text:
Do most chemical reactions happen at constant pressure?
While studying enthalpy $H=U+PV$ and its changes, I realized I am not clear on the following: chemical reactions happen with the external pressure being constant and equal to the atmospheric pressure $P_{atmos}$. But, in general, the system's internal pressure $P_{gas}$ would seem to be different from $P_{atmos}$ and variable during a reaction...
The change in enthalpy $\Delta H= Q$, where $Q$ is thermal energy, only when the gas internal pressure matches the environment external pressure: $P_{gas}=P_{external}=constant$ Under those circumstances, the emitted or absorbed thermal energy $Q$ goes into change #H# while the gas expands/contracts. The work $PV$ done by the gas is perfectly matched by the $PV$ work done on it by the environment.
How common is for the gas internal pressure to be constant and equal to $P_{atmos}$ during chemical reactions to justify $\Delta H = Q$?

What are you talking about? Either reaction is conducted in open reactor - obviously atm. pressure, or closed one with pretty much whatever parameters.

Your gas/diesel engine certainly does not...

If I understand Jon's comment correctly, the gas internal pressure is not constant during the reaction as it can happen in other reactions. Certainly, the external atmospheric pressure is constant if it happens in open air.

What happens during a reaction is important insofar as reversibility is concerned (a reversible path being a special way of performing a reaction in which equilibrium is sustained throughout), but not when computing $\Delta H$. The condition $\Delta H = q$ is derived from the basic law of conservation of energy and the definition of enthalpy. Because H is a state function, to compute the change in its value you need to know only the initial and final states of the system (and of course the values of H corresponding to those states). For $\Delta H = q$ to hold requires that (1) only expansion (pV) work be performed and that (2) the initial and final pressures of the system equal each other.
You can dream up scenarios in which you start at mechanical equilibrium (balanced pressure) between surroundings and system but end up at a constrained system volume (rigid). For instance, if an explosion moves a piston to a constrained setting within a cylinder. In that case it is not true that the enthalpy change and heat are equal. On the other hand, if the explosion is constrained such that the volume can expand until pressures balance (and external constraining pressure is constant), then enthalpy change and heat are equal.


text:
How do I calculate the amount of preservatives one would need for a cleaning solution?
I plan on making a very simple cleaning solution to clean metal contacts on PCB boards. It cannot contain anything that is flammable or hazardous so I've come up with the solution below.

Distilled Water ~92%
Ethoxylated Alcohol (CAS 68439‐46‐3) ~5% "surfactant"
Tetrasodium Glutamate Diacetate (CAS 51981-21-6) ~1% "chelating agent"

I am not making this solution in a perfectly sterile environment so I would like to add a preservative to increase shelf life. I've been researching for weeks and I cannot find anything definitive. I just want something that will easily dissolve into the solution above and not be harmful to metal. The solution will be applied to the metal via a cotton swab. Even if I find a suitable preservative how would I go about calculating how much to use?
My ideas for preservatives:
I think Citric Acid (CAS 77-92-9) would be okay. I think a combination of Methylchloroisothiazolinone (CAS 26172-55-4) and Methylisothiazolinone (CAS 2682-20-4) would be okay as well.
Other concerns are: Do I need an emulsifier? Is there an additional non-hazardous chemical I can add to decrease drying times?

What is *ethoxylatd alcohol" ? Is it ethanol ?

@Maurice some weird ether mixture according to the CAS number https://www.chemsrc.com/en/cas/68439-46-3_907994.html

Most producers/suppliers have recommendations for concentration ranges of their products. If you can´t get it directly, it is often possible to find the information on their websites (e.g. Schülke or BASF).
The ranges are typically around 0,1% - 0,5%, but different products are often mixtures of different preservatives and with different concentrations, so it can be difficult to compare products directly.
Citric acid has some biocidal effect but is not very efficient, so I would not recommend it.
Why do you need a chealant? You use demineralized water and a nonionic surfactant, so ions in the product should not be a big issue.
The CAS-number you refer for the surfactant covers a variety of chemicals, with different C-chains and degrees of ethoxylation. They are generally classified as mostly soluble in water, but it may be necessary to add a hydrotrope (e.g. Cumenesulfonate) or co-solubilizer (e.g. Butylglycol).
It is difficult to suggest the appropriate preservative, as it depends on your exact formulation. I will rather refer to the suppliers/producers. If you want to look through possible chemicals, a place to start could be ECHAs list of biocidal products: https://echa.europa.eu/information-on-chemicals/biocidal-active-substances (see under “In-can preservatives” –PT6). Things to consider is: solubility (and stability) in your formulation, at what pH is it effective and off course any regulatory issues.

I know, or at least think, that the main use for chelants in cleaning solutions is to deal with the metal ions in the tap water they use to make them. I understand my solution uses distilled water but the metal contacts I'm trying to clean come into contact with other metal contacts and they leave fine metal debris on each other over time. Not sure if ultra fine metal debris classifies as metal ions or not but I read surfactants are rendered useless against metal ions without a chelant. Surfactant is Neodol-91-6. Not sure if it's any good. Only thing I could find.

Handling Ca/Mg in tap water is one purpose of chealants. Another could be to bind catalytic ions (eg. Fe, Mg, Cu) which can cause stability problems. Neither seems to be a problem in your case. But a chealant could very well be helpful in dissolving the dirt. Be aware that pH has a great effect on the efficiency of a chealant. Your type of surfactant is used in many different applications, whether it is the best choice for this purpose can only be determined in some realistic tests.

Thank you for the advice. I will have to look into that. My main concern at the moment is determining if I need a preservative, what kind, and how much. My cleaning solution will be worthless if bacteria/mold forms in it. From what I've read, even in a perfect sterile environment it's still recommend to have some, even if it's a minute amount.

You will probably need a preservative. Only solutions with either very low water activity, pH<2 or pH >10.5 are generally considered self-preserving. Production under strictly aseptic conditions is a possibility, but it is not easy to do. And if the product is not distributed in single-dose units, there is a great risk of contamination afterwards.

That's smart. I wasn't event thinking about contamination afterwards. Are there any preservatives you can think off hand that do well with solutions like mine?

@FrankS I have merged your answers into one answer. Please follow this format in the future, i.e., if you have additions to a post please edit in the new information.

source: Stack Exchange

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

sort: votes


text:
Why NaBH4 don't reduce this keto group?
I was solving problems related to Carbonyls. The question was:

What would be the product formed after the reduction of this compound
using $\ce{NaBH4}$?

(This is an isomer of Glutaric Anhydride)

I thought that $\ce{NaBH4}$ will reduce the keto group (not the ester) but my book said there will be no reduction. When I asked my instructor he said "Since there are many Oxygen due to steric and some other reason it won't reduce". Won't $\ce{NaBH4}$ reduce that? if-else Why so?

Which book is this from?

@Waylander It is actually from our coaching material for "JEE".

Your instructor's explanation is unconvincing

@Waylander Yeah.

For what it's worth, there is no evidence in the literature that this won't be reduced by NaBH4. You are being asked to explain a "fact" which nobody has ever observed before.

@orthocresol Yes Actually, I assumed that it will reduce but it is in the book that told the compound won't be reduced by NaBH4. Only LiAlH4 can reduce it.

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


text:
Why is the boiling point of sugarcane juice lower than the boiling point of water?
As the boiling point is a colligative property, one would expect that the boiling point of sugarcane juice should be higher than pure water because Sugarcane juice contains water (75–85%), non-reducing sugars (10–21%), reducing sugars (0.3–3%), organic substances (0.5–1%), inorganic substances (0.2–0.6%) and nitrogenous substances (0.5–1%)1. However, the boiling point of sugarcane juice is only 91-95°C [2]. Are there any chemistry theories that can explain this drop in boiling point?
[3]
1 Sreedevi, P., Jayachandran, L.E. & Rao, P.S. Browning and bioactive composition of sugarcane juice (Saccharum officinarum) as affected by high hydrostatic pressure processing. Food Measure 12, 1962–1971 (2018). https://doi.org/10.1007/s11694-018-9811-7
[2] Tiwari, G. & Prakash, Om & Kumar, Subodh. (2004). Evaluation of convective heat and mass transfer for pool boiling of sugarcane juice. Energy Conversion and Management. 45. 171–179. 10.1016/S0196-8904(03)00143-2.
[3]: Canilha, Larissa & Chandel, Anuj & Milessi, Thais & Antunes, Felipe & Freitas, Wagner & Felipe, Maria & da Silva, Silvio. (2012). Bioconversion of Sugarcane Biomass into Ethanol: An Overview about Composition, Pretreatment Methods, Detoxification of Hydrolysates, Enzymatic Saccharification, and Ethanol Fermentation. Journal of biomedicine & biotechnology. 2012. 989572. 10.1155/2012/989572.

Is there noted atmospheric pressure ? I am not sure it would be boiling, of placed in the boiling water bath. Unless there are volatile components or some decomposition occurs, e.g. decarboxylation. Could you quote broader context from [2] ?

I don't believe the values listed there in reference 2. It was a very very crude set-up and they do not mention what was their criterion of determining the boiling point. Boiling point of filtered sugar cane juice must be higher than pure water.

See this page http://www.sugartech.co.za/bpe/index.php

That page confuses me a little @M.Farooq. Why is the temperature of the sugar solution one of the parameters in determining its boiling point elevation? However, I have seen a source take the boiling point of sugarcane syrup as 105°C but that value wasn't justified. Nonetheless, it does give credence to your idea that source [2] is a little too crude. I have also found the boiling point of filtered sugarcane juice by a very crude method (the best my highschool could offer) and it was about 96.6°C but it had some lime water and SO2 gas passed through it as part of a larger experiment.

Evaporating water from syrup is industrially always done at reduced pressure. The boiling point of sugar solution at ambient pressure is definitely > 100°C.

