text:
The most stable resonance structure
I came across this question:

Which is the most stable resonating structure of p-nitrosobenzene
  among the following?

I have been taught about the priorities for resonance structure stability as follows:

Complete octet $\rightarrow$ more stable
Uncharged $\rightarrow$ more stable
Lower formal charge $\rightarrow$ more stable
Less separation between opposite charges $\rightarrow$ more stable
Negative charge near more electronegative atoms and positive charge near less electronegative atoms $\rightarrow$ more stable

So according to the above rules, (b) and (c) are the among the most stable. Bu then applying rule #4 above, (b) should be more stable, because here each positive charge is in vicinity of negative charge.
However the answer given is (c). Why is this?

A -ve charge on oxygen is better than a -ve charge on carbon, so (c).

@user43496 For future reference, please consider the dimension and orientation of the figures included.  To render the figure more legible, I rotated and down-scaled your photo.

There is no such thing as a most stable resonance structure.

Fyi:  None of these is a structure for $p$-nitrosobenzene.  Properly the name is $p$-nitroso_nitro_benzene.

Because on C; the negative charge should be on oxygen (the more electronegative element) instead of on carbon

The technical terms

There is no such thing as a most stable resonance structure. – Martin

The correct term would be resonance contributor. All these structures contribute, so the charges are spread over the molecule. Also, formal charges are mostly an accounting scheme. For example in the ammonium ion, most of the positive charge is near the hydrogen atoms, but the Lewis structure shows a formal charge on the nitrogen atom.
Here is one explanation of resonance and contributing structures from wikipedia

In chemistry, resonance is a way of describing bonding in certain molecules or ions by the combination of several contributing structures (or forms[1], also variously known as resonance structures or canonical structures) into a resonance hybrid (or hybrid structure) in valence bond theory.

The question
So I edited the question:

Which is the [most stable resonating] major contributing resonance structure of p-nitrosobenzene among the following?

I would agree with @nana: Structures that have a formal charge on nitrogen or oxygen are probably stronger contributors to the resonance hybrid than those with charges on carbon.


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Phosphorus does not form pπ-pπ bonds but how could it form dπ-pπ bonds?
My textbook says phosphorus cannot form pπ-pπ bonds because their atomic orbitals are so large and diffuse that they cannot have effective overlapping.
My doubt is then how does phosphorus form dπ-pπ bonds? I mean dπ-pπ bond is supposed to be more ineffective overlapping. If pπ-pπ cannot happen then how does dπ-pπ happen?

My doubt is then how does phosphorus form dπ-pπ bonds?

Poorly. Influence of the $d$-orbitals on the chemistry of the $p$-block is minor. The current belief is that 'double' bonds are actually dative bonds in most cases.
See the similar cases of chlorine and sulfur
How does chlorine form more than 1 bond?
What is the chemical structure / hybridization of sulfuric acid?
For phosphorus it isn't that bad, but still bad
http://link.springer.com/article/10.1007%2FPL00010730

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text:
How to make a buffer using HClO

If we plan to prepare a buffer with the $\mathrm{pH}$ of $7.35$ using $\ce{HClO}$ ($\mathrm pK_\mathrm a = 7.54$), what mass of the solid sodium salt of the conjugate base is needed to make this buffer? Suppose you want to use $\pu{125.0mL}$ of $\pu{0.500M}$ of the acid.

I know this relates to Henderson's equation, so I do:
$$7.35=7.54+\log{\frac{[\ce{ClO-}]}{[\ce{HClO}]}},$$
which becomes:
$$0.646=\frac{[\ce{ClO-}]}{[\ce{HClO}]}.$$
But I do not know how to go from there, and I don't know how to use the last piece of information in the problem: ("Suppose you want to use $\pu{125.0mL}$ of $\pu{0.500M}$ of the acid").

Hello and welcome to the Chemistry.SE! You should take the tour  to get to know our community and learn how to ask questions. As it stands now, you title is way too long and unclear. Try to state one clearly formulated question.

The information given in the problem, "Suppose you want to use 125.0mL of 0.500M of the acid." tells us that the molarity or concentration of the acid is 0.5M. So, [ACID] = 0.5.
Now, 0.646 = [BASE]/(0.5)
So, [BASE] = 0.646×0.5 = 0.323
So, concentration of conjugate base = 0.323M
       n/V = 0.323
Since, volume is 125.0mL = 0.125L
       n/(0.125) = 0.323
So, n = 0.04
So, no. of moles of conjugate base = 0.04
So, mass of sodium salt of conjugate base i.e NaClO = 0.04×74.5 ~= 3g
(Since, molar mass of NaClO is 74.5)
Thus, your answer is 3g.

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Why alcohol is volatile although it has hydrogen bonds?
We are always told that hydrogen bonds increase the boiling point of a liquid and as a result they decrease the liquid's volatility.
I want to know the reason why, alcohol with hydrogen bonds is still volatile ?

Water has even more hydrogen bonds, and is still somewhat volatile. See, there is no such thing as decrease in general. There is a decrease compared to something.

In general what you have been told is true. Liquids are 'held together' by intermolecular forces (such as due to dipole-dipole and induced dipole-induced dipole (dispersion forces) ) and additionally some have hydrogen bonds. All molecules have dispersion energy between them.
Thermal energy continuously disrupts these forces so that some molecules can escape the liquid thus all liquids have a vapour pressure. What is important is the total of these interactions relative to the average thermal energy RT of the liquid. ($RT=2.49$ kJ/mol at $300$ K). Using the Boltzmann distribution of energies the chance of a molecule reaching, by random collisions, an energy sufficient to exceed a intermolecular interaction of, say $10RT$,  is only $\exp(-10) \approx 4.5\cdot10^{-5}$; very small indeed.
The average intermolecular energies are relatively weak, only a few times larger than RT  i.e. a few times typical thermal energy for small molecules, such as methane, but can be several times this for larger molecules. By comparison chemical bond energies are hundreds of times RT so normally molecules do not decompose before they can boil. Hydrogen bonds have an interaction energy of $\approx 4 RT$ to $\approx 15 RT$ depending on individual types of molecules. Thus if the molecules in a liquid have additional interaction energy due to hydrogen bonding then the vapour pressure will be lower at a given temperature and the boiling point higher. This means that, by random motion, the chance of exceeding this energy becomes exponentially smaller as the intermolecular energy increases.
Comparing three molecules with almost the same molecular weight ($86, 88, 88$ respectively)  shows the clear effect of hydrogen bonding.
n-hexane; b.p. $69$ C, ethyl propyl ether; $64$ C, $1-$pentanol; $137$ C
Finally hydrogen bonding in water is somewhat a special case as there are multiple hydrogen bonds although they are only moderately strong; a water dimer has a single H bond of $\approx 22.4$ kJ/mol $\approx 9 RT$.

I'm going to compare ethanol to water (I think its the obvious choice).
Volatility depends on inter-molecular forces, for example hydrogen-bonds.
The hydrogen bonds ethanol can develop are fewer than in water. It has a alkyl-residue which can only interact via van der Waals interactions.
You could say the hydrogen-bond-density is a lot lower than in water.

You might want to compare also against something else (dimethyl ether?) to emphasize the mentioned decrease in volatility.

@IvanNeretin Yes would maybe be useful. But on the other hand, I would just write #1 to this question again.


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Synthesis of ZnO by wet chemical method and reducing particle size by ball milling
I synthesized bulk $\ce{ZnO}$ which is pure white in colour, but after high energy ball milling colour changes from white to yellow, why?

Figure 1: Zinc oxide (courtesy of Wikipedia).

Are you sure it is not contaminated by the balls  and mill material? What is the nature of the balls and the mill?

How did you proceed in detail, by "wet method" you refer to what liquid?

I suspect non-stoichiometry. Zinc oxide turns to $\ce{Zn_{(1+x)}O}$ which is yellow in color.

I had a parallel experience and noted it in my thesis, 20 years ago.
Was working on an oxide ternary phase diagram (not wet chemistry, traditional shake and bake).  There were some conducting phases (dark) and some insulating ones (white).  I had a ZnO pellet (post firing) and I ground it under acetone (just mortar and pestle).  There was a definite color change to yellow.
I had a lot of other results and never ran this issue to ground.  I did note it and speculated that it had to do with the grain interiors having some conductivity (e.g. from oxygen defects or from impurities) and thus a different color, with the grain exteriors more insulating and white (not having the defects).  This would explain the color change.  There have been papers written (more than the 20 years ago) about differences in defects at grain boundaries and interiors of ZnO.
Very much a speculation at the time and never looked into it more.  But very interesting to see you have a parallel experience.  Maybe you will do more than I did and run it to ground?
P.s.  Consider the appearance of In2O3 and Ga2O3.  Indium oxide out of the jar is yellow and has semiconducting properties, from oxygen defects or the like.  Gallium oxide is much more insulating and is bright white.  Perhaps your ZnO grain interiors are a little In2O3 ish?  After all, both materials are dopable (despite the large band gap) and are TCO materials.  Ga2O3 on the other hand is extremely insulating and undopable, with a wider band gap.


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Chemical equilibrium constant
We define chemical equilibrium as the point in a reaction where the rates of the forward and reverse reactions become equal. We also write
$$K_c = \frac{k_f}{k_r}$$
If at equilibrium the rate of the forward reaction and the rate of the reverse reaction become equal then their rate constants should also be same. If rate constants $k_f$ and $k_r$  become the same, then $K_c$ should always be equal to 1. However, the value of $K_c$ is not always 1. Why is it so?

Welcome to chemistry.SE!
As it looks like you're mixing reaction rate with reaction rate constant.

For an example reaction $\ce{A + B <=> C + D}$
the forward rate is given by
$$r_\mathrm{f} = k_\mathrm{f}[\ce{A}][\ce{B}]$$
and the backward rate is given by
$$r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$
For the rates to be equal one simply needs
$$k_\mathrm{f}[\ce{A}][\ce{B}] = r_\mathrm{f} = r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$
Note that this does not imply that the two rate constants are the same
$$k_\mathrm{f} \neq k_\mathrm{b} \quad \text{(in general)}$$

Although true in principle,  in your example the $k_f,~k_b$   have different units so can never be equal, which weakens your argument.


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synthesis of bicyclic diol from cyclohexanone
I am trying to come up with a synthetic route for the following reaction, my ideas below it.

I think it is a bit of a long shot making this alkene as it would probably be quite strained-any other ideas?

https://en.wikipedia.org/wiki/Pinacol_coupling_reaction would work here

In cases like this, you can almost always exploit the symmetry of the product.


