from the nucleus)
not to mention the heat alone would easily break the bonds between hydrogen and oxygen if you brought water to the sun

Interestingly enough, water has been observed very close to the sun via rotational transitions (at excitation halfway to dissociation).


text:
Can we convert nitrogen from the air into liquid nitrogen?
Is it possible to convert the nitrogen that surrounds us, into liquid nitrogen?
If so, how? If not, why not?
Thx

Yes e. g. http://www.elan2.com/

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed. A poster should not expect respondents would give it more effort then he/she has explicitly spent on it him/herself.

Linde cycle.

@Poutnik this was not a homework question. It was just out of curiosity :)

@cbracketdash  "As homework is considered literal homework, self-study questions, puzzles, worked examples etc."

Yes, why not? If you look at the Wikipedia article of liquid nitrogen, it says:

Liquid nitrogen is produced commercially from the cryogenic
distillation of liquefied air or from the liquefaction of pure
nitrogen derived from air using pressure swing adsorption. An air
compressor is used to compress filtered air to high pressure; the
high-pressure gas is cooled back to ambient temperature, and allowed
to expand to a low pressure. The expanding air cools greatly (the
Joule–Thomson effect), and oxygen, nitrogen, and argon are separated
by further stages of expansion and distillation. Small-scale
production of liquid nitrogen is easily achieved using this principle.

But first, you have to separate nitrogen from other components in air like oxygen, argon, water vapor, carbon dioxide. It is done though either of three methods: Cryogenic Air Separation, Absorption, and Membrane Diffusion. The process by which liquid nitrogen is formed from its gaseous state is more or less the same but the methods become different at how the gases are separated.

Cryogenic Air Separation

The air is sent to a container. Water vapor and carbon dioxide is separated by lowering the temperature in the container and freezing them to the walls of the container which are drained and vented. The left over elements (nitrogen, argon, and oxygen) are sent on to another container which is warmed at $\pu{-185 ^\circ C}$(b.p of oxygen and argon) which forces the two elements to the bottom of the container. Nitrogen is left in a gaseous state which is recovered from the separation Unit by sending to a cryorefrigerator where it is drawn off as a cryogenic liquid at about $\pu{-321°F}$.
The cryorefrigerator consists of a cold head, helium compressor package and a pair of stainless steel flexible helium lines. The cold head is mounted into the top of the liquid tank (dewar). The cold head extends down into the neck of the liquid tank for the purpose of cooling the nitrogen entering the liquid tank to $\pu{77 K}$ ($\pu{-196 °C}$). The nitrogen gas liquefies on contact with the cold head heat exchanger. The liquefied nitrogen drips off the heat exchanger down into the liquid nitrogen inside the dewar.

Absorption process

This method takes advantage of chemical properties of the element and compresses air through absorbent materials like activated carbon  which absorb different elements in the air and leave a sample that is of a majority composed of nitrogen.

Membrane diffusion

This method takes advantage of the fact that different materials diffuse through polymer lined tubes at different rates.  Specifically nitrogen and argon diffuse at a slower rate than oxygen and carbon dioxide.
Here is a homemade liquid nitrogen generator: https://www.instructables.com/id/Homemade-liquid-nitrogen-generator/
References

http://ffden-2.phys.uaf.edu/212_spring2011.web.dir/Benjamin_Lauritzen/Slide2.htm
http://www.amitlabs.com/liquid-nitrogen.html
https://www.mvsengg.com/products/nitrogen/liquid-nitrogen/
https://www.trimechindia.com/psa-nitrogen-plant.html