The experiment in source [2] was done at atmospheric pressure as highlighted in the tables @karl.

I don't have access to that, but the number is obviously wrong, as in unphysical. Find other, better literature. Sugar production is an old industry, there must be dozens of textbooks in your university library. Don't waste time on one weird number from one paper.

@Karl I agree with you but the boiling point of sugarcane juice does seem to have weird properties. There is another paper DOI: 10.17654/HM013040445 whose results show that sugarcane solutions seem to start boiling at about 95-98°C, if you average T2, T3, and T4. Moreover, the experiments I am performing on sugarcane juices for my high school project are also consistently boiling below 100°C. However, I have to admit my methods are very crude, and I am adding lime water and SO2 gas to it.

@Karl: I agree with you. The temperatures reported are obviously wrong. But how? The thermocouples (Cu-constantan) were calibrated. I wonder if the tips were directly exposed to the sugar solution where galvanic corrosion could aid or conflict with the Seebeck effect. Sugarcane juice pH has been reported as 3 to 4.5: https://www.researchgate.net/figure/Evolution-of-the-pH-of-the-sugarcane-juice_fig2_284455766.

@JamesGaidis There sure is some explanation for the numbers. What concerns be more is that the authors don't even discuss it (or so I understand). So either this effect must be basic textbook knowledge for anyone who works with sugar raffination, or these papers were written by a bunch of careless amateur engineers.  ?

@Karl: You hit it on the nose: engineers! I went back to the paper. The word boiling is used many many times, but "pool boiling" is specified, and that term has a specific meaning for heat transfer engineering: it's evaporation from the surface of  a stagnant liquid https://hedhme.com/content_map/?link_id=598&article_id=192 For non-engineers, boiling is boiling, but for engineers, boiling can just be evaporation at near the boiling point. Amazing how difficult it is to communicate across interdisciplinary divisions! Those engineers were not careless; we chemists were ignorant.

@JamesGaidis I change my verdict to "careless pro engineers". ;-))

A more plausible explanation is that it is not water that is boiling.  The  "organic substances" and maybe the "nitrogenous substances" in the juice could include some more volatile components that may pass selectively into the gas phase at a lower temperature than most of the solution.  Such a vapor would not be the pure organic or nitrogenous substance; it would carry off some water vapor as the vapor pressure of the water is a large fraction of an atmosphere.  The temperature at which such selective vapor begins to form is properly called the bubble point.
Of some relevance is the effect of the sugar itself on the boiling point.  In the absence of other solutes this effect is very small, cfailing to reach $5°C$ boiling point elevation until we have nearly 80% sucrose in the solution (by mass).  Sucrose has a large molar mass smd water has a high heat if vaporization, both of which factors attenuate the boiling point rise at concentrations relevant to sugar cane juice.  This lack of an effect from the main nonvolatile solute would enable small amounts of more volatile components to yield a bubble point lower than the boiling point of pure water.

From Ref. [1]
Reference
1.
Willism Heffner and Hinanshi Jin, "Building a Low Cost, Hands-on Learning Curriculum on Glass Science and Engineering using Candy Glass", MRS Online Proceeding Library Archive (January 2009) 1233,
https://www.researchgate.net/deref/http%3A%2F%2Fdx.doi.org%2F10.1557%2FPROC-1233-PP03-06.

Oscar, have you tried sugar cane juice? It is drinkable, in fact it is a popular drink in South Asia road side stalls. Why do you assume there are significant amounts of nitrogenous substances and organic substances there? It would stink or perhaps be bitter if it had those nitrogenous substances in significant amounts.

@M.Farooq See OPs question. ;) And I would assume that means "residual protein". Nothing that would stink, or lower the Bp.

Proteins neither boil nor they are volatile.

Also note that the sugar itself is not very potent in raising the bp to begin with.  See the recejt modification to the answer.  On another comment, proteins could become amino acids.

This also explains the odd occurrence in DOI: 10.2298/TSCI150219213S where the boiling point is written twice. Once the study lists the boiling point as 107°C and once again as "boiling of water in sugarcane juice may start at 91-95C".


text:
Major product in dehydration of alcohol
Consider the dehydration of the following molecule with conc. $\ce{H2SO4}$ at $443 \text K$:

The possible products are:

The first one is the Zaitsev product and the second one Hofmann  product.
Which compound is the major product?
There are 2 factors to consider here:

Zaitsev product is more stable than Hofmann product due to
hyperconjugation
Anti elimination is more favorable than syn elimination due to steric requirements

Obviously both the factors clash with each other, so I couldn't come to a definite conclusion. Can you help me out?

https://chemistry.stackexchange.com/questions/41836/reactions-of-alcohol-with-sulfuric-acid

The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When the OH is protonated and it leaves, we form a secondary carbocation, but that can be further more made stabilized via an hydride shift from the adjacent carbon producing a tertiary carbocation. Now either of the base $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ can capture the hydrogen from the adjacent carbon leading to the desired Zaitsev product. (Also the stereochemistry isn't necessary to worry about here, as the reaction proceeds through E1.)

I would humbly request the contributor to make an edit. The formation of carbocation by removal of H2O is actually a slow step. It is infact the RDS and should proceed slow as is characteristic of any. The addition of the H+ forming the protonated alcohol is the fast step that you've confused with.

but the water is such a nice leaving group that it leaves very fast and forms the cation. That is what I meant.

No, the slow step is carbocation formation. The rest are even faster.

@Desai, please stop using mod flags for everything. Those are meant for exceptional issues. If the answer is wrong, you can comment, or downvote. No need to flag for moderator intervention, or else we'd literally have to fact-check every answer on this site.

@orthocresol the carbocation is formed when the water leaves as a leaving group, but if water leaves fast (as its nature is), doesn't that imply that the carbocation formation is also faster?

OK, let's clarify a few things. (1) The "loss of water" refers to exactly the same step as the carbocation formation. What I'm saying is that that step, whatever you want to call it (water leaving or carbocation being formed), is the slow step. (2) You're still forming an unstable intermediate (a carbocation), which necessarily means that this step represents a large increase in energy, hence higher activation energy, hence slower. Water being the leaving group merely makes that slightly easier, but not to the point where it can be faster than the other steps in the mechanism. [...]

(3) You seem to be confusing "slow step" with a "slow" step. What I mean by this is that, the "slow step" need not even be slow by any means: it can be pretty fast in and of itself. It just needs to be the slowest of all the steps. That means that, even if you disagree with point (2) and say that carbocation formation is fast, it can still be the "slow step", as long as it's not as fast as everything else.

Aside from those two words which I nitpick on, however, your answer is perfectly correct. I think you could also mention that the syn vs anti requirement (as mentioned in the original question) is only relevant for E2 reactions. You implicitly say this by drawing out an E1 reaction, but it's better to leave no doubt.

@orthocresol Oh is it like even though water is a very nice/fast leaving group, the formation of carbocation is a very unstable step and so we say its "Rate Determining" as it generates an unstable product?

Well, sort of, but not directly: it is "rate determining" (or equivalently, the "slow step") simply because it is the slowest step in the entire mechanism. Typically, steps that form unstable intermediates are also the slowest, so there is a connection.

@orthocresol makes sense! I have one more doubt to clarify. When Desai said about this ** The addition of the H+ forming the protonated alcohol is the fast step that you've confused with** . Is this true?

Yes, I suspect that the protonation is probably the fastest step, although the rationale is a bit more involved. That's not really an issue worth worrying about in this mechanism, though. (In all fairness, knowing the slow step is not a particularly huge deal either. The important thing is knowing which mechanism it proceeds via, i.e. E1 or E2.)

Let us continue this discussion in chat.


text:
Calculating phosphorescence quantum yield using the given datas'

A molecular excited state has a fluorescence lifetime of 10 nanoseconds and a phosphorescence life time of 10 microseconds. There are no other processes, what's the phosphorescence quantum yield?

I divided the above two lifetimes naively to get $0.1\%$.  I believe we can use the equation $\phi_f = k_r \times \tau_f$, and similarly for $\phi_p$, then divide the two. But I am not sure at all. Can anyone guide me how should I solve this question and is my answer correct?

Can you explain the different terms in your equation?  The fluorenscence quantum yield? and the fluorenscence lifetime?  It's not clear what you think is actually going on chemically/physically.

$\tau_f$ is the fluorescence lifetime, $\phi_f$ is the fluorescence quantum yield....

The triplet can only be formed by intersystem crossing from the singlet. The fluorescence yield is $\phi_f=k_f^0/(k_f^0+k_i)$ where $k_i$ is the intersystem crossing rate constant to form the triplet and $k_f^0$ the radiative rate constant for fluorescence. The fluorescence lifetime is $\tau=1/(k_f^0+k_i)=10\cdot 10^{-9}$. The triplet yield is $1-\phi_f$. The triplet lifetime is not important and as the triplet only phosphoresces (you mention no other processes, so no, non-radiative, pathway from triplet) all the  molecules formed must phosphoresce (and it does not matter how long this takes as its the only pathway out of the triplet.)  However, you still have one piece of evidence missing, namely $k_f^0$.

