text:
Why is the formation free energy of carbon dioxide almost constant in Ellingham diagrams?
While reading about Ellingham diagram in my textbook, I noticed that for:
$\ce{C_(_s_) + O2_(_g_) -> CO2_(_g_)}$,
the standard Gibbs free energy ($\ce{\Delta_fG^\circ}$) doesn't seem to vary with temperature (I even compared the line's inclination with a horizontal line and they coincided completely!).
Going through the Wikipedia page for Ellingham diagram, I was able to find the following information:

The formation free energy of carbon dioxide ($\ce{CO2}$) is almost
independent of temperature, while that of carbon monoxide (CO) has
negative slope and crosses the $\ce{CO2}$ line near $\mathrm{700 ^\circ C}$.

I agree with this more as I believe that $\ce{\Delta_fG^\circ}$ should vary with the temperature. Upon comparing the $\ce{CO_2}$ line (in the Ellingham diagram in Wikipedia) with a horizontal line, this time the $\ce{CO_2}$ line is inclined slightly upwards.
I still have a doubt as to why the inclination of the $\ce{CO2}$ line is so less. One reason that I am able to think of is:
As there are equal number of gases on each side of the reaction, the entropy of the reaction is going to be less (as there is no net increase of decrease in the number of gases). As $\ce{\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ}$, if $\ce{\Delta S^\circ}$ (whose negative value is also the slope of the lines) is really very less as compared to $\ce{\Delta H^\circ}$, then we can conclude that $\ce{\Delta G^\circ}$ is going to be almost constant. But I seem to have a doubt (more specific to thermodynamics). We know that the entropy of reaction is "less", but "less" as compared to what?, i.e., how are we able to tell that this magnitude of entropy is lesser than the magnitude of entropies of all the metal oxidation reactions?
I think there must be a better way to explain the less magnitude of entropy of formation of  $\ce{CO2}$.
Could anyone please provide a concrete method / reason to prove this?
Also, are there any other factors (other than the entropy factor) governing this?

Your reasoning is quite sound.

Let's subdivide the CO2 standard entropy change reversible process into 3 parts:

Isothermal expansion of one mole of oxygen from 1 bar to the equilibrium partial pressure of oxygen in a reactor operating at equilibrium at a total pressure of 1 bar (a Van't Hoff equilibrium box)

Injection of one mole of oxygen at its equilibrium partial pressure into the equilibrium box reactor through a semipermeable membrane  while removal of one mole of CO2 through a semipermeable membrane at its equilibrium partial pressure in the reactor

Isothermal compression of one mole of CO2 from its equilibrium partial pressure in the equilibrium box reactor to a pressure of 1 bar.

The overall changes (standard changes) in G, H, and S for this 3-step process are -394359 J/mole, -393509 J/mole, and 2.8508 J/mole-K.  Now, let's break this down to the changes in the three steps.  Based on the standard change in G for the overall process, the equilibrium partial pressures of oxygen and CO2 in the equilibrium box are 8.08E-70 bars and essentially 1 bar.
In step 2, since the reaction is operating at equilibrium, the change in G is zero, while the changes in H and S are $\Delta H_2=\Delta H^0=-393509$ J/mole and $\Delta S_2=\Delta H_2/298.15 = -1319.8$ J/mole-K.
In steps 1 and 3, the expansion of oxygen and the compression of the CO2 are isothermal, so $\Delta H_1=\Delta H_3=0$.  In step 1, $\Delta G_1=R(298.15)\ln{8.08\times 10^{-70}}=-394359$ J/mole and $\Delta S_1=-R\ln{8.08\times 10^{-70}}=1322.7$ J/mole-K.  In step 3, the initial and final CO2 pressures are essentially 1 bar, so $\Delta G_3=\Delta S_3=0$
So, it seems that, merely by coincidence, the changes in entropy in step 2 compared to the sum in steps 1 and 3 are of nearly equal magnitude and opposite in sign (-1319.8 vs +1322.7).  This means that their difference is only 2.85, which represents the overall entropy change.  It is coincidental because the mechanistic changes in step 2 compared to steps 1 and 3 are very different.

In the Ellingham diagram in Wikipedia, the slope of CO2 line is slightly positive. As ΔS is coming out to be positive here, does that mean that the slope of CO2 line should have been negative instead?

I don't see a positive slope on the Wiki graph.  At 298, the slope should be slightly negative, and I think that the standard change in the heat capacities is close to zero, so that shouldn't modify the sign as temperature increases.


text:
Starting from any a,ß-unsaturated carbonyl compound, how would you prepare the following Organic compound?
Is this correct so far? Is there any other way to do this? Was it wrong of me to add KCN instead of HCN? Any help would be appreciated, thank you!

This is basically correct. Addition of alkali metal cyanides favours 1,4 addition. Phenyl Grignard should add to the ketone rather than the nitrile. You'll need to be careful with the workup to avoid elimination of the tertiary OH, I would suggest using aq. NH4Cl rather than acid.

Your choice for KCN is good because it commercially available in kilogram scale as a salt. As this salt, the cyanide already is free, contrasting to its undissociated state in the weak acid HCN is. Be careful though when using it because some chemists are unable recognize the scent of HCN.

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text:
Relationship between molar mass and density in solids
In ideal gases, it's pretty clear that 1 mole of gas occupies 22.4 L at STP. By knowing this, it's easy to calculate density given molar mass. However, this isn't true for solids. If I know molar mass of a solid, I cannot derive it's density.
I want to clarify whether my understanding of why we can't derive density in case of solids is correct. For solids, since each solid has its own lattice structure, knowing how the molecules are arranged is necessary, because if there is more space between molecules, it will have a lower density, which isn't really an issue in (ideal) gases where the arrangement of molecules/atoms is uniform.

Density of a solid is given by d= Z.M/a^3.N_A where Z is the effective number of motif present in one unit cell, M is the mass of the motif, a= edge length of a cubic unit cell and N_A is the Avogadro number.

1 mol of an ideal gas does not occupy 22.4 l at STP. Apparently, you are using an old definition of STP that is obsolete since 1982.

If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound.
In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first place.  If you then know the dimensions of the unit cell by the diffraction experiment, knowledge of the chemical formula per unit cell may tell you the number of formula units per unit cell.  See e.g, this demonstration about Ne.
The example above about Ne may be extended to crystals consisting of molecules for that you may assign atomic volumes to each element that add up to a molecular volume. An early approximation for organic compounds[1] was to assign $\pu{18 Å^3}$ for any atom of $(\ce{C, N, O})$ and $\pu{2 Å^3}$ for any atom of $\ce{H}$.  Decades later,[2] backed by much more experimental data, it was possible to refine these assignments, e.g., to average volumes of $\pu{13.87 Å^3}$ ($\ce{C}$), $\pu{11.8 Å^3}$ ($\ce{N}$), $\pu{11.39 Å^3}$ ($\ce{O}$), and $\pu{5.08 Å^3}$ ($\ce{H}$), respectively.  Large deviations from these guesses from the then experimentally determined molecular volume may be caused by solvent-accessible voids in the crystal structure.[3]
The following quotes the averaged atomic volumes (at $T = \pu{298 K}$) about typical elements of organic molecules (table 2 in Hofmann's publication, a selection of the published data):
|---------+--------|
| element | volume |
|         | [A**3] |
|---------+--------|
| H       |   5.08 |
| C       |  13.98 |
| N       |   11.8 |
| O       |  11.39 |
| F       |  11.17 |
| P       |   29.5 |
| S       |   25.2 |
| Cl      |   25.8 |
| Br      |   32.7 |
| I       |   46.2 |
|---------+--------|

References:
1 Kempster, C. J. E.; Lipson, H., A rapid method for assessing the number of molecules in the unit cell of an organic crystal in Acta. Cryst. B28, 1972, 3674; doi: 10.1107/S056774087200857X.
2 Hofmann, D. W. M., Fast estimation of crystal densities in Acta. Cryst. B58, 2002, 489-493; doi: 10.1107/S0108768101021814.
3 Florence, A. J., The use of supplementary information to solve crystal structures from powder diffraction in International Tables for Crystallography (2019). Vol. H. ch. 4.4, pp. 433-441; doi: 10.1107/97809553602060000959.

Buttonwood has given an excellent answer with a nice demonstrational video. However, there is one point I'd like to emphasize. In ideal gases, the molar mass is directly proportional to the density:
$$PV = nRT \ \Rightarrow \ \frac{n}{V} = \frac{P}{RT} \tag1$$
If the molar mass of a gas is $M$ and its mass is $m$, $n = \frac{m}{M}$ and $\frac{n}{V} = \frac{m}{MV}=\frac{m}{V} \cdot \frac{1}{M} = \frac{\rho}{M}$ where $\rho$ is the density of the gas. Thus, applying this in the equation $(1)$ you get:
$$\frac{n}{V} = \frac{\rho}{M} = \frac{P}{RT} \tag2$$
Thus, under known temperature and pressure,  $\frac{\rho}{M} = \text{constant}$ for given ideal gas.
However, for solids (and liquids), this is not true. To calculate density for a given solid, you need more than its molar mass as Buttonwood correctly pointed out (e.g., use Archimedes' principle). You need at least information about crystal packing of the solid (if it is crystalline). My point is some solids pack more tightly compared to others even if they have the same crystal packing structure (e.g., face-centered cubic or fcc). The following example shows how to calculate the density if you know the unit cell dimensions and molar mass, or vise versa.

The molar masses of gold $(\ce{Au})$ and platinum $(\ce{Pt})$ are $\pu{196.967 g mol-1}$ and $\pu{195.084 g mol-1}$, respectively. In solid state, both $\ce{Au}$ and $\ce{Pt}$ consist of the same crystal packing structure called face-centered cubic (fcc). Crystal studies have revealed that the lengths of cubic unit cells ($a$) of $\ce{Au}$ and $\ce{Pt}$ are $\pu{407.82 pm}$ and $\pu{392.31 pm}$, respectively. Prove that $\ce{Pt}$ is denser than $\ce{Au}$.

If you inspect the fcc unit cell closely, you realize that each of the eight corner atoms is shared with eight unit cells while each of the six face-centered atoms are shared with only two unit cells. Thus:
$$\text{Total atoms per unit cell}  = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$
Since the unit cell is cubic, its volume is $a^3$. Based on the value of $a$ and number of atoms per unit cell for each metal, you can calculate the molar volume of each metal ($V_\ce{Au}$ or $V_\ce{Pt}$). Since $a_\ce{Au} = \pu{407.82 pm}$:
$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3 mol-1}$$
and similarly, since $a_\ce{Pt} = \pu{392.31 pm}$:
$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3 mol-1}$$
Now, since we know the molar volume of each metal, we can calculate the density ($\rho$) by using $\rho = \frac{M_\ce{M}}{V_\ce{M}}$:
$$\rho_\ce{Au} = \frac{\pu{196.97 g mol-1}}{\pu{10.21 cm3 mol-1}} = \pu{19.29 g cm-3}$$
$$\rho_\ce{Pt} = \frac{\pu{195.08 g mol-1}}{\pu{9.09 cm3 mol-1}} = \pu{21.46 g cm-3}$$
Therefore, platinum is denser than gold $(\rho_\ce{Pt} \gt \rho_\ce{Au})$ even though their molar masses are reversed $(M_\ce{Pt} \lt M_\ce{Au})$.
Note: If you know the densities without knowing molar masses, you can calculate molar masses using the same strategy (only one unknown).

