text:
Percentage yield of Haber process
The yield of a single pass of the Haber process is around $\pu{10\%}$ to $\pu{15\%}$ of ammonia when operating at $\pu{200 °C}$ and $\pu{450 atm}.$ However, most graphs show that at this condition will yield about $\pu{35\%}$ instead (see graph below). Where did the value of $\pu{35\%}$ come from? Why does it differ so drastically from the $\pu{10\%}$ to $\pu{15\%}$ range?

Chemical reactions may be performed in sequences (equally known as batches), or continuously.  The two differ e.g., by residence time of the reagents and products in this container.
The former form is like what you are used to, e.g., in the kitchen.  You charge your pot with potatoes, close it, cook them, open it; done.  In a flow reactor (the later form), while the reaction is ongoing, reagents enter and products leave the reactor.  If the reagent's time of residence is less than the time necessary to establish the thermodynamic equilibrium $\ce{N2 + 3 H2 <=> 2 NH3}$, then the chemical yield obtained will be below of what could be obtained.
Adjusting the time of residence depends on multiple parameters. Some of them chemical kinetics of the reaction(s) in question, other ones e.g., about cost of production (e.g., management of pressure and temperature prior/after the reactor, and «time is money»).  In the case of the ammonia synthesis, the reactor is a (massive) tube, in other cases it may be a stirred vessel or a cascade of them. Each approach has its pros and cons.  Welcome to an important aspect of chemical engineering.

Your answer is clear and insightful, thank you.

It is the question of equilibrium versus nonequilibrium yield and production rate. Maximizing the ammonia output flow is not the same as maximizing the reaction yield.
Approaching the equilibrium yield would lead to just minimal production. OTOH, the top production rate would be near zero reaction yield and unlimited input flow, what is not technically applicable.
$\ce{NH3}$ output flow is proportional to the reaction yield and the $\ce{H2 + N2}$ stoichiometric input flow.
If there are 2 variants of production, the one with a lower yield but higher production rate is the way to go. 6 times higher input flow of $\ce{N2}$ + $\ce{H2}$ with the 1/3 reaction yield gives 2 times higher output $\ce{NH3}$ flow.
$\ce{N2}$ + $\ce{H2}$, that are left, are recycled for another catalyst pass.

... and maximizing the profit is not the same as maximizing the product output either.

@fraxinus Sure, it is implicitly included in my answer. As for the component flow rates, there is the flow  with the  minimum value of cost/output ratio.


text:
Why is comproportionation thermodynamically favoured for the synthesis of Mn(acac)3?
I'm currently learning about ligands in school and I'm wondering why does comproportionation happen in the overall equation? There are fewer chemical species on the product side and the entropy of the reaction would decrease. Since entropy decreases, delta G would be more positive and the reaction would not be thermodynamically favoured. Is my reasoning correct?
$$\ce{4 Mn^2+ + 8 CH3COCH2COCH3 ->  4  Mn(CH3COCHCOCH3)2 + 8 H+}$$
$\ce{Mn^{II}}$ intermediate reacts with $\ce{MnO4-}$ to form stable $\ce{Mn^{III}(acac)3}$.
$$\ce{ MnO4^- + 7 CH3COCH2COCH3 + 4  Mn(CH3COCHCOCH3)2 + H+ -> \\  -> 5  Mn(CH3COCHCOCH3)3 + 4 H2O}$$
Overall equation:
$$\ce{ 4 Mn^2+ + MnO4^- + 15 CH3COCH2COCH3 ->  5  Mn(CH3COCHCOCH3)3 + 4 H2O + 7H+}$$

Umm, how could Mn(II) not react with MnO4-?

I understand that the reaction can proceed but I'm not really sure how to put it in terms of the thermodynamics behind the reaction.

Mn(II) in acidic environment does react with MnO4-. But SRP Mn III/II is almost identical to Mn VII/II. acac just shifts the equilibrium.

Disproportionation and comproportionation are not special cases to be treated differently than other redox reactions. They just happen to have some element in 3 different oxidation states. The thermodynamics remain the same as if it were not the case.
I assume the standard redox potential $E^°(\ce{Mn^{III}(acac)3/Mn^{II}(acac)2})$ is lower then $E^°(\ce{[Mn^{III}(H2O)6]^3+/[Mn^{II}(H2O)6]^2+})$, because  $\ce{Mn^{III}}$ forms more stable complex with acac and therefore Mn^{II} is more easily oxidized.
It is situation analogical to $\ce{Co^{III}/Co^{II}}$, where  $\ce{[Co^{III}(H2O)6]^3+}$ is very strong oxidant, but $\ce{[Co^{II}(NH3)6]^2+}$ is easily  oxidized by aerial oxygen.
It follows the general scheme that more stable complex with oxidized metal form leads to greater ox/red metal complex ratio than ox/red metal ratio. That means, as both ratios has to lead to the same redox potential, the complex pair has to have lover standard redox potential. And vice versa for more stable complex with reduced  metal form.

ahh I see. Can I also explain that comproportionation is stable due to chelation effect? Manganese would initally form a complex with water and then H2O ligands get replaced by acac to form a more stable complex?

source: Stack Exchange

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

sort: votes


text:
Why is the Br-C-H bond angle in CH3Br smaller than the Cl-C-H bond angle in CH3Cl?
Can someone explain for me why the Br-C-H bond angle in CH3Br is smaller than the Cl-C-H bond angle in CH3Cl?
From what I know, F-C-H in CH3F has a smaller bond angle than Cl-C-H in CH3Cl because F is more electronegative than Cl, which then attracts the electron in the bonding pair closer to F, resulting in a smaller bond angle.
I therefore expected that Br-C-H will have a larger bond angle than Cl-C-H because Br is less electronegative than Cl. However, the data show that Br-C-H has a smaller bond angle. Can someone please help!
Data (taken from CCCBDB)
CH3Br
Br-C-H = 107.7 degree
H-C-H = 111.2 degree
CH3Cl
Cl-C-H = 110.8 degree
H-C-H = 108.2 degree

These are small differences so who really knows why.  That said, the $\ce{C-X}$ bond length is 1.939 A in methyl bromide, but only 1.781 A in methyl chloride (sterics?) (ref), a significant difference.  The longer $\ce{C-Br}$ bond implies that it contains more p-character than the corresponding $\ce{C-Cl}$ bond.  Consequently, there is more s-character in the methyl bromide $\ce{C-H}$ bonds than in the methyl chloride $\ce{C-H}$ bonds.  More s-character in the $\ce{C-H}$ bonds means a larger $\ce{H-C-H}$ angle.

According to VSEPR theory, the bond angles in case of CH3Cl and CH3Br is based on bond pair - bond pair repulsion only as there are no lone pairs in them. The bond pair electrons are spread within the bond distance between the bonded atoms. The bond distance for C-Br bond (1.934 A°) is larger than the C-Cl bond (1.781 A°) and also there is not effective overlapping between the larger, 4th period Br-atom and 2nd period C-atom. It also can be noted that there is only marginal electronegativity difference between Cl (E.N.=3.16) and Br(E.N.= 2.96) atoms. Hence it seems that the spread of electrons in the larger bond distance (C-Br) bond is causing less repulsion between the C-Br and C-H bonds in case of CH3Br which is causing closer approach of C-H and C-Br bonds, i.e., smaller Br-C-H bond angle than Cl-C-H.

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


text:
Reaction of primary amine with mixed acid anhydride
Is this reaction $\ce{S_N2}$ (because attack of nucleophile and removal of leaving group are not taking place in a single step)?

Which part of anhydride should form the substituted amide? If it is an $\ce{S_N2}$, then $\ce{CH3COO-}$ is a better leaving group but a more (alkyl) $\ce{e-}$ releasing group, $\ce{-C2H5}$ would destabilize the T.S.

The reaction between anhydride and primary amine is by the Nucleophic addition-elimination mechanism (NOT by SN2).
First step: Nucleophilic addition step (Rate determining step):
The addition of amine to carbonyl center (at the most electrophilic carbonyl center and it should generate least sterically hindered intermediate) gives rise to a tetrahedral intermediate (4 groups bonded to the carbon center), which is influenced by steric hinderance. Therefore the attack of primary amine will prominently take place at -CH3 end carbonyl center which is more electrophilic and gives less sterically hindered intermediate.
Second step: Elimination step (fast step, not a rate determining):
The internal nucleophile (O- of earlier carbonyl) reverts back to generate the carbonyl center and propionate ion is eliminated.
Note: As the mechanism of reaction is not by SN2, therefore leaving group stability is not the sole criteria.

There's no such thing as a "sterically hindered intermediate". Steric hindrance is a kinetic property of a reaction, not a thermodynamic property of a molecule.

Then  "sterically hindered transition state or pathway" will be correct ?

Pathway seems like a better choice.


text:
calculation of relative atomic masses in the periodic system
If I now want to calculate the relative mass of c-13 based on the definition m(c-12) = 12amu, I need the q/m values of c-12 & c-13, where do I find these (I've looked everywhere on the internet)? Is this how all other amu's occurring in the PS are determined? How did that go before mass spectroscopy?

Masses of near all isotopes can be found on Wikipedia under titles "Isotopes of ", e.g. Isotopes of carbon

@Poutnik Thank you, I can understand that it says that c-12 is exactly 12 amu, but my question was about how you now know that c-13 is 13.0033548352 amu. This cannot be a direct conclusion from the definition of amu, so you need a ratio between c-12 and c-13 (like the q/m ratio I'm looking for)

When you know one, you know the other.

@Poutnik Could you perhaps show me how you calculate m(c-13) using only the definition "m(c-12) = 12 amu"? ^^

I have not said you can do that. You use 12C mass and ratio of q/m ratios for 12C and 13C to get 13C mass. Or you use 12C and 13C masses to get the a/m ratios.

@Poutnik but on the wikipedia page you mentioned i can't find the q/m ratios. There are only: Nuclide, Z, N, Isotopic mass (Da), Half-life [resonance width], Decay mode, Daughter isotope, Spin and parity and Natural abundance (mole fraction).

I have not said those are there. They can be calculated.

@Poutnik yes, but then why did you comment at the first place if you don't respond to my question at all? My Question was "If I now want to calculate the relative mass of c-13 based on the definition m(c-12) = 12amu, I need the q/m values of c-12 & c-13, where do I find these..."

That is why it is called a comment and not  an answer. Comments do not try to provide an answer.

Your definition is not quite correct. You mean m(C-12) = 12 u. The old unit "amu" was based on oxygen and is no longer used.

..that means m(12C) = 12 Da (Daltons)

The exact mass of an isotope like $\ce{C-13}$ cannot be calculated. It is experimentally determined by mass spectroscopy. Why is it $13.003354$, and not $13.008888$, or $13.012345$ ? Nobody knows. The same thing can be said for all other nuclei.

