text:
Why aren't elements with two valence shell electrons stable?
We all know that hydrogen becomes stable by completing duplet and helium is stable since its valence shell has completed the duplet
Still, other elements don't behave like that. I want to know that why aren't elements like magnesium, beryllium etc, with two valence electrons be unreactive like helium atom, or why can't other elements become stable by completing duplet in the valence shell ?

Titles are not supposed to be clear enough, like book titles are not supposed to tell the book story. They should just give a good hint what it is about. OTOH, the question body is supposed to provide good question elaboration, not just rephrasing the title. That includes results of textbook and online research. Otherwise such question can become a closure candidate.

@Poutnik okay I will edit it. Thanks for response.

Stable is a very poor word here. The alkaline earths are all perfectly stable elements. Do you mean "unreactive"?

@Ian Bush yes , by stable I meant unreactive.

@Ankit do you know about orbitals?

@Aniruddha Deb yes I have heard of them.

That's a lot like this earlier question, but maybe this earlier one would better closed https://chemistry.stackexchange.com/questions/140840/why-isnt-just-a-fully-occupied-valence-s-orbital-stable

https://chemistry.stackexchange.com/questions/27739/does-the-magnesium-2-ion-prefer-an-octet-or-more

Short answer
$\mathrm{s}$ electrons of alkali metals and alkali earth metals perceive relatively low central electrostatic charge, because of shielding the kernel by inner electrons. They are also farther from the kernel, compared to 1s electrons of hydrogen and helium, what overall  leads to low ionization energy of such electrons and high reactivity of such atoms.
Helium and hydrogen have no inner electrons to shield the kernel and their  1s orbitals are smaller, compared to respective higher s orbitals  So their  ionization energy, especially for helium, is high, and reactivity reciprocally low.
For these reasons and the ones more described below, there is significant decreasing trend in ionization energy of ns electrons of group 1 and2 elements ( values approximated):
$\ce{H}$ $\pu{13.6 eV}$
$\ce{He}$ $\pu{24.7 eV}$
$\ce{Li}$ $\pu{5.3 eV}$
$\ce{Be}$ $\pu{9.3 eV}$
$\ce{Na}$ $\pu{5.1 eV}$
$\ce{Mg}$ $\pu{7.8 eV}$
$\ce{K}$ $\pu{4.3 eV}$
$\ce{Ca}$ $\pu{6.1 eV}$
leading to respective reactivity.
$\mathrm{p}$ electrons of chalkogens and halogens perceive relatively high central electrostatic force, leading to the high ionization energy of such electrons and relatively high electron affinity and because that also high reactivity of such atoms.
Longer answer
Classical case of electrostatics
There is the well known rule from electrostatics, that if there is a system of charges with perfect spherical symmetry, it acts toward its neighbourhood like if it were a point-like charge with the effective charge equal to the sum of charges.
If we have such a system with point-like central big positive charge and negatively charged spherical layer with spherically symmetric charge distribution, the central positive charge is shielded/screened by the outer negative charge of the layer, so the system neighborhood perceives much lower central charge than it really is, i.e. the difference of both charges.
Multielectron atomic quantum models
Statistical distribution of electron density around atomic kernels follows wave functions called orbitals(*). The electron density and related charge distribution is not therefore spherically symmetric. Orbitals also spatially overlap and go sideways, what is further worsening the screening of the kernel charge.
The overall decreasing of screening efficiency goes in order s, p, d, f. Also, electrons from inner "shells"/"subshells" perform better screening then electrons with the same shell or subshell.
It has the consequence the overall kernel screening and electron energy depends on particular atomic electron configuration. For more you can read about Slater's rules.
Atoms with a new principal quantum level just started being filled, like alkali and alkali earth metals, have the screening by inner electrons particularly good. Therefore these $\mathrm{ns}$ electrons perceive particularly low effective kernel charge.
This has few consequences:

The last occupied s orbital is particularly big - see atomic radii.

The electrons in the last occupied s orbital have particularly high energy ( compared to other atoms ).

High electron energies lead to favourable chemical bonding, releasing energy, for covalent bond cases like $\ce{R-Mg-X}$ or $\ce{R-Li}$

High electron energies lead to easy ionization due low  ionization energy , which can be more than compensated by lattice ionic energy of solid salts, or by hydration energy in water solutions.

Similarly, elements with just few missing p electrons like halogens or chalkogens have for these electrons poor kernel shielding. This leads to a high effective kernel charge and significant amount of energy released  by capturing extra electron to form anion or polar covalent bond. The related term is electron affinity.

(*) Note that the term orbital has 3 major, related but distinguished meanings:

orbital(1) - a wave function as the particular solution of the Schroedinger's quantum wave equation

orbital(2) as a quantum state of an electron, belonging to orbital(1)

orbital(3) as a 3D region(or 3D surface/shape), describing the region of significant probability of electron occurance (or its border threshold value), belonging to orbital(1,2)

I think you got this upside-down: "For such system is said the central inner charges are shielded/screened by the outer charges."

@BuckThorn Perhaps I have misformulated it. I meant the kernel is for outer electrons  partially screened (off?)/shielded by the inner electrons. I will check it.

@BuckThorn I think it is right. Note that that paragraph as about a classical and symmetric scenario. If you have e.g. a central point charge +2e and spherically symmetric charged layer -1e, it would act for the neighborhood as if it was a point charge +e, as +2e is partially screened off by the negative layer.

I'm a tad confused. Isn't your point later in the answer that inner electrons shield outer (valence) s electrons rendering the nuclear charge effectively less, increasing reactivity and lower ionization energy?

@Buckthorn the first paragraph is about classical spherically symmetric scenario: Central pointlike positive, spherical negative layer. For neighborhood it would be like central-layer summary effective charge. the2nd paragrapha is about atoms, with lack of sph. symmetry and orbital overlapping. There is partial screening of the kernel by inner electrons for the outer electrons. this screening is good for   s electrons of K or Ca, but bad for p electrons of O or F.

@BuckThorn Rather: Inner ones shield the kernel, providing kernel shielding for outer ones. then good such shielding leads to lower ionization energy and higher reactivity. Feel free to suggest rephrazing of critical parts.

