text:
How do I interpret characters that are not 1 or -1 in a point group table?
When the character in a point group table is listed as 1 or -1, I understand how an object/orbital etc. transforms to give either itself or the negative of itself.
When the character in a point group table is 2 or -2, what does this mean in terms of transformation? To understand this would be very helpful in terms of assigning the correct irreducible representations to atomic orbitals/LGOs to then determine which pairs mix.
Similarly, I do not understand how to interpret a character of sin or cos of an angle, nor a character of 0.
I am hoping there is a conceptual/visual way of explaining this as I am not up to speed on the matrix math, and the class that this is relevant too does not include that math either.

There are many sorts of characters needed to represent the symmetry behaviour; sine, cosine, complex numbers,  complex exponential, square roots and you have to treat each the same way. The 2 and 3 indicates degeneracy,  so is associated with E and T Mulliken symbols. Sometimes, as in the S4 group E appear as two rows so the degeneracy is clear. There is some information here https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613

When you observe that in the characer table of a symmetry group there are irreducible representations where the character of the identity operation (E) is 2 or higher, you may find that the character of some symmetry operations is 0, +2, -2, etc. This is due to the fact that this IR are degenerate.
To illustrate this idea, consider that you trying to determine the symmetry of the MO of bezene (D6h). You will find that some of these orbitals have the same energy (are degenerate) for example

When you apply the symmetry operations of the D6h point group you can check that what you obtain is a linear combination of these orbitals. If you express this linear combinations in matrix form, the character of the matrix will give you the value that appears in the character table for the symmetry of that orbital. In this case these orbitals have a E1g symmetry.


text:
Nucleophillic addition vs SN2
The following question came in Paper 2 of JEE Advanced 2014-

Here we have $\ce{CH3-}$ as the nucleophile. There are 2 possible attacks:

SN2 type attack on the chlorine group.
Nucleophilic addition on the carbonyl.

The answer given is option D, which will happen by the 2nd pathway(nucleophilic addition).
Why is nucleophilic addition given a preference over SN2? I feel that nucleophilic addition should rather decrease the reaction rate because the carbon goes from sp2 to sp3, decreasing the bond angle and hence increasing the repulsion

Alkyl Grignards do not readily do SN2 nucleophilic attack without catalysis, certainly on on alkyl chlorides

Just wondering, could the coordination of the carbonyl oxygen by Mg in the Grignard reagent play a role? Because one might also ask why the Grignard reagent does not behave as a base and abstract the fairly acidic proton alpha to the carbonyl.

As to @Waylander's point, how could you possibly make CH3MgBr from CH3Br without getting ethane by an SN2 reaction? In addition, experience is worth a lot in answering this question.

@user55119 in theory, even if it did react via SN2, you could add RBr to Mg such that there is never any excess RBr left in the reaction mixture. Analogous to how one would add carbonyl compound to LDA instead of LDA to carbonyl. I'm not saying it does react, I'm just saying that if it did there are ways of getting around it, so the argument isn't complete.

@orthocresol: True enough but TMI for OP.

@user55119 I wonder if we can say something via HSAB?

Importantly "Here we have CH3− as the nucleophile" isn't true and is a source of OP's misunderstanding. @user6376297 That's part of real mechanism and methyl is still coordinated when this happens, only later it's intramolecularly migrated to carbonyl.

I think @Mithoron has got the finger on the core confusion here.

To begin this short discussion one has to differ between Sn2 and addition to the carbonyl group, as the former is driven by orbital interactions while the latter is driven by coulomb interactions.
This can be explained with a closer look at the carbonyl function.
due to the highly polarized π-bond between carbon and oxygen, carbon is rendered quite positive. It results that any nucleophile bearing a negative charge will be attracted to attack it. In addition the reaction of a carbanion and a ketone or aldehyde is irreversible!
there is fundamentally no way in which the carbanion could ever be considered as a leaving group.
On the other hand Sn2 reactions appear on saturated carbon (in the best case methyl or primary) with at least one good leaving group (e.g. a halide or tosylate). Often the sigma bond connecting carbon and the leaving group is not even polarized, at least not significantly. Just consider the electronegativity of iodine, the BEST leaving group for SN2.
That is why orbital interactions are more dominant in thess kind of reactions. It can be easily noticed that a potential nucleophile has to donate its electrons into a sigma antibonding orbital in order to  react with the carbon electrophile. These kind of orbital can be considered as very high in energy as it is the counterpart of the sigma bonding orbital which is quite stable.
Therefore the best nucleophiles for nucleophilic attack on saturated carbon is usually a second or third row nucleophile like sulphur or phosphorus (Wittig reaction).
Therefore for a carbanion the irreversible addition to the carbonyl group happens much faster, resulting in a alkoxid. The new formed alkoxid is in close proximity to the chlorine substituted carbon and therefore likely to form the five membered ring .
Any references that support everything written above can be found in Clayden, Organic chemistry second edition.
Specifically the chapter nucleophilic substitution at saturated carbon will help.
Best regards,
John

Carbonyl addition is certainly not driven by coulomb interactions.

but the attraction of the nucleophile and the electrophile can be considered electrostatic

In that case, you might just say all chemistry is electrostatics.

My big issue here is really that both reactions are strongly dictated by orbital considerations. Otherwise, there wouldn't be such strong stereoelectronic considerations in each case (back aside attack, Bürgi-Dunitz angle).

But nevertheless there is a significant difference in the electrostatic interaction of a carbanion with a carbonyl or a chlorosubstituted saturated carbon. Which should render the attack on the ketone much faster, shouldn't it?

It should not. Consider the reaction of lithium amide with a ketone versus ammonia with a ketone. Depending on conditions, the anion deprotonates instead of adding to the ketone, while ammonia should form the imine.

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Is it true that "Voigt Circuit" would be able to be fitted to various electrochemical impedance spectra?
"Voigt Circuit" which is a series combination of Voigt elements (i.e. parallel RC's), is commonly used for Kramers-Kronig relations checking; and as I imagine it is generally believed to be able to be fitted to various electrochemical impedance spectra?
Is this belief true or not?
And if the answer is "Yes", is there any theoretical basis for that or it is only a practical sense (I mean that it is approved according to fitting results of experimental data to Voigt circuit)?

It is known that an electrochemical impedance spectra can be modelized with series-parallel combination of resistors and capacitors elements. What is more, for a given impedance spectra they are an infinity of different combinations of R-C elements which accurately model the same complex impedance.
Similar observations are commonly done in the field of imperfect dielectrics (non-Debye) for the complex permittivity spectra.
For examples: pages 10-11 , page 16 and pp.27-32 in the paper  https://fr.scribd.com/doc/71923015/The-Phasance-Concept . Each example of complex impedance is modeled with several different networks of R-C elements.
On theoretical viewpoint, all this is related to the energy efficiency of the phenomena in the experimental system, that is the presence of dissipative components (R) and storage components (C) distributed on various manner.  Even if the energy storage and energy dissipation are respectively on different physical nature in different experimental devices, possibly in some cases the complex impedance is likely to be similar due to a similar global energy efficiency behaviour. So, an infinity of different R-C models are likely to be derived.
This can be observed not only in experiments but also in purely theoretical studies for very simplified geometrical configuration : https://fr.scribd.com/doc/23155389/Theoretical-Impedance-of-Capacitive-Electrodes
In addition. To go beyond the question raised, the following quotation is extract from the first paper quoted above :
It has long been recognized that many impedance loci result in well-defined semi-circles with center depressed below the real axis. Linear loci are also commonly observed associated to rough electrode surface, porous electrode, diffusion, etc. "This features mean only that the system includes some part(s) transferring energy with constant efficiency in a range of frequencies. Any affirmation of the nature and mechanisms of the phenomena involved would be pure hypothesis if only based on impedance analysis."

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How can I clean stained garden rocks?
I have some white garden pebbles which  are now  yellow with trace of rust. I have tried to soak them in bleach diluted with water without success.  Could I receive suggestions of an appropriate material to remove the rust?

Bleach is no good against rust. Use acid.

The iron oxide from rust is not soluble in high pH (basic) solutions such as bleach. If you want to remove the the rust, you need an acid such as muriatic acid (wear PPE, follow label instructions). I would recommend an oxalic acid based product such as Bar Keepers Best Friend (I have no affiliation).  In my (pseudoscience) experience, oxalic acid its more effective at sequestering rust (when scrubbed) than anything else I've used including nitric, muriatic, and sulfuric acid. I would be cautious if the white rocks are marble as acid can disslove marble into oblivion.

Marble can also become quite porous with exposure to the elements, so the yellow color may not just be on the surface.

One of the best rust removers is very strong reducing agents such as sodium hydrosulfite and sodium bisulfite. Many commercial rust removers are based on it such as Iron Out. It takes a few minutes and no acid are involved. Read instructions on the label and follow manufacturers directions.

You don’t. You need acid to remove the rust, but since your pebbles are white they are very likely to be limestone or marble. They will dissolve in the acid faster than the rust.
Your best solution is to try and scrub it off mechanically, get new pebbles, or just learn to live with the rust.

H2O2 (3%, from the drugstore) alone might do the trick, but adding a little vinegar should react with the iron oxide and keep it in solution without dissolving much of the pebble.

As mentioned from A.K. something containing oxalic acid should do the trick.
That can be explained by oxalate acting as a chelating ligand. The structure of oxalate is basically two carboxylate functions covalently bound to each other, if deprotonated a bidendate ligand results. As the dianion coordinates to the iron a soluble complex will be formed which should be easily washed off.


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Why does ice crack when a drink is poured on it?
I think most people have experienced this, but while pouring a beverage over ice I noticed the ice cracked.

Before

After
My initial thought was that because the beverage has an antifreeze property to it, that some sort of thermal shock might occur due to the rapid local cooling from dissolving the ice, but I know that this also occurs when other beverages such as various flavored carbonated sugar water products are poured on ice which do not depress freezing as much. So then What would cause the ice to crack?

Not related to the question, just fun fact: declassified KGB memos from 1970s on how to recognize a spy listed that the one is most likely going to add ice to its beverage and prolong the drinking process, whereas soviets wouldn't bother with ice and drink the glass in one gulp :D

If only breaking the ice was as easy as pouring water on people ....    (I think I know why I have issues meeting new people)

This relates to the speed of temperature change between the inside and outside.  I think dropping ice into liquid nitrogen (or dry ice slurry) will also cause it to fracture.

@andselisk Can I have a source to that unrelated fact? I want to learn more.

@MelanieShebel I don't think there is a citeable source, but the memo itself can be found here, and the bullet point 12 (third from the bottom) on the first scan says "Малое употребление хлеба за обедом, разбавление спиртных напитков льдом, водой, употребление алкоголя небольшими дозами, как бы смакуя." ("Low bread consumption at lunch, diluting alcoholic drinks with ice, water, drinking alcohol in small doses, as if savoring." (Google Translate)).

I believe the ice cracked due to residual strains from freezing. Since ice freezes from the outside inward and it expands as it freezes, that as the inner water freezes, it imparts a tensile force on the surrounding ice (like the opposite effect of tempering glass).  As the warm liquid removes ice, the cross sectional area under tension decreases while the tensile force remains the same, causing an increase in stress. Once the stress reaches a certain point, the ice fails to counter the tension mechanically, causing the cracks.