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Calculating amount of alkali additions to get desired pH in (hard) cider
This is linked to my question over on the HomeBrew SE site: https://homebrew.stackexchange.com/questions/18837/reduce-acidity-of-hard-cider
I haven't done chemistry since I was 17/18 but I remember doing molar calculations and so on, and the principle of OH ions binding with H ions to reduce acidity. I would prefer a more scientific method than trial and error.
Given an initial pH, desired pH and volume of liquid, how can I calculate the mass of a given salt to add... and is it as simple as that or are there more 'parameters' needed to generate some equations?
The salts I am likely to consider using are:

sodium bicarbonate (sodium hydrogen carbonate?)
magnesium carbonate
calcium carbonate
potassium carbonate

Is it as simple as working out how many moles of OH each of these contains per gram, and plugging that into a pH equation?


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What difference does a polar protic solvent make in the addition of X2 to alkenes?
After the formation of the cyclic intermediate, $\mathrm{S}_{N}2$ reaction takes place. This reaction would be favored by an aprotic solvent.
How will the products of the reaction be affected if I use a polar protic solvent?
I think that the formation of the cyclic intermediate is independent of the solvent used but I can't figure out what would happen after that.

The $SN_2$ reaction won't take place if you use a polar protic solvent. If you use a polar protic solvent such as water, you would form $\ce{HX}$. This would lead to an electrophilic substitution reaction.

In the presence of a polar protic solvent, the halogen will react with the solvent to form acids ($\ce{HX}$ and ${HOX}$).
$$\ce{Cl2 + H2O -> HCl + HOCl}$$
The high concentration of the cations formed (the hydrohalic acid and the hypohalous acid will exist as ions in a polar protic solvent) can act as electrophiles and initiate an electrophilic substitution reaction which results in the formation of a range of products.
$$\ce{CH2-CH=CH2 + X2 + H2O -> CH2-CH(Cl)-CH3 + CH2-CH(OH)-CH2(Cl) + CH2-CH(Cl)-CH3(Cl)}$$
This reaction is primarily initiated by the $\ce{H+}$ attacking the electron-rich double bond. This causes an electomeric shift of electrons to a carbon such that the carbocation formed is stable (Markonivkoff's rule). As carbocations are formed, rearrangements are possible. Not only are the mechanisms different, the products will be different too. This reaction is entirely different from a $SN_2$ reaction.


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Dehydration of alcohol
I have seen $\ce{P4O_{10}}$ and $\ce{HSO3Cl}$ being used to dehydrate carboxylic acids giving an anhydride and a ketene respectively.
Can  we use them to dehydrate $\ce{-OH}$ (of aliphatic alcohol) group from an organic compound? I have never seen anyone do that. If we can, then please specify the appropriate conditions for the reaction  to be carried out.

$\ce{P4O10/P2O5}$ with some heating dehydrates alcohols.

Look up Eaton's reagent

As you've noted, $\ce{P4O10}$ (or $\ce{P2O5}$ ) is a powerful dehydrating agent. According to the Chemical Book, it can actually dehydrate concentrated sulfuric acid to $\ce{SO3}$. So, it's not surprising that it can indeed dehydrate aliphatic alcohols. Using ethanol as an example, ethene can be produced by refluxing ethanol over $\ce{P4O10}$:
$$\ce{C2H5OH ->[P4O10] C2H4 + H2O}$$
Note that the reaction will proceed with gentle or no additional heating (the reaction itself is exothermic), it will just do so more slowly.


text:
Mechanism of the reaction of an alkene with excess HI in presence of CCl4
Problem

Answer

 2-iodopropane

Question
I thought of Markovnikov's rule and suggested 1,2‐diiodopropane as the the final product:

Apparently, this is wrong. I can not figure out the mechanism of this reaction (will $\ce{HI}$ dissociate in the presence of $\ce{CCl4}?)$ And what is the significance of excess $\ce{HI}$ here?

Probably rather 2-iodopropene.

@aventurin although the answer is an alkane but it's entirely possible that it may be wrong,I don't see how this reaction is any different from an ordinary addition reaction.

Theoretically, it could also be a rearrangement reaction.

what kind of rearrangement might take place here? Please give a hint, I'm stuck with this one for 4 days

I am at a loss to supply the mechanism, but based on the problems I've worked, and based on the theory that my instructor has taught, a general thumb-rule is that iodine on adjacent carbon atoms (vicinal di-iodo, if you will) are unstable, and they are eliminated as $\ce{RCHICH2I -> RCH2=CH2 }$
An idea for this can be picturised (though this may possibly be the incomplete or even incorrect reasoning) by considering that iodine atoms are bulky, and steric factors cause strain in the main chain. Moreover, there are possibilities of van der Waals' forces between vicinal iodine atoms, and also the good leaving tendency of iodine, that may favour the formation of $\ce{I2}$ by elimination.
A brilliant example that lucidly explains this is the reaction of glycerol with excess of HI; after substitution  of all the hydroxyl groups, there is an elimination of $\ce{I2}$ as discussed above, leading to the alkene of your original question. Subsequently, there is an addition of HI, indeed following Markovnikov sensibilities, that leads to the formation of 1-propene. It subsequently adds HI to yield your final product. Thus, 1 mole glycerol consumes 5 moles of HI; 3 via substitution, 2 via addition.
Now, the last example of glycerol wasn't strictly necessary, but I personally find the sequence quite beautiful; a single pathway leading to 3 rounds of substitution, two of addition and all interspersed with a unique type of elimination, and all this when we barely had a handful of reactants to begin with.

According to [1], 1,2-diiodopropane reacts with hydrogen iodide to give 2-iodopropane.
So the reaction path might be
$$\ce{CH2=CH\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH2\bond{-}I}$$
$$\ce{CH3\bond{-}CHI\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH3 + I2}$$
The mechanism of the second part is unclear to me. It may be elimination of $\ce{I2}$ followed by addition of $\ce{HI}$.
1 Th. M. Schmitt, Analysis of Surfactants, Second Edition, p. 67

even if I2 is eliminated, I will be left with a carbocation, how will HI add to it?

$\ce{I2}$ is eliminated from a neutral molecule, 1,2-diiodopropane (see step 2 above). So there's no carbocation left but propene to which $\ce{HI}$ may be added.

2 moles of $\ce{HI}$ are used up in total and in the second step vicinal $\ce{I2}$ are unstable so they get eliminated.After this one more mole has to be used.

According to Markovnikov's rule we would get a vicinal diiodide, which is unstable.
The iodine atoms would undergo elimination and once again form a double bonded structure that is 1-propene.
As to why (excess) has been mentioned, 1-propene reacts with this $\ce{HI}$ undergoing an addition reaction. As per Markovnikov's rule, this would in fact form 2-iodopropene.

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How does black CuO impart a green color to glazes and glass?
I read from a source that cupric oxide (CuO) imparts green to blue colour to glazes and glass. But CuO is black in colour. How is this possible?

Because, bingo, cupric oxide produces different compound in the glazes and glasses.

This is due to the oxidation of cupric oxide.
The main componds believed to cause this colour comprise a mixture of 3 compounds: $\ce{Cu4SO4(OH)6}$ (green); $\ce{Cu2CO3(OH)2}$ (green); and
$\ce{Cu3(CO3)2(OH)2}$(blue).
The following reactions are believed to take place:
Copper (I) oxidised to the black copper (II) sulfide ($\ce{CuS}$)  in the presence of sulfur impurities. Under accelerated conditions, the described process occur at a faster rate.
$\ce{CuO}$ and $\ce{CuS}$  reacts with carbon dioxide ($\ce{CO2}$) and hydroxide ions ($\ce{OH-}$) in water (in presence of air) to form $\ce{Cu2CO3(OH)2}$.
The extent of humidity and the level of sulfur have a significant impact on how fast the compounds develop, (under controlled conditions these reactants are varied to produce a favourable hue) as well as the relative
ratio of the three components.
$$\ce{2CuO + CO2 + H2O → Cu2CO3(OH)2~~~~~~~(1)}$$
$$\ce{3CuO + 2CO2 + H2O → Cu3(CO3)2(OH)2~~~~~~~~(2)}$$
$$\ce{4CuO + SO3 +3H2O → Cu4SO4(OH)6~~~~~~~~~(3)}$$
References

http://www.wskc.org/documents/281621/282063/ENGAGE_E3S_Chemistry_Statue+of+Liberty.pdf/e4f24c7e-3666-425e-9c41-7dbdd0065eb4

Just a couple nitpicks: When you say "over time", I'm not sure what time scale you mean because glazing processes turn CuO green under the right firing conditions. Also, I don't see the relevance of equations 1 and 2 to CuO turning green or blue. I might be missing something, but it seems more like you are talking about long term corrosion type of processes rather than making colored glasses and glazes as the OP asked about.

@airhuff Thanks let me fix it. Over time was basically referring to under natural conditions however under accelerated conditions the reactions are much faster. Let me out that into edit

The compounds you mentioned have nothing to do with color of glasses and glazes

@permeakra Thank you very much, I thought it would have been better if you really understood the question and then link to the answer than just critisize, but its okay if you think that way

@xavier_fakerat Ok, let me elaborate. The compounds you mentioned cause coloration of copper-contaning metal pieces where they are indeed formed thanks to oxidation. But, as you brilliantly noted, the question does not involve coloration by   copper oxide, not metallic copper. This is different. Since you fail to see the difference, I downvote your answer as wrong and misleading.

BTW, none of the compounds you mentioned may survive glassmaking or glazemaking process. They will decompose.

How is citric oxide oxidized.  Copper is known in oxidation states up to +4, but your quoted compounds all have pure Cu (II).

The green-blue color is common for copper compounds, because copper(II) ions in oxide neighborhood usually adsorb light in the red region thanks to relatively low energy of the $d-d$ transitions, that happens to be low because the d-orbitals of copper are small thank to the $d$-shrinking effect (large effective charge affective the $d$-orbitals). The best model for copper glasses/glases would be copper silicates
Here are links to description of two copper silicates
https://en.wikipedia.org/wiki/Egyptian_blue
https://en.wikipedia.org/wiki/Bayldonite
Please, note: these two are mere examples, the exact silicate phases formed in glasses and glazes may have very complex composition and structure or be actually amorphous. The point is that the color pallette is typical for isolated $Cu^{2+}$ ions in oxide neigbrohood, which is the neighborhood of copper in glasses in glazes
So, the actual question is where the black color of copper oxide comes from. Black color is a total adsorption in the visible spectrum, that comes from a family of electron transitions. Indeed, the paper on the band structure of copper oxide says that there are two rather wide bands with band gap ~1.4-1.7 eV (the energy of photons of  deep-red light is ~1.7) Why it has this particular energy is a much harder question with no simple answer.
Bands are the closest equivalent to orbitals known in quantum theory of solids. Bands are formed from individual overlapping atomic orbitals, but since they are formed from infinite number of atomic orbitals (in theory, but in crystals the size of the crystal makes the theory reasonably close to experiment) they form a continuous spectrum. Individual 'orbitals' in a band are known as states. Bands normally have finite width, thus it is possible to have band gaps - energy ranges with no corresponding electron state for this particular crystal. If the highest band is not completely filled with electrons, the crystal(usually) has metallic conductivity as the electrons may freely move from state to state.  If the highest band is completely filled with electrons and there is a gap between it and the next band, the crystal usually is a semiconducor (or an insulator if the gap is large enough). If the gap has wideness near visible spectrum, it may color the crystal. If the gap is larger than energy of violet photons (~3 eV), the crystal is transparent (or white)

I made it a community wiki because it obviously needs heavy editing.