Your comments make sense, it is exactly how I understand it. It's kind of strange, you have seems to me a very logical answer if you only flip that sentence around and write "For such system is said the outer electrons are shielded/screened by the central inner electrons", everything else than makes sense and is consistent.

Let us continue this discussion in chat.

Apparently nobody has ever given an answer to the question, which I repeat here  : why are Mg and Be not as unreactive as Helium ?

@Poutnik I guess I got confused about who you meant was doing the screening/shielding and who was being screened/shielded. It is clearer now.

@Maurice better ?

@ Poutnik. As far as I understand, nobody has really answered the question about why Mg and Be are not at least partly unreactive like He, because the 2s or 3s shell is complete ?

@Maurice They ARE partially unreactive as He, compared to Ca-Ba. with partiality much closer to Ca-Ba than He. It is the quantitative trend, not qualitative-like switch, even if for helium, it practically is, as He compounds are very extremal.

@ Poutnik. How can you say that Mg is partially unreactive ?

You have said it. Also notice their local maximum of ionization energy. https://en.wikipedia.org/wiki/Ionization_energy. Li, Be and Mg are significanly much less reactive then their heavier cousins.

Woah, that answer above me is very complicated. Im sure that it is also right, but let's explore a simpler way to look at this. Instead of thinking of elements becoming unreactive when their octet/duplet is filled, lets think about unreactivity in terms of noble gases. If we look at the noble gases on the periodic table, we can see that they all have 8 valence electrons (except for He). You can verify this by writing the electron configuration or looking at the group number. That is why we say that atoms want to fill their octet to become unreactive, because that would generally give them noble gas configuration. Now, this is not always true. Instead, we should really say that atoms want to become isoelectronic with a noble gas, which means that they have the same electron configuration as the noble gas. Now back to the question, why does only Hydrogen need 2 valence e- to become unreactive? Well, its because it only needs to have 2 valence electrons to become isoelectronic with the nearest noble gas, helium. This wouldn't work with other elements, which need to have 8 valence electrons to become unreactive, since all other noble gases besides He have 8 valence electrons. I am just an AP chemistry student, so I don't exactly have any credentials, so feel free to correct me, but this made sense to me. Have a nice day.


text:
Wavelength and Colour in Coordination complexes
My textbook said that coordination complexes with weak field ligands require a larger wavelength for obtaining enough energy for electron transfer(and hence, the manifestation of a colour), but if E = hv(E is the energy, h is Planck's Constant and v is the frequency), then shouldn't the wave of higher frequency be needed?

Low spin (formed from strong field ligands) complexes result in a larger difference in energy between the $t_{2}g$ and $e_{g}$ orbitals. High spin (caused by weak field ligands) complexes are, evidently, the opposite.
This means thats high spin complexes require lower energy photons than low spin. Since wavelength and frequency are inversely proportional, high spin complexes require lower frequency (longer wavelength) photons to promote transitions.

source: Stack Exchange

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

sort: votes


text:
Ambiguity in definition of primary valency in Werner's theory
In Concise Inorganic Chemistry (5th Edition) on page 196 , the primary valency of $\ce{[Co(NH3)5Cl]Cl2}$ is given to be 2.

In other textbooks and in the question here, we get the idea that:

The primary valency of a compound is equal to its oxidation state.

In 'Inorganic Chemistry(4th Edition)' by Catherine E. Housecroft and Alan G. Sharpe':

Werner referred to the oxidation state of the metal ion as its ‘primary valence’ and to what we now call the coordination number  as  its  ‘secondary  valence’.

This further suggests that that they are in fact exactly identical and just a matter of how they are referred to.
The oxidation state of cobalt in $\ce{[Co(NH3)5Cl]Cl2}$ is +3 and hence would suggest a primary valency of 3.
What is the correct primary valency of the compound? I am aware that these concepts are not in modern use yet it would be useful to know how Werner originally put forward this idea.

Maybe the definition of primary and secondary valences has changed between Werner's time and today.

I had answered it here: https://chemistry.stackexchange.com/questions/126954/what-is-primary-valency/126963#126963

@Maurice . That might be the correct speculation yet it would be beneficial to know exactly how and why it changed.

@M.Farooq Thank you. I had already gone through your answer as it is included in the link I attached in the question. I am all for dismissing these concepts as useless in todays context as you mentioned but it still doesnot explain the disprency in the result.

I believe you are right, that the primary valence here must be 3. But there might be a caveat. If we do a titration of this complex with silver nitrate, do we see two chlorides or three? If we only see two chlorides, then Werner might have concluded a primary valence of 2. I will have to check some historical papers.

Basically we have to check what Werner thought about it not Alan G. Sharpe or J.D. Lee or any other Indian textbook writer. Facts come out from original papers. Check Google books of the 20th century and even 19th century.

We will only get two chloride ions. However almost all recent books say that primary valency is basically oxidation number.  I couldn't find the thesis of Werner in English. I will surely look some more.

I will let you know. Web is full of incorrect info on historical matters and of course on scientific matters too. The work is not in Werner's thesis but rather in his papers in German. They are all online but need subscription. I am basically looking at the book "Coordination Chemistry:A Century of Progress". I am glad that you checked and challenged multiple books.

George B. Kauffman has edited an excellent book, "Coordination chemistry: A century of progress." It was published as an ACS Symposium Series. In the chapter, he writes, "If history teaches us anything, it teaches us that the latest view is not always the best and that change is not always progress." I may add that there is no authority in science. I am glad that you challenged textbooks.
Basically, in Werner's time, the prevalent theory of the complexes was that the complexes are chained compounds, which was not consistent with the solution conductivity and optical isomerism of the complexes. Werner proposed the primary and secondary valence concept.
Contrary to most modern chemistry books, including that respectable Alan G. Sharpe's texts, Werner's primary valence matches the "oxidation" state of the metal in simple cases. The oxidation state matches Werner's primary valence when there is no anion in the coordination sphere. The textbook version of Werner's story fails for the type of compounds you just asked, such as $\ce{[Co(NH3)5Cl]Cl2}$.
Here the oxidation state is indeed 3. However, Werner's primary valency of this compound is 2.
Kauffman mentions

Werner postulated two types of valence—Hauptvalenz, primary or ionizable valence, and Nebenvalenz, secondary or nonionizable valence. According to Werner, every metal in a particular oxidation state, that is, with a particular primary valence, also has a definite coordination number, that is, a fixed number of secondary valences that must be satisfied. Now, whereas primary or ionizable valences can be satisfied only by anions, secondary or nonionizable valences can be satisfied not only by anions but also by neutral molecules containing donor atoms such as nitrogen, oxygen, sulfur, and phosphorus.