Would then, ice that's cooled extremely slowly (in fractions of a degree per day) not exhibit the same tensile stress?

@Valorum I don't think slowly would prevent this, but rather if the ice has at least one side where material freezes last (like in an ice machine) then that would prevent the production of residual stress and thus prevent cracking.

Does the ice from a clear-ice maker not crack?

@Valorum I can't say for certain, but that is what I am implying. I guess an experiment is required.

@Valorum:  I think the shape of the ice cube might also make a difference — thermal stresses can be concentrated at the corners of a material when it heats.  What you really need is a spherical ice-ball maker.

Water freezing into ice EXPANDS. ($\ce{H2O}$ is most dense at $\pu{4^\circ C = 39^\circ F}$) - https://sciencestruck.com/density-of-water. In other words, water expands BOTH as it cools below $\pu{4^\circ C}$, AND as it warms above $\pu{4^\circ C}$.
When ice comes out of the freezer, it is typically well below $\pu{0^\circ C}$. It may be at $\pu{-10^\circ C}$ for example. When you pour a (warmer) liquid over the ice, the SOLID ice CONTRACTS as it warms up from $\pu{-10^\circ C}$ towards $\pu{0^\circ C}$.
Note that this is the SOLID ICE which is contracting as it warms - this is all BEFORE it gets UP to $\pu{0^\circ C}$ (at which point it will melt into a liquid). Since ice is a brittle solid crystal lattice, it cracks as it contracts.
(You can see the expansion of $\ce{H2O}$ as it freezes - notice in an ice tray, or a hose, it expands as it freezes.  If you freeze water in a glass, it will break the glass.)
For more info + details see https://en.wikipedia.org/wiki/Ice#Physical_properties and https://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html

When Ice is warmed from -10 to -1 it very slightly expands.  It contracts as it melts, and continues to contract in the liquid phase from 0 to 4.  Nevertheless, solid ice doesn't show this behaviour, and (like most solids) expands when heated. Your link shows that it is more dense at -10 (919 kg/m3) than at 0 (916)

@JamesK is right, but I don't think it invalidates the answer just changes the explanation..

Yes, I also think the stresses occur between the outer warming (and expanding) ice and the interior cold dense ice.  The outer part is like a metal hoop on a wagon wheel and has to break loose if heated (well shrink tight when cooling in practice).


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Are all ambidentate ligands monodentate?
Several websites define an ambidentate ligand as being monodentate. What about a ligand with say 3 atoms that can act as donor atoms, but due to the sterics of the ligand, only any two atoms can coordinate bond to an acceptor at any one time?

I don't really get your point there. "Ambidenticity" has little usefulness, pretty much only for simplest ligands. There are molecules able to coordinate in so many places that nobody counts.

If two atoms of the species can coordinate bond then the species is bidentate. A ligand is ambidentate if only one of two possible atoms can form a ligand, eg the linear $\ce{NCS^-}$ anion. There is no specific name for a ligand which is physically constrained so that only two of three possible ligand sites can bind.

If the ligand has 3 donor sites and it is able to show multiple denticity in different compounds, then it will be called as flexidentate ligand.
But as per your question, if it is only showing denticity of 2 only it will be called as bidentate ligand.
Actually, this concept arose when arguments started for selecting donor sites for the " monodentate ligands " which were always showing denticity of only 1 in all of its compounds instead they had multiple donor sites.
This ambidentate case arises when a ligand can't use its two donor sites simultaneously due to steric hindurence caused due to small distance between them and when the distance between the two donor sites is more, then the ligand now is able to bond with the metal using its multiple sites making it a multidentate ligand.

When a ligand has multiple donor sites, but uses only say $x$ of them at a particular time, then it's called ambidentate ligand. However, if it uses different number of sites in different compounds (say $x$ in one and $y$ in the other where $x\ne y$), then flexidentate is a more appropriate term as said in the other answer. But this doesn't mean all ambidentate ligands are monodentate.
For example, dithiooxalate is bidentate and ambidentate:

This example is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha).

I guess you should probably avoid mentioning that flexidentate tidbit, because I couldn't find an official definition for the same on the IUPAC website, so the lack of a credible source might make it seem questionable

@YusufHasan: Yes. You're right that the IUPAC Gold book doesn't give a definition for 'flexidentate' ligands. But some papers (for example, 1, 2) use this term and I think its meaning is also obvious from its name. However, if enough opposition arises, I'll definitely remove that piece of text from my answer. Thanks!

@YusufHasan: the IUPAC link in the answer says it's just two reactive centres, of which just one bonds at a time. Doesn't that imply that it's just monodentate ligands that can be ambidentate?


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What happens when chemical equilibrium is restored after being disturbed?
I'm aware that, according to Le Chatelier's principle, a reaction will shift its equilibrium position in order to counteract the disturbance at equilibrium. However, I don't know what that means after equilibrium is re-established.
For example, take the generic reaction
$$\ce{a A + b B <=> cC}$$
At some point, equilibrium is disturbed, causing the forward reaction to be favored. Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

$$K_{eq} = \dfrac{[C]_0^c}{[A]_0^a[B]_0^b} = \dfrac{[C]_1^c}{[A]_1^a[B]_1^b} $$

Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

Let's say it is a homogeneous equilibrium. You have to consider different cases:

The equilibrium was disturbed by changing one of the concentrations (removing or adding a substance). You can't get back to the same set of concentrations because there are too few or too many atoms.
The equilibrium was disturbed by adding solvent. Again, you can't get back to the same set of concentrations (all are lower than at equilibrium, and a net reaction in one or the other direction can't increase all of them at once).
You disturbed the equilibrium by changing the temperature. The equilibrium constant is at a different temperature will be different, so the old set of concentrations will not satisfy the new equilibrium constant.
You disturbed the equilibrium by intermittently changing the temperature (or - more elaborate - by attaching a power source to an electrochemical cell, and then replacing it by a wire shortcutting the cell). In this case, the concentrations change exclusively because of the chemical reaction (no "outside" disturbance), and they can return back to the original concentrations once the disturbance is removed.

In cases 1. and 2., however, the reaction quotient Q will go back to the original one (=K). In case 3., Q will be different.

Equilibrium for a reaction means that $\sum \nu_i \mu_i = 0$
where $\nu_i$ is the stoichiometric coefficient for species "i" and $\mu_i$ is the chemical potential, the full (total) chemical potential, of species "i".
So, the concentrations will change until this summation equals zero. It is that simple.
There is no need to try to explain this with equilibrium constants, but if you wish to, do make sure you do it with proper ratios of activity coefficients, there is little to be gained by approximating activity coefficients with concentrations, as is often done. It leads only to headaches later in life when you don't get thrown simple ivory tower mixtures :)


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How to determine if a mixing process is spontaneous at constant temperature but variable pressure
This is a conceptual question.
Many times we work with the gibbs energy of mixing under conditions of constant temperature but not pressure.
In a problem like this:

Suppose that 2.0 mol H2 at 2.0 atm and 25 ∘C and 4.0 mol N2 at 3.0 atm and 25 ∘C are mixed at constant volume. Calculate ΔmixG.

Is easy to calculate that the Gibbs mixing energy is -9,7 kJ.
But we also know that we cannot use gibbs' spontaneity criterion because it only works on constant pressure and temperature.
Is there any way to determine the entropy of the universe in this case to establish a criterion of spontaneity?
*For example if we know that the total volume is 10L and the division is even.

For an isolated system, the criterion of spontaneity  is an increase in entropy between the initial and final thermodynamic equilibrium states.

So if I calculate system's change in entropy that means I can check spontaneity because equals universe entropy. Ok. Can you show me how to calculate dS?

For determining change in entropy for systems of constant composition undergoing irreversible processes, see https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/.  For systems undergoing irreversible mixing, one uses the same kind of methodology, but with semi-permeable membranes to design and carry out reversible process paths.

Alternatively, for ideal gas mixing, one can use Gibbs' theorem.

You mean Gibbs-Duhem equation?

No.  The partial molar S, H, G, U, etc. in an ideal gas mixture is the same as that of the pure gas at the same temperature and partial pressure as in the mixture.

Mmm... Well the way I understand this example is that "We have two ideal gases in separate compartments at different pressures separated by a divider.If I open that divider the gases occupy the entire vessel acting with its partial pressures. Calculate Gmix = Gmixed-Gseparated dependant only in pressures. But as total pressure changed I cannot use Gibbs criterion." Since dG=-dSu*T only if total P was constant.

Not so.   What are the final partial pressures?  What are the number of moles of each?  What are the changes in partial molar entropy of each?

Volume of H2 = $\frac{nRT}{P}=\frac{(2)R(298)}{2}$
Volume of N2 = $\frac{nRT}{P}=\frac{(4)R(298)}{3}$
Total volume = $\frac{7}{3}(298)R$
Final pressure = $\frac{nRT}{V}=\frac{(6)R(298)}{(\frac{7}{3})R(298)}=\frac{18}{7}\ atm$
Final partial pressure H2 = $\frac{18}{(7)(3)}=\frac{6}{7}\ atm$
Final partial pressure N2 = $\frac{18}{7}\frac{2}{3}=\frac{12}{7}\ atm$
$$\Delta S=-n_{H_2}R\ln{\frac{\frac{6}{7}}{2}}-n_{N_2}R\ln{\frac{\frac{12}{7}}{4}}=-(2)R(-0.8473)-(4)R(-0.8473)=5.08R$$
For ideal gas mixing, $$\Delta H = 0$$
Therefore, $$\Delta G=\Delta H-T\Delta S$$
Another way of doing this is to (a) first allow the divider to float so that the two chambers equilibrate in pressure and then (b) remove the divider and allow the gases to mix.  You calculate $\Delta G$ for each step, and then add them.  In the end, you get the same answer I got.

Do you have any questions about what I did?

No. Amazing. I get it now. Thanks Chester, you rule!


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Which atom bears the negative charge in cyanide? I am confused by the MO diagram

In heteronuclear species, whichever atomic orbital a molecular orbital is closer to in energy contributes more to that molecular orbital, right? In other words, if the orbital is occupied, more electron density is around the atom whose atomic orbital is closer in energy to the molecular orbital. This feature of MO diagrams can predict dipole behavior and which atom bears charge in ions. If this is the case, and we know that carbon bears the negative charge in cyanide (and hence is a carbon nucleophile), why does the MO diagram for cyanide appear to have the HOMO (the "attacking" orbital) concentrated on the nitrogen rather than the carbon atom? Based soley on the diagram and not the accompanying image of the HOMO, I thought, since the HOMO is closer in energy to the nitrogen 2pz atomic orbital rather than the carbon 2pz, the HOMO should have more contribution from nitrogen and the LUMO should have more contribution from the carbon.

There's some negative charge on both atoms. HOMO on C is more diffuse means something contrary to what you seem to say.