Since $\tau=1/(k_f^0+k_i)=10\cdot 10^{-9}$ wouldn't knowing either $k_f^0$ or $k_i$ be sufficient?

it is not possible to get to triplet state without $k_i>0$, so if only fluorescence occurs $k_i=0$ (no intersystem crossing) and so no triplet. If $k_f^0 =0$ no fluorescence and all triplet, so in this case triplet yield is 1 but then no fluorescence can occur but this is not what happens as the fluorescence lifetime is given. The question cannot be answered without more information.


text:
Nernst equations for the reduction of sulfate(IV) and nitrate(V) anions

Write a Nernst equation $(\pu{25 °C})$ for system:
$$\ce{SO3^2- + 3 H2O + 6 e- -> S^2- + 6 OH-}\tag{R1}$$
and system:
$$\ce{NO3- + 3 H+ + 2 e- -> HNO2 + H2O}\tag{R2}$$

I am not really sure if these are correct:
$$
\begin{align}
E_1 &= E^\circ(\ce{SO3^2-}/\ce{S^2-}) - \frac{0.059}{6}\log\frac{[\ce{S^2-}]\cdot \left(\frac{10^{-14}}{[\ce{OH-}]}\right)^6}{[\ce{SO3^2-}]}\tag{1}\\
E_2 &= E^\circ(\ce{NO3^-}/\ce{HNO2}) - \frac{0.059}{2}\log\frac{[\ce{HNO2}]}{[\ce{NO3^-}]\cdot[\ce{H+}]^3}\tag{2}
\end{align}
$$
I don't know how to deal with the $\ce{OH-}$ in the equation. Am I supposed to consider $\ce{OH-}$ at all? Or should I count with it in form of $[\ce{H+}]$—in that case, can I use $K_\mathrm{w} = [\ce{H+}]\cdot[\ce{OH-}]$ (as written above in my solving attempt)?

See Libretexts and Wikipedia. Pay more attention to the prior research before asking.

Consider the Nernst equation as a kind of a wrapper of the expression for the reaction quotient. Water in involved implicitly.

Note that $\ce{SO3^2-}$ is sulfite not sulfate. . .

@Andrew Note that the PIN for $\ce{SO3^2-}$ is sulfate(IV), not sulfite (trioxidosulfate(IV) would be even more correct). "Sulfite" might still be used more often, but, strictly speaking, it is an obsolete name. Given multiple choices, the safest option in my opinion is to follow up-to-date IUPAC nomenclature unless it is a three-lines-of-text name that can be replaced with a short abbreviation or a traditional name.

@andselisk dihydrogen monoxide == water ;)

@NilayGhosh  It may become more convenient to abandon new IUPAC names,  using just formulas.  ;-)

@Merida Why should be $\ce{OH-}$ ion  treated differently than any other ion ? And why do you use $\ce{[H+]}=\pu{e-14}/[\ce{OH-}]$ instead of $\ce{[OH-]}$  ?

@Poutnik We were always told to use expresions for [H+] rather than [OH−]. From our teachers point of view - it is more convenient/useful.

Well, it is case dependent. This is not the case. You can still do it, but properly. Replace [OH-] by Kw/[H+], not by [H+]=Kw/[OH-].


text:
Why is the ionic radius of hydride so large?
The order of ionic radii for halides and hydride is apparently as follows:
$$\ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{H-} < \ce{I-}$$
Why is the hydride ion so large, even larger than bromide which has filled 3d and 4p subshells? Some claim that it is because the ratio of positive to negative charge is 0.5, which is an unusually small ratio. But this explanation doesn't fully convince me.
Nat. Commun. 2014, 5 (1), No. 3515 also makes mention of the unusually large size:

The ionic radius of an $\ce{H-}$ ion is $\pu{134\pm 2 pm}$, which is slightly larger than the radius of $\ce{F-}$(II) ($\pu{128.5 pm}$) and almost the same as the radius of $\ce{O^2-}$(II) ($\pu{135 pm}$). This evaluation agrees with the generally accepted trend that the radius of an H− ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion and increases with increasing electropositivity of the coordinating cation.

Radius of hydride (H⁻) ion.

A weight, for example, is a number that you can trust. A radius isn't.

Weight isn't, mass is. Weight would vary with the planet's gravitational force. But let's leave that for Physics SE. @Ivan Neretin

@Desai Good point!

It is not.
Presenting the order of ionic radii like that is useless.
Ionic radius of a given ion is not constant and mainly rely on crystal structure (coordination environment, more precisely) and used experimental techniques (crystal structure determination and theoretical calculations).
As a consequence, you can only compare homogeneous data.
Proposing the ordering based on some cherry-picked values from random sources or their averaged values is pointless at best.
This very issue is discussed by Messer [1]:

The effective ionic radius of fluoride ion is given as 1.33 Å by Pauling (8) and 1.36 Å, by Zachariasen (9), for coordination number 6. The effective ionic radius of hydride ion varies from 1.27 to 1.52 Å (10), being much more highly sensitive to the particular cation present and to the particular assumptions used in deriving the radius than the fluoride value. For the alkali metal hydrides other than lithium hydride, with the same face-centered cubic structure as the corresponding fluorides, the effective hydride radius is 1.47 Å for $\ce{Na+},$ and 1.52–1.54 Å for $\ce{K+},$ $\ce{Rb+},$ and $\ce{Cs+}.$ For $\ce{LiH},$ $\ce{CaH},$ and $\ce{BaH},$ the hydride ion radius is 1.34–1.36 Å. The closest similarities between hydrides and fluorides should thus appear here.

You can find the order of ionic radii for group 1 hydrides and halides in a more recent work by Lang and Smith [2]:

$$\ce{Li+} < \ce{H-} < \ce{Na+} < \ce{F-} < \ce{K+} < \ce{Cl-} < \ce{Rb+} < \ce{Br-} < \ce{Cs+} < \ce{I-}$$

Again, note that this doesn't only contradict your order, but it is no universal sequence and is only true for these compounds with soft-sphere model used for calculation.
References

Messer, C. E. Hydrides versus Fluorides: Structural Comparisons. Journal of Solid State Chemistry 1970, 2 (2), 144–155. DOI: 10.1016/0022-4596(70)90062-9.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.


text:
Why does tungsten(VI) oxide have higher melting point than tungsten(VI) fluoride?
I'm learning about ionic bonding and polarizability. Larger anions with higher charge should make more polarized ionic bonds. $\ce{O^2-}$ is both larger and has higher charge than $\ce{F^-}.$
So, why does $\ce{WO3}$ have a melting point of $\sim\pu{1473 °C}$ and $\ce{WF6}$ have a melting point of $\sim\pu{17 °C}?$
I should add that ionic compounds usually have higher melting point than the covalent ones, which makes it strange that $\ce{WF6}$ is the one with the lowest melting point, as I would think that to have the most ionic bond.

Both are better understood as mostly covalent.

Same reason why SiO2 is solid and SiF4 gas.

https://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently

Is it really reliable to always consider ionic compounds to have higher melting point than covalent? Diamond is covalent but has a melting point of above 7000°F(in absence of oxygen)[https://www.nationalgeographic.org/media/diamonds-arent-forever-wbt/#:~:text=In%20the%20absence%20of%20oxygen,Celsius%20(7%2C280°%20Fahrenheit).]. I would direct this question to @Ivan Neretin but eager to learn more if anyone else wants to contribute too?

There are molecular compounds, and then they are covalent crystals. Both are covalent, but very, very different in other regards.


text:
CP2K Spherical Cell
I'm trying to run a gas phase cluster calculation in CP2K. I want to create an outside force to constrain water molecules that are in the simulation within a radius of a central ion. I want to create a spherical cell of a given radius $\lambda$, fix the ion in the center and then run QMMD with the waters free to do their motion.
Does anyone know how to create a spherical cell for the simulation with no periodic boundary conditions or a better way to impose this clustering constraint $\lambda$?

Since this is more about how to use a particular modeling program, rather than about the chemical system being studied, I think this would be a better fit on Matter Modeling SE


text:
Why does the cathode in an OLED require a low work function and the anode require a high work function?
I am currently doing some research on organic light emitting diodes and am just trying to get my head around some of the terminology and concepts because I am more from the organic chemistry side don't have a strong material sciences/physics background.
My reading indicates that the work function is the amount of energy required to extract an electron from a material. Electrons are injected via the cathode and electrons are extracted from the anode (i.e hole injection)
I keep reading that for OLEDs, the cathode should have a low work function to match the energy of the LUMO of the electroluminescent layer, and the anode should have a high work function to match the HOMO. I've read this in many journals and review papers but none of them explain why.
Any help would be appreciated.

Basically you want electrodes already near the levels you want to extract / inject electrons to. Therefore the electrode at which the material is reduced, the cathode, must have oxidation potential / work function near the LUMO energy level. Opposite, to work on the extraction of electrons from the HOMO, a metal lining down at higher ox potential is needed. Note that this can reduce the existence of energetic barriers at the interfaces, and in all cases reduces the potential needed for the functioning. If this does not answer your question I'll come back when I'll have time.

Thank you for your response, that does point me in the right direction. I completely forgot about redox reactions (not a part of chemistry that I enjoy that much) but now that you mention it, it makes a lot more sense.

The operation of light emitting diodes involves the electrochemical creation of radical - cations and -anions that upon recombination lead to molecules in their emitting excited states.
The formation of the above charged species happen when the applied forward potential is positive enough to inject (reduction) electrons from the cathode to the LUMO of the molecule and extract electrons (oxidation) from the HOMO to the anode.
Therefore, the electrodes must have work functions about the levels you want to operate onto.
This means that the cathode must have oxidation potential / work function near the LUMO energy level. Opposite, to work on the extraction of electrons from the HOMO, a metal lining down at higher oxidation potential is needed.
Note that this design not only reduces the potential needed for the functioning of the LED, but minimise energy barriers (if any) at the emitting material / electrode interfaces.
Also note that a reverse order can still lead to light emission, but the ON potential to be applied to this "unnatural" configuration is inevitably higher.


text:
Chirality and Optical activity
Do all chiral objects rotate the plane of polarization of EM waves? For example my hand is chiral, will it rotate the plane of polarization of EM waves of suitable wavelength?

"optical activity" is a problematic term if only because it implies EM radiation in the visible range. Unfortunately there isn't a simple general term. Though you could say, "can chiral macroscopic objects rotate the plane of EM waves of a suitable wavelength?"