@xavier_fakerat Cu(I) is usually colorless. Cu(III) is hard to obtain. Other than that this IS the correct answer. But it needs editing to straighten wordings and spelling.

It doesn't need CW and is not really all that bad.

@orthocresol upvote and upgrade if you thingk it's good.


text:
Why does the rate-determining step determine the overall rate of reaction?
I can appreciate that the rate determining step affects the rate of reaction. What I take issue with however is the exact wording given in the definition of the RDS.  From here:

The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds.

The last part about the RDS determining the rate baffles me. The RDS would affect the rate by slowing down a succeeding step, but in what way does it determine it?

Possible duplicate of Is the rate determining step the step with the largest Ea?

@pentavalentcarbon Disagree. Your link asks about how to tell which step is limiting; this one asks why the RDS strictly defines the overall rate, instead of just influencing it in tandem with other steps.

You can imagine this like water flowing through a pipe with a bottleneck. No manner how fast the water flows in the pipe, the time it takes for the water to go through the pipe depends pretty much totally on the bottleneck. So here, in comparison to the RDS the other steps take place quickly, but then they have to go through the RDS. This takes time, and no manner how fast the other steps are, the rate of reaction depends highly on how much time the RDS is taking.

While it looks like a merger of the comment by @Sawarnik (+1) and the answer by @airhuff (+1), maybe an illustration of this quasi-steady state approximation or Bodenstein principle is of benefit.  A brief search on the internet yielded an excerpt of a textbook of biochemistry, made freely available Wiley-Blackwell, with these two figures and a reference to the very original works about this phenomenon:

Please note that the width of the arrow $\ce{A -> B}$ is slightly different to the one relating $\ce{B -> C}$ in left-hand figure 1.12; but not in right-hand figure 1.13.  You may see this as a hint of slightly different reaction rates leading to an increase of B.
According to the author of the book, Chapman and Underhill pioneered the work (DOI: 10.1039/CT9130300496), which was expanded by the one by Bodenstein (literally "A theory of photochemical reaction rates", DOI: 10.1515/zpch-1913-0112) and still is of importance in view of biochemical reactions.

For the purpose of this post, the figures found were rearranged.  The are located on page 13 (first chapter) of said Fundamentals of Enzyme Kinetics, 4th edition (2012), authored by Athel Cornish-Bowden, ISBN 978-3-527-33074-4.

Think of a chemical reaction as a series of processes that must happen in order. If reaction $3$ is the slowest (rate determining) step of some overall reaction, then the intermediates formed in reaction $2$ just accumulate and wait for their turn to participate in the reaction. Depending on the order and kinetics of the reaction, the rate of reaction $3$ can be increased by this build-up of reactants, but there will always be an excess waiting to participate in this slow step. Once the reactants get past step $3$, they breeze on through the remaining faster steps of the reaction. Since reactions $4$ and above are fast compared to $3$, it makes little difference how fast they are.
Note that it is somewhat of an approximation to say that the overall rate is equal to that of the rate determining step. As mentioned in the example above, depending on the order and kinetics of the reaction, a buildup of intermediates from reaction $2$ could increase the rate of $3$, thus the rate of $2$ becomes somewhat relevant. Likewise, it's possible that a buildup of products from reaction $3$ could inhibit its rate depending on the rate of reaction step $4$. Wikipedia's definition of the rate-determining step of a reaction includes this mention that it is an approximation:

In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step. For a given reaction mechanism, the prediction of the corresponding rate equation (for comparison with the experimental rate law) is often simplified by using this approximation of the rate determining step.


text:
Why doesn't fast-scan cyclic voltammetry work with metal electrodes?
Every paper I've seen discussing it uses it with carbon nanofiber or some other carbon material like a diamond. If the physical principles by which it works are based on sweeping through voltages, I don't understand why it can't be used with metal electrodes.
I've tried to search through the literature but I haven't come across any clear answers.

It actually does work. Carbon fiber electrodes are just more useful for the neuro work that fast scans are popular for these days.
Here is an example from the literature of a metal electrode used up to $3*10^6$ V/s , Amatore et. al. JEAC


text:
Determining Structure of molecule based on IR Spectra and HNMR data
I have been given the following data and am asked to determine what the structure is. So far based on the IR spectra given I (think) I have identified a $\ce{C-O}$ bond, $\ce{sp^3}$ $\ce{C-H}$ bond however, I am really thrown off by the 5 hydrogens in a multiplet when there seems to be only one big peak. Any chance I can get help with this?

This is my proposed structure so far (note I am still missing 2 Hydrogens, there's 11 total and my structure contains only 9) however, I am not too sure about it's correctness this is the closest I can get with the information which I think I have interpreted correctly.

Ignoring the multiplet for now, draw out what you have and share it with us.

I went ahead and added it for you but I question how close I am to achieving the actual structure

Two thoughts: 1. Whoever wrote this question couldn't be bothered to actually look up the NMR or run the sample themselves?! They're using a computer generated spectrum... Lazy... 2. If they had given you a real spectrum, you might notice that the multiplet at 7.2 is actually 3 sets of peaks (1H, 2H, 2H), all multiplets. Two of these happen to have very similar shifts and look like 3H but they're not equivalent. Hope this helps.

The 1H and 13C NMR data disagree very strongly with your proposed structure. The fine structure of the multiplet is important in its own right, but you should probably start with the basics: look at the chemical shifts. What does a proton chemical shift at ~7 ppm mean? The IR data should not usually be your starting point unless there is something very characteristic, e.g. carbonyl. But even if you were starting with the IR data, your proposed structure has a carbonyl group, and there's no IR absorption anywhere near 1700 cm-1.

The organic experts here may be able to work with just the NMR, but if you were given an IR spectrum, too, you should also have included it in the question.

@hBy2Py The original photo does contain a few IR data.  Very few, and neither as spectrum, nor as (semi)quantitative listing among "intense", "medium", "weak", "broad".

@Buttonwood Ah, quite right, never even occurred to me to check the text. Weak-sauce IR data, for sure.

@hBy2Py I speculate at the educational level by the OP, providing the IR spectrum like the one at http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_disp.cgi?sdbsno=2670
would rise too much questions.  S/He were likely lost / overwhelmed by the  sheer information and wouldn't see the woods among the trees.  In earlier exercises I had to solve myself (Silverstein and other books...) there was even little plot for MS and UV-Vis.  The Moiré-like pattern may indicate it is a computer-based training here.

In a step wise approach, based on the NMR data provided:

The broadband decoupled 13C NMR spectrum accounts for six magnetically different nuclei.  (2-Pentanone drawn by you is not consistent with that, its carbon atoms would yield at maximum five signals.)  The four signals in the interval of about 120 to 140 ppm could be carbon atoms of an aromatic ring; the two between 40 to 60 could be by an alkyl chain adjacent to some electron withdrawing group.

The 1H NMR spectrum displays a multiplet in the "aromatic region" worth 5 H.  There is a broad singlet around 4.5 ppm; shape and to some extend location of this signal may refer to an exchangable proton by an OH group (this again is not explained by your first attempted attribution).

Still refering to the 1H NMR spectrum, there are two signals like for an alkyl chain adjacent to an electron withdrawing group; yet their chemical displacement ($\delta$) is a bit too high to be just a normal alkyl chain.  This is equally consistent with the two signals in the alkyl region of the 13C NMR spectrum, too.

Mass spectroscopy mentions a molecular ion, correlated to a molecular mass of 122 g/mol.

Hence

How is this consistent with the IR data provided? An absorption in the range of 3600 to about 3200 1/cm is typical for free OH, like in the instance of alcohols.  In this particular example, with (perhaps arbitrary selected) absorptions mentioned by their position (only), there is little of information you may extract.  Maybe for educational purpose in tune of "not every single information adds to knowlegde promptly useable".

Why – on purpose – a different compound like the isomeric p-ethyl phenol does not represent a valid answer?

The 1H NMR spectrum for this compound would be different.  The pattern in the alkyl region were a quartett (accounting for two protons) and a triplett (accounting for three protons).  Very often, the para-subsitution pattern would split the signals in the aromatic region into two dublets of equal intensity, accounting each for two magnetically equivalent protons.


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Friedel-Crafts reaction of anisole?
We know that phenol and aniline, despite being strongly activating, are unable to to undergo Friedel-Crafts reactions because they form complexes with $\ce{AlCl3}$, thanks to the lone pair on nitrogen and oxygen atoms.
However, according to my notes (and Google) anisole does undergo Friedel-Crafts reactions. How can it be possible, why is there no complex formation this time?
Also, I wanted to know how can we say that a certain group like $\ce{OH}$ will surely form a complex? Is containing a lone pair a sufficient condition?

What complex does it form?

@Yashas For eg, phenol forms Ar-(OH+)-(AlCl3-)

Not sure if this correct, but you could use alkyl triflate, instead of alkyl chloride + $\ce{AlCl3}$. Triflate is a good leaving group and can generate carbocations with easily react with the activated ring of phenol or aniline.

Completely a duplicate of https://chemistry.stackexchange.com/questions/31531/friedel-crafts-reaction-of-phenol/31548#31548

Completely a duplicate of https://chemistry.stackexchange.com/questions/72184/friedel-crafts-reaction-of-anisole#autocomment382586

https://chemistry.stackexchange.com/questions/72184/friedel-crafts-reaction-of-anisole#autocomment382587

Friedel-Crafts acylation of anisole with acetyl chloride and aluminium chloride as catalyst is a standard laboratory procedure to synthesize acetanisole.