The keyword is that primary valence is the ionizable valence. The ionizable valence of the complex in question is 2. $\ce{[Co(NH3)5Cl]Cl2}$ would generate three ions in solution $\ce{[Co(NH3)5Cl]^{2+}}$ and two $\ce{Cl-}$. These experiments were easily doable in Werner's time by conductivity measurements. Secondly, as discussed in the comments, argentometric titrations would show only two chloride ions rather than three.

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


text:
Write half-cell reaction for a battery

A primary battery has the reaction:
$$\ce{2MnO2 + Zn + 2NH4Cl -> Mn2O3 + Zn(NH3)2Cl2 + H2O}$$
Write the half-cell reactions for the given equation.

I am having trouble understanding how you know which will be the oxidation reaction and which will be the reduction reaction. From what I have understood I should assign the oxidation states of each reactant and product and from that I assume that the reactions should be:
Ox: $\ce{Zn + 2NH4Cl -> [Zn(NH3)2]^{2+}}$  +  ...
Red:  $\ce{2MnO2 +  ...  -> Mn2O3 + H2O}$  +  ...
And by balancing out the hydrogens and charges I get that:
Ox: $\ce{Zn + 2NH4Cl -> [Zn(NH3)2]^{2+} + 2H+ + 2 Cl- + 2e-}$
Red:  $\ce{2MnO2 + 2H+ + 2e- -> Mn2O3 + H2O}$
Is this correct? Or should the oxidation reaction only be: $\ce{Zn  -> Zn^{2+} + 2e-}$ and then the reduction reaction would consist of the remaining compounds?
All help is appreciated.

Your equation in the question is not balanced by mass, neither oxidation half-reaction.

@Poutnik Oh, I completely missed the Cl$_2$.
So then my oxidation reaction would be: Zn+$2$NH$_4$Cl⟶[Zn(NH$_3$)$_2$]$^{+2}$+$2$H$^+$+$2$e$^−$ + $2$Cl$^-$

@Poutnik Thank you, I didn't know about that notation. But is what I wrote correct?

@ Katara. Now your equation written with the $2$ chloride ions is correct

Please fix the equations in your original question, not in the comments.

The important thing is not us to tell you if what you have written is correct or not. The important is YOU to understand, why it is correct or not. Put in the question why do you think it is or is not correct. BTW equation sides should be either both molecular kind or both ionic kind, not mixed kinds as for the balanced Ox.


text:
How to get equation of state from fugacity
Given that the fugacity of a substance is $f=Pe^{bT}$, how do I obtain the equation of state of the substance?
This is my approach, but I have not made good headway:
$$RT d (\ln f) = VdP \\
d(\ln f) = \frac{V}{RT} dP \\
bdT = \left( \frac{V}{RT} - \frac{1}{P}\right)dP$$
But what next? I see that
$$\left( \frac{V}{RT} - \frac{1}{P}\right)dP = \frac{(Z-1)}{P}dP$$
But I don't see how I can reduce this further.
My end goal is to find derivatives of pressure wrt temperature keeping volume constant. How do I go about this problem?

That first equation is supposed to be at constant temperature.

@ChetMiller If i change it to $VdP - SdT$, and I equate both sides, I get $V = RT/P, -S = RTb$. What can i do next?

You are supposed to do the integration at constant T.

That first equation is supposed to be at constant T.  You can rewrite it as $$RTd\ln{(f/P)}=RT(z-1)\frac{dP}{P}$$or $$\ln{\frac{f}{P}}=bT=\int_0^P{(z-1)\frac{dP'}{P'}}$$So if z behaves like z = 1+aPT, this becomes:
$$bT=aPT$$ or b = aP

Thank you for your answer @ChetMiller. I was trying this problem with $b$ being a constant... Is it possible to for $b$ to be a constant?

Not for real gases.  Check out a corresponding states plot of z as a function of reduced pressure and temperature.


text:
Calculate concentration of 10 µM to ng/ml Cannabidol
I would like to calculate the concentration of 10 µM unit in vitro to blood plasma unit ng/ml?
What formula do I need to use?
Do I need to use molecular weight? In my case, it is 314,47 g/mol for Cannabidiol
Can someone help?

You cannot be serious. I hope it is just a test question at high school level  and not a real lab task.

$10~ \mu$mol is $10^{-5}$ mol, and it weighs $\ce{10^{-5}}~ $mol · $314.47$ g/mol = $3.1447·10^{-3}$ g = $3.1447$ mg.
$1$ mL of this solution contains $3.1447~\mu$g cannabinol, which is also $3144.7$ ng.
So the concentration of your solution is $3144.7$ ng/mL

Be aware of proper taming of precision. 10.000 uM may mean 3144.7 ng/mL, but 10 uM does not. 3140 ng/mL is more than enough

@Poutnik. OK ! Let's admit $3140$ ng/mL, or even $\ce{3.1 10^2}$ ng/mL

You mean $\pu{3.1e3}$


text:
What is the chemical formula of ghee?
What is the chemical formula of ghee (Indian butter)? Also, name the elements present and state the number of their atoms.

There is no formula of "ghee". It is a complex mixture. It should not be called Indian butter because "ghee" itself is an accepted word in English dictionaries.

"The physical and chemical characteristics of ghee extracted from cow's and sheep's milk were studied. Iodine number was lower, but saponification number was higher in sheep's ghee. 1,2-Diacylglycerides were absent in sheep ghee. The range of vitamin A was 315–376 μg/100 g and of cholesterol 252–284 mg/100 g. Fatty acid composition showed a relatively high degree of saturation (53.9–66.8%) with C16:0 (31.7–38.3%) and C18:1 (21.6–33.7) being the predominant saturated and unsaturated fatty acids, respectively." https://www.sciencedirect.com/science/article/abs/pii/0308814693900073

This is a site on chemistry. So all commenters and answering users should tell OP that while the composition of ghee can be discussed, his/her question is not acceptable as it lacks basic knowledge / preparation of the question.