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text:
What is the difference between "primary quantum yield" and quantum yield?
The definition of the primary quantum yield given in Atkins textbook Physical chemistry as "The primary quantum yield, φ, is defined as the number of photophysical or photochemical events that lead to primary products divided by the number of photons absorbed by the molecule in the same interval."
And from Wiki: The general definition of the quantum yield (Φ) of a radiation-induced process - it is the number of times a specific event occurs per photon absorbed by the system.
And both sources use the same equation for Φ:
$$\displaystyle Φ=\frac{Number~of~events}{Number~of~photons~absorbed}$$
Is there any special use for term "Primary quantum yield" (I haven't found the explanation in literature)?

It does not seem particularly problematic. I guess the book wording point more precisely to a specific event/reaction right after absorption. Say, there is a QY for forming a reactive species and a generally small QY that lead to the final product (by a step in which photons are even not involved), etc. Or the QY of luminescence in not the same as that of absorption itself, which is unity.

Thank you, this is very helpful! And I also looked through Braslavsky, S. E. (2007-01-01). "Glossary of terms used in photochemistry, 3rd edition (IUPAC Recommendations 2006)". It gives definition only for quantum yield but distinguishes the integrated, differential  QY's and also photochemical reaction QY. Thus, I don't see the need to use term the primary QY.

Primary quantum yield would seem to be unnecessary. Most,  if not all,  of the literature will use the Wike definition.


text:
Why hydrogen bonding in some acids make them a stronger acid when it is present even before deprotonation?
If we take the example of salicylic acid, hydrogen bonding is present in the acid as follows:

Even after deprotonation, it has intramolecular hydrogen bonding as follows:

My question:
p-hydroxybenzoic acid ($\mathrm pK_\mathrm a = 4.54$) is less acidic than o-hydroxybenzoic acid ($\mathrm pK_\mathrm a = 2.97$), the reason being hydrogen bonding in the latter after deprotonation.
If the same reason accounts for the acidity of salicylic acid (o-hydroxybenzoic acid), then why is the H-bonding before deprotonation not considered (which should decrease the acidity of the compound)?

Almost Duplicate: Ortho-effect in substituted aromatic acids and bases Pretty much a duplicate: Why is ortho-hydroxybenzoic acid more acidic than its para-isomer?

Some of the conformers of salicylic acid are given below, all of which are planar in configuration.$\mathrm{^{[1]}}$

Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) is the conformer with the second lowest energy as described in the published paper as given below:

The molecule can adopt several conformations, although only one of them is highly populated and was assigned in IR experiments. The spectrum was recorded by Fiedler et al., $\mathrm{^{[2]}}$ in tetrachloride solution and indicated a strong intramolecular H-bond. The deviation of the OH
−1
stretching frequency from that in phenol was 395 cm . The theoretically calculated deviation is
359 cm−1 at the B3LYP/6-311+G(d,p) level. The lowest-energy conformer is planar, the phenolic OH is a H-bond donor to the carbonyl oxygen of the syn carboxylic group. The =O...H distance and the =O...H–O bond angle were calculated at 176 pm and 145°, respectively. The second-most-stable conformer is higher in energy by 14.3 kJ/mol, where the phenolic OH is the H-bond donor to the syn carboxylic OH. Similar conclusions were drawn by Yahagi et al.,$\mathrm{^{[3]}}$ by interpreting the gas-phase IR frequencies of the phenolic OH.

Much more convincing is the fact that the lowest energy conformer has a fractional population distribution$\mathrm{^*}$ of nearly 99.7% which further supports the argument regarding the higher acidity of salicylic acid in comparison to its para isomer. The higher acidity of the compound thus is contributed by the first conformer. The lesser interaction of the carboxyl group’s acidic proton and the stability of the conjugate base due to hydrogen bonding thus increases the acidity of the compound.
Regarding detailed mechanism of acidity of salicylic acid you may refer this and this post in chemistry SE. You may further note that the other structures might have decreased the acidity of the compound if their population distributions would have been higher. The same has been mentioned in the published paper as given below:

In a former calculation by Nagy et al.$\mathrm{^{[4]}}$, the two conformers above were found also to be the most stable with MP2/6-31G*//HF/6-31G* energy separation of 13.7 kJ/mol and free energy difference of 12.1 kJ/mol at T = 298 K. All other conformers are much higher in free energy, supporting the estimate of Fiedler that the population of the lowest-energy form is 99.7%.

Notes and References

Nagy, Peter. Competing Intramolecular vs. Intermolecular Hydrogen Bonds in Solution. Int. J. Mol. Sci. 2014, DOI: 10.3390/ijms151119562
Fiedler, P.; Böhm, S.; Kulhánek, J.; Exner, O. Acidity of ortho-substituted benzoic acids: An infrared and theoretical study of the intramolecular hydrogen bonds. Org. Biomol. Chem. 2006, 4, 2003–2011. DOI: https://doi.org/10.1039/B601875K
Yahagi, T.; Fujii, A.; Ebata, T.; Mikam, N. Infrared spectroscopy of the OH stretching vibrations of jet-cooled salicylic acid and its dimer in S0 and S1. J. Phys. Chem. A 2001, 105, 10673–10680. DOI: https://doi.org/10.1021/jp0126199
Nagy, P.I.; Dunn, W.J, III.; Alagona, G.; Ghio, C. Theoretical studies of the 2- and 4-hydroxybenzoic acids with competing hydrogen bonds in the gas phase and aqueous solution. J. Phys. Chem. 1993, 97, 4628–4642. DOI: https://doi.org/10.1039/B601875K
$\mathrm{*}$ Note that here fractional population distribution refers to the population of each conformer in the compound. You may find more information about it on the Wikipedia article.

So, if I got this right, the answer here is experimental fact? Or did I miss the theoretical explanation?

@Buraian Yes here the answer which I have posted only contains experimental facts about the population distribution of the conformers while the mechanism of acidity that has been explained briefly is theoretical.

I don't understand your argument about acidity of the ortho-isomer (which is the main question of OP). What does lesser interaction of the acidic proton have to do with higher acidity of ortho vs. para? Because there is no interaction in para-isomer either. Then you have mentioned the stability of the conjugate base due to H-bonding: does this mean that the conjugate base is more stable than the acid? There is H-bond in the acid as well, as you show in the first conformer. Also, is this part from a paper, or are you concluding it from the paper?

@ShoubhikRMaiti (1)The question of OP is that at the end Why the hydrogen bonding is to be considered(or why not) even before the deprotonation? (not about ortho vs para)(2) The block quoted texts are from a openly accesssable review paper(ref 1).(3) By lesser interaction I mean it is not involved in Hydrogen Bonding as in conformer 2(Refer statement 1 of this comment)

@ShoubhikRMaiti I apologise for not reading your comment properly, I missed the last 2 questions in your comment. (1) ” does this mean that the conjugate base is more stable than the acid?” The first conformer’s high population results from the fact that it is the most stable conformer but I do not know whether the conjugate base is less stable than the acid but I think the data for that should be there in one of the other references which are not openly accessible. (Continued...

(2) ”  Also, is this part from a paper, or are you concluding it from the paper?” The conformer being the most stable is from the paper.

The other answer has given the major conformer of the protonated form of the o-hydroxybenzoic acid, but I believe there is a part of your question that still remains to be answered. You asked why the hydrogen bond makes the ortho-isomer more acidic than the para-isomer, when the hydrogen bond is already present in the protonated form of the ortho-isomer.
For an explanation of this, let's consider the ionization equilibrium of the ortho-acid:

(I am considering only the major conformer as mentioned in the other answer)
Now, there is hydrogen bond in both the protonated and deprotonated form. But, consider the strength of the hydrogen bond in both cases. In the protonated acid, the H-bond donor is a neutral $\ce{>C=O:}$. In the deprotonated state, the H-bond donor is an oxygen of a carboxyl anion. The negative charge is delocalised over both of the oxygen, so the H-bond donor $\ce{O}$ in this case has some extra electron density compared to the neutral $\ce{>C=O}$. So, it can form a stronger H-bond.
This means the H-bond is stronger in the deprotonated state than in the protonated state of the o-hydroxybenzoic acid. So, the presence of H-bond still makes the ortho-isomer a stronger acid (than the para-isomer where there is no intramolecular hydrogen bond), even if the H-bond can exist in both protonated and deprotonated forms.

This answer imo seems more easily understandable and simpler, answers should always strike a balance in accuracy and accessibility. (Not saying this is inaccurate)

It kind of would be fun to look at this with calculations, see if it actually get stronger. Well, "see", more like guess the electronic structure, do some QTAIM, and guess some more... anyway, I've mentioned the number I would look at here: How to identify hydrogen bonds and other non-covalent interactions from structure considerations?


text:
Why does acetate ion react with water?
When NaCl is added to water, it dissociates into $\rm{Na^+}$ and $\rm{Cl^-}$. The newly formed $\rm{Na^+}$ and $\rm{Cl^-}$ do not react with water.
When $\rm{CH_3COONa}$ (sodium acetate) is added to water, it dissociates into $\rm{CH_3COO^-}$ and $\rm{Na^+}$. The newly formed $\rm{CH_3COO^-}$ reacts with water to form $\rm{CH_3COOH}$ and $\rm{OH^-}$ ions.
Why is it that $\rm{Cl^-}$ does not react with water, but $\rm{CH_3COO^-}$ reacts with water?

Search for concept of strong and weak acids.

Chloride ion is stable enough to exist on its own, acetate not so much and thus tries to grab $\ce{H+}$ from water.

According to Bronsted-Lowry Acid-Base theory, stronger acids have weaker conjugate bases and vice-versa.
$\ce{HCl}$, being a strong acid in water, it's conjugate base i.e $\ce{Cl-}$ is weak or it is more stable. Although it reacts with $\ce{H2O}$ but to a very less extent. Whereas in the case of $\ce{CH3COOH}$, it being a weak acid, it's conjugate base is a strong base. So, it reacts with water and forms relatively more stable $\ce{CH3COOH}$.
The degree of hydrolysis, $\mathrm{h}$ of a salt of weak acid and strong base is approximately given by
$\mathrm{h}$=$\sqrt{k_w/Ck_a}$.
If you substitute the $\mathrm{k_a}$ value for $\ce{CH3COOH}$ ( $\mathrm{k_a}$=$\pu{1.8E-5}$) for a $\mathrm{0.1M}$ solution of $\ce{CH3COONa}$, which yields $\mathrm{h=}$ $\pu{7.45E-4}$ where as for $\mathrm{0.1M}$ $\ce{NaCl}$ (although approximate) $\mathrm{h\approx}$$\pu{ 10^-10}$, which is  negligible when compared to degree of hydrolysis of $\ce{CH3COONa}$.

