text:
Does concerted oxidative addition follow first-order kinetics?
So my professor for organometallic chemistry described the different mechanisms for oxidative addition and in talking about the concerted mechanism he said that it had first order kinetics. I find this very odd as no step in the mechanism seems to involve a single reactant, so it doesn't seem like the rate-determining step could possibly be one with only one molecule.
I have included a picture of the mechanism shown on his slides. Is he correct in saying that it is first-order? Or am I missing something?

There isn’t really a single mechanism of oxidative addition (or may not always even be a single step). I’d depends on the oxidation state of the metal, the polarity of the ligand being inserted, the coordination geometry etc.

Did your professor say first order overall, or did he say first order in metal complex reactant ?


text:
Can actinides form diamagnetic coordination complexes?
I am not sure how accurate I can predict the magnetic properties of actinide diamagnetic complexes, due to the fact that they share the F-block electron as Lanthanides, which are already tricky, combined with the fact that the high atomic mass of actinides starts affecting its electron shell in ways that lower mass elements do not achieve.
If there are, could some examples be given or perhaps an accessible page?

Actinium 3+ complexes would be diamagnetic for example, because Actinium 3+ does not have any f-electrons.

Shoubhik's comment mentions actinium, but thorium offers a more common example. The relativistic effects referred to in the OP actually have relatively little chemical impact in the early groups of the Periodic Table, including lanthanides/actinides, and thorium can be ionized all the way to its radon core ($\ce{Th^{4+}}$). Such a core would then be diamagnetic as shown in the negative magnetic susceptibility reported for $\ce{ThO2}$.
Among complexes, Wikipedia mentions halide complexes ans specifically gives $\ce{ThF6^{2-}}$.  From the WP article on the element thorium:

Many polynary halides with the alkali metals, barium, thallium, and ammonium are known for thorium fluorides, chlorides, and bromides.[1] For example, when treated with potassium fluoride and hydrofluoric acid, $\ce{Th^{4+}}$ forms the complex anion $\ce{ThF6^{2-}}$, which precipitates as an insoluble salt, $\ce{K2ThF6}$.[2]

Cited References

Wickleder, M. S.; Fourest, B.; Dorhout, P. K. (2006). "Thorium". In Morss, L. R.; Edelstein, N. M.; Fuger, J. (eds.). The Chemistry of the Actinide and Transactinide Elements (PDF). 3 (3rd ed.). Springer-Verlag. pp. 78-94. doi:10.1007/1-4020-3598-5_3. ISBN 978-1-4020-3598-2. Archived from the original (PDF) on 14 December 2017.

Hyde, E. K. (1960). The radiochemistry of thorium (PDF). National Academy of Sciences. Retrieved 29 September 2017.

I still do not have enough points to comment but I would like to add something to Oscar’s comment on relativistic effects not being present on such early elements in the Periodic Table. I am afraid that that is not the case. You are right that it is not enough to affect the magnetic properties but yes for other physical properties.
https://www.chemistryworld.com/news/relativity-behind-mercurys-liquidity/6297.article

They're present, but they have relatively little effect on Chemistry early in the Periodic Table. The greater chemical effects come in later groups.

source: Stack Exchange

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

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text:
Why does radium form a nitride rather than oxide when exposed to air?
The Wikipedia article states:

Pure radium is silvery-white, but it readily reacts with nitrogen (rather than oxygen) on exposure to air, forming a black surface layer of radium nitride ($\ce{Ra3N2}$).

Why is that?
I was surprised to learn that, given that $\ce{O2}$ is generally much more reactive that $\ce{N2}$:

Dinitrogen is mostly unreactive at room temperature, but it will nevertheless react with lithium metal and some transition metal complexes. This is due to its bonding, which is unique among the diatomic elements at standard conditions in that it has an $\ce{N#N}$ triple bond. Triple bonds have short bond lengths (in this case, $\pu{109.76 pm}$) and high dissociation energies (in this case, $\pu{945.41 kJ/mol}$), and are thus very strong, explaining dinitrogen's chemical inertness.

It is similar to lithium that form lithium nitride when exposed to air at r.t. or at slight heating.

related https://chemistry.stackexchange.com/questions/135167/on-heating-in-the-earths-atmosphere-can-magnesium-react-with-nitrogen-to-form https://chemistry.stackexchange.com/questions/125627/when-aluminium-is-ignited-in-atmosphere-will-it-form-only-aluminium-oxide-or-al https://chemistry.stackexchange.com/questions/34109/why-does-lithium-from-the-nitride-whereas-the-other-alkali-metal-forms-the-azide

So little is said about this process by Wikipedia's article on radium that one is led to wonder whether the formation of a nitride is really known. Nevertheless, a possible mechanism is suggested by the section that describes known compounds.

Radium bromide ($\ce{RaBr2}$) is also a colorless, luminous compound.[1] In water, it is more soluble than radium chloride. Like radium chloride, crystallization from aqueous solution gives the dihydrate $\ce{RaBr2·2H2O}$, isomorphous with its barium analog. The ionizing radiation emitted by radium bromide excites nitrogen molecules in the air, [emphasis added] making it glow. The alpha particles emitted by radium quickly gain two electrons to become neutral helium, which builds up inside and weakens radium bromide crystals. This effect sometimes causes the crystals to break or even explode.[1]

Thus, possibly, the radioactivity of radium could cause nitrogen molecules from the air to become more reactive (by exciting them), and then the nitrogen would dominate over the less abundant oxygen upon reaction with the electropositive radium metal.
Heads up for readers: The WP article gives a link to Ref. [1], but I am not allowed to access it from my device. Perhaps a reader can remedy this. (And then remove this paragraph.)
Cited reference

Kirby, H. W.; Salutsky, Murrell L. (1964). The Radiochemistry of Radium (PDF), National Academies Press, pp. 4-8.

Not to mention, radium oxide hasn't been isolated yet (cc @Petr)

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


text:
Is the SN1 reactions stereospecific or stereoselective?
In the SN1 reaction, the product show retention and inversion. Since one of them is major unless it forms a racemic mixture, does it mean that the SN1 reaction is stereospecific?

products of purely SN1 reactions are racemic in nature, There is no stereospecificity.

From here

The SN1 reaction - A Nucleophilic Substitution in which the Rate Determining Step involves 1 component.

-SN1 reactions are unimolecular, proceeding through an intermediate carbocation.
-SN1 reactions give racemization of stereochemistry at the reaction centre.
-The first step is slower and therefore determines the rate.
-Neighbouring group participation is SN1 reactions can be important.


text:
How can we predict the formula of a coordination compound by just knowing the ligand and central metal?
Question:

On complete reaction of $\ce{FeCl3}$ with oxalic acid in aqueous solution
containing $\ce{KOH}$, resulted in the formation of product A. The secondary valency of $\ce{Fe}$ in the product A is:

Now my teacher told me that in this compound the secondary valency of iron would be 6, and told me to remember the formula of all the compounds in GRB Organic Chem.
I have a rough idea that six is the most common secondary valency, but beyond that I fail to see any trend.
Can someone help me out, so that I can predict the formula of such compounds without rote learning?
Only tetravalent and hexavalent compounds are in my course, I also know how to predict whether a compound would be high spin or low spin. My doubt is mainly about whether they will be hexavalent or tetravalent.

The first thing to notice here would be the fact that you see an acid along side a base.
$$\ce{H2C2O4 + 2KOH -> K2C2O4.H2O + H2O}$$
Now, this reacts with the ferric chloride present in the solution to give potassium tris(oxalato)ferrate(III). Many if not all iron complexes are hexavalent in nature.
$$\ce{3K2C2O4.H2O(aq) + FeCl3.6H2O(aq) -> K3[Fe(C2O4)3].3H2O(aq) + 3KCl(aq)}$$
This is in fact how you would prepare the said complex as explained in this lab manual

Potassium trisoxalatoferrate(III) trihydrate, $\ce{K3[Fe(C2O4)3].3H2O}$ is a green crystalline salt, soluble in hot water but rather insoluble when cold. It can be prepared by the reaction of $\ce{K2C2O4.H2O}$ with $\ce{FeCl3.6H2O}$

Is there any such rule for elements other than iron?

@SunainaDas, in inorganic chemistry there is no fixed rule for all sets of compounds. Instead of worrying about rote learning, just focus on seeing the different patterns. This is a similar such pattern.


text:
Temperature control and side reactions in PPAA (T3P) couplings
In a lot of procedures for amide couplings (acid + amine) using  PPAA, the coupling agent is added to a cooled solution of the acid, amine, and a non-nucleophilic base such as DIPEA. This is done even when the reaction itself is conducted at a higher temperature (e.g. the reaction might be run at 60 °C). Furthermore, unlike with reagents like oxalyl chloride, there is no separate step form formation of an activated acid prior to adding the amine (everything is just mixed at the same time).
On a process scale, this initial cooling step makes sense as there is a slight exotherm to the addition (avoids thermal runaway), but the method is often also used on small scales (less than 100mg) too , even when conducted in higher boiling solvents like DMF or AcOEt.
Unlike HATU where the side reactions are well known, there is little discussion of what can go ‘wrong’ with PPAA couplings (and whether this is related to the temperature).

Is the initial temperature control preventing something from happening before the desired reaction gets going?
What are the most common side reactions? I’m not talking about the PPAA by-products (water soluble P containing molecules) but rather things that can form from the starting materials through competing reactions/mechanisms

As a footnote - I often just mix at room temperature and stirr. 90% of the time this causes no problems, however on occasion the reaction profile (judged by UPLC) gives multiple side products in addition to the product.


text:
How to understand the chemical formula of Prussian blue complex w.r.t its crystal structure?
Note: Just to be clear, I do not refer to "complex molecules" (i.e. "complex" as "complicated"), but to actual complexes.
I am trying to understand the structure of Prussian Blue based on its chemical formula (I ultimately want to explain why it is blue, but this is probably not relevant to this question). This formula is
$$\ce{Fe4[Fe(CN)6]3}$$
where the "first" iron is $\ce{Fe(III)}$ and the "second" is $\ce{Fe(II)}$ but that too is probably not relevant.
Now, having pretty much no experience with complexes, I would read this formula such that for every four $\ce{Fe(III)}$s, there are three $\ce{Fe(CN)6}$s – like this:

(Source)
So far, so good. But I am confused to how this relates to the actual 3-dimensional structure of Prussian Blue which apparently is something like this:

(Source)
I see that from each $\ce{Fe}$, six "bonds" extend to each other $\ce{Fe}$. These "bonds" are, I think, ligands and are actually $\ce{N\equiv C}$. This makes sense because we had $\ce{Fe(CN)6}$ in our original formula. However, there should be some $\ce{Fe(III)}$s, namely four for every three $\ce{Fe-N\equiv C}$. Also, a $\ce{Fe(CN)6}$ shouldn't be able to bond to another because this would mean that there were two $\ce{N\equiv C}$s between each $\ce{Fe}$. This, however, is not the case.
What am I missing here?

https://chemistry.stackexchange.com/questions/16428/carbon-atoms-at-the-edge-of-a-diamond?rq=1 This might be of some use.