The fact that this reaction is possible does not imply that there is no complex formation with $\ce{AlCl3}$. On the contrary, it is quite probable that such complexes do form due to the interaction of the methoxy oxygen with $\ce{AlCl3}$.
Therefore, and to compensate the formation of complexes with the resulting ketone, $\ce{AlCl3}$ is added in more than two-fold excess in the above synthesis.

if complex is formed then does that mean friedel crafts in anisole forms meta product?


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Negative temperature as limiting case for two-level system
For N particles in a two-energy-level system, with energy levels being 0 and $\Delta$, the internal energy tends to $N \Delta /2$ as temperature tends to infinity. At this point, the occupancy of each energy level is equal, and any further increase in internal energy would lead to a decrease in the entropy of the system.
Temperature is defined as $T = (\frac{\partial U}{\partial S})_V$. Using this definition, for a state with equal occupancies described above, the temperature should be negative. Where is the contradiction?

With equal occupancies, the temperature is infinite. It is the population inversion (higher occupancy for a higher-energy state) that makes temperature negative. There is no contradiction.

and inversion is very common, for example in NMR experiments and in lasers

related http://chemistry.stackexchange.com/questions/36885/

I think you would need to describe the derivative $\frac{\partial U}{\partial S}$ at that point, which should be infinity by your logic since the entropy is at a maximum with respect to internal energy ($\frac{\partial S}{\partial U}=0$). This is obtained using $$\frac{\partial U}{\partial S}=\frac{1}{\frac{\partial S}{\partial U}}$$


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How do counter-ions influence the color of coordination compounds?
I've been asking myself for a while now how counter-ions sometimes change the colors of coordination compounds. Like when I use d-metals as counter-ions to precipitate Cyanoferrates.
I looked at some crystal structures and saw that in the case of a cyanoferrate the terminal cyanide-N coordinate in an octahedral-prismatic geometry around the counter-ion. And I know that the cyano-ligand is a bridging ligand like in for example Prussian Blue, where the electron between Fe(II) and Fe(III) is moved across the cyano-ligand, probably through the π-systems. So I expect two ligand field splittings to happen, the one around the iron and the one around the counter-ion. These colors overlap to create a new color. Also depending on how lewis-acidic the counter-ion is it might pull electron density across the cyano-ligand lowering the gap and causing a red-shift.
I couldn't find much about that topic. It might be related to how colors derive in solid state compounds where I have no idea how this works. I know Lithium metal can be formulated as a huge molecule with many Li-Atoms, overlapping their atomic orbitals forming two big bands as the molecular orbitals create overlap so much. And usually salts crystallize in metal crystal structures so for example every second element is exchanged for the counter ion. If this was true then a colored solid state compound (salt) is a huge molecular orbital scheme, too creating two bands and depending on the size of their band gap the color is created (as I said I can't find anything on this topic). Sometimes the thermochromism in ZnO is explained by band gaps so I assume this concept of metals applies for salts as well. Then or course the counter-ion would have a much more complicated role here.
If anyone knows an answer to any of these questions I'd be really thankful.

The characteristic colour in Prussian blue arises due to intervalence charge transfer, which is actually what you described, up until the point where you said "two ligand field splittings" - yes, both Fe(II) and Fe(III) experience different LF splittings, but that does not really influence the colour. The LF splitting of Fe(II) would determine the energy gap between ground Fe(II) and excited Fe(II), for example. However the charge transfer is an electronic transition between Fe(II) and Fe(III).

in other words the LF splitting of Fe(II) determines this wavelength $$\ce{Fe^{II}-C#N-Fe^{III} ->[$h\nu$] ^Fe^{II}-C#N-Fe^{III}}$$ ( denoting excited state) but the charge transfer is actually of the nature $$\ce{Fe^{II}-C#N-Fe^{III} ->[$h\nu$] Fe^{III}-C#N-Fe^{II}}$$

Ah okay, thank you! So in my case where I have an M(II) at the right side it doesn't necessarily turn to M(III) I guess. So in the case of Prussian Blue I will have to descide between pushing the electron into the eg or across the bridge but if the redox reaction is not favored  with the counter-ion it's back to ligand field splitting again, I guess. And if the M(II) pulls electrons stronger doesn't that descrease the electron denisty around the Iron causing a smaller gap?

"if the redox reaction is not favored with the counter-ion it's back to ligand field splitting again" yep. I don't quite get your last sentence, though. Which iron and which gap?

Let's take copper(II)-ferricyanide. Potassium ferricyanide is yellow while the copper compound is dark brown. Well okay Fe(III) and Cu(I) might be likely here but I don't really know the colors of the other ferricyanides. So if there was another metal like Copper, maybe one which isn't reduced that easily as counterion for the Ferricyanide, does this element influence the ligand field splitting for the Fe(II) as it acts as a lewis acid for the same cyano-ligand as well?


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Can photons also work as a catalyst?
A lot of biological reactions need photons to work, like photosynthesis and making vitamin D in our skin. But are there any chemical reactions where photons are really just a catalyst so they are absorbed and emitted the same as the came? Or is there always a change in frequency/wavelength involved which implies perhaps that the photon is 'used'. Or is perhaps the use of the word catalyst in case of photons not very appropriate?

There is a phenomenon called stimulated emission which is used in lasers and which pretty well could have been an example you look for, had not the definition of a catalyst been assumed to include only substances. (Also, it is not a chemical reaction anyway.)

I do not feel that the word catalyst applies to photochemistry.
Photochemistry allows to access much higher energy states and thus reach reaction pathways that are otherwise not possible. Photons do not stabilize transition states (how would you rationalize the interaction?).
Catalysts on the other hand are focused on lowering barriers so that reactions proceed via lower energy transition states.
Because catalysts only affect the transition states, catalysts cannot affect the equilibrium position of a reaction. For example, if you had a catalyst that allowed easier conversion between (E)- and (Z)- double bonds, you could not use the catalyst to selectively provide the one that is less favorable.
On the other hand, with photochemistry, we do not establish an equilibrium because the higher energy state is not necessarily accessible via microscopic reversibility. And indeed, we can selectively use photochemistry to provide the less favorable species as is well know in the case of stilbenes.
EDIT:
Thanks to airhuff for pointing out that the requirement that a catalyst not be consumed also implies that we should not think about photons as catalysts. Hard to imagine a situation where you would absorb a photon and release one that is the same energy...

It's also hard to imagine a photon participating in a chemical reaction without being consumed, or a reaction where another photon of the same energy is re-emitted, which would be required for the photon to truly be a catalyst.


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Non-polar molecules are soluble in non-polar solvents (hexane) but not in water, Why?
My book says that non-polar simple molecular substances can dissove when added to a non-polar solvent, such as hexane due to intermolecular forces forming between the molecules and the solvent.
However, it says simply that
non-polar molecules do not interact with the dipoles in water, hence simple molecular substances tend be me insoluble in polar solvents
Can someone explain why this is?

Well, it is as simple as that: no dipoles, hence no dipole-dipole interactions with water, so water is better off interacting with itself and leaving your substance alone.

You mean no permanent dipole-dipole interactions? @IvanNeretin

Yeah, of course. You can still have induced dipoles, but those are weak.

@IvanNeretin Got it! I answered my own question! Uprate if it may be useful to other people and thanks very much!

For a molecule to be soluble in water, permanent dipole-dipole interactions are required.
Non-polar molecules only have induced dipole-dipole interactions, which are not sufficient for the molecule to dissolve in water.
"The water molecules are better off interacting with themselves"
Credits @IvanNeretin


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N-fluorosuccinimide: any scope?
All halogen derivatives (except fluorine) of succinimide exist and are used in various chemical processes. What makes N-fluorosuccinimde difficult to prepare? Still chemists have not been able to figure out the preparation of N-fluorosuccinimde and don't even bother to synthesize it since electrophilic/radical fluorination is accomplished by  N-fluoro-o-benzenedisulfonimide (NFOBS), N-fluorobenzenesulfonimide (NFSI), and Selectfluor. Moreover, even if NFS is successfully synthesized, can it be used as a fluorinating agent in electrophilic/radical fluorination?
According to Wikipedia, NBS is made by reacting sodium hydroxide and bromine in an ice-water solution of succinimide. The NBS product precipitates and is collected by filtration. I am assuming that NCS and NIS are made in the similar fashion (using appropriate halogens). But, why NFS has still not been made through a similar way (using fluorine)? Is the reaction condition unsuitable for fluorine or is there any other reason?
I was reading a forum discussing the scope of NFS as a fluorinating agent. It says:

Does anyone know why is N-fluorosuccinimide (201281-58-5) such
    uncommon reagent for fluorination? I checked some major suppliers and
    none of them sells it.

In my knowledge, N-Halogenesuccinimide are obtained by reacting of Succinimide with the halogen itself. This is valid for Chlorine, Bromine, and Iodine. Fluorine is too reactive. For the fluorine derivative,  iodinesuccinimide is treated with silver fluoride to exchange the halogen.
The other option is to treat N-Hydroxysuccinimide with Thionyl
  fluoride or Phosphorous trifluoride.

Is this true? I couldn't find any source confirming the above synthesis and I am skeptical about the preparation process. Can NFS be made through the above process?

Maybe NFS doesn't work well?  Bear in mind that bonds from a second-period element to fluorine are generally stronger and less labile than the corresponding bonds to heavier halogens. If that's true, getting fluorine off a nitrogen-fluorine bond requires more work than, say, getting bromine off a nitrogen-bromine bond.
In https://en.m.wikipedia.org/wiki/Electrophilic_fluorination#Mechanism_and_stereochemistry several common fluorinating reagents are given, these being the compounds mentioned by the OP.  The listed compounds all withdraw electrons from the nitrogen-fluorine bond more strongly than the carbonyl groups would in NFS (they use sulfonyl groups or a cationic function), so it seems this attribute is necessary to get the fluorination to go.

Nice. Also, I'd posit that since fluorine is the most electronegative halide, getting a compound $\ce{X-F}$ to act as a source of $\ce{F+}$ is harder than getting a compound $\ce{X-I}$ to act as a source of $\ce{I+}$. Hence, again, a more electron-withdrawing $\ce{X}$ moiety is needed.

The synthesis of NBS and NIS involve adding succinimide, the relevant halogen and NaOH (and AgNO3 in the case of NIS) in water. Fluorine reacts very violently with both oxygen and water forming HOF and FOOF, both highly toxic and explosive compounds. Water is thus an incompatible solvent for running fluorinations, you need a polar aprotic solvent such as MeCN. Secondly, in the presence of base, fluorine could undergo alpha-fluorinations to the carbonyl depending on the pKa of the C-3 and C-4 hydrogens.