Of course it will, provided you have a billion of it. (One molecule does not rotate much, after all.) Also, each should come without the rest of the body, or at least without the other hand, otherwise that would make a racemic mixture. (I'll leave the grisly details to the reader's imagination.) Also, each hand should form the same gesture, otherwise the destructive interference will ruin the effect.
Also, light of suitable wavelength would be microwave radiation, and a cursory search seems to indicate that the methods of polarizing it and measuring its polarization are known.
With all these conditions fulfilled, I don't see any difference from the situation with the molecules and ordinary light.

Slightly off-topic, but I wonder if it's possible to formalise some concept of "maximising chirality", such that a single interaction between a chiral photon and a chirally-maximised object would be significantly different depending on the relative chiralities. With molecules this may be hard to investigate smoothly, but perhaps macroscopic objects could be more freely adjusted and studied.

Screw-shaped things like helicenes come to mind.

and some types of achiral liquid crystal forming molecules have chiral helical phases.

Human hand is a macroscopic object, even if there are billions, they will not rotate light. Optical rotation is a molecular phenomenon not a macroscopic phenomenon.

Microwave radiation is not exactly light.

Still a lot smaller than a macroscopic object.

How so? Its wavelength is rather macroscopic, as to me.

@IvanNeretin, My understanding has be smaller than the wavelength. Let us roughly say microwaves have 1 cm wavelength, in order to see microwave rotation, the objects have to less than that. See this linked paper,  they used springs to show optical rotation with microwaves but the springs were smaller than the center wavelength. https://ieeexplore.ieee.org/abstract/document/7775673https://ieeexplore.ieee.org/abstract/document/7775673

Great! You have found the evidence that the effect does exist and can be measured for microwaves, after all! As for the size, why bother; just take a wavelength 10 times longer (those still would be microwaves).

The OP should not call it an "optical" rotation. Rotation of polarized long waves is more than century old phenomenon. Blocking of polarized radiowaves with wire "polarizer" is a nice demo in physics. Still a human living hand would not rotate microwave radiation. It will simply absorb it quite strongly. I have worked with microwave spectrometers, and trace of water is a headache.

Who said "living"? As for the rest, well, yeah, the OP's terminology needs correction in that the word "optical" typically refers to the visible light and not to the other ranges of the EM spectrum. Ditto for the word "light", BTW.

The wavelength has to be comparable with the object size in order to see those effects. Your macroscopic hand will not rotate ordinary light even if you have millions hands around, because the wavelength of light is so small. Yes, it will scatter light as a large object. This idea of optical rotation by long wavelengths and by large objects was shown more than 100 years ago by J.C. Bose, using twisted pieces of jute (a type of rope made from plants). Bose was in India and jute was a common material there to play around.
See this article.
https://www.jstor.org/stable/115973?seq=1.

I think you got the the 'Bose's mixed up. The Bose whose you're referring to is JC Bose, who was the one who did many experiments with EM waves. The other Bose is SN Bose who made contributions to Quantum Mechanics.(Bose-Einstein)

You are right Milan. I corrected it.


text:
How does one prove Avogadro's law from Gay-Lussac's?
To understand Gay-Lussac's law, Amedeo Avogadro said

In equal volumes of air under constant temperature and pressure, there are equal number of atoms (or molecules).

Let's take a hydrogen and an oxygen molecule. As the latter is larger in size, when stored in two separate equally-volumed containers, doesn't oxygen increase the pressure inside that container, if the same number of molecules occupied each container? (Does this make Avogadro's statement of constant pressure go wrong?) Again,
$$2mvN\cos\theta = F,$$
where $F$ is the force acting on the walls of the container, $N$ is the number of collisions of the molecules with the walls, $2mv$ is the change of momentum of the molecules, and $\theta$ is the angle of collision of the molecules on the walls.
Therefore, if Avogadro was correct, could anyone give me any reason why $N_\ce{H2} > N_\ce{O2}$ or $v_\ce{H2} > v_\ce{O2}?$
If Avogadro was correct, the above conditions must hold, as $m_\ce{H2} < m_\ce{O2},$ right?
Could you please prove the above phenomena happen when Avogadro's statement above holds?

The kinetic theory of the collisions has no connection with Avogadro's law.

@Maurice why do you say they have no connection?

Avogadro's law describes the state of a system, and not a change. Gay-Lussac's law describes what happens when a change happens, when the system temperature is modified

@Maurice yes but, 2mvNcosθ=F is not Gay-Lussac's theory, right? and what I'm asking is how to describe  Avogadro's law when above mentioned conditions (in the question) satisfy?

His name was Amedeo Avogadro, not Amidio nor Amadeo nor Avegadro nor Avagadro. The full name was Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto.
Avogadro's law states that

equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

More exactly, it also includes single atoms, e.g. for a case of noble gases, due the way how IUPAC defines a molecule.
But remember the Avogadro law implies ideal gases, similarly as the Gay-Lussac's law,Boyle's law and Charles' law. For ideal gases, sizes of molecules play no role. Both the size and volume of ideal gas molecules are considered negligible.
Real gases have deviations from this ideal behavior. See  van der Waals equation as the most used real gas state equation.

$$\left(p+a \frac{n^2}{V^2}\right)(V-n b)=nRT$$
$$\left(p+ \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b)=RT$$

where parameters $a$ (addressing cohesive forces ) and $b$ (addresing non zero molecular volume) are specific for the particular gas.
Then the molar amount of a real gas for given $V$, $p$ and $T$ is the solution of the cubic equation for  variable $n$ :
$$pV - n ( pb + RT ) + a \frac{n^2}{V} - ab \frac{n^3}{V^2}  = 0 $$
It clearly shows that the same volumes of different real gases, at the same temperature and pressure, contain ( usually slightly ) different counts of molecules/atoms.

The question lies at the  heart of Avogadro's law and Gay-Lussac's.
The question inquires about the role of "mass" which as a physical unit or parameter does not show up neither in Avogadro's law nor in Guy-Lussac's.
In fact, the principle of Avogadro's lies in mass of the specific gaseous  element (atomic mass)  or molecule  being of no importance, making no difference (it's only coherent not to mention mass in the formula).
How could it be if mass (cp. explanatory text of the question above) determines impulse, and impulse determines pressure?
The more mass a gaseous particle (specific element, molecule) has the slower it is, the lesser its mean velocity as it hits the wall.
Then,
"why NH2>NO2 or vH2>vO"
must, in my opinion, be anwered as follows:
Lesser velocity means lesser number of hits (N for H2 is larger than N for O2 as H2 is lighter, thust faster, hence hits more often).
Again, refering to the example: O2 is the gas that has more mass, as previously said, its velocity is lower which means that the N number of hits is lower - which places O2 at equal footing with H2 the mass of which is less, the velocity and number of hits of which is greater.
This results in the number of hits canceling out the amount of momentums determined by different masses.

At a given temperature, the average kinetic energy of gas particles is the same, no matter what particles make up the gas. That means that lighter particles have a higher average speed than heavier particles.
Average (translational) kinetic energy determines both temperature and pressure of ideal gases, which is why they are proportional for systems with equal volume and amount of substance.


text:
How to name an alkyne with cyclic substituent?

My attempt: 1-(1-methylcyclopentyl)-3-methylpent-1-yne. Looks so messy though.

It gets even a bit messier than that. Also, note that −≡− should be a straight line: whoever drew this structure doesn't know what they are doing.

As far as I am concerned, your attempt was incorrect.
You must use a parent chain with the highest number of carbon molecules. If I am not mistaken, if a cyclic carbon chain has the same number of carbon molecules as a straight carbon chain, the cyclic carbon chain should be the parent chain.
In addition to what I said above, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), rings/cyclic carbon chains are always selected as the parent chain.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

Thus, the name should be: 1-methyl-1-(3-methylpent-1-yn-1-yl)cyclopentane.
Note that an alkyl substituent that is also an alkyne (i.e. 3-methylpent-1-yn-1-yl in your question above) is called an alkynyl group. It is also worth noting that rings/cyclic carbon chains are always selected as the parent chain for the preferred IUPAC name as I mentioned previously.


text:
Identifying the molecular structure of compound with sum formula C8H14O3 formula using NMR and IR spectrometry data
I am looking to identify the molecular structure of a compound with
the sum formula $\ce{C8H14O3}$ using the NMR and IR spectrometry data below.

I was able to figure out the molecule as

My friend asked me why I didn't consider the following molecule:

I couldn't answer that. What inferences from the IR & NMR data can I observe, to explain why the structure my friend suggested is wrong?
Source for these IR & NMR Spectra: https://www.orgchemboulder.com/Spectroscopy/Problems/7.shtml

You have a singlet with 2, so there must be a methylene group coupling with no other protons.

@Martin-マーチン, I drew the structure(That my friend suggested) incorrectly. Could you be kind enough to look at the question again?
I appreciate any help that you can provide.

Look at the two peaks at around 4.2 and 3.5 ppm. The quartet at 4.2 is more deshielded than the singlet at 3.5. If the structure your friend suggested is right, then the 4.2 signal must come from the $\ce{CH3C\textit{H}_2O -}$, and the 3.5 signal would come from $\ce{-O-C\textit{H}_2-C=O}$. This would not make sense, as the CH2 in between two electron withdrawing groups should be more deshielded than the one that is next to one electron-withdrawing group.

@xavitop Instead of adding "(ANSWERED)" to the title, please feel free to post an answer to your question if you think it has been solved.