https://chemistry.stackexchange.com/questions/72198/how-do-counter-ions-influence-the-color-of-coordination-compounds https://chemistry.stackexchange.com/questions/72198/how-do-counter-ions-influence-the-color-of-coordination-compounds

The structure of Prussian blue is not as regular as Jonas thought. It is basically cubic as Jonas has drawn. But Jonas's figure includes the same number of $\ce{Fe^{II}}$ as $\ce{Fe^{III}}$, which is impossible. There must be $\ce{4 Fe^{III}}$ for $\ce{3 Fe^{II}}$. So some $\ce{Fe^{II}}$ sites of the cubic structure drawn by Jonas must be empty. The real structure is described in a publication in French by G. Fornasier, Act. chimique 444-445, Oct. 2019, p.16 - 21
First the Prussian Blue includes $15$ water molecules. Its formula is $\ce{Fe^{III}_4[Fe^{II}(CN}_6]_3$·$\ce{15 H2O}$. It is basically Jonas's image, but with two $\ce{Fe^{II}}$ lacunes (black in Jonas's picture), one in the center of the horizontal upper face, and the second in the center of one of the vertical face of the cube. The CN groups that are binding each $\ce{Fe^{II}}$ atom to the 6 next $\ce{Fe^{III}}$ atoms are replaced around the lacunes by 6 water molecules.
Here is a description of the Prussian Blue, after G. Fornasier. The numbers $2$ and $3$ represent $\ce{Fe^{II}}$ and $\ce{Fe^{III}}$ respectively. The water molecule is like a V shape attached to a $\ce{Fe^{III}}$ ion. The two dotted loops are lacunes where $\ce{Fe(CN)6]^{4-}}$  have been removed in order to save the electric neutrality of the structure. Please observe that every $\ce{Fe^{3+}}$ ion is surrounded by $\ce{6 N}$ or $\ce{O}$ atoms from $\ce{CN-}$ groups, or from $\ce{H2O}$ molecules


text:
Looking for caffeine partition coefficients
As part of a project, I'm looking to carry out a liquid-liquid extraction of caffeine from an aqueous phase to an organic phase, using dichloromethane or  ethyl acetate.
So I'm looking for the partition coefficient of caffeine between water and dichloromethane/ethyl acetate, but I can't seem to find any reliable data.
A number of websites I've seen claim the partition coefficient can be approximated to the ratio of caffeine solubility in the organic solvent to that in water, but I can't find any sources backing these claims up.
Any help finding this date, or studies/document relating to caffeine partition coefficients would be appreciated.
Thanks.

What you’re asking is quite hard, as it depends on concentration, the relative volumes of the phases, temperature, and other factors such as pH of the aqueous media.
LogD values (partition between octanol and buffered water) are common and well tabulated.  iirc, caffeine has a strong preference for remaining in the aqueous phase however repeated extractions into DCM would likely let you achieve good recovery

This seems like an easy enough setup to test? Why don't you run a trial and see what you get?

Consider adding NaCl to the water to decrease the solubility of the caffeine, or perhaps something more alkaline, like Na2CO3 or borax.

Dichloromethane/ethyl acetate are not organic solvents for which the partition of an organic compound between water and an organic solvent are often recorded.  More frequently these partitions are recorded for systems like water/1-octanol (example entry in the Dortmund Data Bank, requires an institutional subscription). Any change in the composition of the organic or/and aqueous phase (e.g., additional components, alteration of the pH value), or temperature obviously may affect  $\log P$.
However, given the significance of these partitions for pharmaceutical compounds for their update, transportation, and excretion within the ADMET profile, there are programs providing a computed prediction as $c\log P$ data (some mentioned here), for example as extra plug-in of the Marvin sketcher, or as an independent application like the OSIRIS Property Explorer.  Their prediction is based on a model about multiple hundreds of experimentally determined $\log P$ data with a wide range of chemical structures, aiming for a high correlation:

(credit, "Calculated Compound Properties" about Property Explorer / DataWarrior)
In the case of caffeine and the system water/octanol, DataWarrior predicts a $c\log P$ of $-0.18$, which is not terrible off from the report of $-0.21$ (median) and $-0.23$ (mean value, $n = 24$) by Harris and Logan.
The extraction of caffeine is an old lab class experiment. Murray and Hansen, for example, report a coefficient K for the partition between 1-propanol, and water/$\ce{NaCl}$ of 3.7.  For a partition between chloroform and water, the same authors compare this with an estimate (based on solubility data listed in the Merck Index) of 8.3.
References:
Harris, M. F.; Logan J. L. Determination of log Kow Values for Four Drugs. J. Chem. Educ. 2014, 91, 915–918; doi: 10.1021/ed400655b.
Murray, S. D.; Hansen, P. J. The Extraction of Caffeine from Tea: An Old Undergraduate Experiment Revisited. J. Chem. Educ. 1995, 72, 851-852; doi: 10.1021/ed072p851.

In the answer elsewhere, Buttonwood has said what OP needs to know about partition coefficients and why they are not recorded for variety of organic solvents. Yet, I found these data (at $\pu{20 ^\circ C}$) in a published article (Ref.1):

Dichloromethane (methylene chloride): $K_\mathrm{caffeine} = 8.0$
1,2-Dichloroethane: $K_\mathrm{caffeine} = 2.9$
Trichloromethane (chloroform): $K_\mathrm{caffeine} = 4.7$
Carbon tetrachloride: $K_\mathrm{caffeine} = 0.12$

The abstract of the paper states that:

The distribution coefficient $(K_\mathrm{d})$ of caffeine between an aqueous solution and organic solvents has been studied, and the influence of $\mathrm{pH}$ and temperature of the aqueous phase on the $K_\mathrm{d}$ of caffeine investigated in the caffeine—dichloroethane-water and caffeine-methylene chloride-water systems.

Since OP is looking to carry out a liquid-liquid extraction of caffeine from an aqueous phase to an organic phase, I thought following grapical representation of extraction would be helpful:

Source of the graphs: Extraction of Caffeine from Tea Leaves - UCLA
Here, $K =$ Partition coefficient or distribution coefficient; $W_o =$ Initial mass of solute; $V_1 =$ Volume of the organic layer in each extraction; $V_2 =$ Original volume of water; and $n =$ number of extractions.
The relationship is is given in:
$$\frac{\text{Final mass of solute in water phase} }{\text{initial mass of solute in water phase}} = \left(\frac{V_2}{V_2 + V_1K}\right)^n$$
References:

G. S. Klebanov, L. N. Mednikova, and A. D. Ovcharova, "Extraction of caffeine from aqueous solutions," Pharmaceutical Chemistry Journal 1967, 1, 221-223 (DOI: https://doi.org/10.1007/BF00770195); Translated from: Khimiko-Khimiko-Farmatsevticheskii Zhurnal 1967, (4), 49-52.


text:
Confusion regarding nucleophilicity
My teacher told me that to compare nucleophilicity, check the size of the attacking atom. Bigger the attacking atom more the nucleophilicity. But while reading Solomon, I found out that the order of nucleophilicity of halides in polar aprotic solvents is
$$\ce{F-}>\ce{Cl-}>\ce{Br-}>\ce{I-}$$
Which is opposite to the order expected from my teacher's explanation. How can we explain this order. Can I conclude that my teacher's explanation is valid only in protic solvents, and in aprotic solvents, more the basic strength implies more nucleophilicity, or this is just an exception?

No. I read the answer before but was not satisfied. In Solomon it is written that this order is for aprotic solvents but the question says that order is different for different solvents, which is making me more confused


text:
What is/was "uraneous oxide"? Was it an accepted name for a compound or mixed oxides of uranium?
In This National Technical Reports Library page for a 1965 report Method for Coating Actinide Particles in the list of keywords both Uranium compounds and Uraneous oxide are listed.
But I don't find other instances at least in a quick search.
However I did run across the word "uraneous" (or "uranious") in the now retired American musician, singer-songwriter, satirist, and mathematician Tom Lehrer song We Will All Go Together When We Go (short, amusing vintage performance in YouTube).
Note, Lehrer is also the lyricist of The Elements Song (YouTube)
Also, see When The Air Becomes Uranious/Uraneous?
Question: Was "uraneous" an accepted word or "uraneous oxide" an accepted name? Did "uraneous oxide" refer to a specific oxide of uranium or perhaps mixed or unspecified oxides?

Could it be "uranous" oxide? Google leads me to uranium(IV) oxide, UO2.

(Cont. to  my above comment) If we talk about mixed oxides, then $\ce{U3O8}$ is a mixed oxide. However, it is named urano-uranic oxide. (FTR I didn't downvote).

One of the oldest reputable sources that I managed to find allowing to define what "uraneous oxide" is, was Roscoe's Lessons in Elementary Chemistry [1, p. 253] (strong emphasis mine):

URANIUM.
Symbol U, Combining Weight 240, Specific Gravity 18·4.
Uranium is a metal which occurs but sparingly in nature, existing combined in two somewhat rare minerals, pitchblende, $\ce{U3O8},$ and uranite.
The metal is of a steel-white colour, and it does not oxidize in dry air at ordinary temperatures, but when strongly heated it burns brilliantly.
There are two oxides which form salts, viz., uranous oxide, $\ce{UO2},$ and uranic oxide, $\ce{UO3}$: the uranous salts are green, whilst the uranic compounds are yellow; and these latter solutions give yellow precipitates with an alkali, in which the uranic oxide acts as an acid, forming a uranate of the base; thus with potash we obtain $\ce{K2UO7}.$
The sulphide is an insoluble salt of a vellowish-brown colour.
The chief application of uranium compounds is for the purpose of glass-staining; the uraneous oxide imparts a fine black, and the uranic oxide a beautiful yellow, to glass: uranium compounds are also now used in photography.

This brief description was extremely helpful in cleaning up the terminology, i.e. "uraneous oxide" is neither pure $\ce{UO2}$ nor $\ce{U3O8}$ nor $\ce{UO3},$ rather a  mix of unspecified oxides.
Another hint was that "uraneous oxide" is used to produce black glass, so it's definitely not uranium(IV) oxide or uranyl, which are used for glass tinting with bright colors (yellow/green/orange).
This could be uranium(IV) oxide which is dark-brown and appears as black powder en masse; however, I doubt its concentration used in glassblowing is enough to yield rich black color.
On the other hand, there are black uranium minerals like uraninite or pitchblende (the name speaks for itself) composed of $\ce{UO2}$ and its oxidized forms.
Oxidation of brown $\ce{UO2}$ at relatively low temperatures yields dark black β-phase nucleated within $\ce{U3O8}$ [2].
An oxygen torch used in glassblowing has the right amount of oxidizing power to partially oxidize uranium(IV) in glass matrix attributing deep-black tint described by Roscoe.
To sum it up, the early definition of "uraneous oxide" was likely a mixture of unspecified uranium(IV–VI) oxides after primary extraction from the ore.
Modern literature doesn't appear to use this term anymore.
However, one can use "uraneous oxide" is an umbrella term referring to any class of oxides containing uranium since linguistics permits usage of adjectival suffix -eous for this purpose:

-eous
suffix forming adjectives
relating to or having the nature of: gaseous. Compare -ious
[from Latin -eus]

on par with more ubiquitous terms "nitrogeneous" or "hydrogeneous".
"Uraneous oxide" is not an accepted name since it's ambiguous, and, of course, doesn't follow IUPAC recommendations for nomenclature.
In the context of nomenclature, it's "erroneous".
As for its use in poetry and songs, I have nothing to add to the following quote by Dickens:

Poetry's unnat'ral; no man ever talked in poetry 'cept a beadle on boxin' day, or Warren's blackin' or Rowland's oil, or some o' them low fellows; never you let yourself down to talk poetry, my boy.
— Charles Dickens, The Pickwick Papers

References

Roscoe, H. E. Lessons in Elementary Chemistry; Macmillan and Co.: London, 1875. ISBN 978-0-266-15565-2 (reprint).
Alberman, K. B.; Anderson, J. S. S 62. The Oxides of Uranium. J. Chem. Soc. 1949, 303–311. DOI: 10.1039/jr949000s303.