Your question is about the measurement of carbon-13 mass. After long efforts of physicists and chemists (more than two centuries), it was decided that everyone will measure the atomic mass with respect to carbon-12, defined exactly as 12 g/mol. Once this value was set, every element was measured relative to this particular carbon isotope.
Before mass spectrometry, there was no idea that carbon has isotopes or oxygen has isotopes so this problem of measuring the relative mass of carbon-13 did not exist! Nobody knew about isotopes before 1913. The mass spectrometers, or the so-called parabola spectrographs were invented in 1912.
It is possible to measure the atomic weights by chemical means (one Nobel Prize) of all the elements. In fact the atomic mass of most of the elements (mixtures of isotopes) were known by pure simple and elegant chemistry, and simple mathematics of ratio and proportion long before MS was invented.

Thanks for this detailed answer :-)
How exactly were the weights measured by chemical means?
I just looked up the isotope B-11 on Wikipedia, there the mass was given with 11.009305167 u.
Now I wanted to calculate this mass myself, so I first thought that 5 protons and & neutrons correspond to about 11 u, the mass of the electron is known since the Millikan experiment in 1909, accordingly also the mass of the proton. Unfortunately, I do not know when the mass of the neutron was determined.
(5x1,672610^-24g + 6x1,674910^-24g + 5x9,1094*10^-28g) x N_a = 11,09094954u which is of (1/2)

course a little bit too much because of the binding energy later.
So assuming mass spectroscopy does not yet exist, mass defect is not known but relative mass is based on c-12 isotope as it is today, would we say boron-11 has a relative mass of 11.09094954u? And also if the cathode ray and oil droplet experiment had never taken place would you say boron-11 has a mass of 11 amu?

Where can I find the q/m values of carbon 12 and any other element? Unfortunately I can't find these values anywhere on the internet.
Because with this I could at least calculate the isotopic mass as given on Wikipedia for isotopes (m(Boron-11) = 11.009305167u) myself by simply calculating: (q/m of c-12)/(q/m of any other isotope) x 12 or?

If you want to know how we got here, go to wikipedia and look up “John Dalton” and “Dalton (unit)”. The history goes further back, but Dalton’s work is a reasonable start. As M. Farooq stated, it took centuries of research and refinement, by chemists and physicists, and we did not get get numerical values, with lots of digits, until late in the story. The whole thing is vastly too big to be answered in a Q and A site, i.e., here. So, good luck with your reading up on this fascinating history!

@iwab See also Semi-empirical mass formula aka Gamow's liquid drop model

@iwab, Can you write a separate question? I appreciate your thought process but refine your next question, read a little bit more because you are self answering many just like the binding energy. You cannot have q/m for Carbon-12. It is by definition.

@M.Farooq I think I have found the answer: https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.15%3A_Measurement_of_Atomic_Weights here, for example, the mass of a c-13 isotope is calculated by initially making a reference measurement with c-12 and then comparing this with the measured value of c-13 in order to determine the mass: m(c-13)=(29.749 cm/27.454 cm)(12.000000)=13.0031u. This should be done in this way with every element, right?

@iwab, I think you are right. Today Fourier Transform Ion Cyclotron Resonance Mass Spectrometer can measure masses of single isotopes to several decimal places. This magnetic sector mass spectrometer was also high resolution.


text:
Acidity and structure of EDTA at different pH
I was conducting complexometric titrations to measure the concentration of magnesium ions in my solution.
But of course, as EDTA is a chelating agent, knowing it’s structure and ionization form at different pH would be relevant.
Wikipedia lists 6 different pKa values for EDTA. What I was wondering is, as it has 4 carboxy groups, which all have pKa around 0-2.6, is how the other two pKa values come into effect (~6 and ~10 pKa).
Am I right in the assumption, that when EDTA is dissolved in water, the four carboxy groups release oxonium ions, which then could protonate the amino groups in intramolecular fashion? Or do the amino groups react themselves with water as a base? Basically, the EDTA structure after dissolving in water (pH ~3) would rather look like:

Then, if the pH is gradually increased to ~6, the first, then at ~10 the second amino group will be deprotonated?
As I said, I am quite unsure about the facts and any help or explanation of this structure at different pH levels would be appreciated.

The premise seems false. As of November 2012 WP lists only the expected four pKa values.

@OscarLanzi Curiously, the sidebar (which is presumably what you read) lists 4 pKa values, but the main text lists 6 (see the section on "coordination chemistry principles"). The main text also states which pKa belongs to which group, which seems like it should address OP's question. But I'm not convinced that the amino group has a lower pKa than the carboxyl groups, as it claims...

... indeed, it seems like the Supporting Information of Belle-Oudry, D. J. Chem. Educ. 2008, 85 (9), 1269. (PDF link) has it the right way round. The first four pKa values (0 to 2) are the carboxyl groups; the 6 and 10 are the amino groups (technically they're pKa values of the protonated ammonium groups). Seems like somebody should edit that Wikipedia page.

@orthocresol IP user introduced this error about a month ago, fixed now.

So is what I said correct? Are the amino groups protonated due to the carboxy l groups, in addition to reacting with water as a base?


text:
Would TFA be a strong enough acid to perform an SN1 reaction on alcohols?
The pKa of sulfuric acid, which is usually used to do this reaction is -3, but would TFA be sufficient even though it has a pKa of 0.5?

Yes. TFA is widely used for alcohol dehydration, for example as described in this review.
TFA is a standard reagent for the ionic hydrogenation of alcohols with hydrosilanes which proceeds via the protonation of the alcohol and the formation of the cation wikipedia article


text:
Why is carbonic acid often ignored?
Water (distilled or not) picks up CO2 from the atmosphere, some of which dissolves to form carbonic acid (H2CO3), lowering the pH by a point or two.
From that it seems like carbonic acid would be the bane of every chemists' existence, yet (in my very limited experience) I rarely see it mentioned and never see it accounted for as a participant in whatever reactions are occurring. Similarly, almost every chemistry experiment or demonstration I've seen was done under air, not some inert gas.
Why is its effect generally ignored? Is the concentration of it usually too low to have any impact? Or maybe it's because it doesn't have a lot of interesting reactions... (no idea)?
Even in experiments where the reactions of interest produce CO2 in an aqueous solution in addition to what is already absorbed from the atmosphere, it still seems to rarely be acknowledged. Why is this?
Are there any (relatively simple or common) examples of processes (in a lab) where it does matter? I.e. when would I need to consider its effects?

One of my colleagues used to devote several weeks of his Quantitative Analysis course to carbonate titration errors. Freshly boiled water is a thing! ;-) As you can well imagine, students loved those lectures!

After boiling the water, would it be at least somewhat effective to have a layer of a liquid immiscible with water (e.g. oil of some kind) on top of the water to keep it separated from the air? Provided that CO2 doesn't permeate through that oil, or is at least hindered.

I'm a geochemist (in addition to other things) and I can assure you that carbonic acid and related species are of major importance in natural systems.

Similarly, almost every chemistry experiment or demonstration I've seen was done under air, not some inert gas.

I don't think it's too much of a stretch to say that the popular chemistry demonstrations are popular at least partly because they are robust towards environmental conditions. The same applies to high-school chemistry labs. If you had to construct air- and water-free setups every time you wanted to show something simple, it would become very tedious and very expensive very quickly.
In research labs which have the equipment for this, it's quite different: a lot of small-scale reactions are done under fully inert conditions.

Why is its effect generally ignored? Is the concentration of it usually too low to have any impact? Or maybe it's because it doesn't have a lot of interesting reactions... (no idea)?

In fact, the issue in such labs typically isn't carbon dioxide; the problematic species are often water (a proton donor and a Lewis base) and oxygen (an oxidising agent). Carbon dioxide is none of that, so is pretty inert compared to those two.
Carbon dioxide is also rather less abundant in the atmosphere. But note that the reactivity is by far the more important point: for example, nitrogen is even more abundant, but is also largely inert, so its presence doesn't matter.

Even in experiments where the reactions of interest produce CO2 in an aqueous solution in addition to what is already absorbed from the atmosphere, it still seems to rarely be acknowledged. Why is this?

Probably because the small shift in pH doesn't actually affect the reaction being shown. However, it's not possible to say anything more than that, unless specific reactions are mentioned.

Are there any (relatively simple or common) examples of processes (in a lab) where it does matter? I.e. when would I need to consider its effects?

Ed V's comment hints at titrations. I think the simplest example is when making a stock solution of NaOH for acid–base titrations (a common experiment at high school level). This will absorb carbon dioxide over time, forming sodium carbonate, and lowering the actual concentration of sodium hydroxide. Titration results will in turn be affected.

Exactly! Every student in the quantitative analysis teaching lab boiled water to prepare their sodium hydroxide stock solutions. It would have been worthless to standardize against KHP otherwise and students were later graded by the accuracy of their subsequent determinations of their primary standard unknowns.

The time scale of contamination is really fast. Half an hour is usually enough.

@M.Farooq Would not a layer of some sort of oil (or generally a liquid immiscible with water) on top of the water buy some time?

@Kaz, Yes, that would work, but gases diffuse even plastic containers. Adding oil will contaminate water.