You can get NFS from a heavier halogen counterpart plus silver fluoride. But that does not actually mean it will work as a fluorinating agent, which I suspect is the real problem.

NFS has a very similar N-F bond strength in MeCN to N-fluoroquinuclidinium (also a pain), so might work. But I think it's just much more difficult to make then NFOBS/ NFSI which do just as good a job.


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Lead(II)chloride solubility in hot water
I just went through my notes from last semester and remembered something my professor told me but only explained it very briefly. I tried to find anything about the topic on the net but couldn't find much. And I want to make sure that this information is nothing he just came up with but based on some actual literature.
So he was talking about the increased solublity in Lead(II)chloride when you heat the water. How he explained it was with the crystal structure of Lead(II)chloride where he said the lone pairs on the lead would take part in the coordination as well and whatever coordination polyhedron might form in that case (he had a selfdrawn picture of that) will have one edge which isn't very strong and likes to break at warmer temperatures which increases solublity.
Now that was all a little fast and I'd like to have some confirmation on that topic. Has anyone ever heard about this before?

Lead chloride is by no means unique in that regard. Most solid salts increase solubility with temperature. This has nothing to do with lone pairs.

I know I found that as well. The most common explanation for Lead Chloride is the same as for all other salts that become more soluble in warm water which is probably why I cannot find anything else on the topic. I meant more like is there really another effect that contributes to this based on the crystal structure of the compound?

See the answer to this question: https://chemistry.stackexchange.com/questions/42698/why-exception-is-noted-in-the-solubility-of-lead-salts


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What products did this aldol condensation create?
I followed a procedure in my organic chemistry lab, but I don't understand what product the procedure created. Can someone tell me what chemicals reacts with which to create the product?
First $\ce{NaOH}$, water, and ethanol were mixed together and cooled. Then acetone was added, followed by benzaldehyde (the reaction mixture then turned yellow/orange). The reaction was stirred for a while. The product was washed with water and ethanol before being recrystallized from ethyl acetate. After crystal formation the product was reduced then cooled.
If I understand correctly the acetone and benzaldehyde reacted together and the ethyl acetate was only used as a solvent?

NaOH is intended to act as base.  The presence of ethanol, beside water, helps to accomodate both inorganic matter (NaOH) and organic matter (acetone, benzaldehyde) in solution. Where, by the way, acetone may be deprotonated?  And, in its then different form could it react with ease on the subsquently added  benzaldehyde?

Straightforward aldol between benzaldehyde and acetone. Enolate of acetone adds to benzaldehyde (which cannot enolise) then eliminates water to give the enone. Product recrystallised from EtOAc.

@JQQ You just need to figure out whether it's single aldol or double aldol.

I think I understand it now, but correct me if I'm wrong. I can take a mp of the product and see whether it's closer to the mp of the single or double aldol to determine what the major product is. If I used a 2:1 ratio of benzaldehyde to acetone, then the double aldol would be the expected product. The only thing left to figure out would be which conformation it has. If sufficient heat was used then I should get the thermo-favored product (E,E).

True. The proportions of reagents used may give you a clue. If xs acetone is used then that would favour the single addition product, if xs benzaldehyde then you get the double


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Effect of pH on positive and negative charges of a zwitterion
I read that the isoionic point is defined as the

pH at which a zwitterion molecule has an equal number of positive and negative charges and no adherent ionic species.

I have two questions regarding this definition.

How does pH affect the number of positive and negative charges of the groups in a zwitterion?
What does "no adherent ionic species" mean?

Consider an amino acid as a zwitterion. It has both a basic amine functionality and an acidic carboxylic acid functionality. At a high pH, the acidic functionality will be largely deprotinated and the ion will have a net negative charge. Oppositely, at low pH, the amine group is largely protinated and the compound will have a net positive charge. Clearly, there is some intermediate pH at which the positive and negative charges are equal (or it's neither protinated nor deprotinated) and the net charge is zero.
To maintain charge neutrality, a charged species must have some oppositely charged ionic species associated to maintain charge neutrality. For example, under acidic conditions when the amine group is protinated, you need some anion present, like chloride for example, to be associated with the compound to maintain charge balance. This counter-ion is also known as an adherent ionic species (adherent meaning adhering to, like when the compound is an undissolved solid).


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pH probe bulb - what is happening within the glass?
I am trying to understand how the glass bulb of a pH electrode of a pH meter works - the glass bulb itself. Not the reference electrode or the rest of the electrode (HCl, Ag/AgCl wire, etc...), the math, or the equilibrium yet. For this question just the glass bulb of the pH electrode.
Here is what I have learned so far. This is my current understanding. I'm not saying it is right - but it's where I am right now.
The thin-walled glass bulb has a conductive solution inside, and the outside should be kept in liquid at all times as well. This hydrates a thin layer of the glass on the outside and the inside. I assume these layers are manufactured differently, otherwise, the glass should be uniformly hydrated after a long enough time. It is important that the middle layer of the glass remains very low conductivity so that a potential difference can be maintained; there are likely other reasons as well.
The glass is amorphous and in this case, the exterior layers are somewhat porous, so there is a large volume of Si-O groups exposed to the solution. Protons will stick to these groups and establish a negative charge on the outside of the glass. The number is related to the pH or hydronium concentration of the solution on the outside of the electrode.
edit: I have just started to read this early discussion, where the idea that the glass itself may behave as a sort of buffer:

Hughes (3) has pointed out that the hydrogen ion concentration in the glass phase may be held relatively constant by the buffer action of the glass which is a mixture of the salt of a weak acid $(\ce{Na2SiO3})$
   with the anhydride of that acid (excess $\ce{SiO2})$.

Note: The hydrated layer is also called a "gel layer", but it is not clear if this is formed naturally as the glass hydrates, or if there is a special gel-enabling material applied to each surface during manufacture.

Do these have to be specially prepared layers of more porous, hydratable glass on the inside and outside of the glass bulb? If so, roughly speaking how is this done? If not, what does limit the depth of the hydrated layer?
When inserted into an acid/base solution, is it just protons diffusing into the hydrated layer of the pH probe bulb by jumping between Si-O sites, or is it the hydronium ions in solution that is diffusing into the glass?
Why is it this often called an ion exchange process? (e.g. not in the Mettler link but in the other two links below, and several random textbooks pulled from a library shelf). Are there Li or Na ions in the glass that are moving? What is being "exchanged"?

below: From A Guide to pH Measurement – the Theory and Practice of pH Applications, Mettler, Toledo.

below: From Theory and Practice of pH Measurement.

below: From The Glass pH Electrode by Petr Vany´sek.

This question is a little complex, but I believe that understanding one or two underlying processes within the glass is all that I need here.

https://www.electrochem.org/dl/interface/sum/sum04/IF6-04-Pages19-20.pdf  This short article gives some useful information for you. In particular, I believe it addresses the third part of your original question. 1st page, right around equation 2: "The exchange of hydronium (or written as proton, H+) between the solid membrane and the surrounding solution,
and the equilibrium nature of this exchange, is the key principle
of H3O+ sensing. " Equation 2 shows that the ion exchange is with the silicon of the glass membrane.

@Tyberius Yep, that's the pdf I've linked to in the question and the source of the figure of the bulb labeled Figure 1, and it is one of the statements there that are bothering me and brought me here to get expert help. I think it is somewhat vague. I am not sure it actually clears up beyond all doubt that hydronium ions do all of the diffusing, and not just the protons, and 3. asks about exchange, and the metal ions within the glass (Li, Na) itself. After all The Chalkboard magazine column is not intended to be a scholarly reference source.

@Tyberius  The glass has four interfaces created by three layers, and I'm also asking about diffusion and exchange within the volume of the two outer hydrated layers.  If I dip a probe in a more acidic solution, is it the hydronium ions that are physically diffusing all the way into the hydrated layer, or in that one second of time as the pH meter updates to the new, correct value, is it really just protons jumping from one SiO site to the next? The dynamics are different.  3. asks about exchange, and the metal ions within the glass (Li, Na) itself - presumably the two hydrated layers.

Have you looked at the Wikipedia page for "glass electrode"? References 8 through 12 cited in the "Metallic function..." section might be worth tracking down.
https://en.wikipedia.org/wiki/Glass_electrode

@J.Ari I'll try - some may be difficult to locate in English.

I remember having read that no H+ ions are crossing the glass membrane, because if Tritium chloride is inserted in the inner solution instead of usual HCl, the radioactivity stays inside the bulb, and never gets out, even after many weeks. Unfortunately I don't remember the reference of this experiment.

@Maurice if not too question-adverse, one might simply ask about it as a new question.

My reference for all information and pictures is Harris' Quantitative Chemical Analysis, 9th ed., pp 347-9. I think it'll be worth your while to consult those pages, but I'll try to summarize the important points here.
An ion-selective electrode is characterized by a thin membrane that, well, selectively binds ions. The glass electrode is an ion-selective electrode for $\ce{H+}$ made of amorphous silicate glass, which consists of connected $\ce{SiO4}$ tetrahedra.

Presumably, no special preparative techniques are required for the glass, and the depth of the hydrated gel layer is mediated by the strength and range of the intermolecular interactions between water and the glass.
Protons are the main ions that bind to the layer, leading to the selectivity of the electrode. They diffuse between the solution and the hydrated gel layer and displace the metal ions originally present on the surface of the glass, which describes an ion-exchange process. Note that they cannot, however, diffuse through the inner glass layer.

A few side remarks.

Equilibrium is reached when the favorable binding of protons to the glass surface is balanced by the unfavorable electrostatic repulsion and chemical potential gradient that result from diffusion into the hydrated gel layer. This provides an equation relating the potential difference to the pH of the solution and allows for $\ce{pH}$ measurement.
Something has to be able to move through the inner layer of the glass membrane to conduct a current and hence allow for a measurement of the potential difference. It turns out that sodium ions can move through this inner layer, but only sluggishly---the resistance of the glass membrane is about $10^8\,\Omega$.

As you've mentioned, the electrical resistance of the glass is of the order of 100 MΩ or more, and is very sensitive to temperature, thus the need for a very high input impedance amplifier. However it does deliver a small current (nA). I had thought that the conductivity of the glass is due to electron (or hole) carriers, not the movement of sodium ions. My thinking was that ionic conductivity might "use up" sodium since its not necessarily replenished by the solutions on the outside or AgCl inside. Perhaps for Na sensitive ionic probes the situation is different.

I'm glad to help. (1) Yes, they are; I've updated my post. (2) I don't know enough about this to comment, but Harris does state that sodium ions are responsible for current: "The $\ce{H+}$-sensitive membrane may be thought of as two surfaces electrically connected by $\ce{Na+}$ transport."