So, it turns out that uraneous is erroneous. sweet!

"Thus with potash we obtain $\ce{K2UO7}$". Typo for $\ce{K2U_{\color{blue}{2}}O7}$, maybe?

@OscarLanzi The source says $\ce{K2UO7}$,  probably a typo in the source material indeed. True, a "double" oxide $\ce{K2O · 2UO3},$ i.e. $\ce{K2U2O7},$ would make more sense.


text:
Is a hydrocarbon necessary for combustion?
Oxygen, heat and fuel are essential for combustion. What counts as a fuel?  Is a hydrocarbon essential for combustion? Is combustion, say, of magnesium really combustion? And what's the difference between ignition and combustion?
I'm new to chemistry. If you could explain the problems above that will be great.

Combustion is a quite technological term (although the minimum technology required is known since the discovery of fire). At the moment I think it defines a reaction with the characteristics you know but the oxidant is air or oxygen. Still X burns  in Y is OK as soon the reaction results in a flame.

A combustion is a chemical reaction of something with oxygen, producing much heat. If it also produces light, it is a flame.
Ignition is the operation required for starting a combustion. Using a match may do the job. Fuel is usually a mixture of hydrocarbons derived from oil or petroleum. But combustion of Magnesium is also a combustion, although magnesium is not usually defined as fuel. Another example : Hydrogen is a gas  burning quite easily, and it is not a hydrocarbon.

Does it really require oxygen ? I have seen titanium burn ( light . heat ) in bromine and reliable sources ( Timet engineer) tell me titanium will burn in a hydrogen atmosphere  when the protective oxide has been removed

You may burn hydrogen gas altogether with chlorine to yield HCl; and this product, dissolved in water, yields hydrochloric acid, performed at large scale (there are HCl ovens, shown, for example on wikipedia). The discern between combustion as a (typically exothermic) reaction of compound x and an oxidizer (often, but not limited to oxygen) and an oxidation (which refers to an increase to the [formal] oxidation number of compound x) - again not constrained to reactions with oxygen - perhaps is that the former uses at least 1 gaseous reagent.

Magnesium (as a solid) is not a typical fuel, agreed on this.  But there are examples of rockets with solid fuel of type «once lit, it burns without stop»; contrasting to those with liquid fuel which may be re-lit and easier moderated in terms of the mechanical momentum they yield.  Though the use gasoline (or an other hydrocarbon, as asked by the OP) as propellant may be not very frequently seen.

@Buttonwood. You are perfectly right. I agree with all you said. But. But the OP said he or she  is new in chemistry. I have adapted my answer to his or her level. It is no use mentioning combustions in bromine or chlorine !!

Taking into account you just started studying chemistry, and to expand Maurice' answer, I would reply your question of

"What counts as a fuel?"

that this depends on the reaction conditions set (e.g., temperature, concentration of the reagents present, pressure).*  Then, any compound may be considered as a fuel provided it

may be at all oxidized.  This often is expressed in the release of energy which may be the emission of heat; yet emission of light, or the potential to perform mechanical work may be other forms. As burning logs of wood, more than one form (e.g., heat and light) may occur simultaneously.  Chemists account for this release of energy (account similar like balancing a bank account) in terms of enthalpy, for example Gibbs free enthalphy.  The more energy released, the better for compound x to be considered a fuel.

and

the oxidation proceeds rapidly.  This is one of the reasons why hydrocarbons are used as fuel in combustion engines, because at set temperature and pressure in the cylinder, it is in a fraction of second that these react and release a lot of energy.
A counter example would be the oxidation of a nail of iron at ambient conditions.  This reaction equally releases energy, but this is way too slow to be useful in an industrial process (by today's standards).  Thus, it is not considered as a fuel.

*) For reactions with gases, there is some conceptual overlap of "concentration of the reagents" and "pressure".  You may return to this detail later in your study.


text:
Can hydrogen peroxide reduce ferric ion to ferrous ion?
I have studied that hydrogen peroxide always oxidizes ferrous ion to ferric ion (source of study : NCERT Chemistry Part II, Textbook for Class XI),
but a question came in IIT JEE 2015 which states:

The answer given is A and B but I doubt that the answer is correct as I have studied the following reactions in my book :

$$\ce{2Fe^{2+}(aq) + 2H^{+}(aq) + H_2O_2(aq) -> 2Fe^{3+}(aq) + 2H_2O(l)}$$
$$\ce{2Fe^{2+} + H_2O_2(aq) -> 2Fe^{3+} + OH^{-}}$$

Where am I possibly going wrong?

Yes, H2O2 is both easily reduced and oxidised.

You are right, and your book is wrong. It is easy to show in a simple test tube. In presence of $\ce{H2O2}$, the $\ce{Fe^{2+}}$ ion is always oxidized, whatever the presence of an acid or a base in solution.

@Maurice I think you mean to say that the question is wrong as my book states that in presence of $H_2O_2$, $Fe^{2+}$ is always oxidized to $Fe^{3+}$.

@sochemistry, Your textbook is right. It is just that question is wrong.

The book example looks like the peroxide is oxidising ferrous to ferric, which is  kind of what I'd expect an oxidising agent to do.      Think someone made a mistake (or two) writing the question.

You are right. This question is "practically" incorrect, although on paper it might appear so if someone is unaware of real chemistry (the question setter). Iron (II) will readily precipitate in alkaline medium and then immediately oxidize to iron (III) in the presence of hydrogen peroxide. It does not proceed backwards, i.e., hydrogen peroxide will not reduce Fe(III). It will start catalytic decomposition rather.
In fact iron (II)+ hydrogen peroxide mixture is a very interesting system and used in environmetal cleaning. It is quite complex chemistry of free radicals. It is called Fenton's reagent (always used in mildly acidic conditions).

Thanks for the explanation so it means that all the options are incorrect for the given question isn't it?

Yes, the whole question is wrong.

@sochemistry See Fenton reagent

Test questions usually require regurgitating something already taught (whether that teaching is absolutely correct or not is not really an issue).
However, while it may be accurate, as the OP stated, that hydrogen peroxide always oxidizes ferrous ion to ferric, it is not incorrect to state that hydrogen peroxide also reduces ferric ion to ferrous.
How does this occur? Consider a beaker of ferric chloride to which is added some hydrogen peroxide:
Fe$^{3+}$ + H2O2 --> Fe$^{2+}$ + $^.$OOH + H$^+$
Fe$^{3+}$  + $^.$OOH --> Fe$^{2+}$  + H$^+$  + O2
These equations are the second and third from the question asked over a year ago (Ref 1). They amount to a hydrogen peroxide molecule going to O2, and 2 protons and two electrons reducing two ferric ions to ferrous. In the Fenton scheme, as the H2O2 runs out, the hydroperoxyl moiety will be the last oxidant in solution, since it is not as active as the hydroxyl radical, which will have been consumed too. So any ferric ions will be reduced to ferrous, but there will be no H2O2 to reoxidize the ferrous to ferric.
So H2O2 can reduce ferric ion to ferrous. Ferrous ions can be reoxidized to ferric only as long as some H2O2 is present:
Fe$^{2+}$ + H2O2  --> Fe$^{3+}$  +  $^.$OH  +  OH$^-$
(the first equation in the question of Ref 1).
While the question may be questionable, with the above analysis, answers A and B could be correct. The pH is an important factor because at low pH, complexation of Fe$^{2+}$ occurs, limiting its ability to react with H2O2 (Ref 2).
Ref 1. Is the effect of hydroperoxyl radicals significant in the oxidation of organic compounds in the Fenton reaction? Why or why not?
Ref 2. https://en.wikipedia.org/wiki/Fenton%27s_reagent


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Calculating the apparent molecule length from density of the bulk
I'm interested to estimate the apparent length of the water molecule assuming it could be approximated as a cube.  This is an intentional simplification.
Calculating molecule length from volume in density equation
I used this equation to find the volume for a mole of water molecules with $\varrho$ (water density, assumed $\pu{0.997 g cm^{-3}}$), $m$ (mass), and $V$ (volume).
For the determination of the molar volume, I substituted mass $m$ by molar mass $M$ and rearranged
$$\varrho = m / V$$
into the equivalent form
$$V_{\mbox{molar}} = \frac{M} {\varrho} = \frac {\pu{18.01 g mol^{-1}}} {\pu{0.997 g cm^{-3}}} = \pu{18.06 cm^3 mol^{-1}}$$
Because $\pu{1 m^3} = \pu{10^6 cm^3}$, the molar volume equates to $\pu{18.06 \times 10^{-6} m^3 mol^{-1}}$.
Given Avogadro's number, I computed the volume of the individual  volume with
$$V_{\mbox{molecule}} = \frac{ \pu{18.06 \times 10^{-6} m^3 mol^{-1}}} { \pu{6.022 \times 10^{23} mol^{-1}} } = \pu{2.999 \times 10^{-29} m^3}$$
Then I took the cube root to find the length, as I want $m^3 \rightarrow m$:
$$\ell(\text{molecule}) = \sqrt[3]{\pu{2.999 \times 10^{-29} m^3}} = \pu{3.10 \times 10^{-10} m} = \pu{0.31 nm}.$$
What I need help with
This isn't taking intermolecular forces into account, but I want to know if it's an ok method for getting a rough idea of the molecule length, and if it's done correctly.

Density often is denoted by $\rho$ ($\rho$) or $\varrho$ ($\varrho$ in LaTeX / mhchem syntax available to you here on chemistry.se), which is different to $p$ then typically used about pressure.  But this is smallish compared to the initial equation, because density is not mass times volume, but mass divided by volume (e.g., water, 1 g per cubic centimetre, or about a [metric] ton per cubic metre.  The length you obtain is about twice as wide as the (intramolecular) H-H distance (about 0.15 nm).

Assuming molecules as small cubes is a simplification and it depends a lot on the circumstances if this is acceptable, or not. It may be a start of a model later to be refined (e.g. limiting the number of interactions modelling crystal growth) for water's bent, or benzene's flat shape. For future reference: if you use an equality sign, then both numbers and dimensions must equate, too. This shows  e.g., when they cancel out (dimension analysis) even if this demands a bit more to type / write.

Over time (and repeated exposure) you will pick up skills new to you.  Once you have enough reputation, for example, you may e.g. access to the edits made by other users.  This equally is a source of inspiration (and eventually, training) by reference for how to format questions and answers in the community of chemistry.se (maybe you know the expression of «when in Rome, do as the Romans do»).