Carbon dioxide isn’t the only compound readily dissolving in water of any kind from the atmosphere. But it really isn’t of much concern. At first, remember that we are dealing with the following equilibrium:
$$\ce{CO2 (g) <=> CO2 (aq) <=>[+H2O][-H2O] H2CO3 (aq) <=> HCO3- (aq) + H+ (aq)}\tag{1}$$
Obviously, we cannot neglect most of this equilibrium – as you mentioned, the pH value does tend to decrease from 7 to around 5 or 6 if distilled water is left standing – but nevertheless the equilibrium is strongly shifted to the left, i.e. the solubility of carbon dioxide is relatively low and carbonic acid is a transient species for most intents and purposes.
Looking at the equilibrium we are dealing with also alerts us to the kinds of processes that could be affected by dissolved carbon dioxide. Evolution of carbon dioxide by some other reaction mechanism (e.g. oxidation of oxalic acid or oxalates with permanganate) would, for example, hardly be affected. In such a reaction, additional carbon dioxide is created which would shift equilibrium (1) away from $\ce{CO2 (aq)}$ – but since the right-hand half of the equation is already near saturation, the strongest effect will be a shift to the left, i.e. liberation of carbon dioxide from the solution, observed by characteristic bubbling. In a sense, using water that is already saturated with atmospheric carbon dioxide would be beneficial in this kind of reaction as carbon dioxide liberation would begin almost immediately.
There are two reactivities of carbon dioxide that might form a problem in certain contexts. The first one is the (weak) acidity of carbonic acid, which could lead to reactions with bases. This is important especially in titrations when a specific concentration of base is desired which should remain stable over time. If basic solutions (such as sodium hydroxide solutions) are left standing, they will not only absorb carbon dioxide but also react as shown in equation (2), leading to a reduction of basicity.
$$\ce{NaOH (aq) + H2CO3 (aq) -> NaHCO3 (aq) + H2O}\tag{2}$$
The second is carbon’s ability to react as an electrophile. This is essentially the reaction that generates carbonic acid from carbon dioxide (nucleophilic attack of water onto the central carbon atom, breaking one of the $\ce{C=O}$ double bonds), but water is not the only nucleophile one has to worry about. Admittedly, this is less of an issue in aquaeous solutions where water is just as good a nucleophile, but carbon dioxide can dissolve in practically any solvent and it can have greater effects in other solvents that are not nucleophilic. For example, Grignard reagents, which can be thought of as carbon-centred anions, can react with carbon dioxide as a nucleophile giving carboxylic acids; see equation (3).
$$\ce{EtMgBr + CO2 -> Et-COO- + Mg^2+ + Br-}\tag{3}$$
This reaction has synthetic utility if one wants to insert a carboxylic acid, but more typically one wants the Grignard reagent to undergo a different reaction and the attack would be considered an undesired side reaction. Grignard reagents themselves aren’t stable in water (as they are strongly basic, they would be protonated by water) but this can be an issue in ether or THF, solvents typically used for reactions with Grignard reagents.
The above notwithstanding, carbon dioxide is often the least of one’s problems in a lab setting. While many reagents and reaction conditions are incompatible with it, there are also many that are incompatible with oxygen gas, water or both. Most of the time, reactions in research labs will be run under inert gas (nitrogen or argon) mainly to exclude water and oxygen from interfering. Sometimes water is the major issue in which case oxygen exclusion often happens anyway as there is ample water in the atmosphere as well. These reactions are performed with dry solvents which already come packaged under inert atmosphere (often argon). Sometimes, only oxygen will have adverse effects in which case water (and all other solvents) will have to be degassed. Degassing will, as a side-effect, also remove carbon dioxide but oxygen will often be the prime target. Thus, not much thought is given to carbon dioxide specificially.
With advancing knowledge of reactions and mechanisms, it will become easy to predict where carbon dioxide might be a problem and where it isn’t – or where it is but so is e.g. oxygen – so you will find that it is rarely mentioned explicitly.


text:
Identifying an alkyl bromide from NMR information
This question appeared on an exam and I translate it into English.

In the NMR $\ce{H^1}$ spectrum of a pure substance, with molecular formula $\ce{C4H8Br2}$, the following chemical shifts (in ppm) are observed: 1.95 (s), 3.95 (s): the proportion between the areas of integration of these signals is 3:1. This spectrum corresponds to the molecule: (A) 2,3-dibromobutane. (B) 1,2-dibromobutane. (C) 1,2-dibromomethylpropane. (D) 1,1-dibromomethylpropane. (E) 1,3-dibromomethylpropane.

Note that no access to chemical shift data or graphs etc was given. The answer provided is (C). I can see that 1,2-dibromomethylpropane has 6 equivalent hydrogens (in the methyl groups) and 2 equivalent ones in $\ce{-CH_2{}-}$ group, corresponding to a 3:1 ratio as in the question.
Answer: 1,2-dibromomethylpropane
However, I cannot see why (A) cannot be an answer. It has a ratio of 6:2 equivalent hydrogens.
Doubt: 2,3-dibromobutane
I know I am not taking into account the chemical shifts. So I am not sure I could do that, therefore asking for help.

In it's present form, I would prefer 1,2-dibromo-2-methylpropane for the first structure shown over the currently used caption of 1,2-dibromomethylpropane.

Note that the names (C) 1,2-dibromomethylpropane, (D) 1,1-dibromomethylpropane, and (E) 1,3-dibromomethylpropane are all not valid.

It is a matter of coupling distance if singlet signals split into multiplets to be resolved as such in the NMR experiment. (Other factors, beyond your question, e.g., the difference between a C(sp³), or C(sp²), too.)
One of the general rules here, the greater the distance between to nuclei which in principle may couple with each other (i.e., a matter of their nuclear spin), the lesser the coupling, and the lesser the split of the signals.  So if you compare two cases

you count the number of bonds between the two types of hydrogen atoms in question.  Assuming coupling by bonding,* you have four bonds separating the hydrogens (case 1, where coupling is denoted as $^4\!J_\text{HH}$), or only three (case 2, $^3\!J_\text{HH}$).
For illustration, compare the 1H NMR spectra of ethyl acetate and propyl acetate. A $^3\!J_\text{HH} = 6...\pu{8 Hz}$ (reference), is resolved today routinely, which includes spectrometers of low magnetic field strength:

($\ce{^1H}$-NMR spectrum of EtOAc recorded in $\ce{CDCl3}$, $\pu{90 MHz}$; published by 1999-03-31 by SDBS)
Now compare an analogue spectrum of propyl acetate, recorded in the same solvent and with the same field strength, where a $^4\!J_\text{HH}$ coupling occurs, with particular attention to the triplet around $\pu{1 ppm}$ about the terminal $\ce{-CH3}$ group:

($\ce{^1H}$ NMR spectrum of propyl acetate recorded in $\ce{CDCl3}$, $\pu{90 MHz}$, published by 1999-03-31 by SDBS)
While the lines of the triplet at least got broader, the split of the signal by $^3\!J_\text{HH}$ coupling is much stronger than the simultaneous one by $^4\!J_\text{HH}$ coupling.
*) A second mechanism is coupling across space, although molecular rotation causes this to be averaged to zero and thus not visible in solution-state NMR.


text:
What is the best way to visualize hybridization
I had learnt hybridization in various complex definitions some of them as "complex overlapping of orbitals (in quantum mechanical model) which leads in changing of shape of orbitals itself.
But these definitions hardly make any sense without understanding it intuitively...
So what is best way to understand this intuitively ?

YouTube

What it worked for me was that given the valence, ie how many bond orbitals shall form, the hybrid ones allow for a more symmetric distribution of electrons and less repulsion and thus an overall lower energy. But I think it is the way that every teacher uses at basic level (by the way, for once a simple description that is accurate).

Hybridisation is a mathematical model. You take the shape of the molecule, then you arrange the orbitals by linear combination to get to that shape. There is nothing intuitive about it.

There is no point in thinking about hybridisation in the sense of a process, where orbitals "change" their form.
You have the basic (atomic) orbitals, and (atoms in) compounds additionally have "hybrid" orbitals. They are called "hybrid" because you can mathematically model them relatively accurately by combining atomic orbitals. To use those "sp", "sp2", .. to describe bonding, you don't have to worry what their wave function looks like, you just remember their geometry.
To form a bond, you anyway have to combine the orbital with one from the other atom. $sp^x$ just tells you what it looks like on one side.


text:
Boiling and melting points of hydrocarbons
What happens while boiling and melting hydrocarbons? Which bonds are broken?
When we compare the boiling point (b.p.) and melting point (m.p.) of certain hydrocarbons the order is inverted. For example: cis-2-butene has b.p. higher than trans-2-butene, but the opposite is true for the m.p.
I am not able to understand why there is a different order. If there is only role of intermolecular forces, then why does the order come out to be different? What are the factors that affect boiling and melting point of hydrocarbons?

The intermolecular forces obviously depend on the packing in condensed matter. Also note the plural: force_s_. Actually a complex multidimensional force field. Why would they all have the same temperature dependence?

Btw. it a higher boiling/melting point. And pls. leave out the ALL CAPS, looks like you're shouting. That's not necessary here.

The entropy change of the two processes can dictate this reversal. Let me see if it can be the case here as well as to find a old answer of mine that address the point. Otherwise I'll write another one in due time.

The comment above is nonsensical in 2-butenes.

I don't understand why this is close. Not really because I have already answered to it, but I think OP couldn't have done much more than noticing the behaviour and correctly argue using reasonable sense that one might expect that stronger intermolecular forces reflects qualitatively in both solid and melt phase. It is in many cases.

The answer involves considering the two main factor that dictates packing or anyway molecular interactions (don't call them bonds!) in alkenes in the two different states.
These are the type of molecular interaction as well as the geometrical shape.
As for the shape, cis-alkenes have a kind of U bent geometry. This interfere with packing as compared to the cis case. Although they have a small electrical dipole moment, the hindered packing overrules dipole-dipole interactions in the solid state. As such, cis-alkenes melt at a slightly lower temperature as compared to the trans isomers.
Opposite, the geometrical requirements for packing are much less important in the liquid phase. Therefore the little polarity of trans-alkenes and the consequent dipole-dipole interaction fully add to the always present VdW forces.
As such, the boiling point of cis-alkenes is slightly higher than that of the trans isomers.


text:
Mechanism for hydrolysis of cyanohydrin
I know that acetals can be hydrolyzed with H2SO4 but I am not sure how benzaldehyde would be formed following hydrolysis of the cyanohydrin. When I searched online it I found answers suggesting that a carboxylic acid functional group will form rather than benzaldehyde but this question indicated otherwise. Can someone please provide a mechanism as to how benzaldehyde can be reformed after reaction with H2SO4?

Cyanohydrins are formed under acid conditions in the presence of cyanide - essentially the addition of HCN across the C=O bond. Mechanism from Wikipedia

Cyanohydrins are stable under acidic conditions, but note that the reactions are reversible. Under basic conditions (eg carbonate work up of the hydrolysis reaction) the reaction runs backwards losing cyanide anion and regenerating the carbonyl compound


text:
Intermolecular Forces and State Change
I know that intermolecular forces determine the state of a substance at a given temperature and that changing state involves the ability of a substance to overcome those forces.  I encountered a question that asks if intermolecular forces change during a state change.  My inclination is that the forces themselves don't change as they are a property of that particular substance, that it is the ability of the particles to overcome forces that drives the state change.  Am I correct, or are intermolecular forces themselves temperature-dependent?

Force does not know temperature when 2 molecules interact. The particular energy state of these molecules can belong to wide range of temperatures.

Look at my answer to this closed as not suitable question. I think this is an example about the subject of your question. The forces that dictate a phase are different, or they overall effect is different. Note that although a molecule is a well defined object, this is not true for the conformation it can assume. So even the reason for the forces might be or not there depending on T (P). https://chemistry.stackexchange.com/questions/159591/boiling-and-melting-points-of-hydrocarbons. Similar thing can happen when geometrical changes are intrinsic to a phase formation.

So...let me see if I understand your last response---because a molecule's conformation can affect intermolecular attractions, and because its conformations are temperature (or pressure)-dependent, intermolecular forces can, indeed, change during a state change?