I agree that Cl- ion is stable because it has octet. Can we give a reason for the lesser stability of CH3COO- ion?

@NikhilKumar Carbon isn't enough electronegative to hold the negative charge


text:
Oxidising power of oxides of 15th group elements
In case of oxides of 15th group elements in form of $\ce{M2O3}$ ($\ce{M}$ is the 15th group element), does oxidising power of oxides increase or decrease down the group?
I tried to check by comparing the standard reduction potential (SRP) values or other related issues but didn't find any reliable sources.


text:
Can aqua regia/royal water dissolve lead?
As known aqua regia can't dissolve $\ce{Ag}$ because formation of $\ce{AgCl}$ precipitate. Similarly will aqua regia precipitate $\ce{Pb}$ as $\ce{PbCl2}$? OR It is not possible to observe this because aqua regia produce heat keeping solution warm ($\ce{PbCl2}$ is soluble in hot water).

The one who wants to reopen the question please tell the reason. If you want to answer the question you may answer linked question or edit Nilay's answer after asking him. I myself has accepted to close the question.

Aqua regia doesn't actually dissolve lead metal. It converts to insoluble lead salt. From here:

When we are working with an acidic chloride solution, like HCl (with
an oxidizer) or aqua regia, most of the lead will become an insoluble
lead chloride which if it is left to settle can be removed by careful
decanting of the solution and filtering, what lead does carry through
the process with this solution can then be removed with a few drops of
sulfuric acid, and precipitated out of solution as a lead sulfate.

Here is a YouTube video for your reference.


text:
Regarding Ellingham diagrams
In my textbook, the following information is given about Ellingham diagram:

The interpretation of $\ce{\Delta_rG^\circ}$ is based on ${K}$ ($\ce{\Delta G^\circ = -RTlnK}$). Thus it is presumed that the reactants and products are in equilibrium:
$\ce{M_xO + A_r_e_d <=> xM + A_r_e_dO}$
This is not always true because the reactant/product may be solid. In commercial processes, reactants and products are in contact for a short time.

I am not able to understand why this is not always true. How does the reactant or product being solid determine if the equilibrium will be reached or not? (Presence of solid doesn't even influence the value of K as it's activity is always taken as 1.)
Also, what does the fact that "reactants and products are in contact for a short time" signify? Do they mean that due to less time of contact, the equilibrium will not be reached (because it takes some time to attain it)?

In the case of "short time," your interpretation is correct.  The rest is a little ambiguous.


text:
Does rearrangement take place in the reaction between ethenylcyclohexane and iodine chloride?

I am stuck at the point if a non classical carbocation should be formed on the 2nd Carbon of the side chain.
One of my friend gave me the following suggestion:
I guess $\ce{ICl}$ will break in the form $\ce{I+/Cl-}$ so considering this if $\ce{I+}$ gets attached to the lower side of the chain then the carbocation formed on the 2nd element of the chain will not be a non-classical carbocation (because $\ce{I+}$ won't give a lone pair to the carbocation as it is electron deficient itself). A simple carbocation will be formed, and the rearrangement will be possible.
After rearranging it will get its most stable position on the carbon joining the ring and the chain so there the $\ce{Cl-}$ attacks and the requires molecule is formed.
I suspect this is wrong, What is the solution to this problem?

https://www.journal.csj.jp/doi/pdf/10.1246/bcsj.56.314 Mechanism in this might help.

Is the reaction correct? It would be great if you would give a source.

You seem to be describing an iodonium ion as a non-classical carbocation. Is this really correct? See Wikipedia: "Nonclassical carbocations are stabilized by charge delocalization from contributions of neighbouring C–C or C–H bonds". An iodonium ion is a carbocation stabilised by an iodine lone pair, which is certainly not a C–C bond nor a C–H bond.

@orthocresol I was actually describing the 2nd carbon in the chain as a non classical carbocation.

A carbon atom is not a carbocation, a carbocation is a molecule (or a molecular ion, to be precise) with a positive charge on a carbon.

I would say the major product is probably the one without carbocation rearrangement - following a mechanism similar to the addition of a halogen (like bromine) across a carbon-carbon double bond

Yes, I believe the rearrangement can happen. However, the product from the rearrangement is probably not the major one.
The closest reaction to what you have drawn that I could find on SciFinder is this:

The paper is mentioned below in the reference. In the paper, the researchers performed similar reactions of alkenes with $\ce{I-Cl}$ and in cases where there is a possibility of a rearrangement to a tertiary carbocation, the by-product similar to $\text{(C)}$ is generated (for example in case of the cyclohexane ring being replaced with iso-propyl, the percentages are $\text{(A)=55%; (B)=2%; (C)=38%}$).
For the reactions drawn above, the researchers did not mention the A:B ratio, however, in most cases A is the major product, which is expected (Markovnikov's rule). Nevertheless, the product formed from the rearrangement is not the major one in any of the reactions.
The mechanism is not mentioned in the paper, but my guess is that it will not be much different from the usual carbocation rearrangement mechanism. Only in this case, the hydride shift will break the iodonium ring, while creating a tertiary carbocation. The Cl can then attack the carbocation and form the product.
This is my proposed mechanism:

To reiterate, this is likely not the major pathway. The usual addition to the double bond will be the major pathway and form the major product.
Reference:
T. Horibe, Y. Tsuji, and K. Ishihara, "Thiourea–I2 as Lewis Base–Lewis Acid Cooperative Catalysts for Iodochlorination of Alkene with In Situ-Generated I–Cl", ACS Catalysis, 2018, 8(7), 6362–6366

I'd nitpick that the iodonium formation requires an additional arrow from iodine lone pair to the internal double bond carbon; but otherwise, I entirely agree with this, and nice find on the paper!

@orthocresol You are right, that arrow should be drawn. I did not draw it because it looked very messy with overlapping arrows.

In Bull. Chem. Soc. Jpn. 1983, 56 (1), 314–317, a paper dealing with the kinetics of addition of $\ce{ICl}$ to alkenes, the following mechanism is proposed

This may lead to the conclusion that the formation of the carbocation is the rate determining step and hence the carbocation can re-arrange to form a more stable carbocation.
Therefore, in this case, the carbocation could rearrange to the tertiary position where the carbocation is more stable.
Another proof that rearrangement does in fact take place in case of $\ce{ICl}$ addition comes in the following review of $\ce{ICl}$, Encyclopedia of Reagents for Organic Synthesis 2013,

Methoxymethyl-protected homoallylic alcohols can be activated by ICl in toluene at low temperatures. The initial adduct is rearranged by a methoxy transfer, through a chair-like transition state, to yield a chloromethyl ether

(emphasis mine)
Therefore, when it comes to $\ce{ICl}$ addition to alkenes, if carbocation rearrangement is possible, I would say it would be possible. Thus, the given reaction is correct.

This first mech. isn't OK. Intermediate cation is cyclic iodonium rather then open carbocation.

The latter example is interesting, but I'm not sure it's relevant; the question isn't about having some intramolecular nucleophile like an OMe group, it's more about whether hydride migration is possible. Mechanistically, it's similar (because hydride migrations are kind of an intramolecular nucleophilic attack on an iodonium), but an OMe and a C–H sigma bond are quite different, so I'm not really sure whether it lends any real insight as to the feasibility of a hydride migration.

The reaction you mentioned is indeed, as you suspect, incorrect.
Iodoniums do form but not as easily as bromoniums though. The formation of a planar carbocation in this case is unlikely. We do get iodonium like in the case of bromonium. It is to be noted that few of the haloniums ions, synthesized/discovered, are stable enough at specific conditions to remain unreacted and X-ray crystallography backs this up. In chemistry, obtaining an X-ray crystal structure of a molecule is considered to be near-definite “proof” of its existence. In fact, recently even fluoronium ions have been synthesized though they are very rare and difficult to isolate.
The point I am making is that electron deficiency of $\ce{I^+}$ is not an absolute base to knock out its formation. Firstly, $\ce{I^+}$  doesn't exist separately; haloniums are of the form $\ce{R-X^+-R}$ where X is a halogen.
In our purpose here, it only acts a reaction intermediate which exists for a very short time. The above reaction proceeds with iodonium ion formation and anti addition of the nucleophile (in this case $\ce{Cl^-}$). Chloride will add at the Markovnikov carbon. In conclusion, rearrangement not possible as no distinct planar carbocation formed.
Notes and References

Stable Bromonium and Iodonium Ions of the Hindered Olefins Adamantylideneadamantane and Bicyclo[3.3.1]nonylidenebicyclo[3.3.1]nonane. X-Ray Structure, Transfer of Positive Halogens to Acceptor Olefins, and ab Initio Studies, R. S. Brown, R. W. Nagorski, A. J. Bennet, R. E. D. McClung, G. H. M. Aarts, M. Klobukowski, R. McDonald, and B. D. Santarsiero
Journal of the American Chemical Society 1994 116 (6), 2448-2456
DOI: 10.1021/ja00085a027
https://www.masterorganicchemistry.com/2013/03/15/alkene-bromination-mechanism/
https://en.wikipedia.org/wiki/Halonium_ion (The tendency to form such bridged halonium ion is shown highest for iodine in the wikipedia extract but that need not correspond with their inclusion in the olefin reactions. That is an in general statement not mentioning about olefins there)

Just found this question. Check this as well: https://chemistry.stackexchange.com/questions/75250/interhalogen-with-alkene

I think a rearrangement can happen without a planar carbocation.

In the formation of an iodonium ion, $\ce{I+}$ does not ever exist, it does not even exist as an "intermediate". For such a strongly worded opening sentence, I find the argument here is just not good enough. I agree that just because an iodonium ion is formed doesn't mean that a rearrangement is impossible. The correct question is not what kind of carbocation is formed, the correct question is whether there is a rearranged product.

Please see also my comment on the question re. usage of the term "non-classical".

I always had the thing in mind of iodonium not being $I^+$.  While typing it accidently skipped out of my mind to share the definition of halonium ion as per iupac. Also, I don't feel that non classical carbocation doesn't fit here. Regarding the rearrangment, I was mentioning about this one particular case where I don't see  why it should happen. There might be other cases where it might happen but certainly not this

Your answer still says that $\ce{I+}$ is "a reaction intermediate which exists for a very short time." That's not true; it doesn't exist at all. "Also, I don't feel that non classical carbocation doesn't fit here." - That's not a valid argument.

"The point I am making is that electron deficiency of I+ is not an absolute base to knock out its formation. Firstly, I+ doesn't exist separately; haloniums are of the form R−X+−R where X is a halogen" are you talking about this @orthocresol ....I did mention that it doesn't exist. Regarding the non classical carbocation, we have been taught in our college that these types are indeed called so. But referring to https://chemistry.stackexchange.com/questions/31732/what-is-a-non-classical-carbocation this question and its answer i get your point now.