The composition of ghee varies depending on the animal whose milk has been used. Ghee is a complex lipid of glycerides (majorly triglycerides), free fatty acids, phospholipids, sterols, sterol esters, fat soluble vitamins, carbonyls, hydrocarbons and carotenoids (only in ghee derived from cow milk).

Milk fat contains at least 500 fatty acids and fatty acid derivatives with 4 – 20 or more carbon atoms in their chain. The fatty acid may saturated or unsaturated and usually contains an even number of carbon atoms. Following table shows detailed composition of buffalo milk and cow milk fatty acid composition:

The vitamin A content ranges from 315 to 375 international units per 100 grams. The saturated fatty acid profile is 53.9 to 66.8 percent, the unsaturated fatty acid profile is 22.8 to 38.0 and the other fatty acids profile is 3.5 to 10.4*. Cholesterol amounts ranges from 252 to 284 mg/100 grams.
*Vaccenic acid (2.18%) and conjugated linoleic acid (0.77% in buffalo and 1% in cow ghee) are the other main ruminant trans fatty acids.
References

http://ecoursesonline.iasri.res.in/mod/page/view.php?id=5793 (table sources)
https://en.wikipedia.org/wiki/Ghee
Parmar, Nirav & Mehta, Bhavbhuti & Aparnathi, Kishorkumar. (2018). Composition of ghee prepared from camel, cow and buffalo milk. Journal of Camel Practice and Research. 25. 321. 10.5958/2277-8934.2018.00046.2.
Composition of ghee (Samn Barri's) from cow's and sheep's milk
AbdulrhmanAl-Khalifah, Hassan Al-Kahtani, Food Chemistry
Volume 46, Issue 4, 1993, Pages 373-375, DOI: 10.1016/0308-8146(93)90007-3

What other fatty acid means, if it is neither saturated neither unsaturated ? Unless it means saturated or unsaturated fatty acid not particularly listed in those 2 profiles.

@Poutnik The 3rd para is from Wikipedia article and while it was not explicitly mentioned what were the other fatty acids, I believe they are long chain polyunsaturated fatty acids.


text:
stability of a negative charge in halogens and that of orbitals
when we talk about stability of halogen anions we say,
$\ce{I-} \gt \ce{Br-} \gt \ce{Cl-} \gt \ce{F-}$, and the reason is that "the negative charge (the electrons) have more space to move about in the larger halogen". But when we talk about stability of a negative charge on an orbital, why do we say that the negative charge on the $sp$ hybridized orbital (which is the smaller one) is more stable than $sp^2$ and then $sp^3$ as the least stable where as the  $sp^3$ orbital is the biggest orbital among all of the three and can allow the electrons to spread out more.

The primary stability factor is energy. Also, what do you mean by stability, particularly what do you mean saying I- is more stable than F- ? It is easy to oxidize I-, but try it (chemically) with F-.

I meant that the iodine anion is more stable than the flourine anion, as there is much more space for the electron density to spread out in iodine anion. @Poutnik

No, that is considered reason of stability, not what you mean by stability itself. Do you mean hydrated or naked ion ? Do you mean acido-basic stability or general chemical stability ? You have put too little context in the question, what usually leads to unnecessary clarification ping-pong.

oh that was a question on checking which compound was more acidic

and that has not been included in the question, because.......    ?  For the basicity of hydrated halogenides part, or acidity of hydrated hydrogen halogenides,  it is much more complex, as the environment interaction and thermodynamics has to be considered. See https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/The_Acidity_of_the_Hydrogen_Halides

hmm my bad for not including it, thank you for your explanation.

When we talk about hybrid orbitals and stability of charge on them, the electronegativity factor plays a more prominent role. The sp hybridised orbital is more electronegative than the others, owing to Bent’s rule which states - Greater the s character of a hybrid orbital, greater is its electronegativity.
Therefore, species having the negative charge on an sp hybridised atom tend to be less basic.
(You might be interested in knowing that, an sp hybridised carbon atom is actually more electronegative than a nitrogen atom.)

Looking at the other comments, I also get suggestions that there may be a slight confusion between basicity and nucleophilicity. Basicity is a characteristic property, and that depends only on how stable the negative charge on an atom is. Nucleophilicity is, however, dependent on a few other things besides that. It’s not a characteristic property, it’s more like a kinetic property.  They do go hand in hand sometimes, but not always.

In the context of my previous comment, for example - the iodide ion is a weaker base than the fluoride ion. And that won’t change no matter what environment (solvent) you want to test their basicities in. But, the iodide ion happens to be a stronger nucleophile in a polar protic solvent, since the fluoride ion is so heavily solvated (or hydrated) that it loses it’s ability to act as a nucleophile to a large extent. The iodide ion, however, is not solvated effectively because of its huge size and a lesser charge density.

may I know why is it that "more the s character for an orbital, more is the electronegativity of it"? You have said it as bent's rule, but could you spare sometime for more explanation. The thing I am actually confused about is, people say "more is the size of an atom, the more stable it is to bear a negative charge", but doesn't it counter to say that "$sp$ orbitals which are smallest, can bear a negative charge and be stable compared to $sp^2$ and $sp^3$"?

The s orbitals are closer to the nucleus than p orbitals, which makes the electron pair in an s orbital more strongly bond. That’s what makes the trend so.

And I totally understand your point regarding the size of the orbitals. But here’s the thing - electronegativity factor and size factor are giving conflicting suggestions. It just happens to be so, that the electronegative factor is more prominent. I’m not saying size factor is incorrect. It’s just that, considering electronegativity gives a better (in this case the totally opposite) suggestion.

If I may add this, it’s just one of those things where we can’t quite understand where we went wrong...one thing being more fit than other doesn’t imply the first thing is unreasonable. Because again, I see no fault in your reasoning.