Compare the quartet at $\pu{4.2ppm}$ and the singlet at $\pu{3.5ppm}$. These quite obviously would, in your friend’s structure, correspond to the protons on either side of the ether-oxygen. However, we should expect the methylene singlet to be more deshielded than the ethyl group’s methylene: It is squashed between an electron-withdrawing $\ce{OR}$ and an electron-withdrawing $\ce{C=O}$. The ethyl group’s methylene is between an electron-withdrawing $\ce{OR}$ and a mostly electron-neutral $\ce{CH3}$ group. Therefore, the order of peaks we observe in the spectrum does not make sense when we consider your friend’s proposal.
Thereafter, one might also argue that a quartet at $\pu{4.2ppm}$ is so obviously an ethyl ester as likely every organic chemist has had way to many NMR spectra after a column that contained a peak exactly there. An ethyl-ether is more shielded (diethyl ether’s methylene proton has a chemical shift of $\pu{3.5ppm}$.

I’m sure someone else can provide an answer that finds ester signals in the IR and a lack of a 1,2-diketone but I won’t use IR if I can avoid it ;)


text:
Using a H NMR spectrum to determine the structure of a protein supplement
This question is about the H NMR spectrum of creatine.
Creatine is a supplement taken by athletes to increase body mass. Recently, supplements containing derivatives of creatine have been marketed. These are usually more lipophilic in an attempt to improve uptake.
The 1 H NMR spectrum of one of these supplements in D2O is shown, with some parts magnified:

This particular supplement exists in an ionised form at pH1 but does not exist in an ionised form at pH 12 (I am unsure about this and will come back to it in my question below!)
Using this information we are asked to suggest a structure for the supplement. I was almost right. The answer is:

The mark scheme says that three things can be extracted from the NMR spectrum and the information given:

My question(s):

I don't understand the relevance of the pH statement. How does that relate to the structure? How does that relate to the ester group? (this is mentioned in the mark scheme which is attached just above this)

How can we use the integrals to calculate that there are 10 C-H protons? I measured the heights of the traces and got, 2.7cm, 4.75cm and 6.9cm but I'm not sure how this leads to '10 C-H protons'...

Would really appreciate help on this question and a breakdown of the key points discussed. Many thanks!

Let me try to explain the point about the 10 atoms and the integral.
The area under the peaks and the ratios of these areas correspond to the ratios of the hydrogen atoms. The blue line adds all these areas up as it goes along. That means at the first peak on the left you simply get the area of that peak. But at the second peak you still have the amount from the first peak plus the new area, which is why you have to look at the difference.
If we denote the three peak areas as $A_1$, $A_2$, and $A_3$, the height of the blue line goes from $A_1$ to $A_1 + A_2$, then lastly $A_1 + A_2 + A_3$.
With your values we have,
$$\begin{aligned}
A_1 &= \pu{2.7 cm} \\
A_1+A_2 &= \pu{4.75 cm} \\
A_1+A_2+A_3 &= \pu{6.9 cm}
\end{aligned}$$
Which allows us to easily determine $A_2 = \pu{2.05 cm}$ and $A_3 = \pu{2.15 cm}$. (Alternatively you could have simply read off the difference of the blue line between two peaks directly when you measured it).
This gives us the ratios of $\pu{2.7 cm} : \pu{2.05 cm} : \pu{2.15 cm}$. If we assume that the signal at $\pu{1.25 ppm}$ (i.e. $A_3$) is a methyl group with $3$ hydrogen atoms, we can divide all values by $\pu{2.15 cm}$ and multiply by 3 to normalize the ratio. We get approximately hydrogen atom ratios of $3.76 : 2.86 : 3.00$. Rounding the result gives a ratio of $4:3:3$ which agrees with $10$ atoms in total.

Where does the ethyl group statement come from? Also, how can we infer that there is an ester functional group?

An carboxylic acid group would be deprotonated at pH 12, yielding the carboxylate anion. A typical ester is not deprotonated at pH 12 and thus forms no anion at  pH 12, which agrees with the statement that the molecule is not in ionized form at pH 12. I am not sure what the functional group on the left with the nitrogens is called. I am not particular proficient in organic chemistry. But i would assume that it also doesn't get deprotonated. One would have to look up how that group behaves at different pH's or how nucleophilic/electrophilic it is.

is there any way I can find out more (somewhere online) about which species will dissociate at what pH and why?

Do you now about acid dissociation constants or $pK_a$ values ? If not you can check out the wikipedia page for a start https://en.wikipedia.org/wiki/Acid_dissociation_constant.

Yep, I know about pKa values, @Hans Wurst

Then your question is strange. A quick google for the $pK_a$ value of ester shows a value of 25. Carboxylic acids are around 5. Water has an $pK_a$ of 14. You should have no trouble deciding which species is found protonated in an equilibrium with water and which is not, if you understand the meaning of $pK_a$ values.

So I can answer the first question.
Esters hydrolyse readily in acidic conditions to the constituent acid and alcohols. As there is a signal in the 4.2 to 4.3 range, which falls in the 3.5 to 5.5 range shown by an alpha Hydrogen to an ether group or ester group.
To decide between an ether and ester, the question mentions the fact that the form is ionised at pH=1 and not at pH=12

I see. Thanks @xavitop! But, how does the fact that an ester easily hydrolyses at low pHs tells us that the structure is an ester? What does it mean that the form is "ionised" - surely hydrolysis is not the same as ionisation? To anybody else who is also reading: Would appreciate any answers to the other parts of the fiendish question! Many thanks, indeed.

About Q1:
The whole thing with information that creatine derivative in question exist in ionised form at pH = 12 and doesn't exist in ionised form at pH = 1 tells us that this compound tends to bind protons and doesn't show tendency to dissociate with proton release (as binding proton makes whole compound to adopt positive charge and dissotiation with proton release results in taking an anion form). Simple conclusion coming from this is that this compound has a basic group and no acidic group (that would be stronger acid than H2O).
Therefore logic conclusion from the information that creatine derivative in question doesn't have an acidic group would be that it has somehow modified carboxylic group of the parent compound. How it is modified? Well H NMR spectrum can be of some help. You can see a quartet from oxygen bonded methylene group (at about 4.24 ppm) that is coupled to methyl protons giving rise to triplet (at about 1.25 ppm) - that is some hint of ethyl ester being present in this creatine derivative.
More, this basic group that binds proton in acidic pH and gives rise to ionised form is that nitrogen-rich group, that sometimes is called in general guanidyl group (as it is related to the structure of guanidine). Nitrogen atoms there posses lone electron pairs and can react as base (similiary to amines).

is there any way I can find out more (somewhere online) about which species will dissociate at what pH and why?

pH related dissociation has to do with acid/base properties. If a compound dissolved in H2O is stronger acid than H2O then it gets deprotonated to the extent that is related to its acidic strength (this compound protonates H2O molecules because its stronger acid than H2O). Analogously, if the compound s stronger base than H2O it gets protonated by H2O molecules (as water is amphoteric, meaning it has acid and base properties). To get an idea of compound's acidity/basicity pKa values are of much help - knowing them you can calculate ratios of different forms of a given compound at given pH.

There are plenty resources online to learn more about it - just type in a serch engine something like "acid/base equilibria".
Something that comes up on the top of a search results list, and at a first glance looks good is:
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.01%3A_Introduction_to_Acid_Base_Equilibria


text:
why the The 15N–{ 1H} spectrum of [Pt(NH3)3 H2O] shows 6 lines instead of 3 if all the NH3 are supposed to be equivalent?
The 15N spectra for [Pt(NH3)3 H2O] shows 6 lines, I don't understand why because the 3 (NH3) are chemically equivalent and when interacting with Pt will generate a double satellite lines and a middle line for a total of three, so why it is giving me 6 lines?. are they not all chemically equivalent? or is there coupling interaction of  the 15nitrogen  with oxygen?. Also there is a note that says there are no  2J(N, N) observed in this spectra.
I was asked to use the method of successive splitting to corroborate the spectrum lines obtained but as I said  don't understand why there are 6 lines....

You may have missed the impact of ligand bonding geometry.  If, as seems likely, the complex is actually $\ce{Pt^{2+}}$ bonded to the ligands, we should expect a square planar geometry.  Thereby, one ammonia ligand is opposite the water ligand and not equivalent to the two other ammonia ligands which are adjacent instead.


text:
Solubility of gases in water in comparison to their liquefiability
My textbook of physical chemistry says

...The following factors affect the solubilities of gases in liquids:
(i) Nature of the gas and solvent: Generally, the gases which
can be easily liquefied are more soluble in common solvents. For example, $CO_2$ is more soluble in water than oxygen or hydrogen. The gases which react with the solvent posses higher solubility. For example, $HCl$ and $NH_3$ are highly soluble in water.
Oxygen, nitrogen and carbon dioxide are much more soluble
in ethyl alcohol than in water at the same temperature and
pressure.
(ii) Temperature: The solubility of most of the gases in
liquids decreases with increase...

Why is it so that the gases which can be easily liquefied are more soluble?
I mean, more liquefiable gases would have a stronger attractive forces amongst the molecule. This would mean that it would become increasingly tougher to dissolve them as their ability to be liquified increase. Then why the opposite is true?

Edit
I found a similar question on Chemistry.SE here asking Why does physisorption (physical adsorption) increase with ease of liquefaction?. I know the above question (and answers) are aimed at adsorption and mine is about solubility, but I think there must be an explanation like that for the aforementioned one.

Make some hand-waving about the strength of the intermolecular interactions.

Because it's a crappy textbook, or you haven't been reading it thoroughly? ;)

@Karl - I feel like you would be happy to know that the textbook is among the standard ones used by students (atleast in my country)

@SteelCubes If this is one of the 10 in One Study Package for CBSE Chemistry Class books, they I have to agree with Karl—the quality of that stuff is way below average from I can see on Google Books. If you must use this sort of literature, then so be it, but I would recommend to take a glimpse of the Resources for learning Chemistry.

@andselisk - The text quoted is from PHYSICAL CHEMISTRY FOR COMPETITIONS by Dr. O.P. Tondon (G.R.B. Publication)

@anselisk - Honestly, even I'd have to agree that these 10 in One Study Package for CBSE Chemistry Class type of books count quite too low in quality

@SteelCubes Isn't it Tandon? Anyway, 10 in One Study Package… is where I found the exact quotation of yours. I don't know who copied from whom, but still a recommendation to get a better textbook holds true.