Strange this question was closed for the reason given but from things you have said @buttonwood my question is answered (about if it can be used roughly to estimate length), would gladly have given you the answer if your stuff wasn’t in a comment.

For me, I struggled to relate your statements of «calculating the total bond energy» and «intermolecular forces» on one hand, with the target the question addresses, the determination of a length of a molecule, on the other.  The later just (implicitly) assumes inflexible molecules, like stacking dies next to / on top of each other. If the molecules come closer to each other (e.g., compression, external pressure, lower temperature [with exceptions like solid water]), then bulk density simultaneously will increase.

That was just a bit of extraneous info I took it out. I think calculating molecule length in this way assumes highest possible density, I ignored intermolecular forces (affects density) as hydrogen bonding can lead to a decrease in density, but as you rightly pointed out, the shape also matters which is being ignored as well, guess I should’ve specified STP

It actually is fascinating how close you get with this absolutely simplistic approach. I#m not sure why this was closed, I wouldn't have. But I'm also not confident enough to overrule this decision. Also (but not entirely sure how helpful this is): https://www.researchgate.net/post/Molecular-volume-of-8-water-molecules-by-DFT-and-connolly-surface-method

Ah you’re right it’s 0.27nm… since water is in a tetrahedron shape, can you convert cubic units to tetrahedral units? I don’t know if that’s even a thing my maths is weak @Martin


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How can I filtrate Pd2(dba)3 and P(o-tol)3? (Celite filtration failed)
I am conducting an organic synthesis involving Stille coupling, using $\ce{Pd2(dba)3}$ and $\ce{P(o-tol)3}$ as catalysts. Reaction solvent is toluene in which the catalysts also dissolve. Many texts said that celite pad can filtrate these catalysts, but I failed. I tried vacuum filtration using buchner filter, and poured $\pu{0.0087 g}$ $\ce{Pd2(dba)3}$ in $\approx \pu{25 mL}$ toluene. $\ce{Pd2(dba)3}$ in toluene has dark red color, but filtrate also has same color. TLC spot was also in same position. Is there any other way than column chromatography?

Filtration is not going to remove the phosphine, or the phosphine oxide which is also likely to be present as they are both toluene soluble. You are going to have to run chromatography.

Thank you for advice. Does filtration work for Pd catalyst? I saw many papers using Pd catalyst employ Celite filtration. (although I failed)

Celite filtration only removes fine solids, so may take out degraded catalyst. You still have to run chromatography to get rid of tin residues.

Your question is actually two: how do I remove residual catalyst, and how do I remove residual ligand.  The situation is complicated yet further by the fact that both catalyst and ligand are capable of being oxidised, for example the phosphine ligand becoming the phosphine oxide during the course of the reaction due to adventitious oxygen.
As pointed out, Celite only really removes particles - it can often be good enough to carry on, but won’t get rid of everything. You can also try filtering through scavengers - these are resins designed to specifically ligate metals.
On a small scale, the easiest option is to concentrate the reaction mixture and purify by silica chromatography. Depending on the polarity of your compound, it’s often possible to elute the molecule of interest whilst retaining the catalyst/ligand.
On larger scales, chromatography becomes challenging and time consuming. Carefully designed work ups, crystallisations, and precipitations can give pure material with incredibly low ppm values of residual metal.
The key question is always whether the material is clean enough for the next step. Often, the easiest and most efficient option is to filter through Celite, evaporate, and carry on as it only really matters whether your final step is clean (for example if you do a Buchwald and a subsequent Boc deprotection, it can be significantly easier to clean the material up after the second step wheee you can make use of a strongly basic centre).


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Cause of reversible reactions in terms of Gibbs free energy
Heading
I recently learned about the standard Gibbs free energy change of reaction, ΔG=ΔH-TΔS, and how its sign indicates whether the conversion of (ALL) reactants and products is spontaneous or not.
I then learned that the equilibrium constant of a reversible reaction depends on ΔG for some reason.

ΔG=-RTlnK

The equation shows that whether a reaction goes is (noticeably) reversible or effectively goes to completion depends on how -ve ΔG is.
Given the relationship between ΔG and K, is there any explanation of why reactions are reversible in the first place?
If a reaction is exergonic and the 100% conversion of reactants to products would increase total entropy, why shouldn't it go to completion?
If a reaction is endergonic and the 100% conversion of reactans to products would decrease total entropy, why should it occur to any extent at all?

It's because of the entropy of mixing.  If you take a look at the picture in my answer here https://chemistry.stackexchange.com/questions/127674/are-all-exothermic-reactions-spontaneous/127676#127676 , you'll see that the free energy of the system isn't a straight line from reactants to products as the reaction progresses.  For the example shown there, if it were, then the equilibrium point would be all reactants.  Instead, there's a dip, such that the minimum free energy, and thus the equilibrium, lies at a point somewhere between pure reactants and pure products.

For the simplest case, $\ce{A <-> B}$, the entropy of mixing is a maximum when you have half A and half B.  In an ideal system, the free energy of mixing is T times the negative of the entropy of mixing, i.e., $\Delta G_{mix} = -T \Delta S_{mix}$.  So the free energy of mixing is a minimum at the half way point.  To get the actual free energy of the reaction mixture, you need to add the free energy of mixing which, being negative, causes a dip.   In non-ideal systems, you need to add the enthalpy of mixing (assuming constant p), which can be either positive or negative.

Finally, in first-order phase transitions (these are the commonplace ones, like ice to liquid water), you have no entropy of mixing, because they're different phases.  That's why, outside of the coexistence point (where the free energies of the pure phases are equal), phase transitions do go to completion, i.e., you have either all of one phase or all of another; and the phase you have is whichever one has the lowest free energy under the current conditions.

Ok so would it be correct to say that endergonic 'reversible' reactions attain equilibrium because the increase in free-energy due to some conversion of reactants into products is more than compensated by the decrease in free energy due to mixing, making the overall free energy change negative?

Also,  from ΔG=ΔG0 +RTLnQ, could i conclude that the entropy of mixing is RlnQ?

Can you describe a reaction that is both exergonic and that doesn't increase total entropy (with 100% conversion or otherwise)? Exergonic refers to $\Delta G<0$ which under constant T and p is equivalent to saying that for the process $\Delta S_\textrm{total}>0$

Βy total entropy I meant entropy of the universe. And apparently, as theorist points out, there's also free energy of mixing that must be taken into account, so  I guess 100% conversion never maximizes entropy(of universe)?

All reactions are reversible and all reactions are spontaneous.  If they told you differently, then they are too stupid to realize that saying that will just confuse you.  If you start out with just pure reactants, combine them, and let them equilibrate, then some products will be present in the end.  And if you start out with just pure products, combine them, and let them equilibrate, then some reactants will be present in the end. So, as I said, all reactions are reversible and spontaneous.  The question is not "yes" or "no," it is only a question of to what extents.
A reaction being spontaneous "if its standard change in Gibbs free energy is negative" is only a very crude rule of thumb.  All it means is that the equilibrium constant is greater than zero.  But a positive $\Delta G^0$ does not mean that reactants will not form products.  It just means that the reaction is less prone to approach completion.  Note that I did not say that if $\Delta G^0$ is negative, the reaction will go to completion; it will just be more prone to approach completion.

But if all reagents and products are at constant activity then are you not guaranteed complete conversion (or no reaction)?

I am aware that all reactions are reversible to some extent. What I want to know is why? Someone mentioned a gibbs free energy of mixing due to increase in entropy when reactants and products mix. Could you elaborate on that?

Not really.  If the Standard Gibbs free energy change is carried out reversibly using compressors, semipermeable membranes, expanders, and a Van't Hoff equilibrium box, then there is no significant  mixing involved.

@BuckThorn I don't understand your question, or how it relates to what I said.

Is it impossible to envision a situation in which $\Delta _r G$ does not depend on composition (that is, where it remains constant during the entire progress of a reaction)?

@BuckThorn Yes, in a Van't Hoff equilibrium box where reactants are added at the same chemical potentials as the mixture in the equilibrium box, and reactants are removed at the same chemical potential as in the mixture in the equilibrium box.  In that case, $\Delta G_r=0$.  Of course, in this case, the reactants and products are not in their standard states at 1 bar.

In the previous comment, I meant that products are removed, not reactants.

But what about a scenario where reagents and products are solids, each at constant activity, and assuming the substances to be otherwise immiscible. I can take an arbitrary amount of these through a cycle (say annealing them and cooling to obtain products) and in principle convert 100%.

@Buck Thorn Is there such a real world situation?  If they are immiscible, does the reaction not take place in the gas phase?  If so, is there not an equilibrium constant for the gas phase reaction?

Well, I meant that the solid reactants reacted in a melt, upon heating, and otherwise formed separate solid phases.

@Buck Thorn So there is a melt phase reaction that occurs?  If so, is there an equilibrium constant and standard free energy change associated with this melt phase reaction?

I am just looking for a ploy to work around the immiscibility, something like the van't Hoff box. Gas phase would work too I guess. If you ignore the mechanism (how to mix the reagents) and focus on the relative stability of reagents and products, assuming these are constant (as in the van't Hoff box experiment) I can't see how not to conclude that you should not drive the reaction to completion in one direction. In such a situation $\Delta _r G $ is constant until completion of the reaction. Sorry about the double negatives.

Let us continue this discussion in chat.


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Action of Fuller's Earth on Marble and Limestone
Using Fuller's earth is a very common and harmless way to treat marble and limestone that has been damaged/corroded by atmostpheric pollutants such as PM2.5/PM10 carbon particles (Soot) , Oxides of sulphur and nitrogen and Acid rain. Such a treatment has been tried on Taj Mahal in India as well as Delphi in Greece.
I know that fuller's earth usually contains Ion Complexes of Si, Al, and Mg. I would like to know what is the action of these ion complexes on the surface in removing the soot particles from limestone and marble.

No chemical reaction but simple mechanical abrasion and adsorbent. It is a good adsorbent of oil (recall face cleaning with so called Multani matti (=Fuller's earth)  in the South Asia.

@M.Farooq I read somewhere that the OH- ions from the Aluminium Hydrosilicate (the main constituent of fullers earth) can attract the carbon particles and other adsorbed compounds from the surface

Yes, this is a good adsorbent but we cannot pin point what is causing what? The question to consider are, what is the surface chage on Fuller's earth? What is the surface charge of carbon particles.


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In benzyne intermiediate of anisole why -I effect of OCH3 is considered not +M effect of it?

Here they considered -I effect of $\ce{OCH3}$ so carboanion is stable at orthroposition;
But $\ce{OCH3}$ also shows $\text{+M}$ effect if $\text{+M}$ is shown carboanion will be stable at meta position(generally in substitution reaction we give preference to $\text{+M > -I}$ of $\ce{OCH3}$).
But for this reaction NH2 in orthro position is not at all formed.
What is exactly happening in this reaction?

carbon already has 8 valence electrons, so -och3 won't donate anymore electrons through +M (since resonance is impossible), we only consider -I here

@Naren I think one explanation is that the carbanion is in the plane of the ring (in an sp2 orbital) while the delocalisation (+M effect) only occurs via p-orbitals perpendicular to the plane of the ring. So, the resonance does not affect the anion.