Thank you for your answer, it helped me to understand.

@AtharvaNavaratne regarding your first message about  Nucleophilicity vs basicity , i have a doubt regarding that .CH3O- is more basic than CH3COO- as the former has a more negative charge density than latter . So ,CH3O - must be more solvated in protic solvent and hence must behave as a weaker nucleophile . But it is the reverse .The nucleophilic character follows the same trend as the basic character in this case . Can you suggest any reason for this ? And what are the other factors except hydration of anion which differentiates nucleophilic character from basic character ?

@AtharvaNavaratne can you give me an example where to decide basic nature , electronegativity of atom bearing negative charge dominates over the negative charge density ?


text:
Is heat a form of energy or is it just a transport of energy, not energy itself?
In every textbook I read, it says that heat is a form/type of energy... but in lecture, my professor said, that there is no such energy as heat and that heat is just a method by which we can transport energy to another system. Do these 2 statements say the same and I'm just reinterpreting the first one badly. If I were to say in my own words what heat is just based on the textbook's definition, I'd say that heat is energy, that... But if I were to say it by the lecturer words, I'd say that heat is a disordered transfer of energy, not energy itself.
What's more correct?

Heat is an energy, because the concept of Heat content makes sense.

No, heat is not the same as thermal energy. Heat is energy transfer other then matter transfer or thermodynamic ( =volume or nonvolume ) work, i.e. mostly, but not limited to, transfer of thermal energy as kinetic and potential energy of molecules/atoms.

So, the latter is correct and the textbooks have it mixed up?

Yes, they mix it up. Heat is a kind of energy, but at the special circumstances. It is energy on the move, the energy flow, aside of that excluded above. Note that we write $\Delta U = Q + A$, not $\Delta U = \Delta Q + \Delta A$. Also, Q can be negative, while thermal energy cannot be.

Small correction: Rather energy transfer than flow, as heat has the same unit as energy [J]. Energy flow would be rather [J/s=W], or [J/s/m2=W/m2], depending on context.

I agree with others. Heat is energy in transit. However, I would not really say the first is totally wrong. It becomes rather semantic. I am more concerned of the opposite, when a Delta over a reaction arrow is taken erroneously as high T instead of the fact that there must be a transit of energy from the surrounding to the reaction vessel. And this transfer, the heat, requires energy, of course.

https://chemistry.stackexchange.com/questions/112043/is-heat-just-a-change-in-temperature

To describe 'heat' as an energy we have thermal energy. This thermal energy is a way by which an increase in kinetic energy of particles occurs.

As you know that this energy gets transfered in the form of infrared rays so heat is more of a layman word

But using it in everyday terms for describing transfer has no problems. Btw Physics Stack Exchange may help you.


text:
If Electronegativity of Cl is greater than H2 then why is the bond angle of Cl2O greater than H2O?
If Electronegativity of Cl is greater than H2 then why is the bond angle of Cl2O greater than H2O?
Cl2 has more EN than H2 and size of Cl2 is more than H2 but if we go according to EN then more angle should be of H2O as electrons in H2O would more nearer to O2 than the electrons in Cl2O near to O2?

Steric strain possibly, due to the atomic radius of Chlorine being significantly larger than that of Hydrogen.

Bond angle is largely governed by the location of electron cloud with regard to atoms in a molecule.
In dichlorine monoxide, the electron cloud is more spread out between oxygen and chlorine while in the water molecule it’s much closer to the oxygen atom. Therefore, the net bond paid-bond pair repulsion is higher in dichlorine monoxide than in water, while the lone pair-lone pair repulsion is almost the same.

Also, chlorine atoms are much bigger than hydrogen atoms, because of which they also undergo steric repulsion causing the bond angle to further increase.
That’s why the bond angle in dichlorine monoxide is greater than in the water molecule.


text:
How can I make a lot of microscopic rods of metal?
I want to make around 40-200 g  of metal cylinders that would ideally have a length of around 1-100 μm in size (smaller preferred). The rods will be made of some ferromagnetic material such as Iron or Nickel.
The reason for this is to make a magnetic field display where the tiny metal rods are submersed in a viscous liquid which is trapped between two clear sheets of material (likely glass or acrylic).
Since these are rods and not balls, this should allow them to stick end to end without bunching up so much and better visualise the magnetic field lines.
Although i'd prefer The rods to be smooth, they don't necessarily have to be a cylinder shape, they just have to have a length to width ratio higher than 2:1 (although i'd prefer a bigger aspect ratio than that).
It is preferred, but not necessary that the rods are all the same length with minimal variation in size.
Last preference is for the metal to look shiny rather than black as this helps a lot with visualising the field lines. I'll take what I can get though.
Haven't found anything online like this but I also don't really know what to search for.
I thought of using thin iron/steel wire and cutting it up but this process would most likely be too difficult and tedious to carry out unless I could construct some machine or jig to do it for me. I also have struggled to find any sort of ferromagnetic wire smaller than 0.1mm which would still be too thick for what I want.
I also thought of long metal shavings but they tend to be twisted and irregular (and possibly too flat) to work well for this.
Perhaps I could use some sort of micro 'cheese grater' to grate pieces off of a metal block?
Could these rods be synthesized via electrolysis somehow?
I'm at a loss so any help or pointers on how to go about this would be greatly appreciated!

You're better off using a commercially available ferrofluid, rather than synthesizing your own. If you want to synthesize your own easily, try experimenting with plain old iron filings and see if the results are satisfactory.

I offer a weird suggestion: find the finest (flexible) iron or steel wire you can get; cut a piece half as long as the distance between the poles of the magnet. Bind a dozen or so pieces of this length of wire together; attach to one pole of the magnet; they should splay out like the mini-particles you are looking for, but without the mess of liquid etc. Attach another bunch to the other pole.