@KaraRowbotham for instance, the dipole electrical moment depends on shape, and the shape of an isolated molecule can be different than that in solid state. Another reason is that the range of various forces isn't the same, as such the predominant one depends on molecular interdistance. The same can be said for pi pi stacking, and so on. The "force themselves don't change" they role yes. As classical gravity is always expressed by the formula you know, but it can even become neglibile if m is too small or r is too big. Back to our chemical case, yes you can also say that there is always a mix.


text:
Best method for molecular mass determination
A number of different methods may be employed for determination of molecular mass of any sample. But I want to limit our focus to the following methods,

Relative lowering of partial pressure
Elevation in Boiling point
Depression in freezing point
Osmotic pressure

All the methods rely on colligative properties of the sample and only work if the sample is non volatile. I want to know which method gives the most accurate value of molecular mass and if there is any explanation for that. Also are these methods more accurate than mass spectrometry?

Even ignoring mass spectroscopy, there is no fixed accuracy ordering of the four methods. Even for a substance that would be suitable for testing via all 4 methods, what are the various measurement errors in each method? The middle two obviously require accurate temperature measurements and the other two require accurate pressure measurements. Basically, this comes down to horses for courses. And mass spectroscopy, where applicable and available, will cross the finish line while those other nags are lumbering to the first turn.

Osmotic pressure method is the most accurate and reliable method among the 4 colligative properties based molecular mass measurement methods.
Due to inverse relationship between the molar mass of solute and the measured colligative property, the highest difficulty is observed for the measurement of molar mass of high molecular weight species like polymers, proteins, etc.
The osmotic pressure is generally measured in terms of atmosphere or torr unit which is experimentally achieved by measuring the height of solution column in the capillary. As we know that 1 atm = 76 cm of Hg = 10.3 m of water column (Note: Density of water = 1 g/mL and density of Hg = 13.6 g/mL). Therefore for any solution we shall get appreciable column height, which can be measured accurately. Data for the same have been given below:

(Image Source: Manas Chanda, Introduction to Polymer Chemistry, A problem solving approach, 2nd Edition, CRC Press, Taylor & Francis, 2013.)
The experimental data comparing the 4 colligative properties based measurement methods have been given in the above table.
Therefore it is the experimental ease and accuracy of the measurement that makes the Osmotic pressure based molecular mass measurement method most suitable.


text:
How high does the base dissociation constant, Kb, have to be for a base to be considered "strong"?
I'm just confused as to how high the Kb constant / how low the Ka constant needs to be for a base to be considered a "strong base." The conjugate base of hydrogen phosphate ion, phosphate ion, has a Kb of 2.083 * 10^-2. This number is less than 1, a number I would consider to be small. And yet phosphate ion is widely considered to be a strong base. Compared to the strong acids, which have acid dissociation constants in the billions, do strong bases require a smaller Kb?

If we can afford to consider the acid/base as fully ionized, we can call it strong one.

A strong base is arguably whatever you want it to be. Ammonia seems pretty strong when it gets to your nose (its biting order comes from acting as a base), magnesium hydroxide looks powerful too when it displaces heavy metal hydroxides from solution.
For a more objective definition, Wikipedia gives the following:

Acids with a p Ka of more than about 13 [in water solution] are considered very weak, and their conjugate bases are strong bases.

This suggests that a freely soluble species with $pK_b<1$ in ambient water solution would be considered a strong base. By this criterion, phosphate ion and calcium hydroxide come up just short with $pK_b\approx2$ (the latter is limited in solubility anyway), whereas all the alkali metal hydroxides and tetramethylammonium hydroxide make it.

This completely answers my question, thank you.


text:
Can a neutral molecule have nonzero formal charge?
I initially thought that yes, there must be a neutral molecule that exists with a nonzero formal charge on at least one atom.
But thinking more into it, if we are given a neutral molecule where the formal charge (FC) is
FC = V - N - B/2 ,
I now think that the formal charge must be zero, because if the molecule is neutral, than valence shells for each atom must be filled, meaning that for each atom, N + B = a full shell, and I think that means this results in a FC of zero.
I feel there are some holes in my reasoning; how would I go about proving this?

Formula charge of each atom in a Lewis dot structure has to be minimized but it does not have to be zero. The sum of formal charges has to be zero. Try SOCl2.

Check formal charges of O atoms in an ozone molecule.

Ok thanks, I tried both, I see now. So sum of FC for a neutral molecule is always zero, but not necessarily for the individual FC's.

For a very simple and intuitive example, there's the ammonia borane adduct.

Look into zwitterions

I feel there are some holes in my reasoning; how would I go about proving this?

No matter whether you have filled shells or not, the sum of formal charges will always be equal to the charge of the molecule or ion.
To show this, we start with the definition you gave:
$$ \mathrm{FC} = \mathrm{V} - \mathrm{N} - \frac{\mathrm{B}}{2} $$
Here, V stands for the number of valence electrons an atom has, N stands for the number of non-bonding electrons of that atom in the Lewis structure, and B stands for the number of electrons in the bonds that atom makes (i.e. 2 for each single bond, 4 for each double bond and 6 for each triple bond).
Now, you can sum over the formal charges using that definition:
$$ \sum_{atoms} \mathrm{FC} = \sum_{atoms} \mathrm{V} - \sum_{atoms} \big( \mathrm{N} - \frac{\mathrm{B}}{2} \big) $$
On the right hand side, the first sum represents the total number of valence electrons of the unbound atoms making up the molecule or ion. The second sum represents the number of valence electrons present in the molecule or ion. This is because the electrons in bonds are counted half for each atom they connect, so every electron in the Lewis structure is counted exactly once (different from checking for octets, where bonding atoms contribute to the octets of both bound atoms).
To summarize, the sum of formal charges compares the number of valence electrons in the atoms (if they were isolated atoms) to the number of valence electrons shown in the molecule or ion. If they are equal, you drew a molecule. If they are different, you drew a cation or anion (depending whether your electron count in the Lewis structure is lower or higher than the total number of valence electrons in the isolated atoms).

Can a neutral molecule have nonzero formal charges?

Yes, individual atoms of a molecule can have nonzero formal charges, as the examples in the comments show. However, the sum of the formal charges should be zero if the Lewis structure is drawn correctly.
Why so complicated?
You could just count all the valence electrons present in the Lewis structure to check to charge of the molecule or ion. However, sometimes the formal charges are already assigned, so then you are dealing with smaller numbers, most of which are zero.
Even if the formal charges are not assigned, you can learn the bonding patterns where the common elements H, C, N, O, F (and other halogens) have a zero formal charge fairly quickly.

For example, nitrogen with a formal charge of zero and a complete octet has 2 electrons that are non-bonding ("lone pair") and six electrons involved in bonding (either as three single bonds or one single and one double bond or one triple bond). If you see a nitrogen with four or with  two bonds, check the formal charge.
Let's take the amino acid alanine as an example. This molecule is zwitter-ionic (Ian Bush mentioned this concept in the comments to the question). You can quickly compare the binding patterns above with the Lewis structure below to exclude most of the atoms from a formal charge calculation because they match the known pattern of a zero-formal charge atom in a Lewis structure. All carbon atoms, for example, have four bonds and no lone pairs, so they all have a formal charge of zero. Through elimination, it is clear that the non-zero formal charges are on the nitrogen (+1) and on the bottom oxygen (-1) in the Lewis structure below:

Of course, you can also learn the common non-zero formal charge patterns, i.e. a nitrogen with four bonds (+1) or an oxygen with just one single bond (-1).

In all this could somebody define V, N and B, please? It's obvious what the essence of the question is, but I have never seen this notation

@IanBush I updated the answer.

Wow, great answer, thanks! The FC summation equation makes a lot of sense, that was the "proof" I was looking for.


text:
Order of basicity for arylamines and ammonia in gas phase
In aqueous solution, the basicity order is: $\ce{NH3 > Ph-NH2 > Ph-NH-Ph > (Ph)3N}$
What will be the order in gas phase?
Following are some considerations:
1- In $\ce{NH3}$, $\ce{N}$ is $\ce{sp^3}$ hybridised but in aryl amines, it is between $\ce{sp^3}$ and $\ce{sp^2}$ hybridised. So, does the steric factor is more in ammonia?
2- In aryl amines, $\ce{N}$ is bonded to $\ce{sp^2}$ hybridised electronegative $\ce{C}$ atom. This would withdraw the lp electrons with $\ce{N}$.
3- $\ce{+M}$ effect of $\ce{-NH2, < -NHPh, < -NPh2}$ will reduce the basicity of aryl amines.
Edit: I did find a source[1] citing standard Gibbs energy change values for proton transfer reactions of various amines. (Table from Ref. [1])
Thus, basicity of $\ce{Ph-NH2 > NH3}$.

Reference
1.
J. P. Briggs, R. Yamdagni, and P. Kebarle. "Intrinsic basicities of ammonia, methylamines, anilines and pyridine from gas-phase proton-exchange equilibriums", J. Am. Chem. Soc. (1972), 94(14), 5128–5130.
https://doi.org/10.1021/ja00769a081

I think you have all you need. Triphenylamines are even flat or rather flat about N, their 3D character is mostly due to a propeller shape.

@Alchimista Mesomeric effect is considered as a very strong effect. Even then, basicity of $\ce{Ph-NH2 > NH3}$

do you mean why in gas phase aniline is more basic than ammonia?  Please make clear what you want to know and what you know already.

@Alchimista Yes, my whole post is about gas phase.

Still it is not clear to me which order do you expect. There is no solvation, the order is dictated by the mesomeric effect, which get stronger with the number of aryl substituents. Stronger to the point of flattering the molecule. In NAr3 there is even no more lp, in a way to say. Unless you have data pointing to a higher basicity of aniline vs ammonia. Still there could be a way to rationalise even that, as for the mesomeric effect can't be assumed independent of the whole structure.

@Alchimista I did find a source. Check the edit part.

I do have a rationale for that like in my last line above. Could you look for Ph2NH as well? I do expect the latter less basic than ammonia....

As per @Oscar Lanzi answ even my last idea does not hold. I just stress that in solid state the hypothetical order might be even another one, as these beasts are not pyramidal....

Reference [1], in the question, indicates that the phenyl group is not really special. In the gas phase substitution onto the nitrogen with either an alkyl (methyl) group or an aryl (phenyl) group increases basicity, and the effect is cumulative when multiple substitution is used. While this reference does not specifically address diphenylamine or triphenylamine, Ref. [2] does. From the abstract:

Using the Brønsted acids $\ce{BH^+}=\ce{NH4^+,CH3NH3^+,(CH3)2NH2^+}$ as reactant ions in the chemical ionization source of [a] mass spectrometer, the general proton transfer reaction $\ce{BH^+ + M -> MH^+ + B}$ was studied. The following order of proton affinities, i.e. gas-phase basicities, was observed: quinineclidine, decylamine, pyridine, 4-N,N-dimethylaminopyridine, N,N-dimethylaniline > dimethylamine > N-methylaniline, m-phenylenediamine, N,N,N- triphenylamine > methylamine, o-phenylenediamine, p-phenylenediamine, N,N-diphenylamine, aniline > ammonia.