Have removed that non classical part @orthocresol


text:
Does the largest known cycloalkane contain 288 carbon atoms?
If you Google "largest known cycloalkane", you will find this:

The above result came from a random HW-QnA site. From where does the result come from?
Similar posts:

Largest alkane having a given alkane as its base name
IUPAC nomenclature of extremely long alkanes

According to Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208[1], Kwang Sup Lee and Gerhard Wegner, attempting the synthesis and study of $n>100$, $\ce{C_nH_{2n}}$ cyclo-alkanes and linear alkanes, were successfully able to synthesize a linear alkane of $\ce{C384H770}$ and cycloalkanes with 288 carbons ($\ce{C288H576}$)

Linear alkanes up to $\ce{C384H770}$, (2k; M = 5378) and cycloalkanes up to $\ce{C288H576}$(3g; M = 4040) were prepared by this strategy.

A brief method of preparation is as follows:

The strategy of the synthesis of long chain linear and cyclic alkanes was developed starting from the work of Schill et al. who had shown that the oligomers obtained by oxidative coupling of 1 following Eq. (1) can be separated by adsorption chromatography.
\begin{align}
\ce{\underset{\textbf{1}}{HC#C-(CH2)20-C#CH} ->[Cu^2+] \\ \underset{\textbf{2}}{HC#C-(CH2)20-C#C-\left[C#C-(CH2)20-C#C\right]_m-H}} \\
\end{align}

A similar strategy was used to synthesize the cycloalkanes as well.

The cyclic oligomers were synthesized by the same reaction according to Eq. (2)
under conditions of high dilution of the terminal diacetylene

Finally, the linear and cyclic oligomers were hydrogenated over a Pd/C-catalyst in a conventional manner to obtain the analytically pure alkanes.

As for the assignment question, I was able to link it to a introductory exercise in Organic Chemistry by Janice Gorzynski Smith.2.

Reference

Kwang Sup Lee, Gerhard Wegner, "Linear and cyclic alkanes ($\ce{C_nH_{2n+2}, C_nH_{2n}}$) with $n > 100$. Synthesis and evidence for chain-folding," Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208 (https://doi.org/10.1002/marc.1985.030060316)
Janice Gorzynski Smith, Organic Chemistry, Sixth Edition, McGraw-Hill book Company Inc.;2020, ISBN: 978-0-07-066720-4

A) Perhaps an accidental error, but for me, the link behind the introductory exercise aims to show Organic Chemistry by Robin Charles Burrel published by McGraw Hill in 1936, but stops midway with the note "No preview available".  Maybe google.co.in differs in this from google.com?  B) I don't have Janice Smith's book in the shelf, but if the content still supports your argument, I recommend to link to the publisher's entry to the current 6th edition by 2020 (incl. an updated ISBN13) here.

@Buttonwood, seems to be an error in google books. The cover of the preview shows Janice gorzynski smith as the author. I've updated the publisher information, but the preview still works for me. I've removed the preview and added the actual image of the question.


text:
At which temperature does a concentrated aqueous solution of sodium chloride begin to boil?
We know, solutes dissolved in water increases the boiling point of water. This question is an MCQ. The options are 96,99,100,104 degree Celsius. The correct answer is 96. Why? If the boiling point INCREASES after adding salt then why not 104?

Have you excluded an error as the option.

The boiling temperature always increases when adding salt into water. It looks as if salt was attracting water molecules and preventing them from quitting the liquid. So that it is necessary to overheat the solution to let water molecules get into the vapor phase. I have often made this measurements in class and always found about $\pu{104°C}$.

In explaining this phenomenon, the key is not an interaction between solute and solvent but instead the inability of the solute to exit into the vapor phase at this temperature.


text:
Thermodynamics Property Tables

Two rigid tanks are connected by a valve. Tank A contains $0.2\ \mathrm{m^3}$ of water at $400\ \mathrm{kPa}$ and $80\ \%$ quality. Tank B contains $0.5\ \mathrm{m^3}$ of water at $200\ \mathrm{kPa}$ and $250\ \mathrm{^\circ C}$. The valve is now opened, and the two tanks eventually come to the same state. Determine the pressure and the amount of heat transfer when the system reaches thermal equilibrium with surroundings at $25\ \mathrm{^\circ C}$.
Assumptions 1: Tanks are stationary and thus the KE and PE changes are zero and there are no work interactions so $W = 0$.

To start this question I know we have to find the specific internal energy for Tank A and Tank B then use this to eventually find the total heat transfer so, I began by finding the properties of water in Tank A at $p = 400\ \mathrm{kPa}$, which are taken from Tables A-5 in Thermodynamics an Engineering Approach
At $400\ \mathrm{kPa}$ the specific internal energies are retrieved to be $$u_\mathrm f = 604.22\ \mathrm{kJ/kg}$$ and then $$u_\mathrm{fg} = 1948.9\ \mathrm{kJ/kg}$$
My question is why are we using the $u_\mathrm{fg}$ or the specific internal energy of evaporation for this question and not the specific internal energy of saturated vapour, $u_\mathrm g$?
Please ask me any further questions you need to help me!

You could use the internal energy for liquid and vapour, or you could use the internal energy for liquid and the energy of evaporation since both pairs essentially give you the same information. What's better depends on how you want to calculate the next properties.

Some data are missing. For example : 1) What is "water 80% quality" ? Is it 80% purity ? What is the residual stuff ? 2) What is the temperature of the water in container A ? 3) What is "saturated water" ?

@Maurice 80% quality means that 80 % of the mass is vapor and 20% is liquid.

I would have used $u_L$ and $u_G$.  I would also have used $v_L$ and $v_G$ .

Ahhh I see, ultimately using either will result in the same values, I too would have used u_L and u_G, Thanks a lot for the responses! How do I proceed from here? How can I upvote your responses or mark as answered ?

"Tank A contains 0.2 m3 of water at 400 kPa and 80 % quality."  What do you want to describe with "water of 80% quality"?

It seems that the OP left, but I liked the question so I will answer it. The notation $_{fg}$ has been addressed in the comments, so I will answer to your "how do  I proceed from here?".
Disclaimer I will use Smith, Van Ness, and Abbott book for the water tables, but it doesn't matter for what we are asked. I think they have the same reference as your book, so answers will be almost the same.
Main Idea If $\ce{H2O}$ is a pure gas or liquid, it has two degrees of freedom, and its thermodynamic state is specified two intensive variables. If saturated, then we need only one.
I will talk about states $1A$ (Tank A initially), $1B$ (Tank B initially), and $2$ (final state).
Initial Calculations
$\ce{Tank 1 A}$: We have that $P_{1A} = 400 \; \pu{kPa}$ and the quality $x_{1A} = 0.8$, so it is saturated water, and the thermodynamic state is specified. They give us the quality $x$ because a general thermodynamic property for a pure fluid in LV equilibrium $M$ is obtained by doing $M = x M_{g} + (1 - x)M_l$.
\begin{align}
  v_{1A} &= 0.3699928 \; \pu{m^3/kg} \\
  u_{1A} &= 2163.0074 \; \pu{kJ/kg}
\end{align}
Of interest will be the initial mass of Tank A
$$ m_{A} = \dfrac{V_A}{v_{1A}} = 0.54055 \; \pu{kg} $$
$\ce{Tank 1 B}$: We have that $P_{2A} = 200 \; \pu{kPa}$ and $T_{2A} = 250 \; \ce{°C}$, thus the thermodynamic state is specified
\begin{align}
  v_{2A} &= 1.1989 \; \pu{m^3/kg} \\
  u_{2A} &= 2731.4 \; \pu{kJ/kg}
\end{align}
Of interest will be the initial mass of Tank B
$$ m_{B} = \dfrac{V_B}{v_{1B}} = 0.41705 \; \pu{kg} $$
The specific internal energy of state $1$ is
$$ u_1 = \dfrac{m_A u_{1A} + m_Bu_{2A}}{m_A + m_B} = 2410.551 \; \dfrac{\pu{kJ}}{\pu{kg}} $$
Balances
For a closed system, where no work is done, the 1st law states
$$ m\Delta u = Q \rightarrow Q = m(u_2 - u_1) \tag{1} $$
where $m = m_A + m_B$.
Also, since matter does not enter or leave the system, our whole mass resides in the tank volumes. Therefore, the final specific volume is
$$ v_2 = \dfrac{V_A + V_B}{m_A + m_B} \rightarrow
   v_2 = 0.73099 \; \dfrac{\pu{m^3}}{\pu{kg}} \tag{2} $$
Final State
With Eq. (2) we are ready to go, since we have our two intensive variables $T_2 = 25 \; \ce{°C}$ and $v_2 =  \; 0.73099 \; \pu{m^3/kg}$. However, upon looking at the water tables, this specific volume is between the saturated liquid and saturated gas at $T_2$. Thus, the pressure is informed immediately
$$ \boxed{P_2 = 3.166 \; \pu{kPa}}$$
And the quality and specific internal energy of this saturated water is
\begin{align}
  x_2 &= \dfrac{v_2 - v_{2l}(T_2)}{v_{2l}(T_2) - v_{2g}(T_2)} \rightarrow
  x_2 = 0.01682 \\
  u_2 &= x_2 u_{2g}(T_2) + (1 - x_2)u_{2l}(T_2) = 143.5728 \; \dfrac{\pu{kJ}}{\pu{kg}}
\end{align}
And Eq. (1) yields
$$ \boxed{Q = -2170.858\; \pu{kJ}} $$
The system has given this energy to the surroundings. Makes sense, because we started with two tanks at high $P$ and $T$, and ended at ambient temeperature.


text:
Why Does Electrical Work Cause Changes in Internal Energy of the System?
If we look at system at constant temperature and volume which is galvanic cell, first law of thermodynamics states: $$ dU = dQ + dW'$$
Where W' is electrical work exchanged with surroundings and Q is heat exchanged with surroundings. I use chemistry sign convention for work.
As far as I know electrical work is work done by electric field when charge moves between certain potential difference. In galvanic cell, potential difference exists between two electrodes.
Internal energy is sum of total microscopic kinetic energy of all molecules and atoms and their potential energy of interaction (energy stored in intermolecular and chemical bonds).
What does work done by electric field have to do with internal energy of the system? Why does electrical work cause changes in internal energy of the system?
In galvanic cells, electrical work is negative as work is done by the system and from the first law it tends to decrease internal energy of the system.
I can't really see the connection between work done by electric field in the galvanic cell and sum of microscopic kinetic energy and microscopic potential energy of interaction.

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

For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell.

Your question seems to be "what does reversibility mean in the context of a galvanic cell?". Is that right?

@Buck Thorn Well, no actually. Question is what does electrical work in galvanic cell have to do with internal energy of the system. What does moving charges between certain potential difference have to do with microscopic kinetic and potential energy A.K.A internal energy?