@Mithoron My original post may have been a bit ambiguous. I have come to associate the position of a MO relative to those its parent AOs with the contribution that orbital has from its parent AOs. As such, I would expect the HOMO in cyanide to primarily reside on the nitrogen and the LUMO primarily on the carbon. The pictures show what I would expect based on my knowledge of the reactivity of cyanide, but I would have predicted the opposite just by looking at the energy diagram.

Carbon bears a negative formal charge, but formal charges do not necessarily represent the actual charge distribution.

As commenters have noted, charge isn't the best choice of a word here, but you're right that the HOMO is more centered on the C, and that's really your question.
The answer is a bit complicated. It relates to those dashed lines on the MO diagram. If we naively construct an MO diagram of cyanide ion, we have two favorable sigma interactions. First, the $2s$ orbitals can interact positively, which I'll call $\sigma_s$. That would be the lowest energy MO on our diagram. Second, we can have an end-to-end interaction of $p$ orbitals, which I'll call $\sigma_p$. Since that $\sigma$ interaction is stronger than a side-by-side $\pi$ interaction, we would expect the $\sigma_p$ MO to be lower in energy than the $\pi$ bonding MOs.
If we look at your MO diagram, though, we see that $\sigma_p$ is actually higher than the $\pi$ MOs, not lower as we predicted. It's the HOMO in the diagram. The reason it is higher than we naively expected is because there is mixing between it and the $\sigma_s$ (and also between their anti-bonding counterparts, but ignore that for now). These two MOs can interact positively or negatively. In the positive interaction, we get the (normalized) sum of the two MOs, which is an $sp$ orbital on C with its large lobe toward N and an $sp$ orbital on N with its large lobe towards C (see the lowest MO in the drawings below). Both large lobes have the same phase, so this is a strong bonding interaction that is the lowest MO on your diagram. Both of the starting bonding MOs are biased in favor of contribution from N, since both the $s$ and $p$ AOs of N are lower in energy than their counterparts on C. Thus, this sum orbital has a stronger N contribution as well.
In the negative interaction, we subtract $\sigma_s$ from $\sigma_p$ and again get $sp$-type orbitals on both atoms, but now the large lobes are facing away from each other. That makes the bonding interaction much weaker, so much so that it moves higher in energy than the $\pi$ bonding orbitals. Furthermore, subtraction takes away more N-character than it does C, and you end up with a greater contribution to the new orbital from C than N. As you get more comfortable with MOs, you can verify this mathematically. Another way to think about it is that the final MO is higher than the either the average of a $Cs$ and a $Cp$ orbital or the average of an $Ns$ and an $Np$ orbital, but it's closer to the C $sp$ average (which is higher than the N $sp$ average), so it has more C character by the same logic you used to predict that it would have more N character (which it would if it were just made up of $p$ orbitals).
Below is a drawing of the five occupied orbitals where you can see that the $\sigma$ ones have become $sp$-like and that the carbon contribution is greater in the HOMO. C is on the left and N on the right.


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Effect of substituents on Ullmann reaction
This question that I recently encountered asked what the most suitable substrate for the Ullmann reaction would be. The answer given is (b) but I think it has to be (d) as the nitro group acts as EWG and thus facilitates the reaction.
However I do not know the mechanism and thus am not sure if (d) is indeed the correct answer.
Any reason why the answer might be (b) and not (d)?

Related: Ullmann's Reaction (similar to Wurtz) - aryl chloride with deactivating groups - stability of free radical

I think that the answer given in your textbook is wrong as they are trying to conduct ULLMANN reaction on a chloride a strong EWD needs to be present at the ortho/para positions for the reaction to be feasible.
There Are generally 2 Mechanisms Proposed for the reaction :

Free Radical Mechanism
Ionic Mechanism

FREE RADICAL MECHANISM
  The Copper Metal (used as catalyst) Generate a Aryl Radical By transferring a electron, and then 2 Aryl Radical combine to form the final product.

IONIC MECHANISM
  Acording to this a Organometalic intermideate of Aryl copper halide is fromed. The copper is added to the Aryl halide by Oxidative addition. then a electron is transfered to the intermidiate to for a organocuppric radical which again perform the oxidative addition on another aryl halide which result in formation of a biaryl copper complex which in turn undergo reductive elimination and make the final product.

You can easily see that we need a EWG to make the Reagent more reactive thus making the reaction feasible thus your answer must be either wrong or you might have checked it wrong. nevethless I also find this paper affirming mine and your belief.
Source:

https://link.springer.com/article/10.1007/s40828-016-0036-2
www.name-reaction.com/ullmann-reaction

I may be wrong,but isn't a free radical destabilized by an EWG,considering that it is an electron deficient species? So why would EWGs facilitate the reaction?

@YUSUFHASAN no I don't  think so as the free readical is formed on a sp2 carbon

@AdvilSell See this It clearly states that a free radical is an electron deficit species,and it's stability is similar to a carbocation

So even if -M effect is not operative then also -I effect surely is via the sigma bonds

@YUSUFHASAN yes i was mad when i wrote it , it has been a bad day till now please accept my apology i have edited the answer the EWG make the intermidate more reactive making the reaction faster as its a radical intermideate reaction the reaction is kinetically controlled

@AvnishKabaj Kabaj Its not so please read this https://link.springer.com/article/10.1007/s40828-016-0036-2 and accept my apology

@AvnishKabaj Read the introduction part around the last para


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What alkali and alkaline earth metal oxides will turn moist red litmus blue and finally white?

How many of the following will turn moist red litmus blue and finally white?
$\ce{Li2O}$, $\ce{KO3}$, $\ce{RbO2}$, $\ce{Cs2O2}$, $\ce{BeO}$, $\ce{MgO}$, $\ce{BaO2}$, $\ce{SrO}$.

Since bleach convert moist red litmus blue and then white, but how we can identify which compound will act as bleach ?

The standard oxides $\ce{Li2O, MgO}$ and $\ce{SrO}$ are water soluble and hydrolyse to form hydroxides and a basic solution. $\ce{BeO}$ is insoluble in water.
$\ce{RbO2}$ is a superoxide which in aqueous solution undergoes disproportionation to $\ce{O2}$ and $\ce{OH-}$ so will give a blue litmus test.
The peroxides $\ce{Cs2O2}$ and $\ce{BaO2}$ react with water to give hydrogen peroxide and the hydroxides, so again a basic solution(source). The hydrogen peroxide produced will bleach the litmus paper.
$\ce{KO3}$(Potassium Ozonide) is highly unstable, decomposes in water to $\ce{O2}$ and $\ce{KOH}$
The oxides which will NOT decolorize litmus are $\ce{Li2O, MgO, BeO}$ and $\ce{SrO}$. The peroxides ($\ce{Cs2O2}$ and $\ce{BaO2}$), the superoxide($\ce{RbO2}$) and the ozonide($\ce{KO3}$) will decolorize litmus.

KO3 is Potassium Ozonide

RbO2 is technically a superoxide.

@OscarLanzi quite correct, my error. I will correct the entry

@Waylander Can you mention Compound Name clearly which finally turn white ? I am asking this because in my book answer of this question is given as 4 I.e. 4 Compound will turn white finally. A/c to your answer only 3 compound will turn white.


text:
Product of oxidation of alkanes
I'm trying to find the product of the oxidation of iso-butane using $\ce{KMnO4}$, I know that alkanes undergo stepwise oxidation with strong oxidising agents to finally give carboxylic acids. With straight chain carbons, its very straightforward - ethane becomes acetic acid etc. But with branched chain alkanes I don't know how to proceed. Can someone tell me if there's a general rule for oxidation of alkanes to get the product?

Are you sure of the premise? KMnO4/H+ is a test for alkenes vs. alkanes. Just the benzyl position undergoes ox. Isn't so? And in general,anyway, if an alkane react than is a multisite reaction. Like in bromination.

Actually, the reactants in the reaction are given as KMnO4/ [O] , so I think oxidation is implied.

Oh ya I remember now, yes KMnO4/H+ is a test for unsaturated hydrocarbons, but KMnO4 is also used as an oxidant according to my textbook, but the only examples given are oxidation of methane and ethane..

You might want to look at the Kuhn-Roth oxidation although it is run with chromic acid.

Cold Neutral KMnO4[Baeyer reagent] is a test for alkenes and also primary and secondary alcohols, aldehydes and probably some amines and sulfides. In acidic or basic solution and at higher T oxidizing agents such as permanganate or dichromate will oxidize almost anything to anything eventually CO2 and water. Look up procedures for chemical oxygen demand [COD] to get an idea. For specific reactions the conditions must be well controlled and many specific reagents have been developed.

Potassium permanganate, $\ce{KMnO4}$, is the most common and probably the most applicable oxidizing agent discussed in organic chemistry teaching classes. $\ce{KMnO4}$ is a very strong oxidizing agent, and is capable of oxidizing a wide range of organic molecules. Usually, it will oxidize organic compounds to appropriate carboxylic acids, but, the final products can vary on the conditions used. If you googled enough to find reduction potentials of $\ce{KMnO4}$, you may find following conditions, which are the most used reduction conditions of $\ce{KMnO4}$ (Wikipedia):
\begin{align}
\ce{MnO4- + H+ + e- &-> HMnO4^-} & \mathrm{E^o} &= \pu{0.90V}\tag{1}\label{1}\\
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & \mathrm{E^o} &= \pu{1.51V}\tag{2}\label{2}\\ \ce{MnO4- + 2H2O + 3e- &-> MnO2(s) + 4OH-} & \mathrm{E^o} &= \pu{1.69V}\tag{3}\label{3}\\
\ce{MnO4- + 4H+ + 3e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &=\pu{1.7V}\tag{4}\label{4}
\end{align}
Other manganese containing compounds other than $\ce{KMnO4}$, which can also be used as oxidizing reagents are listed below:
\begin{align}
\ce{MnO2(s) + 4H+ + e- &-> Mn^3+ + 2H2O} & \mathrm{E^o} &= \pu{0.95V}\tag{5}\label{5}\\
\ce{MnO2(s) + 4H+ + 2e- &-> Mn^2+ + 2H2O} & \mathrm{E^o} &= \pu{1.23V}\tag{6}\label{6}\\
\ce{HMnO4- + 3H+ + 2e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &= \pu{2.09V}\tag{7}\label{7}\\
\end{align}
Yet, as most of the viewers already commented, I also believe $\ce{KMnO4}$ does not oxidize alkanes, except in following situations:
$\ce{KMnO4}$ oxidizes alkane carbon atoms, which contain sufficiently weak $\ce{C-H}$ bonds such as those in benzylic positions ($\ce{C-H}$ bonds in the $\alpha$-positions of substituted aromatic rings). These oxidizable alkyl benzene substituents can be -$\ce{CH3}$, or any $1^\mathrm{o}$-alkyl groups such as -$\ce{CH2CH3}$, or $2^\mathrm{o}$-alkyl groups such as -$\ce{CH(CH3)2}$. All of them will be oxidized to carboxylic acid group, -$\ce{CO2H}$ (See Scheme 1 below):

It is also important to note that when the benzylic carbon is a $3^\mathrm{o}$-$\ce{C}$ (e.g., corresponding alkyl groups such as $-\ce{C(CH3)3}$ group in tert-butylbenzene, see the example given in RHS of Scheme 1), the oxidation reaction does not happen.
In addition to benzylic carbons, $\ce{KMnO4}$ also oxidizes weak $\ce{C-C}$ bonds such as $\pi$-bonds in alkenes and alkynes (Scheme 2):

$\ce{KMnO4}$ also oxidizes $\sigma~\ce{C-C}$ bonds such as that in glycol-moiety (e.g., $\ce{HO-C-C-OH}$) and $\ce{C^{\mathrm{sp^2}}-O}$ bonds such as those in phenolic compounds (e.g., it oxidizes phenol to 1,4-benzoquinone, commonly known as para-quinone).
At last, as mentioned above, conditions of the reaction can be adjusted such a way that desired compound can be achied in high yields. For example, see following example (Ref.1):

Note that no benzylic carbon or -$\ce{OCH2}$- was oxidized here.