text:
Density range of table salt?
Does anyone have a reference for the density range of common table salt $(\ce{NaCl})?$
A pure $\ce{NaCl}$ crystal has a density of $\pu{2.16 g/cm^3}.$  However, the salt granules in table salt don't pack perfectly — there's a lot of air mixed in.
I carefully measured some Morton's iodized table salt at home $(> 99\%$ $\ce{NaCl};$ remainder is calcium silicate, dextrose, and $\ce{KI}),$ and got a density of $\pu{1.40 g/cm^3}*,$ which gives a packing fraction of $1.40/2.16 × 100\% = 64.8\%.$
Interestingly, this is (within my measurement error) essentially the same as the $64\%$ random close packing limit for monodisperse frictionless hard spheres.
But I don't know how much variation there is in the density of table salt, and have been unable to find a reference online.
*Here is how I measured the density:
I started with a metal 1 tbsp measuring spoon.  I didn't trust that its volume was actually 1 tbsp, so I filled it with water and measured the weight of the water (14.25 g) with a calibrated centigram scale, and its temperature (76 F) with a thermometer.  Since water @ 76 F has a density of $\pu{0.997189 g/cm^3}$, the volume of the measuring spoon was:
$$V_{spoon} = \frac{\pu{14.25 g}}{\pu{0.997189 g/cm^3}} = \pu{14.2902 cm^3},$$
as compared with the actual volume of a tablespoon, which is $\pu{14.7686 cm^3.}$
I then weighed a level tablespoon of salt (20.00 g) and, from this, determined that
$$\rho_{table salt} = \frac{\pu{20.00 g}}{\pu{14.2902 cm^3}}= \frac{\pu{1.40 g}}{\pu{cm^3}}$$

Aside of references, I suggest experiments with different table salt sources of different grains. And/or, for a given source of rough grains, try crunching it in grinding mortar, how it affects its density. Sure, the way of grinding and shapes of grains would have their affect too.

@Poutnik.  Nope, not interested in doing more experiments. Note my user name :)!  I don't wish to make this into a project; was simply curious. That's why I wanted a reference.

@Poutnik  The 74% value for dense sphere packing is the maximum packing fraction for a perfectly ordered arrangement  of spheres.  By contrast, the 64% value I quoted is, as I specified, the random close packing limit. Indeed, quoting from the very article you linked:  "For equal spheres in three dimensions, the densest packing uses approximately 74% of the volume. A random packing of equal spheres generally has a density around 64%."

I see, such idea has come to my mind too, but too late.  Random_close_pack - For_spheres

Poured random packing  Spheres poured into bed  0.609 to 0.625
Close random packing  E.g., the bed vibrated 0.625 to 0.641. OTOH, I see no reason to expect equal spheres or spheres at all.

@NilayGhosh  Please re-read my question. The question explictly concerns the density range of table salt, a material that is a mixture of salt and air at a certain packing fraction, not the density of a solid block of salt, which is what you quoted.

There is really no way to exactly determine the packing fraction theoretically. It obviously depends on the distribution of the particle sizes, and the shapes of the particles.

@MaxW  Not asking for a theoretical answer, but actual measured values.

@J... "There is no point for anyone to study this or measure it for salt, specifically...It's also pointless because....."   Nonsense.  What you wrote is demonstrably false.  This was a practical question about the distribution of density for commercially produced table salt, which is in turn determined by the variation in the manufacturing process.  Such information is important to individual table salt producers, so that  they know the variation in salt volume as they fill each container by weight (or alternately, if they sell by weight, but fill by volume.... (continued)...

.... it tells them what needs to be the minimum volume to ensure the weight is achieved. It's also of practical importance for those who use table salt in large quantites (e.g., food processors), and whose formulations are by wt., but find it more convenient to measure by vol. instead. Knowing the variation in density tells them whether density variation would be a significant source of error. You claim this ques. is "pointless" b/c "everyone will have salt made in a different way", but you're not thinking it through – the point of the ques. is precisely to know what this mfr. variation is.

@theorist fair enough.

Theoretical calculations are not going to be very useful here. Salt grains are rarely spherical or, in many cases, regular sizes. Some, sea-salt flakes, for example are irregular flakes which won't pack well. Even industrially produced salt consisting of regular cubes with a regular size distribution, won't pack like spheres. Empirical answers will win for most of these alternatives.

@matt_black  Please note my comment to MaxW:  "Not asking for a theoretical answer, but actual measured values."  Also, as specified in my OP, I'm looking for experimental values specifically for commercially produced table salt (and thus not for salt flakes or other variations). And the experimental values I'm looking for are the range of densities of commerical table salt.