If water is electrolyzed producing $\ce{H2}$ and $\ce{O2}$ gases, a great amount of energy has to be added to this water. The obtained substances are not hotter than the original water. But they possess a much higher potential energy. Their internal energy (and enthalpy, and free enthalpy) is much higher than the same atoms in water. This internal energy may be liberated into heat if these substances are sent in a flame, or into electric energy in a combustion cell.
It is no use thinking about the work done by an electric field on an electron in the atoms. The electric field inside an atom is difficult to calculate, because of all other electrons interfering with the field of the nucleus inside the atom.


text:
What are the similarities / differences between the terms anode / cathode and oxidizing agent / reducing agent?
I'm kind of confused on how to distinguish the terms anode and cathode from oxidizer and reducer or oxidizing agent and reducing agent. Some references prefer to use one over the other in discussions depending on the topic. Are they suppose to mean the same thing in some cases? Like an anode has oxidation taking place so it is an oxidizing agent? The substance used in the anode can be referred to as an anode but it can also be called an oxidizing agent?
I want to be able to know this more clearly and have a distinction of when to use both if I want to use both in the same discussion so that both of them have a specific idea associated with them instead of just allowing interchange between the terms even though in a sense it is also correct.

Anode or cathode are simply the material or the location where oxidation or reductions in an electrolytic or galvanic cell. It means anode or cathode must be electrical conductors. There is no oxidation or reduction when electricity is not passing through the cell.

Also the cathode anode can be part of the redox or merely conducting. They refer to the "place" where the process occurs , but that place can also be involved.


text:
Why are conductance and conductivity additive in nature?
In my book, it is written that:
$κ_{electrolytic solution} = κ_{electrolyte} + κ_{solvent}$
$G_{electrolytic solution} = G_{electrolyte} + G_{solvent}$
Where $κ$ and $ G$ are conductivity and conductance respectively. But why are conductance and conductivity additive in nature? Why is it that we can simply add conductance and conductivity? What is the proof of the above two expressions? Can someone please provide an explanation for the above? I am so confused. Please help.

When we have two resistors in parallel, the same voltage will be across each resistor. However, the total current that flows will be the sum of the currents flowing in each resistor. The conductance is the total current divided by the voltage. This is the same as the sum of the two conductances.

There is expected explicit effort to find the answer before asking, based on thinking, searching and reading.

@Poutnik, but nowhere can I find the solution to my question.

@PranitaBaruah1, This all valid under very very dilute conitions. Search Kohlrausch laws.

@Pranitabaruah1 Thinking is not limited to what you have found. For very dilute solutions, you can derive it yourself  from ion mobility through molar conductivity.

Simple example: suppose you have a large tank of water and you wish to empty the water out. With a single water hose, used as a siphon, you can conduct away, i.e., drain, the water out of the tank. It will drain at some rate. If you now add another hose as a siphon, it will conduct away water at its own rate. The total rate at which water is conducted away, i.e., drained away, is the sum of the two siphon hose rates. The two are independent of one another and their effects are simply additive.


text:
Comparing the basicity of heterocyclic amines

Compare the relative basicity of the following amines.

I know that the higher the electron density on the nitrogen the higher will be its basicity. So here’s my take on each option.
(a)The Nitrogen lone pair undergoes resonance. It’s not good for basic strength
(b) A doubly bonded nitrogen lone pair does not undergo resonance. So, the Nitrogen lone pair are localised.
(c) Both nitrogen atoms are doubly bonded and so their lone pair won’t be delocalised and the electron density is localised here as well.
(d) There’s no question of resonance here so the lone pairs are localised anyway
From the above points my answer is (c)>(b)>(d)>(a) but the answer given is (d)>(b)>(c)>(a)
Is there any mistake in my reasoning?

The correct order is in fact $(\bf{d}) \gt (\bf{b}) \gt (\bf{c}) \gt (\bf{a})$, the reason for this is as follows.
$(\bf{a})$ is definitely last in this order of basicity since its lone pair is delocalised by the phenyl ring. Now we need to compare the other three.
We can split this into two parts, a comparison between

$(\bf{d})$  and $(\bf{b})$
$(\bf{b})$ and $(\bf{c})$

The basicity of $(\bf{d})$ would be greater than that of $(\bf{b})$ because of the fact that in $(\bf{d})$, the nitrogen with the localised lone pair is $\mathrm{sp}^3$ whereas in case of $(\bf{b})$, the lone pair is seen on an $\mathrm{sp^2}$ nitrogen. Since an $\mathrm{sp^2}$ nitrogen is more electronegative than an $\mathrm{sp^3}$ nitrogen, it holds the electrons closer, thereby leading to less accessible electron density. Therefore $(\bf{d}) \gt (\bf{b})$.
Now, moving on $(\bf{b})$ and $(\bf{c})$. In case of $(\bf{c})$, one of the nitrogen atoms has a -I effect on the other nitrogen atom, thereby leading to lower electron density and hence lower basicity. Since we find the basicity of each nitrogen atom, we see that the basicity of each one is less than that seen in $(\bf{b})$. Therefore, $(\bf{b}) \gt (\bf{c})$
Hence, the order of decreasing basicity will be $(\bf{d}) \gt (\bf{b}) \gt (\bf{c}) \gt (\bf{a})$.

How can one nitrogen can impart inductive effect on another nitrogen in (c) if both the nitrogen atoms have identical electronegativities ?


text:
Why do we take two moles of amines in the reaction of amine with an acid chloride?
My textbook states that when reacting an acid chloride (for example, acetyl chloride) with an amine, we need to take two moles of amines because when the $
\ce{-NH-R}$ group displaces the $\ce{-Cl}$ group, the newly formed second degree amide is susceptible to attack by the $\ce{Cl-}$ ion produced and hence the second mole of amine reacts with the $\ce{Cl}$ to pacify it. How is this possible, when $\ce{NH2-}$ is a stronger base than $\ce{Cl-}$?

Actually, it's not a textbook. It's my professor's lecture notes from school.

Another way to put this is that the primary product is not the amide itself, but its protonated form. In this, the leaving group is not a charged amide-ion, but a neutral amine, which is "stable" enough to be displaced by a chloride ion (the reverse of the addition reaction). The second molecule of amine is needed to deprotonate the positively charged species into a neutral molecule, in which the leaving group would in fact be the negatively charged, "unstable" amide-ion.

Oh yeah, it's not the charged ion. That's where I was going wrong. Thanks!

I am not sure what the book is talking about, but I am sure that one amine molecule attacking the acid chloride will form a certain byproduct:
$\ce{R^{(1)}-C(O)-Cl + NHR^{(2)}R^{(3)} -> R^{(1)}-C(O)-NR^{(2)}R^{(3)} + \color{blue}{HCl}}$
The HCl acts as a protic acid and can react with another amine molecule.  To get a stable mixture of products you need one mole of amine to combine with the carbonyl group and another to neutralize the HCl:
$\ce{R^{(1)}-C(O)-Cl + 2 NHR^{(2)}R^{(3)} -> R^{(1)}-C(O)-NR^{(2)}R^{(3)} + [NH2R^{(2)}R^{(3)}]^+ + Cl^-}$

Yeah, that's what the book said. So, if we didn't use another amine molecule, what would happen? Also, when we react esters with amines, why do we not need to use 2 molecules of amine?

If you add $1$ amine molecule for each molecule of $\ce{RCOCl}$, the yield of amide will be only $50$%.


text:
What is a correct primitive hexagonal unit cell to describe a monolayer of molybdenum sulfide?
I searched the coordinates or reduced coordinates of primitive cell of mono layer hexagonal $\ce{MoS2}$, but there are lots of different structures and coordination. For example, material project give this:
\begin{array}{c|ccc} \hline
\textbf{Element} & a &b& c \\\hline
 \ce{Mo} &0.3333 &0.6667 &0.25 \\
 \ce{Mo} &0.6667 &0.3333 &0.75  \\
 \ce{S} &0.3333 &0.6667 &0.362 \\
 \ce{S} &0.3333 &0.6667 &0.138 \\
 \ce{S} &0.6667 &0.3333 &0.638 \\
 \ce{S} &0.6667 &0.3333 &0.862  \\  \hline
\end{array}
but there are another structures too, that is:
\begin{array}{c|ccc} \hline
\textbf{Element} & a &b& c \\\hline
\ce{Mo}&0.0 &0.0 &0.25 \\
 \ce{Mo}&0.0 &0.0 &0.75 \\
 \ce{S} &0.3333 &0.6667 &0.362 \\
 \ce{S} &0.3333 &0.6667 &0.138  \\
 \ce{S} &0.6667 &0.3333 &0.638 \\
 \ce{S} &0.6667 &0.3333 &0.862 \\  \hline
\end{array}
and there more than these. So, which one is correct?

Entries mp-2815 and mp-1018809 differ in their lattice constants ($c = 14.879\ Å$ vs. $c = 13.983\ Å$) reported, thus differ in unit cell volume and density (4.05 g/cm³ vs. 4.31 g/cm³) while sharing the same space group symmetry $P6_3/mmc$.  But only the former is labeled as «stable», the later one is annotated with «Decomposes To $\ce{MoS2}$».  It is possible that the later entry is simply about a metastable polymorph.
About half-way through the pages by materials project you find the entry Electronic Structure with (in these cases) referrals to the ICSD database.  Thanks to a collaboration with CCDC, a sub-set of some data of the databases' corresponding entries, including bibliographic information, may be retrieved on their joint web page by entering the ICSD entry number (result 1, result 2).*
Indeed a search, e.g., on the COD reveal that there are at least two experimentally characterized polymorphs of $\ce{MoS2}$ known where differences justify a description in different space group symmetries:

$P 6_3/m m c$, $a = b\ \pu{ = 3.15 Å}$, $c\ \pu{= 12.3 Å}$, $\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$, e.g., entry COD 1010993.

$R3m$, $a = b\ \pu{ = 3.163 Å}$, $c\ \pu{= 18.37 Å}$, $\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$, e.g., entry COD 9007661.

The $(a, b, c)$ of the lattice constants describe the unit cells' outer dimension.  They are not the fractional coordinates -- better described as $(x, y, z)$ -- to set the atoms position along these vectors.  (Which may [e.g., cubic Bravais class] orthogonal to each other, but need not to be [e.g., monoclinic Bravais class].)
*) Accessing the databases «for real», though, offers you a much easier comparison with other entries of either same chemical composition,  symmetry related space groups, similar unit cells, etc.

Is this true for a true monolayer (2D material) or does this answer only apply to bulk, 3D material. I see 3D unit cell volumes in those links which should be areas in 2D, and there should be no c axis. 1, 2, and 3. I think that the OP may have searched the wrong site if they really want to ask about 2D materials.

@uhoh I don't know for sure if the two entries the answer explicitly names are the two mentioned in the OP.  After all, some polymorphs are re-examined multiple times (e.g., different equipment, different experimental conditions of $T$, or / and $p$, etc.).  Nor did I read the two publications the two ICSD entries refer to. Thus it is not clear if these are about a bulk 3D crystal (and those examined in electron diffraction may be tiny, compared to conventional X-ray diffraction), or a 2D crystal.

@uhoh Depending on the line of thought, true monolayer may be understood differently, too. For self-organized monolayers of thiol-alkanes on Au, or LB films in general, the statement of your reference 1 «The crystal structure of monolayer MoS2 showing a layer of molybdenum atoms (blue) sandwiched between two layers of sulfur atoms (yellow).» already could be conceptually two too many, raising the question «why not a more flat (graphene like) material in first place?» as in e.g., in c2db / 2dmatpedia.org.


text:
Burning of Natural Gas
I would like to compare the burning of Natural gas to heat the external object 1) when Natural Gas gets fired with Pure Oxygen 2) when Natural gas gets fired with Atmospheric Air.