Oops, sorry, forgot to write @uhoh.

Looks great, thank you for all this work! Below my meta question Does anybody here know how a pH probe's glass bulb electrode works? @Martin-マーチン commented "Sometimes it just takes more time. Don't give up just yet." :) +n!

I guess H+ ions are able to diffuse through the glass layer. After all, if the glass would soften at 900 K, then the probability of reaching or exceeding the activation energy for diffusion of an ion (H+, Na+) at 300 K is about exp(-900/300) = 0.05 which is quite high.

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


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How to balance the redox reaction between bismuth(III) sulfide and nitrate anion?
I have to solve the following redox reaction which occurs in an acidic environment, using the half-equation method.
$$\ce {Bi2S3_{(s)} + NO3-_{(aq)} -> BiO3-_{(aq)} + SO4^2-_{(aq)} + NO_{(g)}}$$
There seems to be a lot elements that change their oxidation numbers, except for oxygen:

$\ce{Bi}$ from $\ce{+III}$ to $\ce{+V}$
$\ce{S}$ from $\ce{-II}$ to $\ce{+VI}$
$\ce{N}$ from $\ce{+V}$ to $\ce{+II}$

What should I begin with? How would one go about grouping three redox pairs?

If possible, please try not to use MathJax in titles on Chemistry.SE. Please see this meta discussion for more details.

There might not be a unique answer. Multiple redoxes are difficult because you cannot necessarily add two reduction potentials or two oxidation potentials...

N is from +5, not +3.

While in general a reaction involving three independent redox pairs may not have unique coefficients, this particular one contains one coupled oxidation half-reaction and one regular reduction half-reaction.
The key is to note that both elements in $\ce{Bi2S3}$ are oxidized, and hence we have a stoichiometric relationship between the two oxidation half-reactions. This suggests we write the oxidation half-reaction as follows: $$\ce{18H2O + Bi2S3 -> 2BiO3- + 3SO4^{2-} + 36H+ + 28e-,}$$ and we can proceed with the usual algorithm for balancing half-reactions.


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Will sodium metal stored in mineral oil explode on contact with water?
Will sodium metal stored in mineral oil, like this one offered on amazon.com, explode if it were immersed in water?

https://www.youtube.com/watch?v=LmlAYnFF_s8

In short: the mineral oil will float away, and the metal exposed to air and water may react violently, yielding an explosion. Do not do this.
During shipment and storage, because of sodium's density of about 0.96 g/cm$^3$, the metal will be covered by the mineral oil.  This intends to keep oxygen and water away from the bare metal.  If you find someone shipping it (it is banned from air planes, for example).
If you take the ingots / cubes of sodium out of the mineral oil, and remove adsorbed oil for example with filter paper, the shiny surfaces quickly become dull.  Oxygen and especially moisture present in the air attack the metal, mainly to form a white surface of $\ce{NaOH}$.
Bare pieces of sodium, exposed to water, will vigorously react to yield both $\ce{NaOH}$ and hydrogen gas.  Because of the lower density of the metal, compared to the one of water (1.00 g/cm$³$), it will float on the later.  However this said reaction is highly exothermic, and may ignite the hydrogen generated as shown here.
Disclaimer: Do not repeat this.  This is nothing funny, it is highly dangerous to you (as hot shrapnels of metal may hit you, and you may expose yourself to highly concentrated lye), to others around, and to the environment.  (In addition, he handled the metal without appropriate gloves, nor safety goggles, nor lab coat...). Do not throw sodium into a pond. Think first, than act responsibly.

Totally agree about safety. Safety should be planned into all experiments. Just because you got away with doing something stupid doesn't mean that you'll get away with it next time.


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Electrochemical series - when not at standard conditions

Is there a reason why this question's answer is treating it as if it is at standard conditions? Isn't it using 0.2M aqueous solutions, so the electrochemical series won't be valid for this question? The answer seems to ignore this.
Thanks

While the electrochemical series does assume standard-state conditions, we will show that in certain cases it will also work for non-standard-state conditions. One recalls that the Nernst equation is given by $$E = E^\circ - \frac{RT}{nF}\,\ln Q,$$ and can be used to determine electrochemical cell potentials when the cells are not at standard state. This works directly for the second reaction you list, as $Q = \ce{[Sn^2+]}/\ce{[Cu^2+]} = 1$ and $\ln Q = 0$, so the second term in the Nernst equation vanishes, but will not work for the first reaction, for which $Q = \ce{[Sn^2+]}/\ce{[Ag+]^2} \neq 1$.
In this case, it is reasonable to assume that one knows both (i) that the correction due to the second term is relatively small, and (ii) that $\ce{Ag}$ and $\ce{Cu}$ are relatively inert metals, and strongly favor being in their elemental form relative to other metals. The quoted result then follows.

Thanks. Is this expected knowledge for a final year high school knowledge? Seems rather complex.

Yes---at least this was my experience in adv. chemistry class in high school. Your mileage may vary.


text:
Why color of alkali metal peroxides, superoxides and ozonides deepens down the group?
This is basically a continuation of the question-"Why is potassium monoxide (K2O) coloured?"
I knew that color of alkali metal oxides deepens down the group:

Lithium oxide ($\ce{Li2O}$) is the lightest alkali metal oxide and a white
solid. It melts at 1570 °C.
Sodium oxide ($\ce{Na2O}$) is a white solid that melts at 1132 °C and
decomposes at 1950 °C.
Potassium oxide ($\ce{K2O}$) is a pale yellow solid that decomposes at
350°C.
Rubidium oxide ($\ce{Rb2O}$) is a yellow solid that melts at 500 °C.
Caesium oxide ($\ce{Cs2O}$) is a yellow-orange solid that melts at 490°C.

The reason is due to what @Oscar said in the answer of previous question:

If we go further down the alkali metal oxides the band gap gets
  smaller, and more visible light with longer blue/green wavelengths may
  be absorbed. Then the color becomes stronger and more reddish.

But, it is observed that not only color of normal oxides deepens down the group but the peroxides and superoxides also follow the trend:

Lithium peroxide ($\ce{Li2O2}$) is a white solid that melts at 195 °C.
Sodium peroxide ($\ce{Na2O2}$) is a pale yellow solid that melts at 460
°Cand boils at 657 °C.
Potassium peroxide ($\ce{K2O2}$) is a yellow solid that melts at 490 °C.

Sodium superoxide ($\ce{NaO2}$) is a yellow-orange solid that melts at
551.7°C.
Potassium superoxide ($\ce{KO2}$) is a yellow solid that decomposes at
560°C.

Lithium ozonide ($\ce{LiO3}$) is a red solid which is produced from caesium
ozonide via an ion-exchange process.
Sodium ozonide ($\ce{NaO3}$) is a red solid which is produced from caesium
ozonide via an ion-exchange process.
Potassium ozonide ($\ce{KO3}$) is a dark red solid which is produced when
potassium is burned in ozone or exposed to air for years.
Rubidium ozonide ($\ce{RbO3}$) is a dark red solid which is produced when
rubidium is burned in ozone.
Caesium ozonide ($\ce{CsO3}$) is a dark red solid which is produced when
caesium is burned in ozone.

What is the reason behind the deepening of the color? Is it due to the same reason as @Oscar pointed out? Or is it any other reason? Although superoxides and ozonides are paramagnetic and are colored, the peroxides are diamagnetic but still they are colored. Why?

Are you sure that ozonides are diamagnetic?

@permeakra, yes, my bad, rectified.

Most likely, we are seeing the same band structure effects.
There is an additional trend, whereby the compounds become darker and more reddish as one adds more oxygen.  In oxygen-chain species such as $\ce {O2^{2-}} $, antibonding molecular orbital are occupied.  So The valence band containg these electrons is raised.  That lowers the band gap enabling absorption of more visible light.
Incidentally, the visual effects described here are limited to alkali metal (when oxygen is the electronegative element).  When we move to the corresponding alkaline earth oxide the monoxides are white and the peroxides ($\ce {Mg}$ and heavier) are white too.  Less strongly ionic bonding and a lower cation/anions ratio lead to less negative charge on the oxygen species, thus lower valence band energies and gaps  remaining large enough to remain uv-absorptive only.


text:
Electrophilic halogenation of 4-halo-3-(tert-butyl)phenols
Question

Correct answer
A,B,C
My thoughts
Correct answer should be A,B. The main role is of the steric hindrance between the halogen atom and the tert-butyl group. So why are we even considering the electronic effects of both the groups? They are common in all the cases.
Further, the answer should be A only as its only the halogen atom changing each time, and hence the steric hindrance due to it. Why are we even considering other factors, as they are common in all the cases?

The options ABC are correct answers
Reason:

A: Steric effect of halogens will definitely be taken into account since all 3 halogens are acting on the same compound hence the primary factor which will make the product formed for each case be different is the steric hindrance due to their respective sizes.
B: For steric hindrance to be taken into account both atoms need to be bulky and have equal role to play in the first option we affirmed that due to different sizes among halogens there'll be steric hindrance causing different types of products to be formed however that is only true when we have a bulky group( here, tert butyl) providing steric to the group trying to attach(the halogens) to the adjacent c atoms. Hence for us to take steric hindrance as a factor for different products the steric hindrance caused due to the tert-butyl group and halogens is considered
C: The substitution of the halogens is taking place due to the +M effect of the -OH group thus option C must be true for us to proceed even a bit and it is the first requirement and only after this would we take into consideration the steric hindrance. In other words the reaction is taking place due to the electron donating effect of the -OH group since it is donating electrons to the ring via mesomeric and not inductive. And the steric hindrance serves so as to differentiate and predict which product will be formed

And +I effect of the tert-butyl group?

Inductive effect only strengthens or diminishes the effect of the already existing group which is exerting it's mesomeric effect

In this case the tert butyl group is increasing the charge circulating in the benzene ring due to the +M of -OH

Okay, so wouldn't have it mattered if there had been a mildly deactivating group in place of tert-butyl?

No, only the steric effects of the tert butyl matters, if there has been a -CH3 instead of tert butyl, then yes, it wouldn't have mattered.


text:
How do we know the number of atoms in an inorganic compound?
How do we know the number of atoms in an inorganic compound? If there's a question like 'Write the chemical formula for sulfur dioxide', the only thing I managed to do is answer '$\ce{SO}$'. I don't know how to find out the number of sulfur atoms or oxygen atoms.