@ andselisk - Apologies, yes it is Tandon indeed. Moreover, I'll surely consider your suggestion. But the thing is, I have encountered this statement and its similar variants at multiple occasions. And the above textbook is just one of them. Could you please have a look at this question here

Whenever someone asks here about something that is seemingly sloppily explained in a textbook, I bet myself one buck on the book having something to do with the JEE, and charity isn't doing very well on that. ;) Your country's educational systems is based on textbooks written by undereducated morons, to make sure you never get better than them. Sorry.


text:
Regarding the probability density of finding an electron in between a bond
I'd found the image attached on a website explaining the molecular orbital theory.

My question is, shouldn't the graph have a maximum at the middle of the two hydrogen atoms? since my intuition tells me that the electrons have a great chance of being in the middle rather than anywhere else. hmm, this train of thought has led me to another question, what exactly does this graph even represent? I mean, which electrons probababilty does it even show? or does it show us the probability of finding any electron participating in the constructive bonding.
another question that I have is this, if say, we were to excite one of the electrons into the higher antibonding orbital, what would exactly happen to this orbital, the shape I mean, would it simply cease to exist? since we're literally making the bond cease to exist (bond order=0) or would it assume the shape of this?

I still haven't completely grasped the true nature of such orbitals, I think it's evident from all these questions.

Bonds are in general not maxima of the electron probability distribution. Local maxima are typically centered at the nuclei. This is due to the Coloumb potential. The electron can lower its potential energy by getting closer to the nuclei so they will always try to maximize their probability around the nuclei, where all the positively charged protons are bundled. This is a bit oversimplified but in essence the reason why you see the maxima at the H position or in general at the position of nuclei.
To see bonds, you should look at the difference of the probability density of two lonely  hydrogens and a H2 molecule. I.e. calculate the electronic wavefunction for a single Hydrogen Atom and then add the plots of two such single hydrogen calculations placed at the positions as in H2. Compare this with the plot coming from the H2 calculation and you should see that the H2 distribution has a higher density in between than the two unbonded hydrogens simply taken together.
Bonds as we draw them in simple chemistry are not rigorously defined mathematically and more or less a useful approximation to rationalize reactions without doing difficult calculations. There are ways to obtain a mathematical definition of bonds but these definitions are not neccessarily agreeing with the way that we draw bonds in general chemistry. One such model is Atoms in Molecules which evaluates the electron density and its topology to define bonds. But there are other models, for example Electron localization functions and i am sure that there are more, unknown to me.
Overall it is important to realize that bonds are a useful concept but they have no unambiguous clear cut underlying physical definition. Physics/Quantum mechanics yields no equation that unambiguously defines a chemical bond.

thanks for the answer!
But I also want to ask, where exactly did the idea for an antibonding orbital come from? was it to explain the fact that if we excite the an electron in the bond with energy, the bond will sometimes dissapear? or is there a mathematical intuition behind it as well?
I am aware of the LCAO, but I have the same question for this as well, is it a mathematical consequence? or to explain observational phenomena?

When you determine the bonding orbital, you also obtain the antibonding orbital. Both orbitals are obtained when you search for the best possible orbitals for your molecule. You start with atomic orbitals centered on each nucleus. Then you use a method to obtain better molecular orbitals, typically the Hartree-Fock method. This calculation tells you how to build better molecular orbitals from the original set of atomic orbitals. In H2 you start with 2 x S1 orbitals. You obtain 1 x bonding sigma orbital and 1 x antibonding orbital. This procedure is well defined.

In H2 it appears obvious to identify the bonding sigma orbital with the single bond that we also draw in the structure H-H. But in larger molecules your orbitals often look nothing like a single bond between two nuclei. Instead the orbitals are spread out over the whole molecule. And this is the point where it gets tricky to identify a single orbital with a bond. We can still call them bonding and anti bonding by looking at their energy relative to the energy of the original atomic orbitals. I'd recommend not to overthink orbitals before you had an exposure to the mathematical background.


text:
Elimination vs reduction reactions - how to tell
Here are two compounds:
Compound H:

and compound I:

Compound H reactions with the reagent shown below to form compound I and two other products.
How would you clarify this reaction? I thought it was a reduction, because we are losing the oxygen bonded on the left hand side of the molecule. The actual answer is elimination Any advice for tell-tale signs for this type of reaction, and why my answer was wrong?

The oxygen-bonded group isn't the only thing you're losing. I assume you know what these structures mean -- draw in all the hydrogens explicitly (OK, skip those on the ring) and count them carefully.

If the phosphate group were a bromine, would you even ask the question?

@orthocresol - I'm not sure what you're getting at, unfortunately, I counted the Hs up but how does that help? What tells you that this is an elimination reaction?

You're losing an oxygen-based group (the phosphate), yes, but you're also losing a hydrogen.

By the way - this isn't a criticism of you, but rather I feel I should point it out, as you seem to be taught otherwise - "oxidation/reduction" and "addition/elimination" are entirely orthogonal concepts to one another. An elimination reaction can be oxidative, reductive, or indeed redox-neutral. The reaction you have here falls into this last category: it is an elimination (you are losing groups across a double bond), but there is no redox going on, as the overall oxidation state is preserved.

It is somewhat akin to size and colour. Just because a suitcase big, doesn't mean it can't also be brown. Classifying reactions into "elimination" and "reduction" is like classifying suitcases into "big" and "brown". Most of the time, you'll be OK, but when you get a big brown suitcase this classification falls apart. Likewise, classifying this reaction is fine, but if and when you get shown a reductive elimination, it won't fit into either elimination or reduction: it'll be both.


text:
Why isn't my electrochemical cell producing its potential voltage
For a school project, I am testing how concentration affects the voltage output of three metals. So I am building voltaic cells with 0.1 M copper nitrate and 0.1 M, 0.01 M, and 0.001 M of zinc, aluminum and nickel nitrates as the electrolyte solutions. For the cells with zinc and copper, I found that the voltage produced was generally fairly close to their theoretical value of 1.10 V with the lowest voltage produced being 0.95. However, for aluminum and nickel, I have found that they are very far off from their theoretical values of 2.00 and 0.57 respectively as I have only reached a maximum voltage of 0.50 for aluminum and 0.20 for nickel.
I am using a porous cup and have made sure all of the electrodes are thoroughly sanded before each trial. I am fairly confident the issue is not related to the concentrations, or to the porous cup. At this point, I just have no idea why it isn't working. I read a similar question on here suggesting that the issue might be that aluminum does not easily dissolve in NO3 and I was wondering if the same was true for nickel.

Check the resistance of your porous cup. What is it made of? Internal resistance of the cell can reduce the potential.

Are you also familiar with Nernst equation?

Yep I think I have a pretty good understanding of the Nernst equation. Would the resistance have a different effect depending on the electrolyte? Like could the resistance explain why zinc matched its theoretical value while the others didn’t?

Since the concentration of the electrolyte is changing, I feel the resistance of the internal cell is also changing. I am talking about the resistance of the solution. Can you bring the electrodes closer and see if the potential increases.

Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious.
Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. This oxide is an electrical insulator which makes a barrier between the metal and the solution. Aluminum ions cannot cross this barrier. There is no way of preventing this aluminum oxide formation. If you rub the aluminum surface to remove it, the pure metal so exposed will immediately react with air to remake this layer.
The only way to suppress this layer is to dip aluminum in a mercury chloride solution. Of course today this operation is forbidden by law. But this experiment was done before this legal decision. So let's see what was happening. Some mercury ions may cross the barrier and react with aluminum to produce an amalgam $\ce{Al(Hg)}$ plus some $\ce{Al^{3+}}$ ions getting in solution$$\ce{5 Al + 3 HgCl2 -> 3 Al(Hg) + 2 Al^{3+} + 6 Cl^-}$$ The formula Al(Hg) is not sure, as it is an alloy. Maybe the ratio Al:Hg is different from $1:1$. It does not matter for the present study. The only important thing now is that the aluminum oxide layer does not perfectly cover the amalgamated metal any more, and is partly destroyed. Therefore aluminum atoms are suddenly in contact with water, and they quickly react with it according to :  $$\ce{2 Al(Hg) + 6 H2O -> 2 Al(OH)3 + 3 H2 + 2Hg}$$ or$$\ce{2Al + 6 H2O ->  2 Al(OH)3 + 3 H2}$$This shows that unprotected aluminum metal quickly reacts with water, producing $\ce{H2}$ bubbles, and these $\ce{H2}$ bubbles are responsible of the real anodic half cell according to :
$$\ce{H2  -> 2 H^+ + 2 e-}$$ As a consequence, this "aluminum" electrode would have, at maybe $p$H$ 5$, a redox potential $\ce{E(H2/H+)}$ given by Nernst law : $$\ce{E(H_2/H^+) = E°(H_2/H^+) + 0.0296 V·log[H+]= 0 V + 0.0296 V·log 10^{-5} = - 0.148 V}$$ In your setup, this anode is coupled with an ordinary copper half-cell as cathode, whose standard redox potential is + $0.34$ V. The voltage of this cell is $$\ce{E(Cu/Cu^{2+}) - E(H2/H+) = 0.34 V -(-0.148 V) = 0.488 V}$$ This value is nearly the voltage that you have obtained in your cell ($0.50$ V).

Thank you that is very helpful. In regard to the nickel electrodes would I be correct in assuming that nickel, like aluminum, also forms this instantaneous oxide layer? Also, would zinc then take longer to form this oxidation layer or is there some other explanation as to why its value matched the theoretical?