Late Addition:
Although tertiary $\ce{C-H}$ bond in isobutane (preferred IUPAC name: 2-methylpropane; $\ce{CH3CH(CH3)CH3}$) is relatively weaker than othe alkane $\ce{C-H}$ bonds, I’m highly skeptical of the statement of “Oxidizing agents such as potassium permanganate readily oxidize a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidized to t-butanol.” (vide infra). To my knowledge, the literature evidence for that type of oxidation reactions does not exist. Nevertheless, 2-methylpropane (isobutane) can be oxidized to 2-methylprop-2-ol (t-butanol) by reacting supercritical-phase isobutane with air in the presence of a catalyst such as $\ce{Pd/C}$ at high temperature, Instead of using $\ce{KMnO4}$ (Ref.2):

Abstract: tert-2-Butyl alcohol can be synthesized efficiently on $\ce{SiO2 –TiO2}$ or $\ce{Pd–C}$ catalyst if air is directly introduced to the supercritical-phase isobutane.

tert-Butanol (TBA) is the initial material in methyl tert-butyl ether (MTBE) production and methyl methacrylate (MMA) synthesis.  MTBE is high-octane-number gasoline additive and MMA is a resin material for plastic synthesis (e.g., PMMA or poly-methyl methacrylate). Thus, one would have wondered why scientists did not attempted to synthesize TBA in a such an easy way like using $\ce{KMnO4}$.
However, in a different note, almost exclusive $\ce{C-H}$ bond activation has been observed in the reaction of linear and branched alkanes with $\ce{CeO2+}$ (Ref.3)
Reference:

A. Abiko, J. C Roberts, T. Takemasa, S. Masamune, “$\ce{KMnO4}$ revisited: Oxidation of aldehydes to carboxylic acids in the tert-butyl alcohol - aqueous $\ce{NaH2PO4}$ system,” Tetrahedron Letters 1986, 27(38), 4537-4540 (https://doi.org/10.1016/S0040-4039(00)84997-7)
L. Fan, Y. Nakayama, K. Fujimoto, “Air oxidation of supercritical phase isobutane to tert-butyl alcohol,”  Chem. Commun. 1997, (13), 1179-1180 (DOI:10.1039/A702308A).
A. Gunay, K. H. Theopold, “$\ce{C-H}$ Bond Activations by Metal Oxo Compounds,”  Chem. Rev. 2010, 110(2), 1060–1081 (DOI: 10.1021/cr900269x).

Thanks a lot for your detailed answer. But as you said that alkanes don't generally get oxidised , is the example given by @Anubhab Days right? I haven't a reference of il finars volume 1, so I'm not sure.

@Karun mathews: please see my late addition to the original answer for reasoning.

Thanks for the very researched answer. If I find anything else I'll post it here then..

@MathewMahindaratne Indeed, I had also not heard of oxidation of tertiary hydrogens by KMnO4 until I came across that statement in my answer. Although it is unlikely, Finar may be wrong, as in, it doesn't contain any reference to a real paper to support his claim.

@Anubhab Das: Please post any new finding on this subject. I'm really interested to them.

@MathewMahindaratne I do not expect to find anything new supporting Finar's claim. Your research, seems to me, contains all there is about oxidation of alkanes by oxidising agents :-)

Quoted directly from IL Finar's Organic Chemistry Volume 1 -
"Oxidising agents such as potassium permanganate readily oxidise a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidised to t-butanol."

Ah ok thanks. So could I assume a general rule that 3° H>2° H>1° H for more likelihood of oxidation?


text:
Stability order of carbocations

These are some of the most stable carbocations, but what is the relative stability order between them.
According to me it should be: C>D>B>A........ I came to this conclusion because both C and D have equal number of resonating structures i.e.7.However Bend orbital Bond seems to stabalise more than Resonance so C is assigned higher priority.
However, my teacher says that the order should be C>B>A>D........ and he gives priority to Bend orbital bond above anything else.
However, I seem to disagree with his answer.
Unfortunately,  I don't have the source of the question to verify the answer and couldn't find anything relatable on the internet.
So any help would be appreciated.
(Also, as a side question, which is the most stable carbocation ever isolated, is it tricyclopropylmethylcarbocation or tropyllium ion or any other. If anyone knows the answer might help)
(Found several other questions like mine on cyclopropyl metgy carbocation. However the answers either are using Molecular Orbital Theory Or some other concepts which I haven't learnt(I am still in school). So I wish for a more simplistic approach to the question. Also they are not able to explain the exact position of tropylium ion in the order)

The Cycloheptatriene cation is aromatic. You may want to reevaluate your ordering with respect to D.

Also related: What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?
What is "dancing resonance"? What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

@Mithoron Thanks for the links. Finally got my answer to the question.

@Mithron Any idea about the most stable carbocation ever isolated :)

https://chemistry.stackexchange.com/a/49307/9961

@Mitron The link you suggested shows that C is more stable than D. However which one is more stable among B and D....... I am getting confused

The answer is D>C>B>A.
This can be explained with the circumstance that the fourth structure (D) is an aromatic compound and therefore has extraordinary stability compared with standard carbocations.
The mentioned property arises from: planarity, a circular structure and the fact that the structure fulfills the 4n+2 rule. That carbocation (sp2-hybridized C) is required in order to allow a circular electron flow
The remaining three carbocations are just stabilized by hyperconjugation, which rises with the number of alkylsubstituents bonded to the central carbon.
Here is a reference to the cyclopropenylium cation you mentioned:
Breslow, R.; Yuan, C.; J. Am. Chem. Soc. 1958, 80, 5991-5994. DOI: 10.1021/ja01555a026
And a paper about stable carbocations:
Journal of the American Chemical Society / 94:6 / March 22, 1972
(Here you can find the carbocation D from above again)

I know D is a very stable carbocation due to aromaticity. However, the cyclopropylmethyl carbocation is exceptionally stable deu to its bend orbital bond i.e. a non conventional resonance(popularily called dancing resonance). However you dont seem to have accounted for it in your answer.........

https://en.m.wikipedia.org/wiki/Tropylium_cation

Given the evidence in the linked post, I'd say this is not correct in its entirety.

I inserted a link to the paper I could find, I don't have the time to look for the other 'reference'. I found two candidates be Olah et.al.; DOI: 10.1021/ja00761a041; 10.1021/ja00761a042


text:
Mechanism of the reaction
I came across the following reaction,about which I was ignorant till now.Can anyone guide be regarding it with suitable mechanism.

hello buddy did you try adding the halide resulting in formation of carbocation and the attacking by the oxygen in carboxylic group

This may enlighten you  https://en.wikipedia.org/wiki/Iodolactonization

These reaction are called as Lactonization(Generally) these follow a pretty straightforward mechanism the steps are as follow:

The $\ce{HCO3-}$ ion from $\ce{NaHCO3}$ Deprotonate the oxygen of carboxylic group to give a negative charge on the oxygen atom
Meanwhile the  doublebond forms a non-classical carbocation with $\ce{I2}$
The Oxygen atom attacks the non-classical carbocation forming a six membered ring and a $\ce{I}$ on the adjecant atom

This is the mechanism scheme of the above description :

A similar approach can be followed for the $2^{nd}$ reaction.

this is not the accepted mechanism for halolactonisation. See the reference given  above

@Waylander Sir I have corrected it Please reconsider it

It is worth noting that in the second example the [5] ring is preferred over [6]. This is a general rule of preference for these kind of reactions.

@Waylander I just noticed it can you please explain why it is so....

5-exo-Tet is always favoured over 6-endo-tet by Baldwin's Rules  http://www.reading.ac.uk/acadepts/sd/pharmacy/public_html/staff/andre/BaldwinsRules-RAJ.pdf


text:
Rate of hydrolysis of silicon tetrahalides
Concise Inorganic Chemistry for JEE(Main and Advanced) by JD Lee(Third Edition) writes

If we compare the rate of hydrolysis for $\ce{SiX4}$(where X= F,Cl,Br,I), it is in the same order as that of their Lewis acid strength order,i.e.
  $\ce{SiF4}$ > $\ce{SiCl4}$ > $\ce{SiBr4}$ > $\ce{SiI4}$
  .This is because the effect of π back bonding is less dominant over the negative inductive effect.

My question is, why is the π back bonding neglected here even though 3dπ-2pπ back bonding is strong enough to be evoked in many other cases? Also, Wikipedia lists the bond strengths of $\ce{Si-X}$ as $\ce{Si-F}$ > $\ce{Si-Cl}$ > $\ce{Si-Br}$ > $\ce{Si-I}$ , and I feel that the rate of hydrolysis should be inversely proportional to this order as it will become increasingly difficult to break the silicon-halide bond when bond strength increases,and consequently difficult to hydrolyse the molecule

We can justify the rate of hydrolysis of Silicon tetrahalides if we just think on the statement carefully. The statement claims that the rate of hydrolysis is in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$. So, we are mainly concerned about the rate of the reaction, not on the thermodynamic stability or any energetic criterion for this reaction( I have dealt with it in the next paragraph). If we recall the characteristics for rate of   a reaction, it depends on the slowest step of the reaction. In the hydrolysis reaction of silicon tetrahalides, the first step, i.e the attack of a $\ce{H2O}$ molecule to form the trigonal bipyramidal intermidiate is the rate determining step. Also note that, no cleavage of $\ce{Si-X}$ bond happens in the rate determining step, but it actually happens in the next step, which is a fast step, so it doesn't contribute to the rate of the reaction. So, basically strength of $\ce{Si-X }$ bond doesn't affect the rate of hydrolysis reaction (the situation is exactly same as Nuleophilic aromatic substitution reactions ($S_NAr_2$) also). But if the $\ce{Si}$ atom becomes more and more electrophilic, the attack by a water molecule becomes more easier which subsequently results in the increase of the rate of the reaction, So the electrophilicity of $\ce{Si}$ in the tetrahalides increase in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$, and so the rate of the hydrolysis also.
Here is the mechanism of the first out of the four subsequent substitutions in the hydrolysis reaction of Silicon tetrahalides.