The density of $\pu{2.17 g/cm³}$ refers to the density within a crystal of NaCl.  In chemical engineering, the terms of powder density, tapped powder density and settled apparent density take into account for the air between the grains of a solid. Especially the later recognizes that there may be a difference between the solid simply poured into a container, and after light compression (still with air gaps between the grains) after applying a little pressure e.g. if you shake and knock the tin filling with freshly ground coffee powder.
References like this, this, this, or this .pdf state powder densities of $\pu{1.378 g/cm³}$ (fine table salt), $\pu{1.282 g/cm^3}$ (granulated salt, again from here), and $\pu{1.089 g/cm^3}$ (rock salt). From these values, your estimate of $\pu{1.40 g/cm^3}$ seems plausible.
However, these data lack to state the typical size of the grains (think about the diameter), as well as the dispersion of the grain sizes (presence of larger and smaller grains, equally known as particle-size distribution) of the samples characterized. Both influence the packing of the grains and thus the recorded density.  In this perspective, the softer / more airy harvest of fleur de sel possibly packs much less dense.

Many commercial solid (powdery) products specify a density (usually not specified as to what kind!). If this solid is incorporated into a formulated liquid product which is a solution, or even a dispersion which eliminates air, the bulk density must be used to calculate the final product density. Then the actual product density is measured to provide a degree of quality assurance. Using the wrong density for ingredients causes no end of confusion.

I don't wonder bodies like ASTM and ISO reference density measurements, not only practical ones (coal, or fertilizers for «commoners», but equally materials to deploy under pressure (e.g., coating) up to nuclear fuel ...

"The density of 2.17 g/cm³ refers to the bulk density, i.e. within a crystal of NaCl" unfortunate sort of "typo", just delete "bulk" and it fixes the sentence. Please edit it, it confused me a lot. I'm not knowledgeable enough to 1% of you in this, but I'm positive most stuff online I find "bulk" actually means NOT within a crystal, that's what some people call https://en.wikipedia.org/wiki/Particle_density_(packed_density) (kind of the same as density, but it seems people that work with this use it), quoting the beginning of that article... (1/2)

"The particle density of a particulate solid or powder, is the density of the particles that make up the powder, in contrast to the bulk density, which measures the average density of a large volume of the powder in a specific medium (usually air).", here you can read the bulk density of salt at https://anval.net/Downloads/Bulk%20Density%20Chart.pdf ~1.3g/cm^3 (2/2)

The density of table salt is 1200 mg/cm³ according to the USDA.
The USDA nutrition label for salt lists the density as 18 g/tbsp.  That seems to be rounded to 0 decimals.  However, the nutrition label also state that there are 6976 mg Na in 1 tbsp.  Using that value and the atomic weights of Na and Cl, the calculated density is 17.77 g/tbsp, a more precise value.  The density then is 1200 mg/cm³, rounded to 0 decimal places.
Doing the same calculations with the data for Morton table salt, 590 mg Na in .25 tsp, gives a density of 1218 mg/cm³.  However, since 590 mg seems to be rounded to the nearest 10 mg, I think the USDA number is more precise.

As it reads, you have the density at 1.2 kg/cc. I think you might need to go in and do some editing -

Double Argh.  Proofed it again after correction.  I think I got this time.  Thanks @ToddMinehardt

@WayneGoode - Still no. I will edit it.


text:
Coordination number in π-complexes
I wanted to know what is the coordination number of the metal atom in π-complexes like Zeise's salt, ferrocene and dibenzene chromium.

I found some data on it but its a bit contradicting. Can anyone tell me what's right exactly cause what I found on web is a bit different from what my teacher said. It would be great if someone also tells me the reason and way of finding the coordination numbers in such compounds.

You should link your research here.

I found the answer for coordination number of zeise's salt on this website https://www.sarthaks.com/242855/assertion-in-zeises-salt-coordination-number-of-pt-is-five-reason-ethene-bidentate-ligand , Wikipedia didn't give me any useful result.....it just said 'wherein the platinum is three-coordinate and in oxidation state 0 (Zeise's salt is a derivative of platinum(II)).'

And coming to the other two salts according to Wikipedia the LBN is 5 and 6 whereas the coordination number is 10 and 12. When I read about LAN I found it to be the same. And my teacher says that the coordination number is 5 and 6. So I dunno what's right or wrong.