What are the resultant products in the above two scenarios.
Is there any difference of heating rate (of any external object, if the flame is directed on that external object)
Which one is desirable between the two for achieving required temperature in given time and for achieving maximum combustion efficiency.

The combustion of any natural gas occurs by the same equation with air or with pure oxygen. It is $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$, The amount of heat produced by its combustion is the same with air and pure oxygen. But the final temperature its not the same, as the heat of combustion in air is partly used for heating an important amount of nitrogen. The same amount of heat has to heat $3$ gases with pure oxygen ($\ce{1 CO2 + 2 H2O}$), and $11$ gases with air $\ce{(1 CO2 + 2 H2O + 8 N2}$). So the increase of temperature is about three to four times higher with pure oxygen than with air.

Not counting the  small amount of nitrogen oxides that might be formed, depending on temperature and rate of reaction. See https://www3.epa.gov/ttnchie1/ap42/ch01/final/c01s04.pdf

Such estimate of temp. is just too crude and I wonder why you bother with such questions (rhetoric question).

Mithoron. Of course my estimate of temperature is crude. But I suspect  the author of the question is not interested in a rigorous calculation. He might just need a qualitative approach.


text:
What is the reaction mechanism involved in the ring opening seen in 3,3-diphenyloxiran-2-ide?
I have an epoxide with a negative (-) charge on one of the carbons. In which of the two ways should I open the epoxide? I am not able to understand which one to prefer.

(The general way I think of these rearrangements and breaking and forming of bonds is by doing heterolytic cleavage and placing (-ve) and (+ve) charges on the carbons the bond's between which I am breaking)
I think there's something wrong with the method in which I am approaching this which is causing this issue. Please help me out with the same.
(I know the general theory for opening of epoxide in case of acid/basic attack)

Was this the initial question? or was it something different?

The second way looks good. Note the first way would give you a carbene.

Count your implicit hydrogens carefully in the first option.

@SafdarFaisal an intermediate step in a reaction sequence for a problem. Btw, how do you make these sequences digitally?

@JoshMitchell yeah, that should be highly unstable. Looks like I miscalculated the hydrogens or didn't keep track of them!


text:
Error in the IUPAC Gold Book definitions of exergonic, endergonic, exothermic and endothermic?
All definitions in the IUPAC Gold Book refer to standard values:

exergonic reaction (exoergic reaction) [link]
This expression is often applied to a reaction for which the overall
standard Gibbs energy change ΔG° is negative. Some workers use this
term with reference to a negative value of ΔH° at the absolute zero of
temperature.

endergonic reaction (endoergic reaction) [link]
This is usually taken to be a reaction for which the overall standard
Gibbs energy change ΔG° is positive. Some workers use this term with
respect to a positive value of ΔH° at the absolute zero of
temperature.

exothermic reaction [link]
A reaction for which the overall standard enthalpy change ΔH° is
negative.

endothermic reaction [link]
A reaction for which the overall standard enthalpy change ΔH° is
positive.

Shouldn't all these terms characterize values of ΔG and ΔH at any given condition, independently to any standard state?
Is there another recognized official source for the definition of those terms?
Edit: In view of the discussion starting to build up in the comments to this question, please note that I am not asking about the definition of standard state in thermodynamics.

The word "standard" means that the measurements have been made when the initial and the final state are at the same temperature and pressure (25°C, 1 atm.)

My question was maybe not clear. I have rephrased it. I meant "independently to any standard state".

The entry about exergonic reactions links to the open access publication in PAC. I understand the footnote on p. 164 of this paper as "... [standard refers] to the standard concentration [of the reagents] (usually $1 \mathrm{mol, dm^{-3}}$)" to prevent too large deviation between concentration of the reagents, and their activity.  Maurice' criteria (25°C and 1 atm) might be included, or set on top.  Content in square brackets is my added reading, which may be wrong.

Please don't propose iupac tag. See https://chemistry.meta.stackexchange.com/questions/3801.

The definition of standard conditions does not include temperature. And standard pressure is 1 bar (not 1 atm).

@Long Bye, bye, to T in SATP ?

@buttonwood Despite the names, STP and SATP are not the same as the standard state for thermodynamic quantities. See also Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions (IUPAC Recommendations 1981): “The above definitions of standard states make no reference to fixed temperature. Hence, it is possible to have an infinite number of standard states of a substance as the temperature varies. (…)”

@Loong With this specific reference in hand, I recommend you are going to transfer your comment above (including link to the OA publication) into an answer.  As such, it is easier to recognize (both by visually inspection and appreciating your crawl through the literature to identify the relevant reference).

I agree that IUPAC's decision to tie the definitions of exothermic and endothermic to stanadard pressure is problematic, since it disconnects them from the direction of heat flow.  I discuss this here:  https://chemistry.stackexchange.com/questions/122102/why-are-iupacs-definitions-of-exo-and-endothermic-disconnected-from-the-direct

@theorist Indeed, your post raises exactly the underlying questions I had. So it seems that this question doesn't have a clear answer yet. I have contacted someone from the IUPAC pointing to your question. Let see where it leads...


text:
How to graph a second order reaction with two different reactants

In a laser flash photolysis experiment, $\ce{C6H5NH2}$ $(C_0 = \pu{60 \mu M})$ is oxidized to the corresponding radical cation. The disappearance of radical cation is monitored spectrophotometrically and the results are shown in the table below. There are three possible reaction pathways for the radical cation formed:

Dimerization of radical cation,
Reaction between radical cation and starting material, and
Reaction between radical cation and solvent.
Using experimental data, determine the most probable path of reaction and the rate constant of the reaction.
$$\begin{array}{c|c}
t/\pu{\mu s} & [\text{Radical}]/\pu{\mu M} \\ \hline
0\ & 20\\
5 & 11.12\\
10\ & 6.86\\
15\ & 4.47\\
20\ & 3.00\\
25 &  2.06 \\
30 & 1.43 \\
35 & 1\\
\hline
\end{array}$$

I know how to use the given data to graph a first order and a second order with the same reactants, but I don't know how to do it for a second order with reactants A + B (which is the second pathway: reaction between radical cation and starting material). I know that the graph should be ln$\frac{a-x}{b-x}$ against $t$ but I don't understand how to get $(a-x)$ and $(b-x)$.
From what I understand $a = \pu{20 \mu M})$ and $b = \pu{60 \mu M})$ but what is $x$?

$x$ is the conversion variable.   If the reaction is A+B $\to$ P+Q at time $t$ the concentration of A is $A_0-x$ and of B, $B_0-x$. Then product is $P_0+x$ and $Q_0+x$ where the zero subscript is the initial amount. You have 60 micromol to begin with and 20 is converted instantly by the flash so 40 is left unreacted at time zero. The integrated rate equation is $\displaystyle \ln\left(\frac{A_0-x}{B_0-x}\right)=kt(A_0-B_0)+\ln(A_0/B_0)$. In (3) the solvent conc'n is huge so pseudo first order.

x is also sometimes known as the extent of reaction

@porphyrin I don't quite understand how to calculate ($A_0 -x$) and ($B_0 -x$). I would have assumed that ($A_0 -x$) is calculated as ($60 - 20$), ($60-11.12$) and so on. But that isn't the concentration of the radical. So how exactly is ($A_0 -x$) and ($B_0 -x$) calculated?


text:
Examples of endothermic reactions occuring spontaneously at constant temperature and pressure
Typical spontaneous endothermic processes I can think of (e.g. melting of ice, isothermal gas expansion) involve either an initial temperature difference between the system and the surroundings, or some pressure variation.
I am looking for examples of endothermic processes, if possible molecular, that happen spontaneously at constant temperature and constant pressure, so that the process is both endothermic and exergonic (i.e. $\Delta G <0$).

Are you happy to count dissolving Common Salt in water? https://en.wikipedia.org/wiki/Enthalpy_change_of_solution

Searching here for “endothermic dissolution” turns up more examples.

A practical example is a chemical cold pack. You break the inner bag to allow the chemicals to mix, resulting in an endothermic reaction.

The classic demo example is this. The chemical cold packs suggested by @theorist are excellent choices: I have one of the sodium acetate supersaturated solution ones and they may still sell the ammonium nitrate dissolution cold packs.

The simplest reaction which is endothermic and exeronic is the reaction $$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$
If you add progressively some $\ce{NaHCO3}$ powder in a beaker containing $\pu{150 mL}$ $1$ or $2$ molar $\ce{HCl}$ solution initially at maybe $\pu{20°C}$, you will observe that a huge amount of $\ce{CO2}$  is produced. But the temperature goes down to $15°$, then $10°$ C, if the solution is continuously stirred by a thermometer. Important : stir ! By adding some more powder, the temperature may go down to $\pu{0°C}$ if the beaker is well isolated from the ground.

What happens if one doesn't stir?

The evolution of CO2 means that the volume of the system increases and thus that energy is transferred to the surrounding as work. Does it mean that for any process that is both endothermic and exergonique at constant p and T, the internal energy necessarily decreases? (With your example I realize that this was actually the underlying question that was bothering me when I posted the original question...)

If you don't stir, the powder does not touch the liquid. It is pushed by the bubbles of  $\ce{CO2}$ and stays hanging above the liquid, at least for a while

@TheQuark  No. Suppose (in arbitrary units) delta_H = -3, and the pV-work = +2.  Then delta_U will be -1 (also negative, just less negative than delta_H).  OTOH, if the pV-work is +4, then delta_U will be +1.  I.e., whether the direction of heat flow reverses from positive to negative when you change from constant-p to constant-V is determined by the magnitude of the pV-work relative to the magnitude of delta_H.  At atmospheric pressure, typically the pV-work will be too small to change the sign of q (direction of heat flow).  But at sufficiently high pressures, you could see a reversal.

In the comments @IanBush has mentioned dissolving of salts.
Some salts in particular have a very endothermic dissolution process. IIRC two such salts are $\ce{NH4Cl}$ and $\ce{NaNO2}$. If you can get your hands on any one of them, then just dissolve them in a little bit of water in a test tube and you will be able to see condensation of moisture on the outside surface within a short time.
Oh, and it's not just salts, dissolution of many organic compounds is also endothermic, for example glucose dissolving in water.
(There are exceptions to this: dissolution of NaOH in water is exothermic)


text:
How to compare the basicity of 4-Aminopyridine and 1,4,5,6-Tetrahydropyrimidine
The structures of 4-Aminopyridine and 1,4,5,6-Tetrahydropyrimidine are as shown

Now the nitrogen (I) and (II) of both the compounds have their lone pair of electrons. In both cases the lone pair of electrons is currently not in resonance with the ring or other nitrogen respectively. In both the cases after protonation the (+ve) charge is stabilised by the other nitrogen through resonance. Case (A) has more resonance structures. Also Inductive effect due to nitrogen is lesser in (A) than in (B) which leads to more electron density in (A)(I). It could be argued that (A) is an aromatic amine hence less basic but aromaticity reduces the basic character due inductive effect (or maybe delocalisation of lp aswell). But in this case structure (B) has greater inductive effect (-I) due to the other nitrogen as compared to what the benzene ring would have on (A) and the lp are out of plane of ring in (A) therefore localised too.
All these effects I) Greater resonance structures II) Lesser (-I) in (A) than in (B) make me feel that (A) should be more basic however the case is completely opposite and (B) is more basic.
All that I said above could be completely wrong. Please help me figure out where I am going wrong and what is actually happening in detail. Thanks for your time!
(Note: My statement of the inductive effect of nitrogen being greater than benzene ring Is completely based on my speculation based on the fact that nitrogen is quite electronegative in general. I don't have the data to prove the same. Please correct if wrong)
EDIT: Please see my comments on the question which is linked as a duplicate to this one. I had doubts regarding the specific comparison of these two compounds based on things I've mentioned above, which is why I felt it was appropriate to make a new question to ask this in detail.