Secondly, if we really want to judge the rate of the reaction by numerical calculations, observe that the ultimate product of hydrolysis reaction of all the tetrahalides are same. Also, all the reactions are exothermic in nature. So, $\Delta _r H^0$ (assuming the  hydrolysis reactions are done in standard conditions; it's not a strict assumption, it may not be at standard conditions also, but the changes in the heat of reaction will be almost similar for all the tetrahalides) will be more if $\Delta_f H^0$ is more for the reactants (note that, $\Delta _r H^0 = \Delta_f H^0_{products} -\Delta_f H^0_{reactants} $ and we are bothered about reactants. If $\Delta_f H^0_{reacants}$ are more $\Delta_r H^0$ will be more negative and hence energetically favourable). The values for $\Delta_f H^0$ have the order, $\ce{\Delta_fH^0_{SiF4} < \Delta_fH^0_{SiCl4} < \Delta_fH^0_{SiBr4} < \Delta_fH^0_{SiI4}}$. So, the hydrolysis reaction of $\ce{SiF4}$ will be the least exothermic and that of $\ce{SiI4}$ will be the most exothermic, which is expected also from the bond energy considerations. So, if you have less release of heat energy, i.e thermodynamically less stable product, it will have lesser activation energy and higher rate constant, and thus the rate of the reaction will be highest for $\ce{SiF4}$ and least for $\ce{SiI4}$ as we will get the product along with relatively more thermodynamic stability. So, by thermodynamic consideration also, rate of hydrolysis should be, $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$.
Lastly, the $\pi$-bonding is not important in case of large size difference between $\ce{Si}$ and $\ce{F}$ ($2p$ and $3d$ have course greater size difference though you may not admit). The fluorine atom makes the silicon more electrophilic by its electron pulling ability rather than  back-bonding.

Then why do we consider back bonding in case of the boron trihalides?( The order given for their rate of hydrolysis is fluoride<chloride<bromide<iodide, and although they have not mentioned the reason for this below, in my mind I have always thought that this is due to backbonding,or am I wrong here?)

Also, I just checked in my book, and the author is performing an SN2 reaction for the hydrolysis of silicon tetrahalides,which goes through a single five membered transition state resulting in Walden inversion. So can you please clarify which mechanism is correct?

@YUSUFHASAN We have to consider the back-bonding in case of Boron because it lowers the electrophilicity of the Boron atom, and if electrophilicity is lowered, obviously water molecule will be less prone to attack at the Boron centre which does affect the rate of the reaction. In case of Silicon, only the strength of the bond matters at the end of the reaction. Thus the two cases are diffferent.

So you mean to say that the RDS of hydrolysis of boron trihalides does not concern the addition of water molecule,while it is so for silicon tetrahalides?

@YUSUFHASAN The typical $S_N2$ type reaction occurs always if the central atom is Carbon which can't expand its valency beyond four. But with atoms like Silicon or Phosphorus, where valency greater than four is possible, they can undergo hydrolysis by a trigonal bipyramidal intermidiate. See this Kolodiazhnyi, O. I., & Kolodiazhna, A. O. (2017). Stereochemistry of nucleophilic substitution at trivalent phosphorus. Phosphorus, Sulfur, and Silicon and the Related Elements, 192(6), 621–633.doi:10.1080/10426507.2017.1284842,here they have mentioned both the possibility of intermidiate and T.S.

No, it does. If water molecule's addition is considered then only we will be able to take account of less or more electrophilicity of Boron centre or Silicon centre, right ?

Ohk... But since Boron has an empty orbital,so it seems that we can first form a tetrahedral intermediate by adding water in the RDS,and then removing the halide. So again, it seems like electronegativity only matters here,and the order should be(I feel) same as that of silicon tetrahalides...? Please elaborate on this

@YUSUFHASAN Yes it can form a tetrahedral intermidiate in the r.d.s. But you tell me in which case tetrahedral intermidiate formation will be the easiest ? will it be in case of Fluorine or Iodine? Remember, Boron is already enjoying extra stability by back-bonding from Fluorine. So, shouldn't the tetrahedral intermidiate formation be difficult with fluorides ..

Ok... So just let me summarize what I understood.... The boron trihalides experience a good amount of backbonding(2pπ backbonds) which lowers their electrophilicity in the RDS... While in th case of silicon tetrahalides, the backbonding is not that efficient(due to large energy gap between 3d and 2p orbitals applying (n+l) rule) ,so only electronegativity dominates while the RDS. Am I correct in this,or would you like to add something here?

Yes, Exactly..Nothing more than that..


text:
Molecular Orbital Mixing to get the Molecular Orbitals of a large molecule
I am trying to get the molecular orbital for arbitrarily placed nuclear centres (identical). In other words , I am trying to solve the following one electron hamiltonian.
$$h=-\dfrac{1}{2}\nabla^{2}-\sum_{i=1}^{N}\dfrac{1}{|\vec{r}-\vec{R_{i}}|}$$
I assumed the charge on each nuclear centre is one and to solve the problem I took the following linear combination as the trial function.
$$\psi(\vec{r})=\sum_{i=1}^{N}\alpha_{i}1s(\vec{r}-\vec{R_{i}})+\sum_{j=1}^{N}\beta_{j}2p_{x}(\vec{r}-\vec{R_{j}})+\sum_{k=1}^{N}\gamma_{k}2p_{y}(\vec{r}-\vec{R_{k}})+\sum_{m=1}^{N}\delta_{m}2p_{z}(\vec{r}-\vec{R_{m}})$$
of course, I start with s orbitals and then I proceeded with the above linear combination. My question is that if there are some symmetries in the structure. Does variational principle take care of that and gives the right $$ \alpha, \beta,\gamma,\delta$$, the coefficients of the trial wave functions? Do I need to impose the symmetry of the molecule (if there is any) on the trial wavefunctions? Does it give totally wrong answer if I take the above trial wave functions without worrying about the symmetry of the molecule?Can someone throw some light on it? Also, I am curious to know how do I apply MO theory for polymer like chain molecules? Please suggest some good reference to understand that

Sinmetry always simplify things so I would say use it. With long molecule it will be translational. Not an answer at all but you definitely profit of symmetry for things like molecular hydrogen ion. So impose it.

Thanks a lot for the answer .if the molecule doesn't have symmetry at all, somewhat like protein. Does MO theory work for that? Can I use the above-mentioned trial functions to solve it?

Running the calculation without imposing symmetry constraints should give the same ground energy as one with them because the true electronic ground state of the system (within the Born-Oppenheimer approximation) always has the same symmetry as the nuclei. The variational principle always puts an upper bound on your ground state energy, so in principle it will work just as well. Using symmetry, however, could speed up your calculation significantly because the symmetry operators commute with you Hamiltonian. This gives you an additional set of constraints on the coefficients. It is worth mentioning that single reference methods like Hartree-Fock and DFT will sometimes give lower ground state energies without symmetry constraints because the true ground state is a symmetric superposition of several configurations, but the single reference methods can only examine a single one.
For polymer molecules and other 1D infinite systems you would use periodic boundary conditions. Just make the box really large in the two dimensions you don't want to be periodic (so the system doesn't 'see' itself) and only evaluate non-gamma k-points along the periodic vector.


text:
Can running water carry an electrical current indefinitely, and for long distances?
Okay, not a scientist, but doing a writing project that requires a little knowledge. My question is: If there's a constant source generating enough energy into running water (like a stream, surrounded by non-conductive material) could the water carry an electrical current for an indefinite period? Basically, doing the same job as a wire.
I think the conductors in the water (minerals) may deplete over time because of chemical reactions to this process...but if it's a running stream with constantly renewed minerals, could this happen?
Edit: Clarification was requested for a couple parts. 1) Method of conduction to the water: Not that important, really. I guess a wire capable of conducting the necessary amount of power could work. 2) How fast the water flows: I'm not sure. A strong currant, I guess. But not as severe as rapids.

Water is quite a poor substitute for a wire. Other than that, yes, it will do.

Like the asian carp barrier in Illinois?

Hm...thanks A.K. Could you clarify how the Asian Carp Barrier relates?

Monopol concetion is a long known way to spare the second cable "back" and still is used today on land (e.g. https://en.wikipedia.org/wiki/Single-wire_earth_return) or across see (e.g., https://en.wikipedia.org/wiki/Baltic_Cable between Sweeden and Germany).  To quote: designed for 450 kV DC and up to 600 MW (but operated at a lesser scale).  One entry in the sea "... consists of 40 titanium nets each with a surface of 20 m2", the other entry in the sea is "a bare copper ring with a 2-kilometre diameter". With some problems related to the hydrolysis of water (e.g., hydrogen, pH value), well...

Could you make clear how you want to contact this "wire"? And how fast it flows?

I assume that the fundamental problem with this idea is that due to inherent resistance of the water, heat will be generated. As a result the water will evaporate. If there is no space left for the gaseous water vapor the "pipeline" will just explode.
That is basically why conducting wires are made of copper, it mirrors a remarkably low resistance. Silver and gold would be even better but more expensive on the other hand. Consequently copper is the standard material of wires.
If constant cooling of the system would be permitted, it could basically work.
water itself does not significantly conduct electricity, but with electrolytes dissolved in the mentioned (as suggested by the author of this question) it becomes conducting.
As long as the ions do not precipitate the liquid retains the property of conductance.

Thanks, John! Please clarify: "As long as the ions do not precipitate the liquid retains the property of conductance." Do you mean that the ions could combine or change properties?

I wanted to say that specific ions can form insoluble salts. an example is Ba(SO4) or the well known Ca(CO3). The formed salt is a solid which just sinks to the ground and is not able to carry charge anymore as it is required for conductance.

Ah, thanks. So...basically: if there is a constant source of water, conductive ions in the water, AND an outlet for steam...then this would be a feasible situation?

would say so, yes.


text:
Why do the f-block elements ( inner transition metals) have an oxidation number of +3?
The f-block valence electrons understandably include the outermost two from 6s. But why do the f-block elements have an oxidation number of +3, pulling an electron from an inner shell, not even a d-block electron?

It called f-block nowadays not "inner transition". Also they tend to have +3 cations so what is it with "1 or 2"?

I apologize. I have since researched this question a bit and found out that the oxidation state is indeed 3+ because of the two 6s electrons, which are easily removed and a 4f electron, which supposedly is easier to remove than a 5d. But I'm still questioning why the 4f electron is more easily removed than the 5d? Judging from the nonresponse here, this is a very obvious or unknown answer. Anyone?

Would be good if you edited you question then, to make it mare clear and accurate; also it bumps to homepage.

Related: Why don't we see these lanthanide species?