This ordering is consistent with the favorable effect of increased substitution indicated by the table quoted from Ref. [1], so that the basicity order
$\ce{Ph3N>Ph2NH>PhNH2>NH3}$
is suggested for a solution to the question.
What appears to be happening is that in gas phase ammonium ions electron donation to the formally cationic nitrogen via hyperconjugation is a controlling factor. Highly substituted ions offer much opportunity for such hyperconjugation whereas the parent cation $\ce{NH4^+}$ offers none.
In water solution, however, solvation is strongest with ions that are compact, more or less spherical and with localized charge -- the exact opposite characteristics of the hyperconjugated ions that are most stable in the gas phase; the solvation effect appears to win out.
References
1.
J. P. Briggs, R. Yamdagni, and P. Kebarle. "Intrinsic basicities of ammonia, methylamines, anilines and pyridine from gas-phase proton-exchange equilibriums", J. Am. Chem. Soc. (1972), 94(14), 5128–5130. https://doi.org/10.1021/ja00769a081
2.
Ismet Dzidic. "Relative gas-phase basicities of some amines, anilines, and pyridines. Application of some Broensted acids as reactants in chemical ionization mass spectrometry", J. Am. Chem. Soc. (1972), 94, 24, 8333–8335. https://doi.org/10.1021/ja00779a009

I couldn't understand "...$\ce{NH4+}$ ions electron donation to the formally cationic nitrogen via hyperconjugation..." In anilinium ion, the carbon bonded to nitrogen, does not have any hydrogen.

There can still be hyperconjygation from carbon-carbon bonds.

This is quite amazing giving that in solid state Ar2NH and Ar3N are planar (about N). I was really expecting that delocalisation would have kicked in with at least two Ar groups. You can mention, if I got it right, that PhNH2 shall rather be compared to cyclohexylamine instead, and so on.


text:
Why is the van ’t Hoff relation an approximation?
Lately, I've been looking and thinking about the fact that a professor told us in an advanced physiology lecture. When it came to mentioning osmotic pressure and the van ’t Hoff relationship
$$
\pi = c \cdot R \cdot T\tag{1}
$$
he said it was just an approximation, and if we ever got bored, we should try to prove it mathematically. I remember this relationship from when I took physical chemistry classes but I never looked into it further. Well, since I was bored, I started looking at the derivation of this relationship and possible clues to why it is just an approximation, and it kind of messed with my head.
Is it because in the equation shown in Wikipedia — Osmotic pressure: Derivation of the van ’t Hoff formula
$$
\pi = -\frac{RT}{V_\mathrm m}\ln(1 - x_\mathrm s) \tag{2}
$$
we can apply Taylor series for $\ln(1 - x_\mathrm s)$ in the case of very dilute solutions? Is the result of Taylor’s development a relationship that we can also sometimes see, when instead of concentration $c$ we calculate the sum of the product of concentration and activity for each osmotically active substance in dilute solutions?
Does this part of the relation tell us that it is a simple approximation? And how can I somehow mathematically prove it?

If you go to this Wikipedia page, and follow the derivation of the van't Hoff formula, you will see that there are four approximations: two physical, and two mathematical.  https://en.wikipedia.org/wiki/Osmotic_pressure   Can you tell what they are?

@theorist is it approximation by $-x_s$, $n_s/n_v$ and linear one because of the equilibrium equation shape?

There is a beautiful description of the osmotic pressure here (367-373). It's consistent with my questions above. Thanks for all help anyway.

It uses the approximation that the solution is very dilute, so the mole fraction of the solvent $x_a$ is bigger than the mole fraction of the substance $x_b$ which is dissolved. So for small $x_b$:
$ln(x_a) = ln(1) - x_b \approx -x_b$
$-RTx_b \approx -V_m\cdot p_{osm}$
$p_{osm}\frac{V}{n_A} = RT x_b$
Also we approximate that
$x_b = \frac{n_b}{n_a + n_b} \approx \frac{n_b}{n_a}$
$p_{osm}\frac{V}{n_A} = RT \frac{n_b}{n_a}$
$p_{osm}V = RT n_b$


text:
How to determine if all carbon atoms in a cyclic compound lie on the same plane?
I'm currently doing a-levels chemistry and came across a question asking if all 17 carbon atoms in cyclic rings of a cholesterol molecule lie on the same plane. I've done some googling and it turns out that they do not lie on the same plane, but I could not find an explanation for this. Could anyone provide an explanation?

It's rather opposite, atoms aren't in the same plane unless there's a good reason for that.

This isn't a bad reference. Some suspects are lower-energy conformations of the fused rings (cis- or trans-fused) as well as boat/chair energetic favorability.


text:
What happens in an electrolytic cell when the cations are different from the metal of the anode?
Suppose we have an electrolytic cell with copper(II) sulphate solution as the electrolyte, and lead metal as the electrodes.
Will lead ions be discharged from the anode into the solution? What will be left in the electrolyte after we allow this experiment to run for a while?
My question stems from the fact that during electroplating, either carbon (which is always taken to be inert) or copper is always used but no one ever discusses what happens if the electrode is made of a different metal from the cation. Will the electrodes be active in such a scenario?

For given cathode material, surrounding electrolyte and forced potential, what happens on cathode does not depend on processes on anode and vice versa. // Pb anode  would get oxidized to insoluble PbSO4(s), like in an acid lead battery.

Also, consider that copper ions can oxidize lead metal even without powering the electrolysis cell. We normally do not want that parasitic issue.

@Poutnik Will this PbSO4(s) form a coating around the existing Pb anode or will it sink to the bottom of the solution? Also, just to make sure I understand you correctly, it doesn't matter what metal the anode is made of, it will get oxidized so long as it is not carbon or platinum. Is this correct?

@EdV Yea, perhaps my example was poor. Maybe we can switch it with silver to avoid this problem; so what would happen if we use silver electrodes instead?

Remember oxidation of water is a thing. With silver, water gets oxidized. Likewise with gold.

Yes / Yes / Not exactly, it depends on material and electrolyte.

@Poutnik How does one determine if the metal of the anode will get discharged into the solution? How does it depend on the material and electrolyte?

@EdV This is getting confusing. How do we determine what gets oxidized?

You go by redox potentials, known over voltages, i.e., kinetic issues, and experience. No one said electrochemistry was easy! ;-)

@Tham It is the topic for an independent question, or rather for studying of electrochemistry and chemistry of elements.

As with so many questions in chemistry, the answer to "What will happen if we do x?" is going to be "It depends." unless x is very precisely defined.
While an electroplating setup that contains two electrodes and a solution may seem conceptually simple, there are several intertwined things to consider:

The composition of the electrodes
The composition of the solution (and solvent)
What voltage is applied to the cell

A good starting place to see what might happen is to examine a table of standard electrode potentials.
This is essentially a table of how thermodynamically favourable half-reactions are under standard conditions.
You will quickly see that there are usually many possible half-reactions for any given element (and you must consider also reactions that could happen with the solvent, e.g., the oxidation or reduction of water in an aqueous solution).
Once you have a list of potential half-reactions, the Nernst equation can then be used to adjust the potentials for the actual concentrations present in the cell and you will be left with a set of possible half-reactions and their relative thermodynamics.
The voltage being applied serves to tilt this thermodynamic balance to one side or another.
This approach alone, however is unlikely to give a good impression of what will actually happen in a given cell arrangement because it neglects the other important component of a chemical process: kinetics.
While a given reaction may be thermodynamically favourable, depending on the circumstances, it may proceed very slowly, making a less favourable reaction more significant.
This may depend on many factors, such as catalytic effects by the electrode (e.g., reduction of aqueous hydrogen ion to hydrogen gas occurs at a much higher rate on platinum electrodes versus mercury electrodes), and mass transport (a reaction can only occur if the reagents are present at the electrode surface, so reactions may proceed slower if it is more difficult for the reagent to reach the surface, e.g. an electrode recessed into a tube).
So it should be clear that electrochemical cells can be very complicated, but in the case of electroplating, there is mostly only a single thing we care about: taking the cation in solution and reducing it onto the workpiece without interference, and all the parameters are selected to achieve this condition.
Now, to touch more directly on your question: why is electroplating normally done with a cation in solution and a matching (or inert) counter electrode and what happens if this is not true.
While electroplating, a metal cation in solution is being reduced on the workpiece, and one or more oxidation reactions are occurring at the counter electrode to maintain electroneutrality in the cell.
These oxidation reactions are necessary, but not the intent of the process, so it is desirable to ensure that the reactions that occur do not affect the plating process.
Fundamentally, when you have a counter electrode that matches the metal you are trying to plate, as the counter electrode dissolves, it simply replaces the metal cations in the solution that are being plated on the workpiece.
An "inert" counter electrode is an electrode that will not produce products that interfere with the plating reaction; on something like a carbon electrode this may simply be the evolution of oxygen, which leaves the cell as a gas and (assuming the counter electrode is not so close to the workpiece that the bubbles cling to it) does not interfere with plating.
If you chose a counter electrode that will oxidize and dissolve into the solution (a non-inert electrode), the cations may eventually diffuse towards the workpiece and be plated along with the desired metal, creating an impure coating (and generally adding to the list of possible half-reactions assembled above).
The electrode may also interfere in other ways, e.g., if your counter electrode forms an electrically insulating oxide layer, it may quickly cover the electrode and stop any further plating.
There are certainly materials other than carbon that may be inert in a typical electroplating cell, though carbon electrodes are inexpensive compared to something like platinum, and there is a way to separate the two halves of the electrochemical cell so that the products of the counter electrode do not contaminate the plating solution using a salt bridge, though this adds unnecessary complication and may impose some limits to how quickly the plating reaction can occur.
There may also be an advantage to using a matching metal counter electrode rather than an inert counter electrode in that as the counter electrode dissolves, it is replenishing the supply of metal cations in the plating solution, and there is a mass transport advantage of not needing an aqueous reactant e.g., $\ce{OH-}$ to diffuse to the electrode surface, since the electrode itself is the reactant.
So in essence, the reason to use an inert or matching counter electrode are to avoid adding things into the plating solution that will interfere with the coating process, and while there is no general reason to not use a counter electrode made from a different metal, so long as it is inert, graphite is often chosen because of its cost, especially at large scales.

Such a comprehensive answer; thank you so much!

No problem. I got a bit carried away, but I hope it was helpful. :P


text:
Where are the C and H atoms in 1,3,3,3-Tetrafluoropropene?
I'm learning about refrigerants, and one refrigerant is  R-1234ze, otherwise known as 1,3,3,3-Tetrafluoropropene.
On Wikipedia, it says that the chemical formula is $\ce{C_3H_2F_4}$. But, Wikipedia also has the following diagram, which doesn't have any C or H atoms in it. Is there some sort of shorthand in these diagrams that omits C and H? Or, is this a typo on the Wikipedia article?