You should rather consider conjugate acids - if mesomeric structures are equivalent or not.

I think here mesomeric effect plays a greater role than inductive. There are only 2 resonance structures possible for 2nd case (the one itself) and the other being the one in which (II)N gets a negative charge. If you draw all the resonance structures of the 1st case, you will get to know that in one structure there comes a negative charge on the (I)N. But similarly the charge will be distributed at other 2 ortho positons (wrt -NH2 group). Thus the partial negative charge on (I)N will be lesser than on (II)N. This leads to the 2nd having more electron density than the 1st.

Compare their structures after protonation. The second compound then shows two resonating structures, as given below -

Both the resonating structures are equivalent in this case, which makes the conjugate acid more stable (in the same way in which equivalent resonating structures of COO- anion makes it more stable, increasing the acidity of carboxylic acids).
For the first molecule, the lone pair donated by the top nitrogen is delocalized not only to the bottom N(I) but also to two carbons of the pyridine ring, which decreases the overall electron density on N(I) and makes it less basic.

If I'm not wrong in COO- cross resonance occurs, but in this case, its extended resonance was taking place so I thought it's becoming more stable. Since the usual idea I know is that the more it delocalizes and spreads the better it is, I heard about this partial charge method first time. If possible please provide me with some resources where I can read more about this or some other cases where this happens. Would like to learn more, (+1) for your answer though.


text:
Explanation of the trans-effect (thermodynamic and kinetic) for octahedral transition metal complexes
The thermodynamic trans effect refers to the ground state weakening of the metal-ligand bond in a complex by a ligand trans to it. The kinetic trans effect is the lablisation of a ligand in a complex by the ligand trans to it.
These are commonly known effects, and most of thre resources I find are discussing the trans effect in terms of square planar complexes. The explanation of kinetic trans effect for square planar complexes is simple, the trigonal bipyramidal transition state is stabilised by $\pi$-acceptor ligands.
However, the trans effects of both type are known to be present in octahedral complexes as well.[1] The paper however, does not go into much detail about why the trans effect works. The structural trans effect I can understand as due to trans ligands sharing the same metal d-orbital for overlap, which would result in competitive electron donation/withdrawal. However, it isn't clear how the kinetic trans effect would work. The paper mentions that sigma donation helps to stabilise the transition state leading to the dissociation of one ligand; but I don't understand why $\pi$-donation would not have the same effect, or why $\pi$-acceptors would not have the inverse effect.
So I am looking for an explanation of the trans effect, particularly kinetic trans effect in octahedral complexes. An explanation with orbitals would be very helpful.
Please also feel free to direct me to resources (books, journals) which has these explanations or would be helpful for understanding this.
Reference:

B. J. Coe, S. Glenwright, "Trans-effects in octahedral transition metal complexes", Coordination Chemistry Reviews, 2000, 203(1), 5-80

https://pubs.acs.org/doi/abs/10.1021/ic50180a048 https://pubs.acs.org/doi/10.1021/ic50143a005, have you checked these papers out?


text:
When to consider back bonding and when to not; when will it change hybridisation and when will it not?
This question has been troubling me since my professor taught this concept.
Points that she taught:

pi back bonding can only happen amongst elements of 2nd and 3rd period.
At least one of them needs to be in 2nd period.
Fluorine forms the best back bonding inspite of its electronegativity. (when asked why, didn't get any convincing answer)
Back bonding increases bond angle and decreases bond length.
When there's a clash between decrease of a bond angle according to Bent's rule and increase of the same angle by considering back bonding, one should consider back bonding to dominate.
In some cases, possibility of back bonding drives the hybridisation to change.
According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital (p or d) for allowing the electrons (in p orbital) of the side atom  to delocalize into it .

Trisilylamine should have been tetrahedral in shape but due to back bonding the hybridisation is $\mathrm{sp}^2$ as Nitrogen lone pair is delocalised into the empty d orbitals of silicon. Similarly,  disilyl ether must have bond angle lesser than or equal to $120^\circ$. Less, because there will be a lone pair of oxygen that is not delocalized which should lead to a repulsion between the bond pair and lone pair. But disilyl ether has a bond angle of $144^\circ$.
In conclusion, the main question is as follows:
When do we consider back bonding and when will it change hybridisation?

@RavirajBhosale I recommend this https://chemistry.stackexchange.com/questions/34842/why-is-bcl3-a-monomer-whereas-alcl3-exists-as-a-dimer/64600#64600

$\mathrm p$π-$\mathrm d$π bonding is very weak and normally not considered to be very effective in bonding. Rather, the bonding can be explained via ionic bonding instead due to the large electronegativity differences. This can be seen in siloxanes and trisilylamine.
According to J. Am. Chem. Soc. 1980, 102 (24), 7241–7244

Since our ab initio calculations reproduce the geometry and
dipole moments of the three molecules very well, conclusions
concerning the electronic structure of the siloxane bond derived from these calculations should be at least qualitatively correct. (1) (p-d)π bonding is of minor importance for explaining the unusual properties of the siloxane bond. [...] A (p-d)π bond order of 0.4 has been deduced from experimental bond lengths

Now referring to Inorg. Chem. 1997, 36 (14), 3031–3039, the results section state the following.

The expected number and arrangement of charge concentrations is
not, therefore, observed for ligands less electronegative than carbon, that is, for the molecules that are predominately ionic, and it just for these molecules that the bond angle is substantially larger than the tetrahedral angle.

Further more, it also talks about the assumptions taken in VSEPR and why it doesn't work here.

However, with decreasing ligand electronegativity, each of the four electron pair domains becomes larger in extentsmore spread out or diffusesand so overlaps its neighbors to an increasing extent until, in the limiting case of an oxide ion, there is no electron localization, and each domain may be considered to occupy the whole of the valence shell

This is also the case in disiloxane as explained later in the paper,

That the bond angle in $\ce{H3SiOSiH3}$ ($148.3$°) is somewhat larger than that in $\ce{H2BOBH2}$ ($135.4$°), which in turn is slightly larger than that in $\ce{H2POPH2}$ ($129.8$°), is consistent with the electronegativity values of Si (1.7), B(2.0), and P(2.1). As in $\ce{H2BOBH2}$, there is only one nonbonding maximum in the oxygen valence shell in $\ce{H2POPH2}$ and in $\ce{H3SiOSiH3}$.

Basically the last three paragraphs state that the VSEPR model only works if the attached atoms are sufficiently electronegative enough so that covalent bonds can form. In case of disiloxane, the electronegativity difference leads to it being a more ionic bond than it should be. This is seen also in $\ce{Li2O}$ which has a $180^\circ$ bond angle because of the high electronegativity difference.
For trimethylsilane, see Jan's answer here

Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:
$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$
In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Using backbonding to explain such differences is very non-trivial since it is like trying to build a modern-era supercomputer with vaccum tubes (may be exaggerating here). Chemistry is a science where we observe and then hypothesise.


text:
My question is regarding Periodic Table, Why there are three columns for 8B group in old version of the periodic table?
Why there are three separate columns for 8B group in old version of the periodic table?

Welcome to Chemistry! What efforts have you made to get the answer to your question? Effort is the imaginary currency you use in chem SE to gain answers.

@SafdarFaisal I like that, it's both concise and accurate :-)

@SyedFaisal. You are looking at an old version of the periodic table. Please forget about it. Please consider the periodic table with 18 columns, and not with 8 columns,


text:
What is meant by partial negative and positive charge? How can a charge be partial?
My teacher stated, oxygen has a greater electronegativity than hydrogen. Thus, in water, oxygen develops partial negative charge and hydrogen develops develops partial positive charge.
I am not able to comprehend what is meant by partial negative and positive charge?
(https://i.stack.imgur.com/8T0RO.jpg)                                                      I have always imagined charge as being integral. How can a charge be partial (ie, non integral).

Welcome to Chemistry! What efforts have you made to get the answer to your question? Effort is the imaginary currency you use in chem SE to gain answers.

@Akhil Please have a look at this answer. It might quickly clarify what is exactly meany by partial charge. https://chemistry.stackexchange.com/a/40781/21349

Have you considered probability density of valence electron distribution for polar bonds ?

@Poutnik not really necessary. A distribution of point charges can have electrical dipole. Of course for equilibrium one has to go quantum, but I think this can be answered nevertheless.

@Alchimista Sure. The electrical dipole is other way to say there is biased distribution of electrons.

OP it suffices that electrons are not equally apart from the nuclei. In reality, as Poutnik said, better you think of foggy electrons. Indeed see the answer by Neretin in the linked answer M Das suggested to you.

@Poutnik yes but it can be the distribution of point charges, too. Not need in context to think of electron density. See also my comment to OP. I know what you said is OK, I try to see things in line with OP.

To go to the root of your concern, that doesn't necessarily involve chemistry. You are correct that e is the elementary charge, and electrons cannot be splitted. However, take a positive and a negative charge in space. Now approach this couple with a test charge arriving from almost whatever point and direction. Unless you are far away and feel no charge at all, when you approach the given couple you will feel like there is a partial charge, because one attract you and one repels you (or viceversa) but with different forces. This should let you see why the idea of partial...

... charge isn't that exotic.

If two identical atoms are bound in a covalence, the center of gravity (barycentre) of the bonding electrons is exactly in the middle of the distance between the two nucleus. The atoms are not charged.
If two different atoms are bound in an ionic bond, the center of gravity of the electrons are exactly situated on one of the nucleus. This atom is charged, or has a total charge.
If two different atoms are bound in a covalent bond, the center of gravity of the electron is somewhere between one nucleus and the middle of the distance separating the nucleus. In that cas, we say that this atom has a partial charge.

While I see the analogy, I don't see why one should use the gravitational field or the gravity force instead of electrical field or electric force. The need of quantum level, that can be omitted here, isn't removed using the gravitational analogy, so why? It just dangerously mixes different facts. Indeed, (almost) nothing change in the baricentre of a hypothetical covalent NaCl and the real ionic one. And a novice can be confused. Downvote.


text:
Why is the ortho product major in the nitration of anisole with nitric acid and acetic anhydride?
There is this weird information I found in my organic chemistry book ("Essential Organic Chemistry" by Ranjeet Shahi)
The question asks us to give reason as to why the following occurs.