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Why is a conformer of hydrogen peroxide rotationally inactive?

Hydrogen peroxide may exist in four different forms depending on the dihedral angle, $d = \angle\,\ce{H-O-O-H}$. The four conformers are: Conformer I with $d = 0^\circ$, Conformer II with $d = 60^\circ$, Conformer III with $d = 180^\circ$, and Conformer IV with $d = 300^\circ$. Which conformer is rotationally inactive?

The answer given to this question is that conformer III is rotationally inactive.
However, what does it mean to be rotationally inactive? $\ce{H2O2}$ doesn't seem to have any steric effects that prevent it from rotating about the $\ce{O-O}$ bond. What would cause the conformer with $d = 180^\circ$ to become rotationally inactive?

"Rotationally inactive" may mean several totally different, unrelated things. Sure, nothing prevents the molecule from rotating about the O−O bond. Now what about rotating the plane of pol$\rm\tiny{ar}$......

@IvanNeretin, do you mean to say, the reason for it being called rotationally inactive is because the dipole moment is zero? I understand that the definition to be applied here is that the compound has no rotational spectrum due to symmetry.

@SafdarFaisal It seems probable to me. The question context would help, but the original may provide no context either.

That would be the third possible meaning, and yes, it might be relevant too.

Rotationally inactive surely means that the discussed conformer has no rotationally active modes. Nothing is about barriers or hindered rotation. The language used is perfectly sensible in spectroscopy. That certainly comes from a section on spectroscopy.

BTW H2O2 doesn't  have any steric effects that prevent it from rotating about the O−O bond, but has stereoelectronic. That makes rotation barrier quite high, but I guess isn't really relevant here.

I think you mean the rotational energy barrier is highest for d=180 since it is the most stable conformer of all (minimising lone pair interactions which govern the bond strength of N,O,F )

"Rotationally inactive" is probably about the rotational spectrum of the molecule
I'm assuming the (unsaid) context of the question is that it is about rotational spectroscopy as this is where the term "rotationally inactive" makes sense.
Rotational spectroscopy–like its better known cousin infra-red spectroscopy–measures absorptions of electromagnetic wavelengths corresponding to transitions in molecules. IR spectra mostly involve vibrations in molecules that have energies corresponding to wavelengths of IR light. Rotational spectroscopy (usually gas phase, as molecular rotations in solids are not usually easy) measures rotational modes in molecules in a similar way (strictly speaking for both techniques absorptions happen when the wavelength/energy of the radiation matches a transition between two rotational or vibrational modes).
But for an absorption to happen the radiation has to have a mechanism to interact with the molecule. For both vibrational and rotational spectroscopy, this can only happen if the molecule has a dipole moment (or, strictly, the relevant vibration or rotation has a changing dipole). While different vibrational models in the same molecule can be "active" (eg the asymmetric stretch in carbon dioxide) or "inactive" (the symmetric stretch in carbon dioxide), in pure rotations there is no way for a molecule without a dipole to have an "active" rotational spectrum as rotation doesn't change the dipole. So only molecules sufficiently asymmetric to have a dipole moment can interact with the radiation. Conformer III of hydrogen peroxide is the only conformation with no net dipole so is the only one which is "rotationally inactive".
The terminology is confusing outside this context as it could be taken to mean something to do with the interconversion of the different conformers by rotation around the central O-O bond (which is likely fairly easy).

(Disclaimer: not an expert in the field.) While the terms 'IR-active', 'Raman-active', 'NMR-active', etc. are common, I've never seen 'rotationally active', not when I did this in undergrad, nor in some of the common textbooks. Hollas, for example, writes: "All regular tetrahedral molecules, which belong to the $T_\mathrm{d}$ point group, may show such a rotational spectrum. However, those spherical rotors that are regular octahedral molecules and that belong to the $O_\mathrm{h}$ point group do not show any such spectrum."

@orthocresol I suspect few have seen much of rotational spectroscopy as it isn't that common. And I'm not entirely sure that textbook is strictly correct (unless it is talking about 2nd order effects which are dependent on vibrations as well as rotations) as Td molecules don't have permanent dipoles so have no pure rotational spectrum.

Oh, I quoted it a bit out of context, sorry: it's describing instantaneous deviations from Td geometry due to centrifugal distortion. IIRC this is not exactly the same effect as rotational/vibrational coupling, but similar idea in that it's not a first-order effect. So Td molecules do have very weak (but nonzero) rotational spectra; though I must confess I've never actually seen one myself.


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Do ambident nucleophiles form functional isomers?
I came across cyanide and isocyanide in an example for functional isomerism. As they are also ambident nuclephiles, I wanted to know if other ambident nucleophiles would also exhibit functional isomerism.
Thanks

Yes, ambident nucleophiles show functional isomerism. Two examples are:

Cyanide ($\ce{CN}$) and isocyanide ($\ce{NC}$)
Nitrite ($\ce{ONO}$) and Nitro ($\ce{NO2}$)

Nitpick: Cyanide is already mentioned in OP's question. Here is a different example: "thiocyanate $\ce{-SCN}$".


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Effect of d-orbital electron shielding on atomic radius
In a standard book it is given:

"Atomic radius of Gallium is less than that of Aluminium. This can be understood from the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm)."

In another chapter, it is given:

"As we move along the period in 3d series, we see that nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner sub shell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly."

In the first para, it is given that the d-electrons shield poorly, but the opposite is given in the second but it is unlikely that the book is wrong. So what am I missing?

See Slater's rules for an empirical evaluation of shielding.

Both sections do not contradict themselves, as they address different shielding aspects.

3d electrons give worse shielding of 4s/4p electrons than 1-3s and 2-3p electrons.
3d electrons give better shielding of 4s/4p electrons than 4s/4p electrons themselves (mutually).


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Mechanism of Wiliamson Synthesis
I was studying about Williamson's Synthesis, and my textbook had the following statement:

Williamson's Synthesis is a nucleophilic substitution reaction and proceeds via $\ce{S_N2}$ mechanism.

Then there was this reaction shown as an example:

It says that the reaction proceeds through a slow $\ce{S_N1}$ mechanism.
Now I searched on the internet and saw that Williamson's synthesis proceeds through $\ce{S_N1}$ mechanism. In that case, what about the shown example? Is what my textbook says correct? Or is there any finer detail?

It depends on the alkyl halide substrate

That means, reactions involving primary alky halides proceed through SN2, while those of tertiary alkyl halide substrates proceed through SN1 mechanism?

Yes you are right

@Nilabja  and secondary halides proceed by both hence the example given

If you got your answer, feel free to self answer it. Self answering is appreciated and it also help future readers.

@Waylander I have read in my textbook (Organic Chemistry for JEE Advanced, Cengage Publications) and in the standard class 12 textbook of India (NCERT) that alkoxide being a strong nucleophile proceeds only through SN2 mechanism. Is there a disparity?

@Waylander in the same context the textbook also mentioned that Williamson synthesis was not a good method for secondary alkyl halides (as in secondary substrate elimination competes with SN2) and also not for tertiary alkyl halides (which don't give SN2 and hence elimination being the only product). Is what I have learnt wrong?

@TRC Saying that Williamson proceeds only by SN2 is going too far. It certainly goes better when SN2 is the dominant mechanism. There are plenty of examples of ether formation between tertiary species and primary alcohols. Search "tritylation of ethanol" for examples, though whether these reactions classify as Williamson is moot.

@Waylander In the general case, does it prefer SN2 to an extent where yields with secondary and tertiary halides are low enough to favour elimination as the major product? The textbook mainly emphasized on questions asking for the best method of ether synthesis, given the ether - and for secondary or tertiary halides the best method mentioned would be alkoxymercuration-demercuration.

You always get some elimination with Williamson conditions but yes, with secondary and tertiary it can become significant.  Williamson works best for systems like methyl or benzyl halides where there are no alpha hydrogens. No-one in the real world uses mercuration - the hazards are too great.  The trichloroacetimidate chemistry is a good alternative.


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While adding simultaneous equilibria , when is the dissociation constant of both reactions the same?
I have a doubt regarding chemical and ionic equilibrium .In reactions involving simultaneous equilibria , instead of analysing the equilibrium simultaneously we sometimes add/substract both the equilibrium and then find net equilibrium constant and solve it easily .But then we assume that both the reactions have same degree of dissociation  (alpha) which is not always true ,Thats why we sometimes get wrong answer while adding the simultaneous equilibria.
SO basically how do we get to know when the degree of dissociation of both rexn being added will be same ?

Adding equilibrium equations has little to do with chemistry, but it is mathematical operation, that may be advantageous at some scenarios. If L1 = P1 and L2=P2, then L1+ L2 = P1 + P2. but there is no assuption of the same dissociation constant. Can you provide more context and particular illustrative examples ? BTW, alpha is not dissociation constant, as it is not a constant at the first place. It is the dissociation degree.

yes sorry its degree of dissociation\

how can i attach images in comments ?

Why not to attach them to the question body ?  // Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it  can be challenge to decipher.  Consider  copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations.


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Is there any condition where alcohol oxidizes in presence of chromium trioxide to give carboxylic acid
My text book says that alcohol in presence of $\ce{CrO3}$ will only give aldehyde. But I then saw this question in which $\ce{CrO3}$ oxidizes alcohol directly to carboxylic acid.
I need to know in which conditions it will give aldehyde and in which conditions it will give carboxylic acid.

https://chemistry.stackexchange.com/questions/43283/mechanism-for-oxidation-of-primary-alcohols-to-carboxylic-acids/43294?r=SearchResults#43294 ... https://chemistry.stackexchange.com/questions/104277/is-there-a-way-to-oxidize-salicyl-alcohol-to-salicylic-acid-without-using-chromi?r=SearchResults

Generally you need anhydrous conditions to get the aldehyde because aldehydes that can form hydrates are further oxidised to the acid. Anhydrous conditions with a complexing agent such as pyridine to aid solubility (e.g. Collins reagent 1) will give good yields of aldehydes.
Aq. acid conditions (known as Jones reagent) 2 give the carboxylic acid
See this answer  How does PCC not oxidize aldehyde?
The use of Chromium reagents has greatly diminished due to their toxicity and waste disposal problems.

Sir, is chromic anhydride in glacial acetic acid or PCC a better reagent to convert alcohols involving C-C double bond to ketones without affecting the double bond? On internet, I found that PCC is not good as compared to first because it can oxidise double bond also . Please tell.

@An_Elephant If this relates to a reaction you actually need to do I would not consider using a chromium reagent at all. If this is a theoretical exercise I would use a very precisely measured amount of PCC in DCM as it will preferentially oxidise the alcohol. I am not aware of any references on PCC oxidising alkenes.This reference says it will not  https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/The_Oxidation_of_Alcohols/Oxidation_by_PCC_(pyridinium_chlorochromate)   Do you have a reference for this?