@John Purely by the standard redox potential for $\ce{Ni(s)/Ni^2+(aq)}$, nickel should be quite reactive. The fact that it is not and is used in corrosion resistent alloys or metal platings says something. $\ce{ZnO}$ is probably not  so compact and is quite soluble in acidic solutions. Note also that some standard redox potentials are more or less just theoretical equivalents of related standard reaction Gibbs energy and cannot be ( or just  with difficulties ) directly measured, at least in simple scenarios. I have once seen redox potential for C/CH4 and would like to see its measurement.

The operation is forbidden by law? I'm surprised to hear that it's illegal to mix two chemicals. I would expect that it's legal to mix them but then the law would be very picky about how you dispose of the mercury compounds - including the mercury dissolved in the aluminium, if it does that.

@User253751 In my country you cannot use mercury in chemistry lectures and in the laboratory.

@Maurice - Mine also. There used to be a demonstration of a simple electric motor consisting of a star shaped sheet of metal rotating above a dish of mercury, between the poles of a horseshoe magnet. The points of the star dipped in the mercury. The current flowed from the battery to the centre of the star to the tips, via the mercury, and back to the battery. The star rotated and successive points dipped in the mercury. And out again, making a little spark. Not enough to hurt the students much, but sufficient to hurt the teacher during his or her career.

@Maurice can I see that law? The law that says you aren't allowed to use mercury in your laboratory, ever? Or is it only when teaching?

@user253751 - many things are banned in school labs but not others.

@User253751. I first would be pleased to know to whom I am writing. Anonymous name like yours does not prompt me to give you any answer. Second : My answer will last longer than allowed for an answer here. So tI will split my answer in two parts. let's go ! In my country try Switzerland, there is a federal law, dated 15. 12. 2000, named 813.1. and called "Law on chemicals" or LChim.  The Annex 1. 732 of the Article 3, is Called "Mercury". Its Chapter 2.b states :" Is forbidden the use of elementary mercury, of compounds of mercury, and of preparations containing mercury".

@User 253751...The following Chapter 3 is called Exceptions. In §3.1.c, it is stated that "the preceding interdiction does not apply if no substitutes of mercury is known for its use". When the law was introduced, our Society of chemistry teachers wrote to the authors of the text to be allowed to heat mercury oxide $\ce{HgO}$ in a test tube for showing in our lectures that both $\ce{O2}$ gas and droplets of elemental mercury are produced, and that nothing remains in the lower part of the tube at the end of the experiment. We got the answer that we should replace $\ce{HgO}$ by $\ce{Ag2O}$.

@MichaelHarvey then you should have specified school labs.

@Maurice article 3 specifies a definition of "dangerous substances". It does not appear to list forbidden substances, nor does it contain an annex? The annex of the complete law merely specifies that it repeals a previous law. The law itself does not appear to ban any substances, either; it says they require approval before being sold.

@user253751. I was the President of the Swiss Association of Chemistry Teachers at the time where the law was started. Before this new regulation we could decompose $\ce{HgO}$ in our lectures. And we did it without problems. I had personally gone to the central office in Bern, obtained the support of the Swiss Chemical Society, discussed with clerks in charge of the new regulation. We have repeated the same argument that you are giving now to me. And still others. We had long discussions. There was nothing to do. No mercury in high schools (Gymnasium in Switzerland) !

@Maurice It was not clear that the sentence "the operation is forbidden by law" was only talking about high schools


text:
How to completely remove Iron II Acetate from a steel surface through a chemical reaction?
Is there any suggestion to completely remove iron II acetate from a zinc-alloy steel surface? It can't be something abrasive, or if it is, it must be completely soluble.
Background:
My bike chain, which is made of zinc-alloy steel, got some bits of rust here and there after I cleaned it and left it wet for a while. I wax my bike chain, so the there's no protection against rust. I did some research and decided to sink it in white vinegar in order to get rid of the rust, since I figured between the plates and the pins must have got a bit rusty as well. I sank the chain in pure white vinegar for 10-15' and couldn't see any visible reaction. I guess I had to leave it there for some hours/days. I decided then to put the vinegar and the chain in an ultrasonic cleaner. After some 10-15' I checked it back. The rust was gone, but to my surprise (I'm not a chemist of course :) ), the chain turned completely black/dark brown-ish. I guess the ultrasonic cleaner catalyzed the process.
I guess the brown-ish layer is iron II acetate from what I researched. Basically, the acetic acid reacting with the steel.
My problem is, I want to get the chain completely free of the iron acetate. I was able to scrub most of it off with an iron brush, but only on the outside area. There is residue between the links and inside the pins. I already tried hot water and denatured alcohol (left the chain sank for hours), but none worked.
I wax my bike chain and I do it for a few reasons. It's environmentally friendly as no chemicals are needed to clean it, just hot water. Cleaning is fast and easy and I like to keep the drive train pristine. The issue with the iron acetate is that although the chain looks clean after each waxing. After the first run, the wax turns black-ish and the chain starts to look dirty again.
I read somewhere that NaCL would turn iron III acetate soluble. Maybe that would work for II as well? In any case, I'm not sure dumping the chain into a solution of water and table salt would solve my problem.
Any comments would be very welcome. Thank you very much.


text:
How can potassium hydroxide be prepared from available potassium chloride and quick lime
I live in a small town where Potassium hydroxide is not readily available. (I need above for preparing liquid soap.) But Potassium chloride ($\ce{KCl}$) and Quick Lime ($\ce{CaO}$) are available as fertilizer and White-wash ingredients respectively. Can $\ce{KOH}$ be prepared from above ingredients? What will be reaction conditions?

Any crudely prepared KOH will pose a lot of problems. Calcium forms scum.

Mixing $\ce{KCl + CaO}$, or $\ce{KCl + Ca(OH)2}$, will never produce pure $\ce{KOH}$ without $\ce{Ca(OH)2}$. And this Ca(OH)2 will prevent soap from being synthesized out of oil, as $\ce{(Ca(OH)2}$s destroy soap in case a little bit of soap has been synthesized. The only way of producing $\ce{KOH}$ out of $\ce{CaO}$ or $\ce{Ca(OH)2}$ is to mix it with potassium carbonate $\ce{K2CO3}$. Of course $\ce{Ca(OH)2}$ is not very soluble in water. But $\ce{CaCO3}$ is still more insoluble. So the following reaction is possible :$$\ce{Ca(OH)2 + K2CO3 -> 2 KOH + CaCO3(s)}$$ And the insoluble $\ce{CaCO3}$ can be filtrated to produce a KOH solution

Isn't potassium carbonate available as fertilizer? It's a lot more common. Then it would be a simple matter of mixing $\ce{CaO}$ with water and adding $\ce{K2CO3}$. Eventually you'd be left with solid $\ce{CaCO3}$ and $\ce{KOH}$ in solution.

A big limitation with this method is calcium hydroxide is not readily soluble in water.  You might get a gram or a few grams per liter into solution at thd start, but if you allow hydroxide ions to accumulate in the water then the solubility equilibrium shifts and the calcium hydroxide dissolution is choked off.

historically potash lye was made by leaching the ash containing these two reactants, continously collecting the koh solution. Perhaps this works by preventing the equillibrium shift as hydroxide ions don't accumulate


text:
What does the salt actually do during electrolysis?
Pure water does not conduct electricity.
When you dissolve a salt like $\ce{Na2SO4}$ in water the solution becomes a better conductor.
If electrolysis is done on this solution, oxygen and hydrogen gases form at the electrodes.
The ions $\ce{Na+}$ and $\ce{SO^{2-}4}$ don't seem to be doing anything further.
The electrons from the cathode don't involve the ions:
$$
\begin{align}
\ce{4 H+ + 4 e- &-> 2 H2} \\
\ce{4 OH- &-> 2 H2O + O2 + 4 e-}
\end{align}
$$
How are the salt ions helping here or what are they doing?

Does this answer your question? Electrolysis with two similar metals

An inert salt here is basically reducing the resistance of the electrolytic cell. Anything which is thermodynamically feasible gets reduced/oxidized at the electrode, in this case it is water. Note that sulfate ion is not always inert. It can be oxidized to peroxydisulfate ion under certain conditions.

This is most certainly not a duplicate of the linked question. Did you even read the two questions?

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's see what's going on at each electrode.
At the cathode (negative pole), water is decomposed according to : $\ce{2 H2O + 2 e- -> H2 + 2 OH-}$. This is only possible if there are some positive ions available near the cathode to compensate the negative charges appearing in the solution with these newly created $\ce{OH-}$ ions. Without positive ions like $\ce{Na+}$ near the cathode, these new $\ce{OH-}$ ions would repel further electrons coming from the outer circuit and prevent them from arriving to the electrode and from reacting with water. This would stop the electrolysis.
At the anode (positive pole), water is decomposed according to : $\ce{4 H2O -> O2 + 4 H+ + 4 e-}$. This is only possible if there are negative ions arriving near the anode to compensate the positive charges appearing in solution because of these newly created $\ce{H+}$ ions. Without negative ions like $\ce{SO4^{2-}}$ in the solution near the anode, these new $\ce{H+}$ ions would make a positively charged solution. This would prevent further electrons to be ejected from water. This would stop the electrolysis.
At the end, the composition of the solution is modified around the electrodes. Near the cathode, the solution contains more and more $\ce{NaOH}$. Near the anode, the solution contains more and more $\ce{H2SO_{4}}$ Of ourse if you stir and mix anodic and cathodic solutions, the acid and the base neutralize each other, which regenerate the solution of the original salt.


text:
Can we dissolve salts with higher lattice enthalpy by providing heat?
It is well-known that

A salt with a higher lattice energy will not dissolve at all, because then $\Delta_{\text{sol}}H^\ominus$ will be negative.

So, the reaction should be endothermic, right? My question is:
If then we provide heat to the reaction, then we should be able to dissolve such salts. Is this true?

Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt solubility product would increase with higher temperature.
But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.
The effective result depends on the particular case. The typical cases where significant solubility increase is observed are  potassium salts  $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the solubility table.
Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by van't Hoff equation

$$\frac{\mathrm{d}}{\mathrm{d}T} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

For solubility products, being equlibrium constants, is valid the following reaction:
$$\Delta G^{\circ}_\mathrm{dis} = \Delta H^{\circ}_\mathrm{dis}  - T \Delta S^{\circ}_\mathrm{dis} = RT \ln K_\mathrm{sp}$$

Can you elaborate on why heat is not a factor? Negative $\Delta H$ implies we need to supply energy, which we can, in the form of heat.

Okay, one final time: is this reasoning correct? Since by Hess' Law, $\Delta_{\text{sol}}H^\ominus =\Delta_{\text{lattice}}H^\ominus+\Delta_{\text{hyd}}H^\ominus$. Therefore, if $\Delta_{\text{lattice}}H^\ominus$ is big enough, $\Delta_{\text{sol}}H^\ominus$ will be largely negative, and the salt may not dissolve at all. Can this be improved upon more?

Heat is only a factor as it is needed to increase temperature, or to keep temperature during the endothermic process. Negative $\Delta H$ means the equilibrium constant increases with increased temperature. But as the process of practically insoluble salts causes negligible heat consumption, it does not play much role.

For c = a + b, saying if a is big enough, c is largely negative does not make sense. Note that for the solibility product value, the decisive is the value of the Gibbs energy of dissolution, so the entropy change can affect the raw value of the enthalpy.

So, what's the proper reasoning?

Try to think about any of your question for some time, before asking. It gives you more than if you get the answer by a cheap, easy and fast way.

Can you please, as the last time, let me know the proper reason for saying that salts with higher lattice enthalpy may not dissolve at all?

Let us continue this discussion in chat.

The point is, my book simply says that without any reason, and I want to conform my guesses.


text:
SN2 on compound containing halogen and carbonyl
I was given the following question on a sample exam for my organic chemistry course:

I am quite certain that it would happen in a SN2-type reaction considering that

the nucleophile is $\ce{CN-};$
the solvent is DMF.

However, I'm not sure what electrophilic center would be attacked first. I'd appreciate any explanation about why the nucleophile would choose one carbon over the other (if that were the case). The question also states that there are 2 product formed, so could it be possible that an elimination reaction is happening as well?

What do you think would happen if it attacked the carbonyl? Ok, you add into the carbonyl group, then what happens?

The $\pi$ electrons would go up to the carbonyl oxygen, giving it a negative charge?

If it attacks the carbonyl carbon, then you form an unstable tetrahedral intermediate with a negative formal charge on the oxygen. Unless you treat the compound with water so that the oxygen can remove it's formal charge, what will happen is that the reaction goes backwards where the CN group is kicked off.
However, if the CN attacks the Cl group, then you can have a substitution where you get a stable compound finally.

Can the elimination reaction also occur?  The question on the exam also stated that there are two products (A & B) formed

If the elimination occurs, the product is a Michael acceptor and cyanide will add giving the same product as the direct nucleophilic displacement of chloride. The other product is likely the cyanohydrin


text:
Why is +3 the oxidation state of Indium?
The three more energy electron layers for Indium(49) are, sorted from lower to higher energy:
[Kr] $5\text{s}^24\text{d}^{10}5\text{p}^1$
From this configuration, it could be said that is easy to subtract the $5\text{p}^1$ electron (oxidation state +1), next start subtracting electrons from the 4d layer. However, the more common oxidation state of Indium is +3 (as in $\ce{In2Se3}$). It seems electrons are subtracted from layer 5s before to the ones of 4d. This could break the rule of filling the lower energy layer first.
How to explain the +3 oxidation state of Indium taken into account electron configurations?

Indium is hardly the first element which exhibits similar behavior.

@IvanNeretin: yes. Question and answer can be extend to some others. Could be Copper +1 is the first one.

https://chemistry.stackexchange.com/questions/27662/what-is-the-electronic-configuration-of-feii-ion

When electrons are removed from an atom, those in the outermost orbital is removed first. The p orbitals are most away from the nucleus, then the s orbitals, and finally the d orbitals. In fact, the closeness of d orbitals to the nucleus accounts for higher energy of the 4d orbitals than the 5s orbital. Because electrons are closer, there is a stronger repulsion between them, increasing the energy. In the case of indium, electrons in the 5s orbital are farther away from the nucleus and are removed first.
I've also found a relevant discussion here from a chemistry community at UCLA.


text:
Verify heat produced by flame using enthalpy of fomation
I'm simulating using a CFD solver a pure oxygen/methane flame. However, I doubt that the combustion model does not produce enough heat. The heat from reaction is defined by $S_h=-\sum_{j} \frac{h_j^0}{M_j}R_j $ with $h_f^0[J/kgmol]$ the enthalpy of formation, M$_j$ the molar mass and $R_j$ the reaction rate of species $j$. The integral of the heat is my simulation is around $50kW$.
I tried to calculate analytically the heat prodruced using the equilibrium equation $CH_4+2O_2\rightarrow 2H_2O+CO_2$. With the non-stochiometric molar flux $\dot{n_{CH4}}=0.9625mol/s$ and $\dot{n_{O2}}=1.059mol/s$ I know I should be able to calculate:
$\dot{n_{CH4}}h^0_{f,CH4}+\dot{n_{O_2}}h^0_{f,O_2}=\dot{n_{O_2}}h^0_{f,H_2O}+\frac{1}{2}\dot{n_{O_2}}h^0_{f,CO_2}+(\dot{n_{CH4}}-\frac{1}{2}\dot{n_{O_2}})h^0_{f,CH4}+\Delta H$.
There I'm getting $\Delta H\approx472kW$, which is a factor 10. Either my simulation is really wrong or my ethalpy calculation are nonsense...
Is this way of chemical calculation correct?

For formatting of chem/math formulas/expressions, use can use MathJax with mhchem extension \ce{}. Note that chemical formulas should use upright fonts, what is honoured by mhchem \ce{}.

Never heard about these formula ! Where are they coming from ?

Enthalpy equilibrium,https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation#Use_in_calculation_for_other_reactions

Your equation looks OK.  What values did you used for the heats of formation?

@ChetMiller, thanks good to know.
$ h_{CH4}=-73.4e3;J/mo$l
$h_{O_2}=-0;J/mol$
$h_{H2O}=-285.83e3;J/mol$
$h_{CO_2}=-393.5e3;J/mol$
I'm more and more certain that my simulation calculation is wrong.

Shouldn't you be using the heat of formation of water vapor (rather than liquid water)?  I don't know whether that will change things much.


text:
Mechanism of Hypervalent Iodine Phenolic Coupling
I need some help understanding the hypervalent iodine phenolic coupling reaction taking place in this synthesis of morphine, the reaction uses $\ce{PhI(OAc)2}$ as the hypervalent reagent and trifluoroethanol as a solvent.

What is the mechanism taking place here? From my understanding the hypervalent compound contains a 3c-4e bond in which the iodide bears the positive charge and therefore is what the lone pair on the $\ce{-OH}$ attacks, what takes place after this?

This reaction results in the dearomatisation of one of the phenol rings, isn't a dearomatised ring a thermodynamically more unstable product, what is driving this reaction?

The paper describes the trifluoroethanol solvent as crucial, what makes the use of this solvent crucial? Is it the lack of nucleophilicity preventing a reaction with the iodine reagent?

The paper I am looking at is this one for reference.

Welcome to ChemSE! Your question will likely be considered as homework. You need to make some effort to answer the question yourself . Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange. Try searching "phenolic coupling".

I don’t think this is homework.


text:
Why does the infrared spectrum of HCl only have one peak?
Suggest one reason why HCl has only one peak?
Does this have something to do with fact that peaks (in IR spectroscopy) represent areas of the spectrum where specific bond vibrations occur and therefore since HCl has a single bond it only has one peak to show only one bond vibration?

Sorry had the wrong spectra - Gas phase HCl

Please, don't use all caps in your title. It comes off as very rude, even though that may not have been your intention. Please also try to choose appropriate tags for your question. This is not about organic chemistry or reaction mechanisms.

Do you recall that a linear diatomic molecule has only 3(2)-5=1, i.e., one degree of freedom.

@orthocresol it only comes off as rude to you please don't comment things that don't make sense and that don't help me answer my question. Also thank you to everyone who left a comment sorry I didn't reply sooner but I did end up figuring out the answer

@VENDINGMACHINE I'm afraid there is only your complaint that doesn't make sense. Before flagging a comment faithfully suggesting a change for better readability as rude, you could at least do a little research. Feel free to read about all caps usage on Wikipedia or check community guidelines on SO.Meta or CSE.Meta.

A molecule's vibrations are quantised making a ladder of vibrational energy levels. At very, very low resolution a single wide line (or peak) is seen from the $n=0 \to 1 $ transition. At high resolution rotational levels are also resolved as shown in the link in the comment by MaxW above. Additionally, v. weak overtones can be seen from $n\to 2$ and $n\to 3$ transitions (and so at higher energy) due to what is called anharmonicity in the bond's vibration. See a text such as 'Modern Spectroscopy' by Hollas (publ Wiley) for details but many Phys. Chem. textbooks will also explain this.


text:
IUPAC Nomenclature for cyclic compounds without principal functional group
I am struggling naming cyclic compounds with multiple substituents and no principle functional group . If anyone can please elaborate the rules, it would be appreciated.

Nomenclature gets variegated there. And hybrid nomenclature is tolerated (at least it was..).

I have provided example, here I don't understand where to start numbering from.

I would call this 2-chloro-4-nitrotoluene.

@AravindSuresh And you would be correct if the question were about any name. PIN though is 2‐chloro‐1‐methyl‐4‐nitrobenzene.