Here is a relevant 2013 paper: Synthesis of pure iron magnetic nanoparticles in large quantity, to quote:

Free nanoparticles of iron (Fe) and their colloids with high saturation magnetization are in demand for medical and microfluidic applications. However, the oxide layer that forms during processing has made such synthesis a formidable challenge. Lowering the synthesis temperature decreases rate of oxidation and hence provides a new way of producing pure metallic nanoparticles prone to oxidation in bulk amount (large quantity). In this paper we have proposed a methodology that is designed with the knowledge of thermodynamic imperatives of oxidation to obtain almost oxygen-free iron nanoparticles, with or without any organic capping by controlled milling at low temperatures in a specially designed high-energy ball mill with the possibility of bulk production. The particles can be ultrasonicated to produce colloids and can be bio-capped to produce transparent solution. The magnetic properties of these nanoparticles confirm their superiority for possible biomedical and other applications.

So, to answer your question, it actually may be feasible based on more recent work. Note: the article appears in the applied physics section of Journal of Physics, reflecting the likely significance of this topic in this area.

Thanks for the reply! Ultrasonification + ball milling sounds like a really cool concept and I might play around with that idea myself!
What exactly is bio-capping the particle? Is it the same as forming a micelle around an iron particle?
A transparent liquid that is also magnetic would be really amazing to make and I can see it being used in a lot of different applications!


text:
Glucose and related aldohexoses are interconvertible with a little base. Does anybody know how much of each sugar is present at equilibrium?
If you add NaOH to D-glucose in water, a mixture of almost all aldohexoses will arise over time, including mannose and fructose. I wonder how much of each will be present and what could explain the distribution. My assumption is that the amount will match stability and that the most stable D-beta-glucose will be most abundant. I'd love to see better quantitative data on this. Do you know how much of each aldohexose is produced?
I found a paper using sodium titanate nanotubes as catalyst but not sodium hydroxide. There, Kumar et al. show in figure 3a(CC-by) that glucose is converted to mostly fructose (blue) and to a lesser extent mannose (red).

The paper does not really answer my question as it only touches upon 2 aldohexoses and doesn't give a molar % (or I didn't find it). I do wonder how much of each sugar would be present in the basic reaction of glucose described above. Does anybody know of experiments to answer that?

Maybe look at this: https://pubmed.ncbi.nlm.nih.gov/11559169/

Thanks, Buck, for the link to the paper by Zhu et al. It's an interesting NMR study, teasing apart how much of which acyclic and cyclic form of various aldohexoses including glucose is present. It does not look at the interconversion of those. That's what peaked my interest; the fact the glucose can be converted to the other aldohexoses.


text:
Specific Heat Derivation
The specific heat capacity at constant volume and the specific heat capacity at constant pressure are intensive properties defined for pure, simple compressible substances as partial derivatives of the functions u(T, υ) and h(T, p), respectively,
$$c_v=\left ( \frac{\partial u}{\partial T} \right )_v$$
$$c_p=\left ( \frac{\partial h}{\partial T} \right )_p$$
Can someone explain why this is? The book from which I got this from doesn't derive these expressions.

Those can be regarded as definitions. Do you have a more specific question about the definitions?

@BuckThorn Yes. I was trying to find out how to obtain this expression through differentiation.

Well, for $c_p$ for instance, start with the enthalpy function h(T) and take the derivative with respect to T while holding the pressure constant. It is not more mysterious than that.

@BuckThorn  I'm just confused as to what h(T) would be.

Consider a gas which was subjected to two different processes: isochoric and isobaric.

Isochoric: If a certain amount of heat $dq$ was given to this gas, then this heat will be completely used up in raising it's temperature since $dw$ is zero. So, $$dq_v = dU$$ Also $$dq_v = nC_v dT$$ where $C_v$ is constant volume molar heat capacity. Therefore we can say that $$nC_v dT = dU\implies C_v =\frac{1}{n}\frac{dU}{dT}$$ So for 1 mole of gas $$C_v =\left(\frac{dU}{dT}\right)_v$$

Isobaric : Now in this process a certain amount of heat $dq_p$ is given to the gas and this will raise it's temperature and a part will be used up in doing work against external pressure (if any). Change in Enthalpy $\Delta H$ as we know is defined as heat tranfered in an isobaric process. This process is isobaric so $$dq_p = dH$$ Again writing heat transer in terms of molar heat capacity $$dq_p = nC_p dT$$ Therefore $$nC_p dT = dH\implies C_p = \frac{1}{n}\frac{dH}{dT}$$ For 1 mole of this gas $$C_p = \left(\frac{dH}{dT}\right)_p$$

I don't think this is an answer because you have only illustrated the equations for an ideal gas instead of having a discussion on the actual quantities involved.

Actually these equations are applicable to real gases as well since I haven't used the ideal gas equation anywhere

Oh you are right, I again misunderstood. On a lighter note, I changed places of isobaric and isochoric since you explain isochoric first

I think it'd be easier to undestand if you wrote from where you got each definition i.e: one where $ dq_v = nC_v dT$  and $dq_v = dU$

I believe he already knows about this since he is talking about heat capacities in the question.

@Eyyboss Thanks for the answer. I can see that this does apply to real gases as the ideal gas law is not used anywhere.

If a body absorbs a quantity of heat $q$ its temperature will normally rise by a value $\Delta T$. The average heat capacity  over this temperature range is defined as $C_{av}\equiv q/\Delta T$. The instantaneous heat capacity at temperature $T$ is $C\equiv dq/dT$. This definition is not exact enough, however, until the path of heating is specified. From the first law $q$ could be replaced by $dU+dw$ so it is necessary to fix another parameter which is $V$ or $p$ giving rise to the definitions you quote.
Note (a) that at a phase change $C$ becomes infinity because heat is absorbed with no change in temperature. (b) As the heat capacity is the gradient of internal energy with temperature it describes how the internal energy increases. For example when $C$ becomes constant at high temperatures the molecules are absorbing heat a constant and maximum 'rate' with increase in $T$. (c) Using statistical mechanics the heat capacity can be calculated from a knowledge of a molecule's energy levels.

Can you explain what the function h(T,p) and u(T,v) would be mathematically?