P.S. This was closed due to being a "homework question." I can see why people might think I am a freshman undergraduate trying to get a leg up on homework. However... I asked this because I am trying to get a handle on refrigerants and why they are so damaging to the climate, and what better refrigerants might look like. I already have a PhD in a different field, but I am a noob at chemistry.  I wish I had time to go back to school and build up a lot of chemistry knowledge!

It is the common convention C and H atoms are not drawn.  It is $\ce{F-CH=CH-CF3}$

Ah, that makes sense! I thought it must be something like that. If you post that as an answer, I will accept it.

https://en.wikipedia.org/wiki/Skeletal_formula

It is the common convention C and H atoms (attached to C) are not drawn.  It is $$\ce{F-CH=CH-CF3}$$
It comes particularly handy for complex structures of sacharides and other molecules in biochemistry.


text:
Saturated vapor and liquid densities (VLC curve) from Helmholtz free energy equation of state
I have been reading about the statistical associating fluid theory (SAFT) which computes the Helmholtz free energy ($A$) from the SAFT EoS (a hell lot of equations) for a molecule of interest, and is then used in the estimation of intermolecular potential parameters by minimising a relative least-squares objective function ($F$):
$$\min_{\varepsilon,\sigma} F=\sum_{i=1}^{N_{P_{sat}}} \Bigg[\frac{P_{sat,i}^{exp}(T_i)-P_{sat,i}^{calc}(T_i;\sigma,\varepsilon)}{P_{sat,i}^{exp}(T_i)} \Bigg]^2 + \sum_{j=1}^{N_{\rho_{satL}}} \Bigg[\frac{\rho_{satL,j}^{exp}(T_j)-\rho_{satL,j}^{calc}(T_j;\sigma,\varepsilon)}{\rho_{satL,j}^{exp}(T_j)} \Bigg]^2  $$
$P_{sat}$: vapour pressure
$\rho_{satL}$: saturated liquid density
$\varepsilon,\;\sigma$: say, Lennard-Jones parameters
I know that the vapour pressure can be computed from the derivative w.r.t $V$, i.e. $P_{sat}=-(\partial A/\partial V)_T$ (although I have not tried to actually do it yet, by hand or code).
But I am struggling to understand how is the saturated liquid (or vapour) density computed from a Helmholtz free energy derivative (similar to that of vapour pressure). I read that ancillary equations have been used to compute saturated properties (https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=918040). However, SAFT EoS have been developed to predict properties below the critical temperature.
Can anyone please help me to understand how this works (theory, calculation, equation, code, etc.), or point me to any reference paper that explains about this (because I have not been able to find any...)?


text:
What reaction does happen when molten sodium hydroxide reacts with aluminium?
The reaction between aqueous sodium hydroxide and aluminium  follows this equation:
$$\ce{2NaOH + 2H2O + 2Al = 2NaAlO2 + 3H2}$$
But it's obviously not possible to balance this equation without the water included and so I'd like to know what is the reaction of only sodium hydroxide and aluminium? (in this reaction the sodium hydroxide/aluminium has to be molten to react). I tried searching for any info on this reaction but i couldn't find anything.
I've then tried melting a small piece of aluminium and putting a single pellet of sodium hydroxide on top of it (i had to melt the aluminium instead of the sodium hydroxide because molten sodium hydroxide attacks almost everything including glass). The reaction produced a flammable gas that immediately ignited which was probably hydrogen. I'd like to know what the equation of this reaction is and what are all the products.
I'm once again saying that this reaction is not the reaction of aluminium and aqueous sodium hydroxide solution, but the reaction of aluminium and molten sodium hydroxide which means that the water is absent in the equation.
The equation is probably $\ce{2Al + 6NaOH ⟶ 2Na3AlO3 + 3H2}$ but I'm not sure if this is right.

Proposition :  $\ce{2 Al + 6 NaOH  -> 2 Na3AlO3 + 3 H2}$

Does this answer your question? How does aluminium react with bases to form aluminates?

@NilayGhosh  The referred question  addresses the aquaeous scenario, what is not the case here.

@Maurice I agree. According to one source it reacts with water following this equation: Na3AlO3 + 3 H2O → 3 NaOH + Al(OH)3 but i'm not sure if that's true. Could be also Na3AlO3 + H2O → NaAlO2 + 2NaOH     Edit: the outcome is gonna be sodium aluminate (NaAlO2) in both cases because the sodium hydroxide will react with the sodium hydroxide.

There are multiple cases of anhydrous sodium aluminate

@Poutnik well anhydrous sodium aluminate is NaAlO2 and hydrated sodium aluminate is NaAl(OH)4 but both of them are made from the reaction of aluminium and sodium hydroxide solution and not aluminium and sodium hydroxide reaction.

Other related compounds, sometimes called sodium aluminate, prepared by reaction of Na2O and Al2O3 are Na5AlO4 which contains discrete AlO45− anions, Na7Al3O8 and Na17Al5O16 which contain complex polymeric anions, and NaAl11O17, once mistakenly believed to be β-alumina, a phase of aluminium oxide.

Do you know how dry your NaOH was?

Still another possibility: with excess sodium you could form sodium hydride, which is stable up to 638C (WP) and thus compatible with molten NaOH. Such a reaction could take the form  2NaOH +Al -> NaAlO2 + NaH +(1/2) H2.

@Snijderfrey the melting point of sodium hydroxide is just above 300°C so all the water should've just evaporated.


text:
Why pyrolysis of C6H10 doesn't give C5H10 + CH4
In my book it's given pyrolysis of C6H14 gives C6H12 + H2 Or C4H8 + C2H6 Or C3H6+C2H4+ CH4 ( as it gives atleast one alkene) but there are other possible alkenes  as well such as C5H10.

They may not have listed all the products. Also a typo in the title, should be C6H14.

"My book" is meaningless. Please always cite your sources properly. Besides, what are the conditions? Why the compound from the title doesn't match the one from the body? Is it a branched alkane (which one?), their mix or just n-hexane?

Yeah it's C6H14 ! Thnx for the ans.. And I wrote my book because there is no name of that book it is a question bank given by our school and we have to do particular questions after every class.

Indeed $\ce{C5H_{10}}$, specifically pent-1-ene, has been documented[1]:

The pyrolysis of n-hexane has been investigated in the ranges 723–823 K and 10–100 Torr at up to 3% decomposition. The reaction is homogeneous and free from the self-inhibition by olefin products observed for several other alkanes. The products of the reaction are hydrogen, methane, ethane, ethene, propene, but-1-ene, and pent-1-ene, with smaller amounts of propane.

Most likely, the book simply did not list the products exhaustively.
Reference
1.
Freddy E. Imbert, Roger M. Marshall. "The mechanism and rate parameters for the pyrolysis of n-hexane in the range 723–823 K". International Journal of Chemical Kinetics 19 (2), 81-103. https://doi.org/10.1002/kin.550190202


text:
What is the condition to have a schiff base precipitate for sulfonamides?
In the sulfonamides group we observe that some category (like sulfanilamide, sulfamethizol) gives products of condensation with carbonyl so a precipitate, whereas other category (like sulfaguanine, sulfapyridine) don't give a precipitate in reaction with carbonyl. Why is there that difference although these 2 categories have the amino-benzene substituent?


text:
Balancing complex redox reaction in a basic solution
I'm incredibly lost right now. I understand fully how to balance redox reactions in acidic/basic solutions, but the problem my teacher posted is not letting me use the method taught to us. I've searched online for help too, but can't find any.
The equation to balance with the solution is:
$$\ce{6 Ag + 3 NaSH + 2 K2CrO4 -> 3 Ag2S + 2 Cr(OH)3 + 4 KOH + 3 NaOH}\tag{R1}$$
The solution he provided doesn't show work, but it includes an $\ce{SH-}$ and I have no clue where that can come from.
I wrote separate half-reactions:
$$
\begin{align}
\ce{6 \overset{0}{Ag} &-> 3 \overset{+1}{Ag}_2S}\tag{R2} \\
\ce{2 K2\overset{+6}{Cr}O4 &-> 2 \overset{+3}{Cr}(OH)3 + 4 KOH}\tag{R3} \\
\end{align}
$$
This doesn't work unfortunately, I'm just so stuck.

These "half rxns" are anything but half-reactions. Where are the electrons and what happened to the mass balance?

Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown.

The Ag is oxidized since its oxidation number changes from 0 to +1, and Cr is reduced since it changes from +6 to +3, which is why I set up my reactions as such. I really don't see why they're wrong to begin with.

Half-reactions show electrons on one side or the other. They are also mass balanced, as @andselisk also said. And you are kidding yourself if you believe you “understand fully how to balance redox reactions in acidic/basic solutions”.

From the context of what I learned in class, I understand everything that I was taught. It's really not helpful when I'm coming here for help and just get talked down upon, especially by someone probably much older than me. I'll seek help elsewhere.

The oxidation numbers (which I appended into formatted reactions) are fine, it's electrons that is the crucial part of the half-reactions method. And there is problematic sulfur in R2 (mass balance) and potassium has no place in R3.

I was able to work through a bit by looking at the solution guide. I have 6Ag + 3SH- -> 3Ag2S as one, but the other one I'm still a bit stuck. I have H2O + 2CrO4 -2 -> 2Cr(OH)3 + 2OH-

The "solution" is missing the water reactant. Hydrogen and oxygen atoms are clearly unbalanced. Standard redox methods, properly applied, reveal the water as a reactant.

This equation is the superposition of two equations : 1) oxidation of $\ce{Ag}$ by $\ce{K2CrO4}$ producing the ion $\ce{Ag+}$, 2) precipitation of $\ce{Ag+}$ by $\ce{NaSH}$. These two equations are :
$$\ce{3 Ag  + K2CrO4 + 4 H2O -> 3 Ag+ + Cr(OH)3 + 2 KOH + 3 OH-}$$
$$\ce{2 Ag+ + Na+ + SH- + OH- -> Ag2S + H2O + Na+}$$
Multiplying the first equation by $2$ and the second by $3$, and adding the results gives, after simplification :
$$\ce{6 Ag + 2 K2CrO4 + 5 H2O + 3 NaSH  -> 3 Ag2S + 2 Cr(OH)3 + 4 KOH   + 3 NaOH}$$
Hopefully you have followed my development. I have forgotten intermediate calculations, which are usually well treated in classroom. In case of necessity I may develop some further steps of this reasoning.