Sir, unfortunately I don't. In our country there is a entrance examination in which it was asked and the answer key says that first one is best . The reason provided by all the solutions website is first and they say that it is because PCC have chances to reduce alkenes also . Here is a list of those ( they are not official website of exam organiser) : https://www.toppr.com/ask/question/the-best-reagent-to-convert-pent3en2ol-into-pent3en2one-is/

https://byjus.com/question-answer/the-best-reagent-to-convert-pent-3-en-2-ol-into-pent-3-en-2-2/

https://www.vedantu.com/question-answer/the-best-reagent-to-convert-pent-3-en-2-ol-into-class-12-chemistry-cbse-5f626a8896466665ae05aa12

Sir, similarly there are dozens of more. However, the organisers only release answers and nor full solutions. So these cannot be considered as very authentic/reliable. Thanks sir

That sums up the difference between exam chemistry and real world chemistry. Nobody should be using chromium reagents.

Means in theoretical terms, first one is better? I think such questions are just non-sense and should never be asked in exams.


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Molecular orbitals energy under pressure
Suppose we have a hydrogen molecule $\ce{H2}.$ At room temperature the average distance of the nuclei is $r_0$ (without applying pressure).
If we put $n$ hydrogen molecules in a box with volume $V_\mathrm b < nr_0^3$ and the box  isn't compressible, then the hydrogen atoms will be under pressure and will be forced to come closer than $r_0.$
Does that change the energy of the molecular orbitals, and if so, how?

Yes, by complicated way, due complicated potential field.

Think about the other extreme - if enough pressure is applied you will get solid H2, and so the individual orbital energies of the isolated molecules have morphed into the bands of the solid. In between you will get something in between, so as pressure is increased the orbital energies will start to spread under inter-molecular interactions into wider and wider bands as the pressure increases

You would have a particle in a box with infinitely high walls inside which is the potential corresponding to the H2 molecule (H$_2^+$ would be simpler) the walls come into the calculation as boundary conditions on the integrals. There are many text-book examples of PIB's with potentials inside.


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Relation between common ion effect and the temperature of the solution
Adding extra Na+ in a saturated NaCl/water solution decreases the solubility of NaCl. How does this exactly happen? AFAIK, the solubility of a solid only decreases with the temperature of the solution. Am I to assume that when I put Na+ in a saturated NaCl/water solution the temperature of the solution somehow drops? If that is what happens, how does it actually happen? If I add extra Na+, the extra Na+ will react with Cl- and will create precipitated NaCl. AFAIK, forming new bonds is exothermic. According to that, the temperature of the system should actually increase and the solubility should increase as well. What exactly is happening here?

"solubility of a solid only decreases with the temperature" - for solubility in pure solvent that's a decent approximation. If you add another salt that's no longer pure.

You are mixing up different effects. The effect of temperature is determined by the enthalpy of solution of the salt. When you examine the effect of another salt (common ion effect) assume the temperature is constant unless told otherwise.

@aNoNyMoUs When you say "the solubility of a solid only decreases with the temperature", it is only true for solution of pure compounds. It is not valid for solutions containing more than one compound. For such mixtures, specially if the solutes have a common ion, a different calculation must be done. The solubility has to be calculated by the solubility product.
For example, at $25$°C, $\pu{1000 mL}$ water can dissolve $\ce{370 g NaCl (6.33 mol}$). So $1$ Liter of such a saturated solution contains $\ce{6.33 mol Na+ + 6.33 mol Cl-}$. We'll see later that the so-called solubility product $\ce{[Na+][Cl-]}$ is important and has the numerical value $\ce{K_s = [Na+][Cl-] =  6.33^2 = 40.1 mol^2 L^{-2}}$ It should be stated that the dissolution of this amount of NaCl is supposed not to have modified the volume of the solution, which is of course a rough approximation.
If now  $\pu{1.00 mol}$ of a more soluble salt like $\ce{NaNO3}$ is added to this solution, the concentration of the ion $\ce{Na+}$ per Liter increases up to $\pu{6.33 + 1 = 7.33 mol/L}$. In such a case, the thermodynamical laws of equilibrium require that the concentration of the other ion (here $\ce{Cl-}$) must decrease in such a way that the solubility product $\ce{K_s = [Na+][Cl-]}$ remains constant. So $x$ moles par liter of $\ce{Na+}$ and $\ce{Cl-}$ must appear as a deposit of $\ce{NaCl}$ in the container. The solubility of the ions $\ce{Cl-}$ must go down to : $\ce{[Cl^-| = 6.33 - x mol/L}$, and $\ce{[Na+] = 7.33 - x}$, in such a way that $\ce{[Na+][Cl-] = K_s = 40.1 mol^2 L^{-2}}$. This a $2$nd degree equation whose solution is : $\ce{x = 0.48 mol (28 g)}$.
Si the final concentration of the ions are : $$\ce{[Na+] = 7.33 - 0.48 = 6.85 mol L^{-1}}$$ $$\ce{[Cl-] = 6.33 - 0.48 = 5.85 mol L^{-1}}$$ $$\ce{[NO3^-] = 1.00 mol L^{-1}}$$

And what will be the solubility after all of that?

@aNoNyMoUs. The solubility of $\ce{NaCl}$ is $\pu{5.85 mol/L ; 342 g/L}$). This is less than in pure water ($\pu{370 g/L}$)


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Quantum chemistry Python package to solve the Coupled-Perturbed Hartree–Fock equations
Two of the most well-known Python quantum chemistry pakcages, PySCF and Psi4, can solve the Hartree–Fock equations. However, I am interested in finding analytic derivatives of the electron integrals:
$$h_{pq}(R) = \displaystyle\int dx \phi_p(x)^{*} \left( -\frac{\nabla_r^2}{2} - \displaystyle\sum_{I} \frac{Z_I}{|r - R_I|} \right) \phi_q(x)$$
$$h_{pqrs}(R) = \displaystyle\int dx_1 dx_2 \frac{\phi_p(x_1)^{*} \phi_q(x_2)^{*} \phi_r(x_2) \phi_s(x_1)}{|r_1 - r_2|}$$
with respect to the nuclear coordinates. To do this, one must solve the coupled-perturbed Hartree–Fock equations. It doesn't seem like Psi4 and PySCF have this functionality, so I was wondering if there was any other package I can use to do this?

Hello, and welcome to Chemistry! I see you've already participated on Matter Modelling, so you'll be aware that this question would also fit there. That doesn't mean it's off-topic here, it's perfectly fine; but in the event that you ever want to move it there, feel free to let us know. (But please don't double-post on both sites.)

Just checking the obvious: What order derivatives do you want? If it's only first order you don't need a CPHF solver.

Related question on Matter Modeling. As mentioned there, any program that can do geometry optimizations probably has a way of generating the gradient/hessian (or at least relevant matrix-vector products that are needed for the optimization). So I suspect Psi4 and PySCF have these quantities, its just a matter of it they provide an interface to access them.

@IanBush First and second order. What can I do other than CPHF to get the first-order derivatives?

@Tyberius Thanks, the thing that I'm interested in isn't just getting the Hessian of energy gradients, but rather the derivatives of the Hamiltonian itself.

@JackCeroni But I mean if these programs computing the geometric Hessian of energy analytically, they would need the 1st and 2nd derivatives of these integrals as part of that process.

Are these packages calculating the gradients/Hessian analytically? I thought that it was done with finite difference.

@JackCeroni For first order there's a standard trick involving differentiating the orthonormality relation to relate changes in the coefficients to changes in the overlap matrix. Any good book on quantum chemistry should cover it, for instance section 3 in appendix C of Szabo and Ostlund. Also see https://chemistry.stackexchange.com/questions/89831/analytical-gradient-of-the-hartree-fock-nuclear-electronic-repulsion-term

Thanks! And as for the second order derivatives, I assume there is no “trick”?

Yes, for second order or greater you need a CPHF solver of appropriate order.

I see, and do you know of any quantum chemistry packages that have such solvers implemented?

I'm a co-author of CRYSTAL (https://www.crystal.unito.it/index.php) which has one, but it's not open source. Most large scale packages will have something. Anyway, bed time.

Are you looking for only python based softwares? And do you need only the hessian matrix (i.e. force constat matrix for atoms) or do you also need the values of the derivatives of electron integrals calculated to get the hessian? (What I am asking is—do you need just the end result or also the intermediate steps?)

Python-based software would be ideal, but anything that gives me the derivatives I'm looking for works (for instance, something like GAMESS). As for the second point, I'm looking for the derivatives of the electron integrals, not the Hessian itself. Specifically, I'm doing a research project, involving new methods of computing the Hessian, and one of the things I require is the first and second derivatives of these integrals.

I think a lot of electronic structure software packages will have what you need, but since people don't typically ask for derivatives of the electron integrals individually (they just want the end result, like the Hessian or the geometry optimization), the codes won't let you output these by default. You might have to go into the code and add some print statements. I also think you should consider the worth of working on new methods to calculate the Hessian, because quantum chemistry has been around for a century and people have been working on ways to calculate Hessians for about as long.

Makes sense, thank you @user1271772. There definitely must be these capabilities in quantum chemistry packages, but I've done some digging (jn PySCF, in particular), and can't seem to locate the specific part of the code where these derivatives are calculated. Do you have any suggestions on how to find this functionality?

I typically use the grep function. I think it's doable, but the effort-to-value ratio is immense. Someone may help you with this if the Bounty is large enough. Quantum chemistry codes involve millions of lines. In PySCF, Qiming Sun alone has added content to 1.5 million lines and removed content from 1.6 million lines. Be very careful about quantum computing projects, especially ones that involve quantum chemistry.

Related: https://mattermodeling.stackexchange.com/q/3787/5

As many of the comments here and on your question at MMSE suggest, pretty much all major electronic structure software can solve the "coupled-perturbed Hartree–Fock equations" as Ian's comment points out that it's needed for the second derivatives, and most (if not all) mainstream electronic structure packages use the Hessian matrix (involving the second derivatives) to do geometry optimization.

Ian Bush mentioned here that CRYSTAL can do it, but is not open source.
My comment here mentioned that OpenMOLCAS can do it.
Susi Lehtola's answer here mentioned that PySCF and Psi4 can both do it, and that Psi4 allows you to print out the Hessian matrix by default (i.e. without adding your own print statements) and that the libcint library that PySCF users, or the libint library that Psi4 uses, can offer you the derivative integrals that you want.
The fact that the above "constellation" of software packages all offer this functionality, makes me convinced that all major electronic structure programs can calculate those derivative integrals and solve the CPHF equations (though the ease of accessing them may vary).


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Which step is the rate determining step in Benzyne mechanism - Elimination or Addition?
I have this doubt for a while now:
How can we compare the reactivity of two compounds towards benzyne mechanism, which have different leaving groups (E.g. chlorobenzene and bromobenzene)?
To know this, I think it is necessary to know if the elimination step is the rate determining step, or if it is the addition step.
I did some research about this but wasn't able to find any satisfactory information. Could anyone please explain which step is the rate determining step?

I would believe elimination.