These functions are in any thermodynamics book, for example $H=U+pV, \Delta U=q+w$

The heat capacities arise when you define various differential forms of U and H. The differential forms define how the energy responds to changes in thermodynamic variables such as pressure, temperature and volume.
For instance, one form for U reads:
$$\begin{align}dU = \left(\frac{\partial U}{\partial T} \right)_V dT +  \left(\frac{\partial U}{\partial V} \right)_T dV \end{align}$$
Here we can now use the following definition (symbolized with ":="):
$$C_V:= \left(\frac{\partial U}{\partial T} \right)_V$$
Think of "$C_V$" as shorthand for "$\left(\frac{\partial U}{\partial T} \right)_V$".
Use of shorthand forms leads to the following equation:
$$dU = C_V dT +  \left[ \frac{\alpha T}{\kappa}  -p\right] dV $$
Similarly
$$C_p:= \left(\frac{\partial H}{\partial T} \right)_p$$
Use of this and other shorthand forms leads to
$$\begin{align}dH = C_p dT +  V\left[ 1-\alpha T\right] dp \end{align}$$
There is more to heat capacities than that, of course. For one thing, we can attach a physical interpretation to heat capacities in the context of particular experiments.
Regarding how a functional form of the enthalpy (or energy) may be obtained as a function of T, that is another can of worms.  One strategy is to use the van't Hoff equation. Another is to use calorimetry.
Also, internal energies and enthalpies are not reported as absolute values. Values are reported as differences relative to a reference value (not "$H$", rather "$\Delta H$"). Once you define the enthalpy as a function of temperature at a particular value of the pressure, computing $C_p$ just involves taking the derivative wrt T. For $C_v$ we start from internal energies as functions of T at constant volume. Finally, as mentioned in another answer, you should keep in mind the difference between heat capacities as extensive properties, and molar or specific heat capacities, which are intensive properties.

Principally, definitions cannot be derived. They are chosen.
What you have described in the question are heat capacities without the adjective specific,  being heat capacities of a whole system.
Note that specific heat capacities are related to the substance unit mass amount, mostly 1 kg, so I would expect the mass in denominator, like $$c_V = \frac 1m . \left(\frac{\partial U}{\partial T}\right)_V $$ $$c_p = \frac 1m . \left(\frac{\partial H}{\partial T}\right)_p $$
Similarly, molar heat capacities would have there the molar amount instead of mass.
$$C_V = \frac 1n . \left(\frac{\partial U}{\partial T}\right)_V $$ $$C_p = \frac 1n . \left(\frac{\partial H}{\partial T}\right)_p $$


text:
Should temperature and pressure conditions for two reactions be same for them to combine for enthalpy calculations?
I was solving a question in which I had to calculate Enthalpy of formation of $\ce{CS_2(l)}$ using the given Enthalpy of combustion (not in standard state) of $\ce{C(s)}$, $\ce{S(s)}$ and $\ce{CS_2(l)}$.
To calculate the $\Delta H_f$ of $\ce{CS_2(l)}$, we combine the corresponding equation of combustion of these chemicals  in appropriate order to form the equation :
$$\ce{C(s) + 2S(g) -> CS_2(l) }$$
Should the temperature and pressure conditions of the combustion equations be same for me to combine them mathematically?

Yes, strictly speaking, they must be the same.

yes they should be for a simple reason: when you do make C and S react together  you are going to assume that the temperature and the pressure of the box you're using for the reaction is the same. In particular you can even think of putting for example the hot solid in the box with the cold gas, when you're talking of entalphy of formation you're basically speaking of the equilibrium situation. You will reach equilibrium only once al the macroscopical quantities are perfectly equal in all the box you're working with. If this was not the case, the equations you would have to solve are much more complicated than those you are usually seeing .


text:
Why is Standard heat of reaction equal to difference between standard heats of formation of products and reactants?

The knowledge of standard heat of formation of various substances can be used to calculate the heat of reactions under standard conditions. The standard heat of any reaction is equal to the difference between the standard heat of formation of the products and the reactants

Why is that so? I'd prefer physical derivation for this:
$$\Delta H^\circ = \sum\Delta H^\circ_{products} -  \sum\Delta H^\circ_{reactants}$$

Technically, I prefer the form where the equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Now, the physical
(as in measurable) verification becomes problematic when the energy released includes things like light. Note: Hess's Law applies to heat.

It is the consequence of the Hess's law and that is the consequence of the energy conservation law. Overall it says the enthalpy change does not depend on the path, but only on the start and final point.
Comparing to the real reaction path, you can formally get there also

via the path of conversion reactants to pure elements in their standard form
and then via forming products from those elements.

If the result were not the same, you could create perpetuum mobile switching both states via 2 different paths, gaining the nonzero net energy.


text:
Industry Standards of Polymer Leeching Measurement
I am a research assistant currently researching the leaching of polymers the byproducts of the reaction of ethanol and wood coating from wood billets into water and I'm at a crossroads with my colleagues.
Do the the somewhat niche nature of the research, it is difficult to find a standardized method of testing leeching from wood. Is there a format approved and tested by NIST or is it wisest to search the literature for a tested method and build from there?

If you want to do scientific research, the answer must be search the literature! in any case. ANY scientific research must start with an exhaustive literature study.

Leaching may be a better word than polymer leeching. You have to find target molecules of interest in order to be reasonable and meaningful literature search.


text:
Best Way to Store Copper Acetate Solution?
I am preparing to copper electroplate an ABS plastic sheet in my home and wanted to know what the best container/environment would be to store the copper acetate solution.
First, I created the solution using copper metal, vinegar and hydrogen peroxide.
Can I simply store this solution in a plastic container?
Should it be closed with an air-tight lid?
Would food-grade LLDPE plastic container be acceptable?
Does the color of container matter? Is black color better than clear color as it reflects more light?
Is it OK to leave it in my home at room temperature or should it be refrigerated?
Thanks!

Far too many questions at once. You need to hire some industrial chemistry consultant!

Instead of hydrogen peroxide, aerial oxygen and time can be used. Many people consider air less expensive but time more expensive than hydrogen peroxide.

Electroless copper plating requires a copper salt solution, of course (https://en.wikipedia.org/wiki/Electroless_copper_plating), but there are many other considerations, like the reducing agent and surface preparation.
I think that storage of the homemade copper acetate solution is a minor issue in your project and a plastic container with an air-tight lid should be perfectly fine, without refrigeration, for a few years.
However, for the actual electroplating, your copper solution may not be adequately characterized. There may be excess acetic acid, other stuff from the vinegar (which could be perfectly safe to consume, but not for sensitive chemical reactions). Commercial copper salts are available inexpensively and with high purity, if the project you are working on has significant value.