Also, don't bother giving reasons for the first (above) reaction. I would like to know the reason for the below one. Why would ortho be major in presence of acetic anhydride ?

Can someone also let me know how to create such images of reactions ? It was very cumbersome to gather images from web and paste and move them

ChemDoodle. Also, see this: https://chemistry.meta.stackexchange.com/questions/3245/how-to-draw-structures-of-compounds-while-writing-questions

It is complicated, as this paper from Olah et al. here shows, but this passage offers a possible explanation:

With the reactive nitronium salts, the isomer distribution of the nitration of anisole shows the highest ortho/para ratio (2.7-2.4), reflecting the "early" (i.e., starting aromatic-like) nature of the transition state of highest energy. Indeed, the calculated electron density in the ground state of anisole is the highest around the ortho positions, even higher than in the case of toluene. As the reactivity of the nitrating system decreases (e.g., with $\ce{CH3ONO2}$/$\ce{BF3}$ or acetylnitrate), the amount of para-substitution increases. Because the para-methoxy substituent can stabilize the arenium ion intermediate better than a para-methyl group, these reactions seem to show increasing intermediate ion-like nature of the transition state.

$\ce{HNO3}$/$\ce{Ac2O}$ has a similar ratio to the formal salts of $\ce{NO^+2}$ so may be classified as a reactive nitronium source.
A later passage in the same paper suggests that the nitrating species in the $\ce{HNO3}$/$\ce{H2SO4}$ may be the less reactive $\ce{NO+}$

If this is the case, it reflects the much weaker electrophilic nature of $\ce{NO+}$ (as compared to $\ce{NO^+2}$) and its reaction through an intermediate areniumion-like transition state of highest energy.

It would also be worth reading some of the papers referenced by Olah for further discussion.


text:
Why same metal salt is required in electrolytic refining?
Why we use same metal salt in electrolytic refining?
Ex. For $\ce{Cu^2+}$ if we take suppose $\ce{FeSO4}$ so there are 2 ions ready to react on cathode ie $\ce{Fe^2+}$ and $\ce{Cu^2+}$.
But from a series (who will react first) we know $\ce{Cu^2+}$ is more likely to react with cathode instead of $\ce{Fe^2+}$.
Then why we need only same metal salt we can take any metal salt that will not react with cathode first according to series?

Answers on CH SE site are figuratively paid by the user's own effort. When you ask, you are supposed to provide explicit compact summary of partial answers or at least ideas you have got until then. Effort not shown may be  considered as effort not done.

Let me suggest you look up the meaning of «refine».  This time, it need not be a reference related to crystallography, nor chemistry, because its meaning is so old (in terms of etymology) that a good English dictionary plus some thought may rapidly clarify the situation.

I have literally no idea what you are asking here. In which processes is the same metal salt "required"? There are a lot of different processes.

You see, in electrolytic refining, the goal is to get pure metal on the anode. Let us take a specific case of electrolytic refining of $\ce{Cu}$. Normally we would use $\ce{CuSO4}$ solution, so let us take that case.
Impure copper is set as the anode, $\ce{Cu^{2+}}$ ions are released. But you see, the $\ce{Cu^{2+}}$ ions from the anode don't directly go to the cathode.
It is a sort of chain reaction if you will, the $\ce{Cu^{2+}}$ ions make the solution positively charged (i.e. there are more $\ce{Cu^{2+}}$ ions than $\ce{SO^{2-}4}$ ions), so the positive ions near the cathode will try and neutralize themselves to maintain the solution's neutrality.
Now, because we have taken $\ce{CuSO4}$ solution, there will be only $\ce{Cu^{2+}}$ ions near the cathode, so they will proceed to the cathode become $\ce{Cu}$ atoms.
But if we take $\ce{FeSO4}$ solution, there will be some deposition of iron too on the cathode, implying impure metal which is the opposite of what we want.

Welcome to the site.  Note, chemical information may be advantageously formatted using on ChemSE with mhchem.  Take moment to familiarize with this.  You are encouraged to use it in the body of questions, answers, and comments.   Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Yah that i wanna to ask why there is some iron deposits, because there is a series like reactivity series that tell if there r 2 or more ion than which would react on amode and cathode. And according to that series copper should react first iron is late in series. Or can little iron could also react????

Iron deposits are there because Fe2+ ions also neutralise or reduce to give Fe, iron atom. Yeah iron can also react, the reactivity series is supposed to be basically how much can a metal react (in very basic terms). It tells that copper will react faster than iron to give copper atom. But it doesn't say, copper will only react and iron won't, but the copper reaction will be faster. Also, another thing to consider is that at first, there will be a lot of Fe+2 ions. So, the Fe+2 ions are more prone to react that the lower amount of Cu+2 ions.


text:
How can I prove ascorbic acid degradation with Uv-vis spectroscopy?
I have to prove scientifically why a sodium ascorbate solution in water turns yellowish - reddish over time. I know ascorbic acid is supposed to oxidize into dehydroascorbic acid, and the latter then oxidizes into more subproducts... But I have to prove this by using UV-vis spectroscopy and whenever I analyze the yellowish solution the only peak I see is the ascorbic acid peak (around 262 nm). Why don't I see the dehydroascorbic acid peak (210 nm) or even other peaks related to the other subproducts ?
Update: I found this graph on a published article and I was expecting to obtain something like it:

Below are the peaks of a fresh sodium ascorbate 10% w/w solution (orange peak) and a solution with the same concentration that has been aged for 50 days and has acquired a deep yellow color (green peak). Both of the solutions have been diluted to 1:80.000 in water for the UV-vis analysis. The ascorbic acid in the aged solution has obviously degraded in some way since the peak is lower, but why are there no new peaks regarding the degradation products ?

Since the solution turns “yellowish-reddish” over time, that means it is absorbing light in the visual region of the spectrum, i.e., between roughly 400 and 700 nm. So what happens when you obtain spectra from 200 nm to 700 nm? How about running experiments with fresh and aged solutions and put those into your edited question?

Also something may diminish.

Thank you for your answers, I edited my question with your suggestions ! Any more thoughts ?

How about not diluting the (deep yellow colored) aged solution: put 3 mL in a cuvette and run the visible spectrum from 400 to 900 nm? Use 3 mL of water as the blank. If no significant difference, then the spectrometer might be in need of repair or re-calibration.

The way the baseline of the green sample increases with decreasing wavelength suggests you've got some light scattering going on. Possibly aggregation or precipitation.

If from literature, add to the black and white plot spectrum. Check if it records Abs for the same sample concentration, or a.u., or normalized in intensity to display all three simultaneously, too. Otherwise, the Abs by ascorbic acid and dehydro asccorbic acid will be literally in the shadow of the Abs of decomposition products and difficult (if at all) to record.

Your spectra:  1) You have an issue with baseline correction as the yellow curve drops do $\approx -0.4$). Some accidental underpass might be fine, but 0.4 is like the sample acts like a lamp to create new light.  2) The number of tics along Abs (after all, about the interval $-0.1(0.1)1.0$)) does not match the expected number of tics. 3) The step at $\approx \pu{320 nm}$?  This looks like (for example) either the halogen, or deuterium lamp wasn't yet stabilized when recording the blank. Check how many hours the lamps already worked (they have a meter), and replace when exceeding this limit.

Thank you for your comments, I have had the lamps replaced and will restart the analysis tomorrow. The number of hours was indeed exceeded... However, I do not understand when you talk about the "tics"... What are "tics" ? Kind regards

Some of the experimental details are missing, but if you recall
$$\mathrm{Abs} = \varepsilon \cdot c \cdot d$$
about the absorption depending on the molar extinction coefficient $\varepsilon$, the analyte concentration $c$, and the optical path length $d$, it could be due to a low concentration of the analyte.
Independent of the former, given the $\pi$ electron systems of the carbonyl groups are not in conjugation with each other, nor with a $\ce{C=C}$ double bond (as in ascorbic acid) it is quite possible that $\varepsilon$ is low; equally attenuating the recorded signal.

(composite from here and here)
You mention to be able to record a signal at $\lambda = 262\,\mathrm{nm}$.  Thirdly, given you now equally look for a signal around $\lambda = 210\,\mathrm{nm}$, either the material of your sample cell, or / and the solvent of your analyte no longer is optically transparent enough for the given path length $d$ for this wavelength.  (Did you check that the spectrometer's second lamp, the deuterium lamp, equally is running and stabilized, too?)

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Than you very much for your detailed answer. None of those things had occurred to me. Nevertheless I recently edited my post with more details so feel free to add a coment if you have some more thought on it, thank you again !!


text:
How is this reaction not a decomposition reaction? And are all disproportionation reactions decomposition reactions?
When I asked my teacher whether all disproportionation reactions are decomposition reactions, he stated the following reaction:
$$ \ce{4 KClO3 -> 3 KClO4 + KCl} $$
and said that this isn't a decomposition reaction and that it is a hydrolysis reaction. How is this not a decomposition reaction as the definition of a decomposition reaction is the splitting of a single reactant into multiple products. I see the same happening here, the singular reactant ($\ce{KClO4}$) is splitting into multiple products ($\ce{KClO4}$, $\ce{KCl}$). And also how is this a hydrolysis reaction as I see no $\ce{H2O}$ molecule taking part in it. It would be great if you answer with some proper explanation to help me understand.

decomposition implies a more complex thing falls apart into more less complex things. kclo4 is kclo3 plus another atom so it's is the opposite of decomposition. but kcl is less complex so overall it's neither decomposition nor synthesis. that's what we call disproportionation

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Unless IUPAC says something else, that can be a decomposition reaction, at least the decomposition of the solid lattice of the starting salt. But could also be hydrolysis or whatever. I would say one named it in the more useful way giving the context. In fact, you know it is a disproportionation. An internal example of the latter? Perhaps in organic chemistry. I would wait other users for this point, but I am pretty sure the hypothetical reaction(s) would have a better representative name, again.

$\ce{KClO3 + O -> KClO4}$ does not seem to be to be decomposition.

(1) If you talk about decomposition, then this is a decomposition reaction: $\ce{2KClO3 -> 2KCl + 3O2}$. (2) The reaction you mentioned is not hydrolysis because there is no water involved.

First, this is not hydrolysis. Word hydrolysis originate from Ancient Greek hydro- 'water', and lysis 'to unbind'. By definition  it is "any chemical reaction in which a molecule of water breaks one or more chemical bonds". Second, from IUPAC gold book - "decomposition is the breakdown of a single entity (normal molecule, reaction intermediate, etc.) into two or more fragments". Nothing said about complexity of products, so this definition can be used for the given reaction.

all right @NilayGhosh I'll do it

As the comments say, in the reaction, $\ce{KClO_{3}}$ turns into $\ce{KClO_{4}}$ which is not decomposition as $\ce{KClO_{4}}$ has an extra $\ce{O}$ molecule. Water is not involved so, it is not hydrolysis either. So, we can conclude that all disproportionation reactions are not decomposition reactions.