You can do it using the standard redox half-reaction method, but because chromium does not form free ions and  the products do not include free silver ions you need to incorporate additional species to match the reactants and products. This would then enable the electron-balanced sum to reflect the actual reactants and products.
Tarnishing silver
Consider first the silver oxidation reaction. With free silver ions it would simply be $\ce{Ag -> Ag^+ + e^-}$, but here the bisulfide ion must be added to provide a sulfur source for $\ce{Ag2S}$. So we render
$\ce{2Ag + SH^- -> Ag2S + 2 e^-} \text{ (unbalanced)}$
where the two electrons are determined by the change in oxidation state of two silver atoms, not by charge balance. So charges along with the hydrogen must be balanced in a second step. In basic solution, we use hydroxide ions to balance the charges and then water molecules to balance hydrogen and oxygen atoms. This gives
$\color{blue}{\ce{2Ag + SH^- +OH^- -> Ag2S +H2O +2 e^-}}$
Chromium gets down to basics
Chromium reduction is rendered as reduction of chromate to chromium(III) hydroxide. Before balancing this would be rendered as
$\ce{CrO4^{2-} + 3e^- -> Cr(OH)3}\text{ (unbalanced)}$
We need five hydroxide ions to balance charges and then four water molecules to balance hydrogen and oxygen atoms:
$\color{blue}{\ce{CrO4^{2-} + 4H2O + 3e^- -> Cr(OH)3 + 5 OH^-}}$
Adding it all up
Now add three times the silver oxidation to twice the chromium reduction to balance electrons:
$\ce{2 CrO4^{2-} + 6 Ag + 3 SH^- + \color{blue}{5 H2O} -> 2 Cr(OH)3 + 3 Ag2S + 7 OH^-}$
Note the water molecules that were missing from the given equation.
All that remains is to add the spectator ions. Since the hydrosulfide is added specifically as $\ce{(3) NaSH}$ and the chromate as $\ce{(2) K2CrO4}$, there will be three sodium ions and four potassium ions, which is reflected in the ratio of the two alkali products.

IMO, it would help if you made it clearer that the "reactions" you've drawn in black are deliberately unbalanced, since that's easy to miss at a glance and since the OP here specifically seems to have a poor and/or incorrect understanding of how reactions should be balanced. Maybe something like "$\ce{2Ag + SH^- -> Ag2S + 2 e^- + \dots}$" or "$\ce{2Ag + SH^- ->T[unbalanced] Ag2S + 2 e^-}$" might work? (Alas, I can't figure out how to properly typeset "$\not\to$" using mhchem.)

Well, somebody didn't like my latest edit I guess.

@ilmari I just wrote the word "unbalanced", but \not\to seems to work for your symbol.


text:
Converting dissolved chromates to solids for disposal
I have 4 liters of electrolysis waste that was used to process stainless steel cathodes.
About half of it was oxalic acid based and the other half was sodium bicarbonate based (currently stored separately, pH 1.7 and 8.4 respectively). It is all mostly liquid with some precipitates currently. My assumption is that all of it contains dissolved (?) chromate complexes/compounds or maybe other chromium compounds like oxalates, hydroxides, oxides, etc.
Since chromates are controlled hazmat here I have to take them to a disposal facility. It's straightforward enough for me to take the liquids there, but I'd rather find a way to get all of the chromates and chromium compounds out of solution so that I can neutralize and dispose of the liquids as household waste and just store the solids to take in one big trip to disposal services later.
Since my ultimate goal is to just dispose of them, I'm not trying to extract chromates with any sort of purity. It's more important that I just get them all out of the liquid phase.
Is there some process by which I can do this using other relatively safe and easy to obtain chemicals?
Currently on my shelf I have a decent collection of filter papers, a vacuum filter, a pH tester, oxalic acid, sodium bicarbonate, hydrogen peroxide, table salt, hydrochloric acid (muriatic), distilled water, and a wide range of various shop solvents (ethanol, isopropanol, xylene, acetone, mineral spirits, MEK, and I'm pretty sure there's some toluene somewhere). Also I have iron, mild steel, aluminum, and copper stock, and a DC power supply. Not sure if any of that is useful but, that's a list of stuff that I already own.

Simplest thing, which does not generate additional waste: let the 4 liters of solution evaporate down to dryness, or close to it, and take it to the hazardous waste folks.

@EdV Sweet. That was my first thought too, and that's what I was hoping someone would suggest. I wasn't sure if there was any chromium stuff that liked to exist as a gas. Can I boil it down, or give it at least a little extra heat, instead of waiting for it to evaporate?

Sure, but gently and with good ventilation. It will likely corrode or strongly stick to the vessel, which is why I would just let the solution evaporate in disposable plastic containers. The less you have to mess with it, the better.

Sounds like a plan. I'll filter first to get the easy stuff out (or not if I'm lazy) then just stick the rest outside in a wide container in the sun. I can neutralize the acid with the bicarbonate solution too. Post it as an answer if you want some bonus points, heh. Thanks!

One thing: if you put it outside, please ensure that critters cannot drink that toxic brew. A shed or such is ideal for this.


text:
Oxidation of tert-butyl alcohol with potassium permanganate
I was doing a high school lab, and according to my expectations for organic chemistry reactions, the following reaction of potassium permanganate and tert-butyl alcohol (t-butanol) should occur:
KMnO4 + C4H10O -> No Reaction
This is because, as far as I understand, while primary alcohols produce aldehydes and secondary alcohols produce ketones, tertiary alcohols do not react in standard oxidation.
My issue is that when I combined 3mL of t-butanol and 1mL of potassium permanganate, plus an H2SO4 (sulfuric acid) catalyst, a visible reaction did occur, turning the solution Orange.
For reference, the initial colour of t-butanol is clear, and KMnO4 is a clear but pink colour.
My initial thoughts are there must be some sort of contaminant. That being said, the only other chemicals nearby that COULD have possible contaminated the solution if someone mixed pipettes werea sec-butyl alcohol and 1-butanol, and I already know the reaction for those with potassium permanganate produces a clear product.
This is my first time using the chemistry StackExchange but I'm used to using StackOverflow, so  please feel free to let me know if there is anything missing in my question and I will put it in an addendum in my post. All help appreciated.
Idea
Maybe it formed formaldehyde?

Normally, the tertiary alcohols do not react with potassium permanganate. However, you added a strong acid (H2SO4). So, the sulfuric acid can react with the alcohol, generating an alkene, and then the alkene could oxidize with potassium permanganate.

I think I understand: The acid generated isobutylene, which in turn was oxidized by the potassium permanganate originally intended to prove that tert-butyl alcohol doesn't react. This is then oxidized into water, carbon dioxide, and 2-propanone (acetone). This makes sense. But my understanding of acetone is that it's clear (as is water and carbon dioxide obviously). What's the cause of the colour change?

It is probably $\ce{MnO2}$.

When I ran reactions with KMnO4, the reaction media became slightly yellow due to the MnO2... I agree with M.L. It could be the MnO2. Or it's other impurities, I guess.

Given time and heat acidic permanganate will oxidize almost anything even water! The Baeyer reagent, cold alkaline permanganate was originally a test for alkenes, isolated double bonds; it oxidizes a double bond to a glycol in decent yield giving a copious MnO2 precipitate. As a test when run properly the end result is a mess and obvious. Unfortunately, other moieties such as primary-, secondary-alcohols, amines, sulfides react. Confirmatory tests or analyses are necessary.
Acidic permanganate [or an intermediate MnO2] is reduced to Manganous ion. The yellow color is not likely to be a manganese compound, unless a complex is formed or a reaction step is inhibited. What is needed here is a more complete stoichiometric analysis. It seems that you reacted an excess of alcohol with a very dilute [pink] permanganate solution and an indefinite amount of acid and got either a pale yellow or orange solution. Possibly a small amount of impurity was reacted with the permanganate and the reaction halted at some intermediate stage. You may be on to something.
First be certain of the purity of the alcohol and the cleanliness of the glassware. Second do a careful account of the amounts possibly a titration with the permanganate instead of a gross addition; control the temperature; an ice bath might be appropriate. The reductions of ions such as MnO4- and CrO4= are stepwise and complicated involving intermediates and intermediate reactions that are sensitive to acidity and concentrations. Add in a secondary reaction such as carbocation formation, makes it more interesting. Careful observation is essential. Have fun and good luck.

I suspect the color us MnO2. Sulfuruc acid was added as a "catalyst", thus a small amount, and permanganate reductions consume acid. So you msy have gotten the neutral reduction product instead of the acidic one.

MnO2 is "insoluble" in water.  You are suggesting the amount of acid was such to neutralize the solution to give a colloidial suspension of MnO2. The info given is insufficient to make conclusions; it is a poorly thought out experiment altho the results are intriguing and invite some serious experimentation not specious conclusions. Also a catalyst is not consumed so the use of the word is amusing at best. The PINK permanganate[have you ever seen pink MnO4- solution?] raises question if there was enough to do anything.

So I said "may". The kay mistake may really have been made by tge book saying there is no reaction with the tertiary alcohol. Some sort of reaction evidently did occur.

True something happened but not what happens with primary and secondary alcohols at least in the end product. We might be seeing a first step in a sequence or just an impurity. Without more data specific speculation is futile.  Permanganate and chromate reductions are complex and varied. I see a possible research topic here.


text:
What is the difference between PEF and PTF bioplastics?
I am trying to research single use plastics replacements, and wanted to explain how PEF and PTF are a good choice. Although as I am researching I can't find clear information about them, also how they differ as article label them as the same. What are they? And how do they differ in an environmental standpoint?

Can you spell out the acronyms? Or is that part of your question?

Polyethylene furanoate (PEF) and polytrimethylene furandicarboxylate (PTF)

Looks like only one more methylene group in monomer, so not much of a difference. FYI better add info you did find to the question and make it more specific, if you don't want it getting closed.


text:
Why do radical inhibitors inhibit a reaction by creating a "more stable" radical, but for monohalogenation, we WANT to form the most stable radical?
According to my notes, radical inhibitors tend to turn a more reactive radical into a less reactive radical by making it more stable. For instance, the hydroquinone reacts with a radical species, forming a more resonance stabilized radical, which is less reactive.

However, when deciding to abstract a proton to create the major mono-halogenated product, we must choose a proton that has will create the most stable radical. In another words, the weakest C-H bond is broken first to form the most stable radical, which reacts to form the major mono-halogenated product. To me, this implies that greater radical stability increases rate of the monohalogenation reaction. And this contradicts the idea of the "less reactive, more stable radical" in radical inhibitor mechanism.

What am I missing? Any help will be appreciated.

Just take the more stable radical formed in a reaction as an inhibitor for the formation of the other possible ones and the conflict should disappear. Also don't take this type of assertions as full 0/1. Reactions can give messy/mixed products mixtures. Anyway, you don't need a "stable" radical, just one that is more stable of the others. It can be rather unstable & reactive, but it gives the majority product.