First of all, lets define which elements we're discussing as f-block.  Generally the lanthanides are treated as f-block elements, though this introduces some ambiguity as La (which starts the lanthanides) has 1 d electron and no f electrons while Lu (which ends the series in some definitions) has 1 d-electron and all 14 4f electrons.  So one of these two is a d-block element but its not entirely clear which one.  Recall that in the f-block there can only be 14 elements due to the 14 electrons that fill f orbitals (just as the d-block can only have 10 elements).
Not all of the lanthanides have 5d electrons, only La, Ce, Gd and Lu do!  The rest fill the 6s and 4f only.  The 6s are heavily shielded and easily lost.  Only one 4f electron is lost because the 4f closer to the nucleus and more effected by losing electrons, ie once one is removed its difficult to remove another because the rest are far more attracted to the nucleus.
So the real question should be why are La, Ce and Gd exceptions to the filling order?  For La and Ce this is because they have a lower Z effective, they're not splitting the energy of the 4f and 5d as effectively.  For these two the 5d is lower energy and fills first, then the 4f fills.  For the later Ln with a higher Z effective the 4f (being in a lower shell) is pulled to significantly lower energies and fills first.
The reasons for Gd having a different filling order is slightly more complex but can be related to whats seen in the transition metals, recall that Rh and Ru don't actually fill the 5s despite not being at a half-shell closure, they fill one 5s and the rest in 4d.  Lu is not an exception.

How are there 14 lanthanides if you include lanthanum?  The lanthanides are supposed to end with lutetium so starting with lanthanum gives 15.  Then do you mean 3 out of 15 or 2 out of 14 manganese having d electrons?

@Oscar Lanzi The number of lanthanides depends on the classification system, but by an orbital perspective (which is what we're discussing here) there can only be 14 elements in the f block (like there is only 10 d block elements) because there are only 14 electrons in the 4f.  Both La and Lu have 1 5d electron, but Lu has the 4f completely filled and La has the 4f completely empty, so which is truly a 4f elements is depends on the perspective.  La is an exception to the filling order which introduces this ambiguity which is not present in d-block elements.

Then not 14 lantanides, but 14 f block elements in Period 6.

@Oscar Lanzi As a whole people often conflate "lathanides" with "f-block" though the two are not really the same.  Some people count La and Lu (making 15 lathanides), but some systems don't count Lu (making 14), though it seems more common now to count all 15.  It's really a matter a semantics only


text:
Is it considered safe to sonicate with flammable solvents?
I use a sonicator at work in a cleaning process. It involves sonicating acetone, NMP, and IPA. We would like to move forward with other flammables as well. We currently do the work in a hood, but no other precautions are taken. OSHA has no guidance. What is your experience with this issue?

Done this many times without problem

In general it should be relatively safe considering you take all the precautions required when working with volatile flammable liquids.
Specifically for sonication I can think of two safety aspects:

Sonification promotes formation of the aerosol/fog consisting of the atmospheric gases and solvent vapors.
Depending on its composition, it might or might not be fire hazardous or even explosive.
That's why keeping an ultrasonic setup in a ventilated area — a fume hood, ideally — is a good idea.
Keep the power supply and all controller circuits outside, if possible, in order to prevent spark vapors ignition.
Be careful when sonicating heterogeneous solutions.
Some materials would absorb the ultrasonic energy and this can result in local overheating.
To prevent this, use mechanical stirring.

+1 especially for keeping an eye on temperature control since -- due to the cavitaion noise it is easy to close the hood, "forget it" and return surprised how warm the water bath got in the meantime (e.g., running an us-assisted acetalization with aceton).

A very good point. I've run Zn/NH4Cl reductions in MeOH under ultrasound and after an hour needed to change the water because it had got significantly hot.


text:
C5v E1 SALC production using Full Projection
I have another SALC question. @Andrew and @user5713492 did a phenomenal job clarifying my previous confusion with Methane T2 SALC production, so I attempted to apply this to a new point group, and have hit another wall.
Utilizing the full projection operator, I can perform the following analysis to get one of the e1 SALC orbitals in C5v (assuming a purely sigma framework through 1s orbitals as basis on the ligands):
The e1 matrix representation can be generated from {x,y}, resulting in:
$$\begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'\\
\begin{bmatrix}1&0\\0&1\\\end{bmatrix}&
\begin{bmatrix}-1&0\\0&1\\\end{bmatrix}&
\begin{bmatrix}Cos(72)&-Sin(72)\\Sin(72)&Cos(72)\\\end{bmatrix}&
\begin{bmatrix}Cos(-72)&-Sin(-72)\\Sin(-72)&Cos(-72)\\\end{bmatrix}
\end{array}$$
$$
\begin{array}{cccccc}C_{5}^2&C_{5}'^2\\
\begin{bmatrix}Cos(144)&-Sin(144)\\Sin(144)&Cos(144)\\\end{bmatrix}&
\begin{bmatrix}Cos(-144)&-Sin(-144)\\Sin(-144)&Cos(-144)\\\end{bmatrix}
\end{array}
$$
$$
\begin{array}{cccccc}\sigma_{v}'&\sigma_{v}''\\
\begin{bmatrix}-Cos(-144)&Sin(-144)\\Sin(-144)&Cos(-144)\\\end{bmatrix}&
\begin{bmatrix}-Cos(72)&Sin(72)\\Sin(72)&Cos(72)\\\end{bmatrix}
\end{array}
$$
$$
\begin{array}{cccccc}\sigma_{v}'''&\sigma_{v}''''\\
\begin{bmatrix}-Cos(-72)&Sin(-72)\\Sin(-72)&Cos(-72)\\\end{bmatrix}&\begin{bmatrix}-Cos(144)&Sin(144)\\Sin(144)&Cos(144)\\\end{bmatrix}
\end{array}
$$
Using the labeling scheme below (ignoring the $\psi$1 1s orbital at the top of the pentagonal pyramid) I can also generate the transformation series of each of these basis functions. Let's do $\psi$2 first:

$$
\begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'&C_{5}^2&C_{5}'^2&\sigma_{v}'&\sigma_{v}''&\sigma_{v}'''&\sigma_{v}''''\\
\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}
\end{array}
$$
Now, when I choose to project using the matrix elements (1,1) and the basis set $\psi$2, I get the following:
$$
SALC_{e1}'\propto C(72)(\psi3-\psi4)+C(-72)(\psi6-\psi5)+C(144)(\psi4-\psi6)C(-144)(\psi5-\psi3)
$$
$$
\propto \psi3+\psi6-\psi4-\psi5
$$

Where C($\theta$) and S($\theta$) are cosine and sine of theta respectively.
This SALC is obviously transforming as $y$ in our chosen coordinate frame (+y up the page, -y down the page, +x to the right and -x to left).
My confusion starts when I take any other combination of the $\psi$ projections and multiply by any other (row,column) combination of the matrix elements. Intuition tells me that I should either get the complementary SALC which transforms as x, or get the y-like SALC again. This is not true.
For instance, if I choose $\psi3$ as my starting point, and do (row,column) = (1,2), I get:
$$
\begin{array}{cccccc}E&\sigma_{v}&C_{5}&C_{5}'&C_{5}^2&C_{5}'^2&\sigma_{v}'&\sigma_{v}''&\sigma_{v}'''&\sigma_{v}''''\\
\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}&\begin{bmatrix}\psi6\\\end{bmatrix}&\begin{bmatrix}\psi2\\\end{bmatrix}&\begin{bmatrix}\psi3\\\end{bmatrix}&\begin{bmatrix}\psi4\\\end{bmatrix}&\begin{bmatrix}\psi5\\\end{bmatrix}
\end{array}
$$
and
$$
SALC_{e1}'\propto S(72)(\psi2+\psi3-2\psi4)+S(144)(\psi6-\psi2)
$$
$$
\propto a\psi2+b\psi3-2\psi4+c\psi6
$$
Or

Which does not, at least at first glance, appear to satisfy our requirements of orthogonality, at least at position $\psi$5.
Can anyone indicate what I've done wrong here? Or show where I strayed off the correct path? Because I am quite confused. I would like to stick to the full projection method if at all possible, as I understand that that method should work, meaning I've just messed up somewhere.
Thanks!

You are doing the right thing, but I think it was Bones who said that Evil usually triumphs unless Good is very, very careful. And you just weren't being careful enough. You seem to be going wrong with your geometric interpretation of your vertical reflection matrices. Not surprising since you don't offer an obvious physical description of which is which. I am going to say that $\sigma_{vj}$ is the reflection through the plane containing $\psi_j$. Then it should be obvious that
$$D^{(E_1)}(\sigma_{v2})=\begin{bmatrix}-1&0\\0&1\end{bmatrix}$$
To get to $\sigma_{v3}$, we employ a similarity transformation that moves $\psi_3$ into where $\psi_2$ was, then reflects with $\sigma_{v2}$, and then moves $\psi_3$ back from its position where $\psi_2$ was back to where it started, so it will be a reflection that ends up leaving $\psi_3$ invariant:
$$\begin{align}D^{(E_1)}(\sigma_{v3})&=D^{(E_1)}(C_5)D^{(E_1)}(\sigma_{v2})D^{(E_1)}(C_5^4)\\
&=\begin{bmatrix}c_1&-s_1\\s_1&c_1\end{bmatrix}\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}\\
&=\begin{bmatrix}-c_1&-s_1\\-s_1&c_1\end{bmatrix}\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}=\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}\end{align}$$
Where we have adopted the notation that $c_j=\cos(2\pi j/5)$ and $s_j=\sin(2\pi j/5)$. This is your $\sigma_v^{\prime}$ matrix, but you have $\sigma_v^{\prime}$ swapping $\psi_2$ and $\psi_3$, which is what $\sigma_{v5}$ does. Just as a check, note that a vector in the direction of $\psi_3$ should be an eigenvector of $D^{(E_1)}(\sigma_{v3})$ with eigenvalue $1$:
$$\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}\begin{bmatrix}-s_1\\c_1\end{bmatrix}=\begin{bmatrix}s_4\\c_4\end{bmatrix}=\begin{bmatrix}-s_1\\c_1\end{bmatrix}$$
So it works! I offer a table as in my previous post:
$$\begin{array}{ccccc}E&C_5&C_5^2&C_5^3&C_5^4\\
\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}c_1&-s_1\\s_1&c_1\end{bmatrix}&\begin{bmatrix}c_2&-s_2\\s_2&c_2\end{bmatrix}&\begin{bmatrix}c_3&-s_3\\s_3&c_3\end{bmatrix}&\begin{bmatrix}c_4&-s_4\\s_4&c_4\end{bmatrix}\\
\psi_2&\psi_3&\psi_4&\psi_5&\psi_6\\
\psi_3&\psi_4&\psi_5&\psi_6&\psi_2\\
\psi_4&\psi_5&\psi_6&\psi_2&\psi_3\\
\psi_5&\psi_6&\psi_2&\psi_3&\psi_4\\
\psi_6&\psi_2&\psi_3&\psi_4&\psi_5\\
\hline\end{array}$$
$$\begin{array}{ccccc}\sigma_{v2}&\sigma_{v3}&\sigma_{v4}&\sigma_{v5}&\sigma_{v6}\\
\begin{bmatrix}-1&0\\0&1\end{bmatrix}&\begin{bmatrix}-c_3&s_3\\s_3&c_3\end{bmatrix}&\begin{bmatrix}-c_1&s_1\\s_1&c_1\end{bmatrix}&\begin{bmatrix}-c_4&s_4\\s_4&c_4\end{bmatrix}&\begin{bmatrix}-c_2&s_2\\s_2&c_2\end{bmatrix}\\
\psi_2&\psi_4&\psi_6&\psi_3&\psi_5\\
\psi_6&\psi_3&\psi_5&\psi_2&\psi_4\\
\psi_5&\psi_2&\psi_4&\psi_6&\psi_3\\
\psi_4&\psi_6&\psi_3&\psi_5&\psi_2\\
\psi_3&\psi_5&\psi_2&\psi_4&\psi_6\\
\end{array}$$
Using this, we can see that the $(1,1)$ projector annihilates $\psi_2$. You knew something was going wrong here because the $(j,k)$ projector with $k=1$ has to spit out something that transforms as $x$, not $y$. Operating with the same projector on $\psi_3$ I get
$$(1-c_3)\psi_3+(c_1-c_2)\psi_4+(c_2-c_1)\psi_5+(c_3-1)\psi_6=-2s_1\left(-s_1\psi_3+s_3\psi_4-s_3\psi_5+s_1\psi_6\right)$$
And this is $x$-like in that $\sigma_{v2}$ turns it into its additive inverse. In this case it seems easiest to get the $y$-like partner by operating on $\psi_3$ with the $(1,2)$ projection/transfer operator and the result I get eventually is
$$-2s_1\left(\psi_2+c_1\psi_3+c_3\psi_4+c_3\psi_5+c_1\psi_6\right)$$
You may check for yourself that these transform like an $\{x,y\}$ pair.