For all compounds, the systematic name of that compound should provide information on the composition and structure. Your case is an example of the nomenclature of binary covalent compounds. Consider the following example:

dinitrogen tetroxide

The first word in this name tells us which element is first in the formula and how many of them there are. In this example the "di-" prefix tells us we have two of something.
$$\mathrm{dinitrogen = }\ \ce{N2}$$
The second word in the name tells us which element is second in the formula and how many of them there are. In this example the "tetr-" prefix tells us that there are four of something. By convention, we usually use the "ionic" form of the second element in the name even if it is a covalent compound, so "oxide" means "oxygen".
$$\mathrm{tetroxide=}\ \ce{O4}$$
Therefore, the full formula for this compound is $\ce{N2O4}$.
The prefixes are as follows (up to 10, with larger numbers being rare). Note that the trailing "o" or "a" of a prefix is dropped with oxygen to increase ease of pronunciation.
$$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$

What about names like iodine monochloride? How many iodine atoms are there in the formula?

The "mono-" prefix is often left out if there is no compound out there that have more than one of those atoms in it. Thus, iodine monochloride is $\ce{ICl}$. Why not iodine chloride? There is another compound out there "iodine trichloride," which is $\ce{ICl3}$.


text:
Exchange of electrons in f orbital
If an f orbital has 6 unpaired electrons then how many electrons can be exchanged.
According to me if 2 electrons are added then a half filled f orbital would be formed which is stable (not as much as a completely filled orbital but still quite stable) however, the answer is supposedly 15.
If one knows the reason please do tell.

There are six unpaired electrons, and we can swap any two of them, so we have:
$${6\choose2} = 15$$

However doesn't exchange imply in out transfer??

Also I'm not able to view your answer properly. I'm seeing a yellow box below your text 'so we have: ' and the yellow box has in it written in red font Undefined control sequence \binom.' I'm not sure what this popup signifies as I'm viewing it for the first time, if you know, please do tell.

I am not familiar with your definition of exchange. The issue you're experiencing may be a MathJax issue, but I am not sure about the specifics. In different notation, my answer is $_6C_2 = 15$.

@a-cyclohexane-molecule for some reason, \binom doesn't currently render in the Android app of SE. I put in a suggestion to edit it to \choose, which uses different syntax, but works in the app.

Ok I get it now. What I was talking about would've been valid if it bad been specified that transfer of electrons is happening. Thanks.


text:
Alkenes v/s Alkynes in two different reactions
Alkynes are less reactive towards electrophylic addition than alkenes. However, in case of catalytic hydrogenation, alkynes are said to be more reactive. Why? Also, if I want to add H+ as an electrophile, then will alkenes be still more reactive than alkynes?
Does the reason lie in the mechanism of both the reactions (that was obvious, I guess)? If so, please explain, I cannot seem to justify for the later one.

The first stage of electophilic addition is addition of a cation to alkene/alkyne with formation of a carbocations. Vinyl carbocations are much less stable then alkyl cations, so the path is strongly hindered for alkynes.
The first stage of catalitic hydrogenation is addtion of hydride-ion on the metal-surface to unsaturated hydrocarbon, forming a carbanion bound to metal surface. Vinyl-anion are more stable than alkyl-anyons for the same reasons vinyl-cations are less stable than alklyl-cation. Acetylenes also are much less hindered sterically than alkenes and adsorb on a metal surface much easier.
Please, note, than nearby substituents may affect these properties dramatically. Deactivation of alkenes in electrophilic addition by strong $\pi$-electron-withdrawing groups is well known.

As suggested, vinyl carbocations are "much less stable" than alkyl cation and are probably NEVER intermediates in protic additions to simple (unconjugated) alkynes.  Strong acids form pi complexes with alkynes and these are attacked by the resident nucleophile to produce an anti addition product.
The chemical reduction of alkynes involve a sequence of electrons and then protons being added to an alkyne which initially puts a negative charge in an sp2 orbital where it is relatively stable compared to being in an sp3 orbital if an alkene is used.


text:
Is ethanol acidic enough to lead to elimination?

The question is asking for the reason of obtaining an acidic solution. The only reason can be when ethanol follows a substitution reaction approach to substitute the bromine atom and form HBr as a sidr product. This will indeed form an acidic solution.
However, if we had gone for Elimination pathway, we can see that it would require

Lesser role of steric hindrance by adjacent methyl groups.
Resonance stabilised carbocation (if we choose E1 mechanism).
The aquous solvent will too stabilise the carbocation (although both 2,3 points also valid in SN1 reaction).

The product form will be in conjugation with the benzene ring.

Aren't these enough for ethanol to act a bit more acidic and lead to elimination than substitution?

Couldn't it also be an elimination reaction? In this case the other end products will be $\ce{Br-}$ and $\ce{CH3CH2OH2+}$ (protonated ethanol). Protonated ethanol is also acidic.

Other organic chemists out there, please correct me if I'm wrong: I would say it's mostly water (rather than ethanol) that reacts with the organic molecule, probably by substitution, forming HBr as a side product. Even if there weren't an inherent difference in nucleophilic reactivity between water and ethanol, the big difference in molecular weight between them means that even a solution where there is more ethanol in mass % may actually have a higher molar fraction of water. But OK, my main point is that it's probably a substitution that we're looking at rather than an elimination.
As for elimination, I am not sure why the acidity of ethanol should favour it. I would say elimination is facilitated in acidic conditions when there is a group that must be protonated to become a good leaving group (e.g. OH). Here we have Br, which is an 'OK' leaving group most of the time, no protonation involved.

Would the person who downvoted my answer be so kind as to explain what they think is wrong with it? As far as I can tell, I answered the original post. If I am right, there is no reason to downvote what I wrote. If I am wrong, but I'm not told why, how am I supposed to improve my contributions in the future?

ethanol does play a significant role by reacting with the organic molecule


text:
Confused about excercise with w/w%
I found super simple excercise which is killing my mind. I should prepare 500 mL 1M HCl. I should make it from 36 w/w% HCl. I do not understand how I can make from less concentrated solution something more concentrated. Please, could you help me?
Thank you.

Welcome to chemistry.SE. 36% HCl is "concentrated HCl", i.e. the highest concentration possible in aqueous solution. Have you tried googling "concentrated HCl"? You will find its molarity, then let us know if you are still having trouble.

:) Hello and thank you very much! That was the problem. Stepan

You will solve this question, step by step.  Elements to bring together are

starting with the concentrated solution, the concentration of 36 m% refering to 36 g of HCl per 100 g solution (acid) may be equally expressed in a molar concentration $c$:

$$ c = \frac{n}{V} $$

The number of moles $n$ is the fraction of mass $m$ and molar mass $M$

$$ n = \frac{m}{M}$$
(You may check your result, the concentration of the concentrated hydrochloric acid should be similar to here.)

The relation of concentration $c$, molar quantity $n$, and volume $V$ mentioned above may be rearranged.  For example, keeping the molar quantity the same, but doubling the volume of the solution will half the concentration. Keep in mind, the relationship described by the first equation is true for the concentrated solution, as well as for the diluted solution to be prepared, individually. By the same token, you know the solution to be prepared has to meet a certain volume, and a certain concentration.  So you could rearrange the first equation to determine the molar quantity of HCl necessary to prepare the dilute solution.

$$ n(\ce{HCl}) = \frac{c\,(\text{dilute acid})}{V\,(\text{dilute acid})}$$

Now, you rearrange and subsitute the equations:
$$ c = \frac{n}{V} \Leftrightarrow{} n = cV \\
V(\text{concentrated acid}) = \frac{n\, (\text{molar quantity in dilute solution})}{c\, (\text{concentration in concentrated solution})} \\
V(\text{concentrated acid}) = \frac{c\,(\text{diluted solution}) \cdot{} V\,(\text{diluted concentration})}{c\,(\text{concentrated solution})}
$$

Try this on your own, then compare.

 By this calulation, you need 42.9 mL of the concentrated HCl to prepare 1 L of 0.5 mol/L dilute solution.

Let me suggest to you to try additional calculations of this type with different acids, different concentrations, etc., to get used to such calculations.  A molarity calculator like this one may help you to check if you did correctly. Deviations in the results will occur, depending if (or if not) the calculation accounts for a density of the concentrated solution equal to 1 (like water), or not.

Thank you very much for your help, the problem was, that I forgot about CONCENTRATED Hcl molarity. Stepan


text:
Dangers of Colbalt Chloride Paper Burning
I am a science technician and foolishly decided to heat a small amount of cobalt chloride paper on a glass petri dish over a Bunsen. Unfortunately the petri dish broke and the paper hit the metal gauze under and burnt. I immediately put the rest of the paper in water and opened the window.
The cobalt chloride paper was white (not pink) and approximately 3 x 1 cm in size. I am very concerned that I have released very toxic material and would welcome any comments.

LD 50 80 mg/kg  i.e. over a gram for a grown man and the content in the paper isn't all that big... I wouldn't be very concerned. Copper chloride dissolved in a large amount of water will produce insoluble cobalt carbonate over time, so it isn't all that environment unfriendly, and it is a vital microelement, so it will be eventually recycled by environment. That said, work in a fume hood when possible in the future.

Not a good story to share ...

It is not surprising that a rapid heating of a Petri dish provided by a Bunsen burner cracks the former.  By application of the heat, the glass wants to expand, to dilate; hence you swirl test tubes over a Bunsen, provided your lab still has access to natural gas.
A more gentle heating of the glass, to dry the paper impregnated by $\ce{CoCl2}$ were gradual heating on a heating plate.  By this you equally would obtain the colourless, waterfree form, in a more safe manner (no open flames, no ignition).
Yes, $\ce{CoCl2}$ is considered toxic, and dangerous for the environment.  It is suspected carcinogen. You set yourself unnecessarily in danger, not only because of the flames, but that the dry salt, too: finely dispersed in air, it is then easily inhaled.  You won't die on the spot, however, do not repeat this. And ChemSE is not a subsitute for a physician.
Depending where you did this, disposal of heavy metal, and aqueous solutions of heavy metals (like $\ce{CoCl2}$) is regulated by law, and hence the rinse with water was probably fine to extinguish the fire, but the waste water generated is not normal household waste.  Especially if it were on a larger scale.
For future reference: look up safety informations of chemicals you intend to work with, chemical suppliers include them in MSDS sheets. In chemistry lab classes, students are requested to get familiar with these information prior to performing the experiments.


text:
What is the criteria for determining structure of a molecule?
Does the shape of the molecule depends on symmetry?
Why is the shape of dinitrogen trioxide ($\ce{N2O3}$):

(oxidation number of nitrogen atoms: +4, +2)
Instead of my proposal, which is much more symmetrical in terms of shape and oxidation number (of nitrogen atoms: +3, +3)

As seen from the case of $\ce{N2O3}$, which is unsymmetrical, then molecule of pyrophosphorus acid $\ce{(H4P2O5)}$ should be unsymmetrical too.
But the actual structure is symmetrical:

Instead of my second proposal

Another interesting thing is: the central atom of the two previously mentioned molecules has the tendency to have have highest oxidation state. That is: +5 in case of phosphorus atom of pyrophosphorus acid and +4 in case of nitrogen atom of dinitrogen trioxide.
Is my above mentioned conjecture is valid?