I agree, given that the benzyne is a highly unstable structure I think the addition will be fasr

@Waylander Oh, well that does make a lot of sense; Thank you very much for your explanation

For a benzyne mechanism formation of benzyne intermediate is always the slow step or rate determining step but unlike normal nucleophilic addition elimination reactions formation of final products in benzyne mechanism is very different.
Here the nucleophile simply doesn't go and attack the leaving group but instead the nucleophile of the base obstructs the proton adjacent to leaving group which then triggers the benzyne formation and simultaneously the leaving group leaves.
This kind of mechanism is only shown by halo benzenes and it is an elimination-addition type mechanism not an addition-elimination type.
The reason we consider the elimination-addition mechanism is because the addition of NH2 group takes place after elimination of leaving group.
Interestingly this reaction was one of the main cause for discovery of benzyne, even when this mechanism was first suggested, benzyne's existence wasn't confirmed until the use of spectroscopic methods came into existence but there was another reaction in which benzyne intermediates were formed, that is a kind of sigmatropic reaction known as Diels alder reaction, it suggested the existence of benzyne, as it formed an entirely different compound from what was expected to form, that is called Diels alder adduct where the triple bond of benzyne acted as dienophile.
Here's a quick mechanism for formation of benzyne intermediate

Another important thing to remember is that this reaction occurs at very low temperature approximately -33 C which is why the suggestion of addition elimination mechanism wasn't accepted, as those kind of reactions occur at relatively high temperatures.
So yes if we have to compare chlorobenzene and bromobenzene the rate of reaction occurs faster with bromobenzene ,as it is a better leaving group.

1)When there is a ch3 group with halo benzene the major product is meta product.

2)An electron withdrawing group gives para as major product.

The reason ortho product dominates is because the nucleophilicity of the carbon at that position increases due to the presence of electron withdrawing group, which makes it harder for NH3 to attack at that position and this can be easily seen by drawing the resonance structures.
I hope this helps !

Please edit your post for grammar and style. MathJax is preferred for typesetting. If you can clean-up your figures and make them more legible, that will help.

Can you please consider using some punctuation?


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Are oxides of sulphur and carbon adsorbed into limestone/marble?
Oxides of sulphur and carbon are commonly present in the atmostphere in areas close to industries, can such oxides get adsorbed into the surface of limestone and marble and cause discolouration and corrosion ?
If yes, then what amount of such oxides will get adsorbed into the surface over time as the amount of avalible surface for adsorption varys for every material.

Study the concept of acid rain. There is not adsorption but these oxides react with moisture in the air and form acids. Yes, limestone gets damaged.

$|ce{CO}$, $\ce{CO2}$ or $\ce{SO2}$  are not absorbed on limestone or marble. But if $\ce{CO2}$ or $\ce{SO2}$ is dissolved in water, their solutions can react with marble or limestone, but these reactions are rather slow, according to $\ce{CaCO3 + CO2 + H2O -> Ca^{2+} + 2 HCO3^-}$ or $\ce{CaCO3 + SO2 + H2O -> Ca^{2+} + HCO3^- + HSO3^-}$.


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Looking for a compound that when mixed with sugar will burn when a drop of water is added
Years ago, I watched a Johnny Carson episode where a guest on the show mixed a compound with ordinary table sugar and added a drop of water and the mixture burst into flames.  I went and bought the stuff at a chemical supply and we had a lot of fun with it.  Over the years I have tried to remember what the compound was.  I think it was a sodium or potassium compound with a trioxide $\ce{O3}$ at the end but not sure about that.  We would mix the compound $50-50$ with sugar and adding 1 drop of water caused a reaction that flamed up pretty good. Does anyone know this trick?  I would love to try this again.

You can mix sugar with potassium chlorate ($\ce{KClO3}$) and then pour a few drops of sulfuric acid on it to create a fire: https://www.angelo.edu/faculty/kboudrea/demos/instant_fire/instant_fire.htm . This is a very violent and unsafe reaction though.

@ShoubhikRMaiti I remember using these mixtures with metal salts as "bengal fire" effects, when I assisted my professor to prepare effects for TV, when I was studing at a chemical high school  in early 80s. Not sure about mixtures with just water drops. It is applicable for iodine + aluminium dust, but it produces iodine vapours, not flame. Perhaps mixture sugar + KClO3 + solid organic acid ? But it would be unsafe to keep.

@Poutnik The website says sulfuric acid is a catalyst. My guess is that sulfuric acid would carbonify some of the sugar, which would then be oxidised by KClO3. In that case, would an organic acid work? I am not really sure though. The iodine and aluminium idea seems more promising, this video shows that a flame can be seen in that reaction: https://www.youtube.com/watch?v=RclM2S2YIBg

@ShoubhikRMaiti Hm, I would rather say $\ce{H2SO4}$ partially converts $\ce{KClO3}$ to much more reactive $\ce{HClO3}$. Even $\ce{HClO4}$, which is less reactive than  $\ce{HClO3}$, ignites organic matter. // I2 + Al, the amount of mixture of water temperature may matter.

No, this was a simple mixture of sugar and the granules and 1 drop of water would set it off. No acids or anything like that. The one time we put in a pvc pipe, it nearly blew the guy's hand off. I've seen sugar rocket engines on YouTube - it is similar to that type of reaction. I made this post almost 3 years ago and was googling and came across it.

It is probably a mixture of sodium chlorite $\ce{NaClO2}$ and ammonium thiocyanate $\ce{NH4SCN}$. It takes fire with the least drop of water. But the mixture is difficult to do, because these two products have to be totally dry, and ammonium thiocyanate is rather hygroscopic. If they are even a bit wet, they react and take flame by simple contact. It is even safer not to try to mix the products. Just add some sodium chlorite to a pile of ammonium thiocyanate in a crucible, without touching anything. And add one drop of water. The fire lasts one or two seconds, producing some $\ce{NaCl}$ smoke and leaves a yellow deposit of sulfur. Mind ! Don't confuse sodium chlorite $\ce{NaClO2}$, with sodium chlorate $\ce{NaClO3}$, which is more common. The reaction is :$$\ce{2NaClO2 + NH4SCN -> 2NaCl + S + 2 H2O + N2 + CO2}$$ If you add some more sodium chlorite to the hot mixture, the sulfur will burn according to : $$\ce{S + NaClO2 -> SO2 + NaCl}$$


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Identifying products
How to determine the major and minor products in a reaction? What factors are useful to do so?
For ex: Nitration of phenols, gives the ortho nitro phenol as the the major product and para nitro phenol as the minor, while Nitrosation of phenols give para nitroso phenol as the major product and ortho nitriso phenol.

This is a very broad question as the major product of a reaction may depend upon multiple factors each different for each reaction.

It is almost like asking How to learn organic chemistry from a 1 page answer ? The  right answer is even shorter: Study many textbooks, books and articles and gain years of experience in chemistry. // I guess the 80-20 rule is applicable even here: Understanding 80% of cases takes 20% of effort.

You won't know until you run the reaction. You may be able to predict the likely course of the reaction based on pre-existing literature and you may be wrong.

The main factors used to determine the major and minor products are:

Stability of products
Kinetics of the reaction
Reaction conditions
Catalyst used , etc.

I just gave out a few examples but, there are hundreds of other factors which you can only learn by practice.


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Calculating pH of aqueous ammonium hydrogen sulfide solution
I was doing the following question yesterday:

Calculate the pH of a $\pu{0.1 M}$ solution of $\ce{NH4HS}$, given $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$, and $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ of $\ce{H2S}$ are $7.04$ and $11.96$, respectively.

My approach
The following equilibria exist in the solution
$$\ce{NH4HS -> NH4+ + HS-} \tag{Eqn:1} $$
$$\ce{NH4+ + H2O <=> NH3 + H3O+} \tag{Eqn:2}$$
$$\ce{HS- + H2O <=> H2S + OH-} \tag{Eqn:3}$$
$$\ce{HS- + H2O <=> S^{2-} + H3O+} \tag{Eqn:4}$$
$$\ce{H2O + H2O <=> H3O+ + OH-} \tag{Eqn:5}$$
Setting up the equations

Material Balancing:

$$\pu{0.1 M} = \ce{[NH_4 ^+] + [NH_3]} \tag{Eqn:M1}$$
$$\pu{0.1 M} = \ce{[H2S] + [HS-] + [S^2-]} \tag{Eqn:M2}$$

Charge Balancing:
$$\ce{[H3O+] + [NH4+] = [OH-] + [HS-] + 2[S^2-]} \tag{Eqn:C}$$

Writing $K_\mathrm{eq}$ expressions:

For $(\text{Eqn:2})$, $$\frac{K_\mathrm{w}}{K_\mathrm{b}} = \frac{\ce{[NH_3}][\ce{H_3O^+}]}{\ce{[NH_4^+]}}$$
For $(\text{Eqn:3})$, $$\frac{K_\mathrm{w}}{K_\mathrm{a1}}  = \frac{[\ce{H_2S}][\ce{OH^-}]}{[\ce{HS^-}]}$$
For $(\text{Eqn:4})$, $$K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H_3O^+}]}{[\ce{HS^-}]}$$
For $(\text{Eqn:5})$, $$K_\mathrm{w} = [\ce{H_3O^+}][\ce{OH^-}]$$
These equations are driving me crazy and any approximation(s) that can be used to reduce number of variables here are appreciated.
Answer

 $\ce{[H+] =\sqrt{K_{a1}.[\frac{K_\mathrm{w}}{K_\mathrm{b}} + K_\mathrm{a2}]}}$
$\implies$ pH =$8.14$

I have seen this question but in that case, the $\ce{NH4+}$ doesn't undergo hydrolysis.

Actually, in given example, $\ce{HSO4-}$ has under gone hydrolysis to give the final pH. The hydrolysis of $\ce{NH4+}$ is the one in question.

Consider simplifications based on assumed strong inequalities. E.g. the trivial formula for pH of weak acid solution pH=1/2*(pKa - log c ) assumes  c >> [H+] >> [OH-]

Second pKa of H2S is much lower than that and even if this value was correct, it should still be ignored.

@Mithoron The answer has Ka2

I did some calculations and we just have to prove $\ce{\frac{[NH_4^+]}{[NH_3]} = \frac{[HS^-]}{[H_2S] - [S^2-]}}$

@M.L How to prove it?

I don't know. I was just writing down what I was able to do.

You can neglect HS- as an acid, as the Ka2 is too low. With some simplifying assumptions, you can consider the system consisting a weak acid NH4+ and a weak base HS- and the hydrolysis balance [NH3] = [H2S] >> max( [H+],[OH-])

From here:
$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b})$$
Plugging in $\mathrm{p}K_\mathrm{a} = 7.04$ and $\mathrm{p}K_\mathrm{b} = 4.74$, you get $\mathrm{pH} = 8.15$. As mentioned by Poutnik and Mithoron in the comments, $\ce{HS-}$ can be neglected as an acid due to a low $K_\mathrm{a2}$ value. So this is just the classic case of the salt of a weak acid and a weak base.


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How to estimate temperature loss of a mixture when we know the heat loss due to vaporisation of one of the component?
So I am modelling in Chemkin and want to include phase change phenomenon in my model. But there's no capability of phase change in Chemkin so I have to think of some out of the box method for this.
So I have a mixture of H2O(g), O2(g) at 1080 K and this mixture mixes with C2H5OH(l) at 298 K. Now in the real world, at this temperature C2H5OH(l) will boil and form C2H5OH(g) as ethanols B.P is 351 K. The process is happening at 10 atm. I know the latent heat of vaporization of ethanol, so how do I calculate the temperature drop of this overall mixture?
So I will know the total amount of heat loss caused by 'm' amount of ethanol(l) boiling by applying
Q=m*hvp, where hvp=latent heat of vaporisation.