Radical inhibitors like dihydroquinone ($\ce{H_2Q}$) are quenchers which inhibit the radical reaction by quenching an active radical ($\ce{R^.}$) to $\ce{R-H}$ and itself gets converted into a resonance stabilized non-reactive monohydoquinone ($\ce{HQ^.}$) radical species. Once the $\ce{R^.}$ gets converted to $\ce{R-H}$ the reaction stops as the formed radical, $\ce{HQ^.}$ is not reactive enough to participate in the reaction further.
But in the case of mono-halogenated product formation, we are making a reactive intermediate species, $\ce{R^{.}}$ in the first step which will react further with the halogen ($\ce{Cl_2}$) in the second step to give the mono-chloro derivative ($\ce{R-Cl}$). The first step, i.e., $\ce{R^.}$ formation is the slowest and rate determining step, therefore activation energy for the radical intermediate formation gets lowered when more stable radical is formed (See: Hammond's postulate). Hence the rate of mono-halogenated product formation will increase when more stable radical is formed in the first step of the reaction which will give rise to major product if other possibility of hydrogen abstraction exists.

Hello, please have a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?; in particular, you should use $\ce{H2O}$ $\ce{H2O}$, not just H_2O $H_2O$, which is incorrect as chemical element symbols are not supposed to be italicised.

@orthocresol :  Noted your suggestions.


text:
Justification for using valence bond theory and molecular orbital theory together?
Elementary gen chem books will make the careful distinction between the two theories and then proceed to claim that chemists can use both to complement each other to explain the behavior of certain electrons. The classic example is Benzene, held together by a $\sigma$ bond framework with delocalized $\pi$ orbitals. In general, resonance uses MOs and VBT describes the framework to hold the molecule together.
What is the justification for utilizing both theories at once? The construction of the wavefunctions are dramatically different. My only guess is that you need the concept of hybridization from VBT and afterwards, you are constructing molecular orbitals that are superpositions of various hybridized orbitals and other atomic orbitals, is this the right idea?

This is quite the can of worms you're opening. At the level of your textbook, the justification is pretty much just following the time-proven tradition in chemistry of "I don't care if it's wrong as long as it looks simple." The "orbitals" you see in your textbook are so far away from the actual maths and physics behind them that it's hard to bridge that gap without going way beyond the elementary level. For an actual discussion of how this mixing of wavefunction approaches is or isn't wrong, you'll need to go a lot deeper into a discussion that is still ongoing in theoretical chemistry.

This might be beyond the scope of an elementary general chemistry textbook, but it is far away from a can of worms. It is quite simple actually. The two theories are essentially equivalent and once you take them to their respective mathematical limit within the same framework of approximations they produce the same wave function. They simply approach the problem from different perspectives. However, the way these theories are used and abused by some textbooks is a different story. Suffice to say, neither theory can be a justification of the other, but neither can one refute the other.

https://pubs.acs.org/doi/pdf/10.1021/acs.jchemed.1c00919 might be of interest

@Martin-マーチン The overall wavefunction stays the same, but only for MO are the constituent orbitals actually static and have energies. In VB, you can't just pick one orbital and investigate its shape and energy, because the former isn't fixed in time and the latter doesn't exist. Thus, displaying a bonding electron pair using a localized orbital is extremely misleading. But then you have to put the caveat that this discussion is confined to a one-electron picture anyway. And then you have to discuss whether static one-electron orbitals are actually a useful physical thing or not. Can of worms.

They are both approximations that look at the problem from different angles. MO looks at the molecule as a whole, VB looks at it as a collection of bonds. In their theoretical limits, both are the same. As with all things in chemistry (and life) it is good to look at problems from different perspectives. AND combining them is more powerful than either one alone (ie. the sigma and pi systems of benzene). Both theories get much more complicated as they try to refine their respective approximations and get closer to "truth". At the introductory level, it is enough to know that they compliment each other and neither is better than the other. Chemistry is a layering process, once you get the basics you can dig deeper and deeper.


text:
What is the effect of temperature on the enthalpy of reaction?
In the study of reaction mechanisms, enthalpy of formation for intermediate compounds at various temperatures is studied. I understand the reason to do that.
How much (and what) effect can temperature have on the reaction enthalpy? Is it too different for different reactions, or is there some trend?

The temperature trend of reaction enthalpy is determined by the difference of heat capacities of reagents and products, following the Hess's law. As total enthalpy change of heating reactants and cooling products back is nonzero.
$\Delta H_{\mathrm{r,}T_1}=\Delta H_{T_1 \rightarrow T_2} + \Delta H_{\mathrm{r,}T_2} + \Delta H_{T_2 \rightarrow T_1} $
$\Delta H_{\mathrm{r,}T_1}=
\int_{T=T_1}^{T_2}{[\sum_\mathrm{reagents}{(n_i \cdot C_{p,i})} \cdot \mathrm{d}T]} +
\Delta H_{\mathrm{r,}T_2} + \int_{T=T_2}^{T_1}{[\sum_\mathrm{products}{(n_i \cdot C_{p,i})} \cdot \mathrm{d}T]}$


text:
Does fluorine have 5 or 7 active valence electrons?
In a paper on chemical periodicities of elements, Cao et al. show the number of valence electrons in a periodic table (https://doi.org/10.1515/pac-2019-0901). Apart from Boron and Aluminum shown as group 3 and 13, the table matches what is taught about main group elements in textbooks except for the entry for fluorine.

Fluorine is listed as 5 "active" valence electrons, implying perhaps that the 2s electrons do not participate in bonding. Why is fluorine treated differently than oxygen (or does oxygen make compounds where the 2s electrons are more involved in bonding than those of fluorine)?
If I look at the effective nuclear charges experienced by the 2s and 2p electrons in nitrogen, oxygen and fluorine, there is no jump in the energy difference that would place fluorine in a special category (values found on Wikipedia from Clementi et al 1963 and 1967):

Basic beryllium acetate, maybe?

This "periodic table" looks so bad I wouldn't put it in a post here, much less any paper. Not to mention it's wildly inaccurate. For starters all noble gases can be protonated - somehow their electrons are "active" enough.

@Mithoron The only way possible to discuss a bad figure is to post it. It is what I am asking about, not an illustration where I have a choice whether to use a high-quality figure or not.

I was taught many years ago that the F 2s orbital is so low in energy that it does not contribute meaningfully to bonding. Eg in SF6 or BF3, the ligand orbitals are essentially completely 2p. But I don't have any actual calculations to back that up.

I've also heard of the statement that the 2s orbitals of oxygen/fluorine are essentially core-like; Google does seem to show some open-access articles talking about "oxygen 2s semicore orbitals".

And why 11 in column 11 ?

Here is an example where the fluorine 2s orbitals are described as too tightly bound to the nucleus to participate in covalent bonds:

This is for the $\ce{SF6}$ molecule, discussing why the 2s orbitals are non-bonding. The full text with figures is e.g. here and the open access MO diagram is e.g. here.
Here is another, simpler example (image credit: Nick Greeves' Chemtube3D, also used in Libretext document), the MO diagram of hydrogen fluoride:

The OPs question shows a table with effective nuclear charges for the first and second period. If instead you look at ionization energies and include the third period elements, you do find that the s electrons in oxygen and fluorine hold the record when comparing outer shell electrons, and it might make sense to set them aside when making molecular orbital diagrams:

Source: https://www.colby.edu/chemistry/PChem/notes/AOIE.pdf

I really hate this classification of bonding and non-bonding and anti-bonding the image implies of using. In my point of view is stupid to base something on a reference that cannot be true any more. And it totally interferes when you later try to understand back-bonding. In MO theory there is no such thing as non-bonding, it can't be, it is using canonical (akin) orbitals, which are delocalised, therefore you cannot unambiguously define bonds. Anyway, in that regard, orbitals are the worst tool to try to capture bonds.


text:
Numbering of carbon chain in 3-bromo-2-methylpropene
I have a specific confusion in numbering the carbon chain for 3-bromo-2-methylpropene.

I found two carbon chains of equivalent length (three atoms) and equivalent side chains (one atom).

For both carbon chains the double bond of propene lies in first position.

Hence, I find it hard to name this structure according to IUPAC system as both points are satisfied. Particularly, I get 2-(Bromomethyl)propene considering the alternative case, but obviously this compound doesn't exist.

Case 1
Case 2

3-Bromo-2-methylpropene
2-(Bromomethyl)propene

Correct
Wrong

Please enlighten me on wherever are my concepts lacking.

Then why 2-bromomethylprop-1-ene is incorrect? It does follow the rules of IUPAC and accounting for bromine attached to methyl group at C2, it seems perfectly valid..

Because you would have"propene" as the root (noun) and "bromomethyl" as the modifier (adjective). The root must include the bromine with methyl as a substituent.

Case 1 is 3-bromo-2-methylprop-1-ene and Case 2 would be 2-(bromomethyl)prop-1-ene. So your actual question is why 3-bromo-2-methylprop-1-ene is preferred over 2-(bromomethyl)prop-1-ene and not about the other mistakes you made?

Yes Loong. That is precisely the question. I have edited not included the "1" in prop-1-ene as per my book, prop-1-ene can be considered propene.

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

The compound that is given in the question doesn’t have a principal characteristic group that is expressed as a suffix; so, Rule P-44.1.1 doesn’t help.
The principal chains in 3-bromo-2-methylprop-1-ene and ‘2-(bromomethyl)prop-1-ene’ have the same length; so, Rule P-44.3.2 doesn’t help.
The principal chains in 3-bromo-2-methylprop-1-ene and ‘2-(bromomethyl)prop-1-ene’ have the same length; so, Rule P-44.3.2 doesn’t help.
3-Bromo-2-methylprop-1-ene and ‘2-(bromomethyl)prop-1-ene’ have the same number of double bonds; so, Rule P-44.4.1 doesn’t help.
Finally, Rule P-45.2.1 is looking for the structure that has the maximum number of substituents cited as prefixes to the parent structure. Therefore, the preferred IUPAC name is 3-bromo-2-methylprop-1-ene (two substituents) rather than ‘2-(bromomethyl)prop-1-ene’ (only one substituent).


text:
How can I calculate the density of a gas mixture using the ideal gas?
I am trying to calculate the density of the ideal gas mixture given the following data:

The gases are helium and oxygen. There are $\pu{8 g}$ $\ce{He}$ and $\pu{16 g}$ $\ce{O2}$. The temperature is $\ce{300 K}$. The pressure is $\pu{100 kN m-2}$.

So I know density is mass per volume, but I don't know where to get the volume from. I thought I would get it from the Ideal Gas law, but I can find $p\cdot V$ for both gases after finding the quantity for both gases, but I don't know where to go from there. I only know that the pressures add up to  $\pu{100 kN m-2}$.

The volume is the same for both gases, and the sum of their partial pressures is equal to 100 kPa.  Do it algebraically, and solve for V.

Your sample is made of $2$ moles $\ce{He}$ + $0.5$ mol $\ce{O2}$. So the total gas amount is $n = 2.5$ moles. Its total mass is $m = \pu{ 2 mol · 4 g/mol + 0.5 mol · 32 g/mol = 24 g}$  Now the volume of $2.5$ mole ideal gas can be calculated from : $V = nRT/p$, with $p = 10^5$ Pa, and $T = 300$ K. Divide $24 $ g by this volume, and you get the answer. Go ! Do the calculation yourself !