Your teacher must have told about the hapticity which is 5,6. Also, the answer provided on sarthaks is wrong . if hat were the case the EAN of ferrocene would be 22 and that would be epic. In reality the EAN of ferrocene is 36, considering the two rings of hapticity five which have an effective contribution of 10e

Does this mean that the coordination number of ferrocene and dibenzene chromium is 10 and 12 respectively? Then will the coordination number of Pt in zeise's salt be 5?

yes, since it has hapticity 5 and two rings, the total number of atoms coordinated is 10 ; EAN= 26 + number of electron taken from ligand which is 10 in this case. If you go by the definition "the number of atoms or ions immediately surrounding a central atom in a complex or crystal." , then CN would be 5


text:
Dissolution of SO2 in water
Majority of texts (at high school level) say that when $\ce{SO2}$ is dissolved in water sulphurous acid $\ce{H2SO3}$ is formed. However my text book states that it forms clathrate $\ce{SO2.6H2O}$ when dissolved in water, and can significantly exist as $\ce{H+ + HSO3-}$.
I found similar finding on Wikipedia:

Sulfur dioxide is fairly soluble in water, and by both IR and Raman spectroscopy; the hypothetical sulfurous acid, $\ce{H2SO3}$, is not present to any extent. However, such solutions do show spectra of the hydrogen sulfite ion, $\ce{HSO3−}$, by reaction with water, and it is in fact the actual reducing agent present:
$$\ce{SO2 + H2O ⇌ HSO3− + H+}$$

However there's no mention of clathrate on the whole page. Do what's the actual product on dissolution of $\ce{SO2}$ in water?

which textbook?

Clathrate appears only at low temperatures, near $0$°C, and relatively high pressures.

Thanks for bringing up this topic, and I would have appreciated it a few years earlier, however!
The important topic I am referring to is the apparent exclusive gas-phase formation of the molecule H2SO3, as correctly noted in Wikipedia on H2SO3, to quote:

There is no evidence that sulfurous acid exists in solution, but the molecule has been detected in the gas phase.[1] The conjugate bases of this elusive acid are, however, common anions, bisulfite (or hydrogen sulfite) and sulfite. Sulfurous acid is an intermediate species in the formation of acid rain from sulfur dioxide.[2]

The implication for acid rain formation has previously been noted, for example, in an MIT article, with cited  Reactions (1) to (3) below:

$\ce{SO2 + .OH + M → .HOSO2 +M (1) }$
$\ce{ .HOSO2 +O2 → SO3 + .HO2 (2) }$
$\ce{SO3 + H2O → H2SO4 (3) }$

However, in this recent 2019 work: A New Mechanism of Acid Rain Generation from HOSO at the Air–Water Interface, some important chemistry:

The photochemistry of SO₂ at the air–water interface of water droplets leads to the formation of HOSO radicals. Using first-principles simulations, we show that HOSO displays an unforeseen strong acidity (pKₐ = −1) comparable with that of nitric acid and is fully dissociated at the air–water interface. Accordingly, this radical might play an important role in acid rain formation.

where the net photolysis of gaseous sulfurous acid (in addition to SO2) likely proceeds as follows:
$\ce{H2SO3 (g) + hv -> .OH (g) + .HOSO (g) }$
Supporting source: See Page S6,Table S2, Eq (1), Eq (2), Eq (5) and Eq (12) in this available supplement. Also, related results for the photolysis of nitric acid, to quote:

Here we present both field and laboratory results to demonstrate that HNO3 deposited on ground and vegetation surfaces may undergo effective photolysis to form HONO and NOx, 1–2 orders of magnitude faster than in the gas phase and aqueous phase. With this enhanced rate, HNO3 photolysis on surfaces may significantly impact the chemistry of the overlying atmospheric boundary layer in remote low‐NOx regions via the emission of HONO as a radical precursor and the recycling of HNO3 deposited on ground surfaces back to NOx.

Note, there is thus an implied similar possible acceleration in the rate of photolysis of gas-phase SO2 after becoming H2SO3 at the air–water interface.
The implications of the above chemistry is that in addition to the cited Reaction (1) above (which is a sink for the removal of the hydroxyl radical, that otherwise could be involved in an ozone depletion cycle), the UV photo-induced decomposition of also gaseous H2SO3 likely leads to more problematic radicals cited in the acid rain formation and even ozone depletion.


text:
In metallic bonding, what causes a metal to lose its outer electrons and become an ion?
In metallic bonding, what causes a metal to lose its outer electrons and become an ion? I am trying to understand metallic bonding, atoms and ions.

When a metal looses its outer electrons (at least out of the transition elements region), it looks like a noble gas. And apparently, it gains in stability.