Thank you! And I especially appreciate the color commentary with the Star Trek reference. I will double check for myself.


text:
What is the oxidation state of bromine in BrO3?
I came across this compound $\ce{BrO3}$ in my textbook, and was trying to find the oxidation state of bromine.
This is how I proceeded:

Since this is not a superoxide or a peroxide, or any other case where the oxidation state of oxygen is not −2 according to this, so the oxidation state of oxygen should be −2.
For oxidation state of Br as $x$, I can have $x + 3(-2) = 0$, which gives $x = +6$.

But, according to Wikipedia, the +6 oxidation state of Br is not listed.
I was confused because my book (NCERT Part I for class XII) says that "The bromine oxides Br2O, BrO2, BrO3 are the least stable halogen oxides (middle row anomaly) and exist only at low temperatures."
Please tell me where I am going wrong.
P. S. This is not an ion $(\ce{BrO3-})$ but a neutral compound.

Bretherick’s Handbook of Reactive Chemical Hazards lists bromine trioxide $\ce{BrO3}$ structure as the one composed of cationic and anionic parts [1, p. 116]:

The solid produced at $\pu{—5 °C}$ by interaction of bromine and ozone is only stable at $\pu{—80 °C}$ or in presence of ozone, and decomposition may be violently explosive in presence of trace impurities [...]. The structure may be the dimeric bromyl perbromate, analogous to $\ce{Cl2O6}$ [2].

so that formula written as $\ce{BrO3}$ is fine being a formula unit, but more correctly bromine trioxide should be written as $\ce{[BrO2]+[BrO4]-}$, implying that bromine is in two oxidation states: $+5$ and $+7$.
References

Bretherick, L.; Urben, P. G.; Pitt, M. J. Bretherick’s Handbook of Reactive Chemical Hazards, 7th ed.; Elsevier: Amsterdam, The Netherlands ; Boston, MA, 2007. ISBN 978-0-12-373945-2.
Pflugmacher, A.; Rabben, H.-J.; Dahmen, H. Zur Kenntnis der Bromoxyde. Über das Bromtrioxyd. Zeitschrift für anorganische und allgemeine Chemie 1955, 279 (5–6), 313–320. https://doi.org/10.1002/zaac.19552790507.


text:
I left a bottle of water in the fridge for the past three months. I took it out today. There was a block of ice in it. How?
How is this even possible?
I know my fridge is not freezing items. The bottle was sitting at the back and I totally forgot about it. I went to empty it and heard a clank and noticed there was ice in it. How does this make any sense? I don't think that the fridge is disproportionately cooling some regions to below zero. And if so, why would the bottle have a mix of that block of ice and water in it then?

Obviously, it is cooling some areas more than others. Look fro an air vent from the freezer.

I'm voting to close this question as off-topic because it's a question about fridges not chemistry ;)

See answer below. Looks like it IS chemistry-based. :)

About fridges? How do you ask a chemistry question that involves empirical results? Won't it involve something?

The piping for refrigerant circulation is in the fridge walls, so obviously it's colder near the back as there is no heat from the front door reaching that area and heat exchange with compressor/coolant circuit is at maximum.
As for the partially frozen water, this happens to mineral water and water containing any salts in general. As water freezes out, the salinity increases, resulting in freezing point depression for remaining solution. These processes are the basis for fractional freezing.


text:
Need some help proving the following statements in Thermodyamics?
So in my second year thermodynamics class we are currently talking about entropy and deriving its definition from the 1st law.
$$\mathrm{d}q_{rev} = dU - dw_{rev}$$
$$= C_{V}(T)dT + PdV$$
$$= C_{V}(T)dT + \frac{nRT}{V}dV$$
And then it just says since the mixed second partial derivatives aren't equal, i.e.
$$(\frac{\delta Cv(T)}{\delta V})_T \neq (\frac{\delta(\frac{nRT}{V})}{\delta T})_V$$ dq is not an exact differential (no proof of this is given). It then stipulates by dividing both sides of the equation by T
$$\frac{\mathrm{d}q_{rev}}{T} = \frac{C_{V}(T)}{T}dT + \frac{nR}{V}dV$$
suddenly it becomes an exact differential because
$$(\frac{\delta Cv(T)/T}{\delta V})_T = (\frac{\delta(\frac{nR}{V})}{\delta T})_V$$
Can anybody give me a proof of this with calculus because my textbook gives 0 proof whatsoever and it irks me because I want to understand the theory underlying this. Thanks!

These statements can be proved in general i.e. irrespective of whether you take your system as ideal gas or not. As you have used in the question $PV = nRT$, you inherently assuming that your system of thermodynamic interest is an ideal gas. So, let's first prove this for an ideal gas and then do the general proof.
$\textbf{Case I : Ideal Gas}$
For an ideal gas, $C_V(T)$ is independent of volume $V$, which means that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. (We will prove the general relation in the next section, and this statement will be obvious from that proof. ), But on the other hand $\big(\frac{\partial (nR T/V)}{\partial T}\big)_V = \frac{nR}{V}$  by simple partial derivative. So, thus you get the proof that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T \neq  \big(\frac{\partial (nR T/V)}{\partial T}\big)_V$. Thus $\delta q_{rev}$ is not an exact differential.
On the other hand if you divide both sides by $T$, you still have, $\big(\frac{\partial (C_V(T)/T)}{\partial V}\big)_T = \frac{1}{T}\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. And also on the other hand you have, $\big(\frac{\partial (nR /V)}{\partial T}\big)_V = -\frac{nR}{V^2}\big(\frac{\partial V}{\partial T}\big)_V = 0$ (obviously, as we are taking partial derivative of a constant term. Thus, by the equality of mixed partial derivatives of $\frac{\delta q_{rev}}{T}$ it becomes exact differential.
$\textbf{Case II : General System}$
For general system we have, $\delta q_{rev} = dU +PdV$. But remember we can't write, $dU = C_V(T) dT $ and $PV =nRT$. We have to write $dU = C_V(T)dT + \big(\frac{\partial U}{\partial V}\big)_T \  dV$ .
So, now we have $\delta q_{rev} = C_V(T)dT + \Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big)dV $.
Also, from the definition of Helmholtz free energy ($A = U - TS $ and $dA = -SdT -PdV$) and by using Maxwell's relation it can be shown that $\Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big) = T \big(\frac{\partial P}{\partial T}\big)_V$
Now, it can be shown properly that
$$\left(\frac{\partial C_V(T)}{\partial V}\right)_T = \frac{\partial^2 U}{\partial T \partial V} = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \qquad(= 0\text{ if } PV=nRT)\tag{i} $$
$$\left(\frac{\partial (\frac{C_V(T)}{T})}{\partial V}\right)_T = \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V} =  \left(\frac{\partial^2 P}{\partial T^2}\right)_V \tag{ii}$$
and for the right hand side
$$\frac{\partial}{\partial T}\Bigg(T\Big(\frac{\partial P}{\partial T}\Big)_V\Bigg)_V = T\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V + \left(\frac{\partial P}{\partial T}\right)_V\tag{iii}$$
and if you divide the right side by $T$ and then check the partial derivative criterion, you will ultimately get $\Big(\frac{\partial^2 P}{\partial
 T^2}\Big)_V$ on the r.h.s also (exactly same as $(ii)$ ) , But note that, $(i) \neq (iii) $. Thus $\frac{\delta q_{rev}}{T}$ will become exact differntial but $\delta q_{rev}$  will remain inexact.


text:
Concept of enthalpy at a given temperature
In thermodynamics, we always quote a fixed temperature, whenever we mention enthalpy of a reaction. For instance, one can determine the enthalpy of combustion of methane at 25 °C. Now almost all reactions involve a change in temperature, i.e. they are either endothermic or exothermic.
How should we physically interpret the meaning of enthalpy change at a fixed temperature? I recall from one textbook on experimental physical chemistry (perhaps it was Shoemaker's Experiments in Physical Chemistry stating that these state functions; still how can physically interpret enthalpy change at a given arbitrary temperature? I have not seen any textbook explicitly addressing this point.

Well, you run your reaction, then you let the products cool down back to the temperature where you started, and that is your final state.

Interesting, but I never heard of this approach before. You mean T_initial is the same as T_final, eventually? Have you seen this written somewhere? I am in the field of chem for a while and searched scores of books but never saw this explanation.

Er... I don't mean to be rude, but this is pretty much equivalent to asking how do I know that after stepping with my left foot when walking, I should step with my right foot, then again left and so on, and whether I have seen that written somewhere. Yes, that's how things are done in this field, and have been for a while.

Once I get Shoemaker's book, I will share the exact quote. Till then let us wait for other opinions which can be referenced. I mean we all know matter is made from atoms, yet we still find it written in textbooks :-) I feel you are bringing the DSC (differential scanning calorimetry) working principles in here. https://en.wikipedia.org/wiki/Differential_scanning_calorimetry

Not really. See, thermodynamics is full of imaginary processes. We talk about the things like enthalpy of formation all the time, even though the corresponding reaction (that of formation of our compound from the elements) is often impossible. Same thing here. Whether the process can be actually performed or not is irrelevant. With that in mind, yes, in some cases it can be done, and that's what they do in DSC.

This is Hess law really elementary stuff.

@Mithoron, that is the same thing as saying that enthalpy is a state function, which was already stated in the original post. As Sherlock Holmes would have said, "it is elementary my dear Watson".

The enthalpy change at a fixed temperature is equal to the amount of heat you need to add in order for the final temperature to be equal to the initial temperature.  If the enthalpy change is negative, this means that the reaction is exothermic, and you need to remove heat for the final temperature to equal the initial temperature.
If the reaction is carried out adiabatically (no heat addition or removal), the temperature of the products will be either higher or lower than the temperature of the reactants, depending on whether the reaction is exothermic or endothermic, respectively.  To determine the temperature change, you can follow a two-step procedure:  1.  allow the reaction to take place isothermally, by adding or removing the heat of reaction as necessary; then, 2.  reverse the heat addition or removal that you did in step 1, by applying it to the products of the reaction from step 1.  The net effect will be the adiabatic temperature rise (or fall) produced by the chemical reaction.