I would argue that a molecule being symmetric doesn't make it more likely to form. Symmetry doesn't always lead to a more favorable for any number of reasons (charge distribution, sterics, angle strain, etc). As a side note, I found an article that shows the symmetric form of dinitrogen trioxide can be formed under the right conditions.

No, not symmetry per se but generally the lowest energy arrangement of  the atoms.  What structure this is is not always obvious by just combining atoms according to the valency rules; if it were, experimental (analytical) methods would not be needed. This is particularly true of 'organic' molecules where one formula can give many similar structures, e.g. isomers, but also of the small molecules you list.
The rules we use are the result of numerous experiments over the last 150 years or so reacting one species with another and trying to make a consistent set of rules to explain things. But ultimately structural studies are needed, the definitive method is x-ray diffraction from a crystal of the material but of course this is a not always possible. In this case infra-red, Raman and NMR spectroscopies have been used as well as mass spectrometry.  These methods do not give complete structures directly (bond lengths and angles as does x-ray diffraction) but allow confirmation of one structure (as topology) over another, for example $\ce{N2O}$ is  NNO rather than  NON. More latterly computational methods are being used to predict structure but ultimately these will have to be backed up by experiments.


text:
What is happening in this video of solvated electrons donated from sodium in ammonia?
I just saw the Periodic Video Liquid Electrons - Periodic Table of Videos where sodium is added to liquid ammonia. The demonstration shows that even if electrons are solvated, if you have a high enough electron density, the substance will turn shiny and reflective and metallic. Sodium is dissolved in liquid ammonia and donates the electrons to the solution.
There are a number of things happening in the demonstration that I don't understand.
There is supposed to be a separation - once a high enough electron density is reached, the solution is supposed to separate into a high electron density and low electron density layer. Why the separation instead of a uniform concentration? And why would the high electron concentration layer float on top?
Also, at concentrations below the metallic appearance, the sodium in ammonia solution starts out very strongly blue colored. Why? What is it about adding sodium to liquid ammonia that instantly produces such a deep blue, almost black color?
It seems that the demonstration did not go quite as planned, so isopropyl alcohol was added, producing sodium isopropoxide. Why?

The isopropanol is added to quench the highly reactive solution and would be required whatever the result of the reaction was. This is a common way to (gently) destroy sodium metal that you don't want to have to throw down the sink.

@matt_black perhaps "destroy sodium metal" is not the best way to describe what is happening? Watching the video, it seems to have been added early on in the demonstration, and the reason is explained. I just don't understand their explanation.

They say in the video they add the isopropanol to "quench" the reaction: this means to destroy any remnants of unreacted sodium metal as you can't dispose of the metal easily (unlike sodium isopropoxide). This has the beneficial effect of destroying the dark blue low-concentration sodium/ammonia solution making it easier to see the metallic sheen of the high concentration solution. It wasn't that the experiment didn't work just that it was hard to see the result on video.

@matt_black ok I see, to consume the remaining unreacted sodium metal so it can be disposed of later more safely, and to help viewers focus on the metallic color and not the strikingly blue layer. Thanks!

I'll try an answer to this question because I watched this video a while back and did a bit of reading on it at the time and I think I understand the big picture. The problem is that these solvated electrons are very complicated things, and do not lend themselves to the traditional ways that chemists would like to think about things. For that reason, there is quite a lot of literature coming from physicists which is very complicated but nonetheless valuable. Additionally, solvated electrons show up in more than just this rather unusual metal-liquid ammonia solution. For instance, a recent paper in Nature chemistry by Sieferman et. al. [1] showed that there are transient solvated electrons at the surface of water which can have major implications for natural electron transfer reactions. So, understanding these systems is truly quite important.
The Blue Color:
In order to get at the deep blue color which is seen in this sodium-ammonia solution, we must understand what it is that is absorbing (reddish) visible light and hence leading to the transmission of a lot of blue light.
As it happens, at low concentrations of sodium (I'll address this in a bit), the solvated electrons exist in a bound state. This is not too dissimilar to ordinary electronic energy levels, except that there is not really a molecule to be found. Rather, the electron exists in a bound state which it creates for itself. That is, the electron polarizes the surrounding solvent such that it is contained in a bound state. Sometimes people refer to this electron as being a "polaron". One important difference between this and energy levels in a molecule is that electronic excitations generally take place from one bound state to another bound state. In the case of a solvated electron, however, the transitions are from a bound state to a continuum state.
Obviously one question that would be nice to answer is why is it such a deep blue solution? Well, this sort of question is always hard to address because the color of things is more or less an accident. The fact that it is blue just tells you that the electron is relatively weakly bound because it is excited by reddish light.
For more details on the bound-continuum transition, see Aulich's paper [2] which uses photo-electron emission to study solvated electrons as a function of concentration and photon energy. For a theoretical treatment of the bound states at dilute concentrations of metals, see Jortner's paper [3].
The Transition to a Bronze Color:
One reason that people have been very interested in this system is because of this transition that takes place as the concentration of metal increases. Let me first try to give some intuition as to why this transition takes place at all. First, going off of our model that the electrons polarize the solvent and exist in a low-energy, bound state, we should expect that as more electrons are added to the solution (i.e. as more sodium is added), these cavities which the electrons occupy will become closer and closer to each other. This means that electron-electron interactions become quite important. This means two things: first, we have to begin worrying about satisfying Pauli exclusion, and second, we have to be concerned that at some point the electrostatic repulsions will become larger than the binding energy of this electron.
These two points have been studied by some physicists. Alavi and Frenkel's paper [4] develops a model and then performs simulations on ideal fermions (I honestly don't know what makes a fermion ideal and don't understand this paper), demonstrates that the low-concentration bound states are very unstable as the concentration of electrons increases. Thus, some transition must take place.
The best paper on the theory of this transition can be found here [5]. The reason this transition is so interesting is that it is a "metal to non-metal" transition. Remember when you were first introduced to metals and the characteristic description is that they are a "sea of electrons"? That is, the electrons are free to move around, and this leads to high conductivity and other observable properties, one of which is that they absorb and reflect nearly all wavelengths of light, which leads to the silvery color of most metals.
Well, the exact same phenomenon is being observed here, but it's very peculiar because this is all happening in a liquid. For instance, Schroeder et. al. wrote a paper [6] which measures the conductivity of several metal-ammonia solutions as a function of concentration of the metal. I unfortunately couldn't add the relevant figure, but the conductivity basically increases linearly with concentration of the metal until it hits a plateau and does not increase any further.
This gradual increase explains what is going on when we see a separation into two phases. The explanation goes like this. There is only so much space which allows for the existence of the bound states we described above because these bound states occupy a cavity of relatively large volume in the solvent. As more metal is added, more electrons are free in the solution, but the solution is already saturated with these bound electrons. Thus, the electrostatic and exclusion effects become such that any additional electrons added can only exist in a metallic state. This is peculiar because this metallic state is in the liquid phase and is actually fairly dense (I read this somewhere but can't remember the reference... Will update later). If one continues adding electrons, they always become incorporated into the metallic state because the bound states are saturated. Eventually, enough electrons are present that the destabilizing effects due the presence of other electrons is large enough that no possible bound state can exist and the whole system becomes metallic. Hence the plateau in the conductivity plot I described above. At this point, I believe the sodium ceases to dissolve and just stays as plain old metal sodium.

I do not have a good explanation for why the metallic phase is specifically bronze colored as opposed to silvery. I suspect this is a very subtle feature and would certainly be temperature dependent. I haven't described the temperature dependence of all of this, but there quite a few papers that address only this point and it seems to be quite complicated.
I also don't know why the metallic phase forms on the top. I suspect this may just be an artefact of the experiment in that video. That is, it is entirely feasible that solid would sink to the bottom if it didn't just freeze to the side of the test tube. I could be wrong about that though.

References:

Siefermann, K. R., Liu, Y., Lugovoy, E., Link, O., Faubel, M., Buck, U., ... & Abel, B. (2010). Binding energies, lifetimes and implications of bulk and interface solvated electrons in water. Nature chemistry, 2(4), 274-279. DOI
Aulich, H., Baron, B., Delahay, P., & Lugo, R. (1973). Photoelectron emission by solvated electrons in liquid ammonia. The Journal of Chemical Physics, 58(10), 4439-4443. DOI
Jortner, J. (1959). Energy levels of bound electrons in liquid ammonia. The Journal of Chemical Physics, 30(3), 839-846. DOI
Alavi, A., & Frenkel, D. (1992). Grand‐canonical simulations of solvated ideal fermions. Evidence for phase separation. The Journal of chemical physics, 97(12), 9249-9257. DOI
Jortner, J., Cohen, M. H. (1976). Metal-nonmetal transition in metal-ammonia solution. Physical Review B, 13 (4), 1548-1568.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298. DOI
Thompson, J. C. (1968). Metal-nonmetal transition in metal-ammonia solutions. Reviews of modern physics, 40(4), 704.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298.

This is stackexchange at it's finest - thank you so much for the substantial, thorough, and thoughtful answer! You've sent me to the library today which is always a very good thing. I'm marking this as accepted because you've addressed the optical properties with my main focus here. I will break out the question of the separation of layers as a linked but separate, stand-alone question in a few hours, since it is probably not strongly related.

First off, awesome +1 answer. I do still have one point of confusion regarding the phase of the bronze colored material. At one point you said "this is all happening in a liquid" and shortly thereafter "a solid is formed (I believe) and the electrons move more or less freely around". I'm probably mixing up just what you are referring to, but my question about this is simply, is this bronze colored sea of solvated electrons precipitating out as a solid or is it a liquid (as it appears to be), or is it something more complex and kind of in between the two?

Good point @airhuff . This is a detail which was unclear to me but I updated that sentence because indeed it is still a liquid from what I can tell. This would definitely be relevant to why it stays on top.

I'm not sure a full answer is 100% ironed out here, but I'm going to click accept because there is enough insight and linked material to help me on my way to understand what's happening. Once I'm finished reading I'll ask some additional follow-up questions. Thanks for the very helpful answer!

Just to let you know, I've just asked the follow-up question How to think of solvated electrons?