Then again, the electrons are not quite lost; they are still around.

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the topic in the question by at least basic own topic review and writes what he/she understands and what is the eventual stumble stone. The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed.

A metallic bond doesn't really involve ions. Are you talking about an ionic bond?

For any reaction to happen/any bond to be created, the process has to be thermodynamically favourable, meaning the materials involved must acquire a more stable state via losing as much energy it possibly can. It is a fundamental characteristic of any spontaneous process in nature.
Here, while forming a metallic bond, the metal atoms lose their valence shell electrons to form the electronic configuration of a noble gas, which requires energy. After losing the electrons, the cations of the metal form a metal lattice, in which the electrons lost previously float. Now, the interaction between the electrons and the ions help stabilize the entire metal, releasing much energy. Thus released energy more than enough compensates for the required energy. And thus, the energy of the metal reduces, resulting in the formation of metallic bonds.
In short, metallic bond is formed because it is thermodynamically favourable. For more, read https://en.m.wikipedia.org/wiki/Metallic_bonding

That's a common, but highly incorrect thinking about metallic bonding. In reality there's no ions in there whatsoever.


text:
How is the hydration state of a chemical related to its adsorption of water?
I am trying to understand adsorption and hydration states better, and I was thinking about the different hydration states of calcium sulfate - anhydrous ($\ce{CaSO4}$), hemihydrate ($\ce{CaSO4*1/2H2O}$), or dihydrate ($\ce{CaSO4*2H2O}$).
I would assume that $\ce{CaSO4}$ and $\ce{CaSO4*1/2H2O}$ would adsorb more water than $\ce{CaSO4*2H2O}$, and I am suspecting that $\ce{CaSO4*2H2O}$ barely adsorbs any water at all since it is "saturated" in a way. But I cannot find any information on this. In fact, I can't seem to find any sources mentioning adsorption in the context of hydration, which is surprising to me - I would intuitively assume that these phenomenons are closely related.
I would therefore appreciate an explanation of the relation of adsorption and hydration. Basically, does a lower hydration state generally imply stronger adsorption? Is adsorption in fact the dominant process by which a chemical changes its hydration state? And how does desorption work for hydrated chemicals?

The three are different compounds with different crystal structures, you should not assume such relations. Of course given time, they should turn into one another, but how long that takes, how much energy each step releases?

@Karl, that is why I am asking the question! I want to know whether there is a general rule that fits my assumption, but it is entirely possible that no such rule exists. However, bear in mind that the crystal structure breaks down at the surface, which is where adsorption occurs.

Adsorption is a surface process, and, by its definition, not a chemical reaction. Hydration however is.

Surprisingly enough, hemihydrate reacts quickly with water to produce dehydrate (gypsum). It is due to the fact that hemihydrate is produced by heating gypsum to a high temperature. This heating produces a structure with plenty of holes or voids corresponding to the previous positions of the water molecules. Water molecules can easily fill these holes if hemihydrate is mixed with water. Anhydrous calcium sulfate is a mineral (anhydrite) that does not react easily with water. The structure of anhydrous calcium sulfate is compact and does not contain any holes between its atoms to include easily water molecules.


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Does HI-6 form a covalent adduct with acetylcholinesterase?
According to sources provided below, the oxime HI-6, or Asoxime Dichloride, as it is also called, besides being a reactivator of acetylcholinesterase that has been inhibited by organophosphorus compounds, can act as an AChE inhibitor itself.In the process of inhibiting AChE, does HI-6 form a covalent adduct with the enzyme (Similarly to how TMTFA forms an adduct) or is the inhibition due to HI-6 simply occupying the active site without covalently binding with the enzyme?
References:

Soukup O, Pohanka M, Tobin G, et al. The effect of HI-6 on
cholinesterases and on the cholinergic system of the rat bladder.
Neuro Endocrinol Lett. 2008;29(5):759-762.
Pohanka, M. (2012). Alpha7 Nicotinic Acetylcholine Receptor Is a
Target in Pharmacology and Toxicology. International Journal of
Molecular Sciences, 13(2), 2219–2238. doi:10.3390/ijms13022219
Pohanka, M. (2013). HI-6 modulates immunization efficacy in a BALB/c
mouse model. Environmental Toxicology and Pharmacology, 36(3),
801–806. doi:10.1016/j.etap.2013.07.012