One approach I can think of is, we know total enthalpy of mixture at say t=0 sec just before mixing and know the amount of heat loss, Q, so we know total enthalpy at say some t= x sec after mixture has come to equilibrium state, by applying
'total enthalpy at x sec = total enthalpy at t=0 sec - Q'.

So do I just use total enthalpy = sum of individual enthalpies of each component and try out different temperatures till I get the desired total enthalpy at x? Will it be as simple as this or there's some catch in between? Any help will be appreciated.
P.S-( Chemkin assumes every species is a ideal gas so enthalpy is only a function of temperature, neglecting effect of pressure and entropy).
Also can someone put out a link for syntax while writing chemical symbols, I searched in the help section but couldn't find anything.

Treat ethanol liquid and ethanol vapor as separate chemical species, with different heats of formation that undergo a reversible chemical reaction with the liquid species producing the vapor species.  Heats of formation of these items are available in tables.

Hi Chet, thanks for replying. So I already have tried that step. I added a phase change step with the activation energy as the latent heat of vaporization, and found the pre-exponential factor by matching the output with experimental data for vaporization of that liquid species. And add heat loss to the reactor equal to the  heat of the  endothermic reaction. But chemkin is having convergence issues with it's solver when I add this step to the detailed mechanism of the species in the gas phase.  So I am trying the approach in the question. Do you think it will work?

I think I was too hasty when I gave you the advice that I gave.  After further consideration, what I would do would be to take the total enthalpy of the liquid ethanol and the gas mixture in the initial state, and, for final state, assume that all the ethanol has vaporized while the enthalpy of the new mixture is the same as the initial.  That sounds pretty much what you did.

How will the enthalpy of the new mixture be same as that of the old mixture?. Shouldn't it less by the heat loss due to latent heat of vaporization?

The increase in ethanol enthalpy by vaporization and temperature increase is equal to the enthalpy decrease of the water-oxygen mixture by temperature decrease, so the change in enthalpy of the combination is zero.


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How Does Internal Resistance of the Galvanic Cell Affect Voltage Drop of the Cell?
When galvanic cell operates or when current starts flowing in a circuit, voltage drops from electromotive force of the cell (EMF) because of internal resistance of the battery.
There are three sources of internal resistance or sources of polarization:

Ohmic polarization
Activation polarization
Concentration or Diffusion polarization

Working voltage of the cell is given by following equation:
$$ E = EMF - I(R_o+R_a+R_c) $$
where R represents polarizations mentioned and I is current in the cell.
EMF of the cell is created because of charge separation in electrochemical double layer on anode and cathode due to electrochemical processes of certain species or redox systems present. More charge is separated, bigger the absolute value of electric potential developed on electrode. EMF of the cell can be calculated from Nernst equation if equilibrium is established on both interfaces/electrodes.
My question is how do these three types of polarizations mentioned affect amount of charge separated on interfaces when current starts flowing in a circuit physically speaking? What processes happen on electrodes so that voltage of the cell drops from EMF value?
For diffusion polarization that is quite clear since if activity of some species like ions is different close to electrode compared to bulk of electrolyte if diffusion is slow process, from Nernst equation electrode potential is different compared expected value calculated from activity of species in the bulk.
If we look at ohmic polarization it seems that electrode potential developed is affected by ohmic resistance of conductors in the cell. Since voltage of the cell drops, this would mean that electric potential of cathode decreases while of anode increases compared to Nernst equation values. Why is it that potential of cathode drops while of anode increases?
Activation polarization represents resistance due to kinetics of electrochemical reactions. It also decreases voltage of the cell. Slower the reaction, bigger the voltage drop of the cell compared to EMF. What does activation resistance have to do with charge separation in double layer and why does it decrease cathode potential and increase anode potential?

Lots of questions, but in regard to internal cell resistance, you can model the system as an ideal voltaic cell in series with the internal resistance. The open circuit voltage is just the ideal cell voltage, as per Nernst. But if significant current is drawn, then you have a de facto voltage divider: the cell resistance and the external load resistance. So your measurement gives the voltage across the external load. No need to worry about the individual electrodes: only the difference matters.

After Ed V, you should simply determine  the tension measured on the cell before ($\ce{E_0}$) and after connecting the cell to an external resistance $\ce{R_1}$. The cell produces  now  a tension $\ce{E_1 < E_0}$ and a current $\ce{I_1}$ . The inner resistance $\ce{R_0}$  is given by : $\ce{\frac{E_0 - E_1}{I_1} - R_1}$

@Ed V, yes, however why is it that voltage drops from EMF, how does activation resistance for example affect voltage of the cell and why would it drop?

It has to. Every voltage source, such as a Daniell cell, has an internal impedance that is, in general, complex. Even ignoring capacitance and just looking at the internal resistance, that resistance comes from the solutions in the cell reservoirs and the salt bridge. There must be a complete circuit, including the resistance of the solutions and that is not accounted for in the Nernst equation. Think about two cells with different salt bridges: one short, with large cross section, and the other very long and narrow. Then the latter cell has high cell resistance and cannot supply much current.

Yes, that holds. What I am interested is what physically happens so that voltage of the battery drops since EMF is created due to charge separation in double layer on both electrodes. Why is amount of charge separated in double layer affected by activation and ohmic resistance? What processes and how do they affect this?

This is just a circuit problem! The electrical resistance of the solutions results in a voltage drop as current passes through the cell solutions. With no current flow, there is no change at the electrodes. In any event, you never measure the ideal voltage, only the voltage of the ideal source in series with the series impedance. You need to get a good grad level book on electrochemistry and read it.

I do know what you've written only it didn't really answer my question. I do have some good textbooks in electrochemistry. Any suggestions regarding this problem?


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Can alkali-doped exohedral fullerene molecules show superconducivity?
Alkali doped fullerenes are one of the few organic materials that show superconductivity. However, when attaching a new molecular system to a fullerene doped with an alkali metal, will it still show superconductivity?

Attaching what system? Most of stuff you'd attach would react with an alkali metal.

These things are probably best formulated [M+]3 [C60]3- , at least according to Housecroft and Sharpe. There is also says the K, Rb and Cs salts are superconducting at low T, but not the Na case. So it seems perfectly reasonable to think about what happens if you add something onto the fullerene - though my guess is that as this is addition rather than substitution you are too radically changing the electronic structure to see the same effect. Fullerenes are not benzene.

Superconductivity is a tricky thing, you can't expect to predict it with qualitative considerations.


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Removing hydrogen bubbles by reacting them away
I have a microfluidics problem where hydrogen is forming tiny (nanometer-scale) bubbles which cause problems. The hydrogen is a reaction byproduct that comes from n etch between silicon and TMAH. The bubbles essentially cause etch masking.
We are trying physical/mechanical means of removing the bubbles (lowering pressure, vibration, etc.) but I wonder if we could add something to the etch solution which would react with the hydrogen and form something soluble, thus prevent formation of the gas-phase hydrogen bubbles.
So far I found suggestions for forming various metal hydrides, but metals are considered contaminants. Also I guess sodium will react with it, but sodium isn't allowed either. We could add oxygen to the solution to create H2O but free oxygen in the solution would probably cause oxidation and impact the process. A lot of things will react with hydrogen though...I wonder about citric acid or IPA? Any chemical which would react away the hydrogen.

Hydrogen isn’t quite as reactive as you might think - mixing it with oxygen for example doesn’t instantly create water.

Adding to what @NotEvans. said, no go for hydrides, sodium, citric acid or either IPA (India Pale Ale or isopropyl alcohol). Maybe an ultra thin palladium membrane to let the hydrogen diffuse though and be safely removed.

I don't see how a membrane would be compatible with etching. All in all, @BetterSense, this is a genuinely tough problem, so don't expect an easy shortcut. Very few things react with hydrogen at room temperature (in particular, none of those you've mentioned).

How about increasing pressure to dissolve the hydrogen? Perhaps there is a solvent compatible with the etchant that could help dissolve the bubbles... see https://www.bre.com/PDF/Solubility-of-Hydrocarbons-in-Physical-Solvents.pdf for some solvents.

Various methods for improving the alkaline etching of silicon are given in this open source by Pal et al. [1]. In the section on chemical additives they describe the effects of hydroxylamine, $\ce{NH2OH}$, which produces an oxidizing radical that reacts quickly. In the case of a TMAH solution they report from Swarnalatha [2] that 5% (not higher) TMAH plus 10% hydroxylamine gives a maximal etching rate.
The authors also quote results from Schriffrin et al. [3,4] demonstrating that suppression of hydrogen bubbles by adding oxygen or isopropanol (the latter acts as a surfactant causing the bubbles to detach) is effective in producing a smoother etched surface.
References
1.
Pal, P., Swarnalatha, V., Rao, A.V.N. et al. High speed silicon wet anisotropic etching for applications in bulk micromachining: a review. Micro and Nano Syst Lett 9, 4 (2021). https://doi.org/10.1186/s40486-021-00129-0
2.
Swarnalatha, V (2019). "TMAH based high speed silicon bulk micromachining", PhD thesis, Indian Institute of Technology Hyderabad, India.
3.
Campbell SA, Cooper K, Dixon L, Earwaker R, Port SN, Schiffrin DJ (1995). "Inhibition of pyramid formation in the etching of Si P(100) in aqueous potassium hydroxide isopropanol."  J Micromech Microeng 5:209–218.
4.
Baum T, Schiffrin DJ (1997) AFM study of surface finish improvement by ultrasound in the anisotropic etching of Si <100> in KOH for micromachining applications. J Micromech Microeng 7:338–342.


text:
SN1 reaction of alkyl halides
I was going through Peter Sykes organic chemistry where I came across a section where the author is saying:

For halides that do not readily undergo SN1 attack ,this can be promoted by use of silver salt of halogen as $\ce{Ag+}$ promotes carbocation formation by precipitating silver halide, $\ce{AgX}$.

So I am confused what is the role of $\ce{Ag+}$ ions in enhancing and why does precipitation of silver halide occurs? Also can this reaction be utilized to obtain primary carbocation? Lastly we know that primary  carbocation is very unstable and thus has a high energy of activation so is this energy provided by the lattice enthalpy of silver halide?

During $\ce{S_N1}$, there is an equilibrium step where the leaving group leaves, in this case, a halide. This is the mechanism for an alkyl bromide substitution.
What $\ce{Ag+}$ allows is the precipitation $\ce{Br-}$ anions produced during the equilibrium step and therefore shift the equilibrium more to the right (the side with the carbocation) according to Le Chatelier's principle. This will promote $\ce{S_N1}$ reactions.

So can we obtain a primary carbocation by this procedure?

In solution, Sn1 and Sn2 reactions occur pretty much simultaneously I believe. In general, for primary carbocations, Sn2 will predominate, about 1000 times the rate of sn1. Adding Ag+ may increase the rate of Sn1 but I would say not too much. Again, there are many factors to this: solvent, ksp of Ag-Hal, any other functional groups, etc. However, I think Sn2 would still be predominant for primary halides even with Ag+ added. I could be wrong though.