Thanks. I got hold of Shoemaker's Experimental Physical Chemistry (reading it after 10 years again), and this is how he describes it as well.

First, in order to compute state functions we often devise idealized paths to get from initial to final states, since we know that the value of a state function at each extreme is independent of path. The path we pick is a reversible one usually, and where the intensive variables describing the system can be related functionally in some simple way that allows us to compute the difference in the state functions, using for instance the ideal gas law. These idealized paths have nothing to do with the actual course of the experiment. Still, as long as the initial and final points in the calculation and experiment match, the computation should (within the approximations of the model) be an accurate descriptor of what happened experimentally to the state functions.


text:
Freezing trend of water
In this experiment, I used a digital temperature probe to measure the freezing point of water and some other salt solutions.
I was just wondering if there was any explanations for 1. The first plateau and 2. The dip in temperature right before the freezing point. (2.) can be observed for other salt solutions.
I think that the dip before the freezing point is caused by the formation of ice needing a 'nucleation point', which requires the temperature to be much lower such that a regular structure can be formed. I have no good explanation for 1. but it could be due to uneven cooling or some sort of issue with the probe.
Please give some suggestions that explains 1. and 2. if possible, even if you are unsure. Thank you for the help!

What is your experimental set-up? You are freezing water and other salt solutions, please add some details. Are you stirring the system? Salt solutions should not freeze at zero degrees. I feel this is rather an uneven cooling issue.

I agree with your nucleation explanation for the dip prior to freezing.

What are the two differently colored lines? Agree with M. Farooq: are you stiring? What kind of probes are you using, what previous calibration did you do, how are you cooling the sample? etc

@M.Farooq@Try Hard, The coloured lines are just different trials; I am using a plastic container with 100ml of each solution (water, NaCl, FeCl3 etc.) and the temperature probe is here: https://www.pasco.com/prodCatalog/PS/PS-3201_wireless-temperature-sensor/index.cfm. I am not stirring the solution, but simply setting the container with no lid into the fridge until it freezes. The image is only for distilled water, other salt solutions produces a similar graph with a different freezing point.

Is the water deionized? (as in, not the sample you added salt to - obviously)

@TryHard It is not.

First, it looks like you supercooled water. This at least explains the dips in temperature below zero degrees Celsius. It is not difficult to supercool water up to a few degrees below the normal freezing point. It remains metastable provided you avoid perturbing it, say by shaking. See for instance: http://www1.lsbu.ac.uk/water/supercooled_water.html. And yes, availability of nucleation points (small crystals) accelerates freezing.
The plateau observed around $\pu{2^\circ C}$ may be due to density differentials across the sample. At about $\pu{4^\circ C}$ water reaches its maximum density. Above this temperature, the colder water at the surface sinks, creating convection currents and accelerating the cooling process. As you dip below about $\pu{4^\circ C}$ the convective currents cease (around 3000 s), since the density gradient is stable: colder, less dense water is at the top. The water at the top of the sample continues to cool and decrease in density without sinking. If the thermometer is placed away from the air interface, it should at this point measure a slowly changing temperature as the temperature gradient toward the interface builds up. The temperature behavior will be a response to heat diffusion instead of heat convection.
You may gain additional information if you add a little dye near the thermometer and attempt to follow its movement, either before or after the convective behavior is presumed to have stopped. Convection should result in mass flow, mixing the dye more rapidly. In the diffusion regime dye should disperse more slowly and locally.
This discussion makes various assumptions of course with regard to the purity of the water and the geometry and insulation of the container it's in. It also assumes that the water has not frozen at all before about 4800 s. It also makes assumptions about the location of the thermometer.


text:
How to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?

Carbon dioxide reacts with calcium hydroxide as shown by the equation:
  $$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$
  A solution containing $\pu{0.1 mol}$ of calcium hydroxide reacts with
  excess carbon dioxide at $\pu{298 K}$ and $\pu{100 kPa}$. Which statement about the reaction is correct?
  (a) It consumes $\pu{4.4 L}$ of carbon dioxide gas
  (b) It produces $\pu{24.5 L}$ of liquid water
  (c) It consumes $\pu{0.5 mol}$ of carbon dioxide
  (d) It produces $\pu{10.0 g}$ of solid calcium carbonate

I have assumed that if there are $\pu{0.1 mol}$ of calcium hydroxide would there also be $\pu{0.1 mol}$ of carbon dioxide (since they both have a stoichiometric coefficient of 1?), please correct me if I am wrong.
Under this assumption, I have calculated the volume of carbon dioxide gas using the equation:
$$n=\frac{V}{V_\mathrm{m}},$$
where $n = \pu{0.1 mol}$ and $V_\mathrm{m} = \pu{24.79 L/mol}$, as this is the volume  one mole of particles will be contained in at a temperature of $\pu{298 K}$ and a pressure of $\pu{100 kPa}$.
Solving for $V$, I got:
$$
  V(\ce{CO2}) = n \cdot V_\mathrm{m}
  = \pu{0.1 mol} \cdot \pu{24.79 L//mol}
  = \pu{2.5 L} \text{(1 dp)}$$
However, from the statements above, $\pu{2.5 L}$ of carbon dioxide gas is not an option. I am not sure whether I have calculated the volume wrong or if I am supposed to use another approach for this question.

OK, you eliminated (a). How about (b), (c) and (d)?

I know that it can't be (c) either, since the reaction consumes 0.1 moles of carbon dioxide, not 0.5. Just to clarify, would the number of moles of CaCO3 and H2O also be 0.1 (since all the reactants and products have a stoichiometric coefficient of 1)?

Correct (c) is wrong too. // Yes, the coefficients in  balanced equation give the molar ratios.  So 0.1 mole of calcium hydroxide and 0.1 mole of Carbon dioxide yield 0.1 mole of calcium carbonate and 0.1 mole of water.

Okay, I got the answer, it's (d). Thank you for your help.

Correct, the answer is (d). // I thought that this was an interesting question. I like the mixing of the units and components.

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OP (or @MaxW) I think the question is quite interesting for a homework problem. Maybe one of you could provide an answer with the rest of the way to the solution. (Self-answers are encouraged.)

(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas

(wrong)
You've correctly calculated the volume of carbon dioxide $V(\ce{CO2})$ via the molar volume of ideal gas $V_\mathrm{m} = \pu{24.790 L mol-3}$ at SATP (Standard Ambient Temperature and Pressure):
$$V(\ce{CO2}) = nV_\mathrm{m} = \pu{0.1 mol}\cdot\pu{24.790 L mol-3} \approx \pu{2.5 L}$$
I personally wouldn't recommend to remember molar volumes at STP, NTP, SATP etc. as the numbers are close, but still different, and it brings up a lot of confusion.
Instead, only remember the ideal gas constant $R$, and remember it well as it's being used ubiquitously.
Pretending you forgot the molar volume for these conditions, you can use ideal gas law directly:
$$V(\ce{CO2}) = \frac{nRT}{p} = \frac{\pu{0.1 mol}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}}{\pu{1e5 Pa}} \approx\pu{2.5 L}$$
Here, keep in mind that $\pu{1 Pa} = \pu{1 J m-3}$ and $\pu{1 m3} = \pu{1e3 L}$.

(b) It produces $\pu{24.5 L}$ of liquid water

(wrong)
Liquid water and SATP both imply that the density of water $ρ(\ce{H2O})\approx\pu{1 g mL-1}$, and its volume can be found via molecular mass $M(\ce{H2O})$:
$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{ρ(\ce{H2O})} = \frac{nM(\ce{H2O})}{ρ(\ce{H2O})} = \frac{\pu{0.1 mol}\cdot\pu{18.02 g mol-1}}{\pu{1 g mL-1}} \approx \pu{1.8 mL}$$

(c) It consumes $\pu{0.5 mol}$ of carbon dioxide

(wrong)
The simplest out of four.
As you've already ruled out, there are equimolar amounts of both reactants and products according to the stoichiometry, so $n(\ce{CO2}) = n(\ce{Ca(OH)2} = \pu{0.1 mol})$:
$$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$

(d) It produces $\pu{10.0 g}$ of solid calcium carbonate

(correct)
Mass of calcium carbonate $m(\ce{CaCO3})$ can be found trivially from its molar mass $M(\ce{CaCO3})$:
$$m(\ce{CaCO3}) = nM(\ce{CaCO3}) = \pu{0.1 mol}\cdot\pu{100.09 g mol-1} \approx \pu{10.0 g}$$


text:
Markovnikov Rule of Alkenes
$$\ce{CH3CH(CH3)CHCHCH2CH3 + HBr ->}$$
According to Markovnikov Rule, "the rich get richer". But, as we can see the 2 carbons have the same amount of $\ce{H}$. So, what's probably the answer?

In the general case of a mid-chain alkene you get a mixture. In this case you should consider the possibility of a hydride shift to give a tertiary cation intermediate.

The rich get richer" only applies to unsymmetrically, alkyl substituted double bonds where the reagent HX has H more electropositive than X. But what of the hydroboration of 2-methyl-2-butene where H adds to the more substituted end of the double bond and boron adds to the less substituted end? This is a Markovnikov addition because boron is more electropositive than H. To allow Markovnikov's Rule to be updated, consider the electronegativity of the atoms of the reagent undergoing addition

My suggestion is to stop trying to predict the product by rules like “rich becomes richer” or “proton goes to carbon having greater number of hydrogens attached to it”, etc. for Markovnikov addition and opposite for anti-Markovnikov addition.
These are subject to many exceptions.
Instead, draw the direction of dipole in the molecule at the site of the double bond, (order of precedence: +R effect, +H effect, +I effect for considering direction). After doing this, $\ce{H+}$ will attach the carbon with δ− charge (as electrophile will attack first, and carbocation formed on δ+ carbon which is more stable), and nucleophile going to other carbon atom in case of Markovnikov addition.
The opposite holds for anti-Markovnikov addition, i.e. $\ce{H+}$ will attack carbon with δ+ charge.
Now coming to your question. After you figure out on which carbon $\ce{H+}$ will attack, the carbocation can further be stabilized by Hydride shift (always check for rearrangement possibilities like hydride, methyl, ethyl shift, ring expansion, etc.). After rearrangement only does the nucleophile $(\ce{Br-}$ here) attack.

It would make sense to post the direct answer to the question (see Hint answers revisited).

Markownikoff's rule states that with the addition of a protic acid HX to an asymmetric alkenes, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets attached to the carbon with more alkyl substituents. Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greatest number of hydrogen atoms while the X component is added to the carbon with the least number of hydrogen atoms.
Take a look at this image, this summarises the formation of major product.

After protonation a second degree carbocation is formed. This intermediate undergoes 1,2-hydride shift to give a comparitively stable three degree carbocation intermediate. Finally addition of $\text{Br} ^-$ yields racemic mixture of 2-bromo-2-methylhexane.