Thanks so much for the reply. I was actually thinking of doing electroplating rather than electroless plating, but perhaps I should consider electroless?
I have seen copper salts (such as copper sulfate pentahydrate). However, my intuition was that it would be cheaper to create the copper solution myself rather than buy pre-prepared copper salts.
Can I simply dissolve the copper sulfate in water to create the electrolyte solution?
Going back to my method above, it seems vinegar and hydrogen peroxide create a toxic paracetic acid. Are there better solutions to create the electrolyte?

It's fun to experiment, but sometimes frustrating if you try too many untested things all at once. Find an example to imitate, then do something different, one thing at a time. Googling electroless copper plating  (and videos) will bring up many videos. Try reproducing one to get the feel of it, then adapt to your materials. Peracetic acid is in equilibrium with acetic acid and H2O2, and your concentrations will be very low because the copper metal will likely catalyze the decomposition of the H2O2. Good luck; be careful.


text:
Regarding change in temperature in Bomb Caloriemeter
In a Bomb Calorimeter, we measure the change in temperature of water $\Delta T$. But when it comes to calculating the change in internal energy we use $$\Delta U = C_{cal}\Delta T$$ Where $C_{cal}$ is the specific heat capacity of the whole calorimeter. Why is that? What I mean is that we're measuring the change in temperature of the water and assuming that heat from the reaction transferred completely to water, so shouldn't we apply $$\Delta U = C_{water}\Delta T$$ since we're calculating the change in temperature of the water?
Or are we assuming that the change in temperature of the water is the change in temperature of every element in the calorimeter?

Can you give a reference i.e where you learned this?

This isn't where I learnt but it says the same thing https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Calorimetry/Constant_Volume_Calorimetry

Oh btw I think you should change your title because right now it's really broad

Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings), and ignition circuit connected to the bomb. (*)

The whole device known as calorimeter consists of a bomb and a water bath, they are not two separate entities. So, $ C_{cal}$ is actually referring to the heat capacity required of the steel vessel and also the water. The construction of the calorimeter is such that the thermal mass of the calorimeter system completely absorbs the heat of reaction (**).

*: This Wikipedia
**: Refer 29:27 of this video

So the change in temperature is actually tge change in temperature of whole apparatus?

Yes, the calorimeter is the whole thing not just the bomb

It's even said so in ncert, see page-163 under (a) $ \Delta U$ measurements [ I'm assuming you study in cbse]

In your first equation, the $\Delta U$ is not the change in internal energy of the gas reaction.  It is the change in internal energy of the surroundings (the calorimeter and water), which is the heat transferred from the reaction mixture to the calorimeter and water.  Since the calorimeter/water and the reaction chamber are both of constant volume, the $\Delta U$ in this equation is equal to minus the change in internal energy due to the reaction.

But that means that change in temperature of water and the calorimeter is same. Is that true? Also what about the second equation $\Delta H = \Delta U + \Delta n_g RT$

Yes, the change in temperature of water and calorimeter are the same at final equilibrium.  The second equation is used for a reaction involving ideal gases to get $\Delta H$ once $\Delta U$ is known.  It follows from $$\Delta H=\Delta U+\Delta PV=\Delta U+(\Delta n_g)RT$$


text:
What is the relationship between the stability constant of a copper (II) electrolyte complex ion and the output voltage Ecell?
As a part of a school investigation I constructed a voltaic cell experiment where for each run I added a different ligands to the copper (II) sulfate half-cell electrolyte in a daniell cell to see if there was a change in the output voltage Ecell. When plotting a graph of this, it seemed that Ecell increased as the stability of the ligand increased (indicated by the stability constant Kc) however I am struggling to explain why. I presume its because the standard potential at the cathode Erhe increases with ligand stability due to greater oxidation.

what are you talking about exactly? The electrolyte is already dissolved in water. Which ligand? are you talking about H2O? The Nernst Equation gives us the Open Circuit cell potential (are you measuring your cell potential at zero current?) It is expressed as $$ Ecell = RT/(nF)*(ln(K/Q))$$ so it is a function of the concentrations and temperature

The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual $\ce{Cu^{2+}}$ ion in solution. And in any solution of $\ce{CuSO4}$, adding a ligand produces a decrease in concentration of the free ion $\ce{Cu^{2+}}$. So, according to Nernst's law, the potential $\ce{Cu^{2+}/Cu}$ becomes less positive.
For example, the standard redox potential of $\ce{Cu^{2+}/Cu}$ is + $0.34$ V. Let's consider the ligand chloride $\ce{Cl-}$. If enough chloride ions $\ce{Cl-}$ is added to a $1$ M $\ce{Cu^{2+}}$ solution so that at the end the excess of $\ce{Cl}$- ions is $1$ M, the complex $\ce{CuCl4^{2-}}$ will be formed, with a concentration $1$ M. The measured residual [$\ce{Cu^{2+}}$] concentration is $2.5 ·10^{-6}$ M. In this case, Nernst's law gives the following $\ce{Cu^{2+}/Cu}$ redox potential : $$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$ This +$0.174$ V is lower than the standard redox potential (+$0.34$ V).
The stability complex of the $\ce{CuCl4^{2-}}$ ion is : $\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$
If another ligand is used, which makes a stronger complex with copper $\ce{Cu^{2+}}$, the stability constant is higher, and the redox potential will be still lower.

Your first two sentences create false impression you may think standard redox potential of copper depends on Cu^2+ concentration.

@ Poutnik. You are right. I will edit the coprrectkion.

My speculation starts with a comment by ChemEng, who noted here that the Nernst Equation is essentially a function of concentration and temperature.
In this regard, I start with an example of a side-reaction:
$$\ce{ Cu (s) + Cu(II)L <=> 2 Cu(I)L }$$
where the above reaction proceeding, given the general low solubility of Cuprous (absence a ligand) critically depends on the presence/stability of the highly soluble ligand L itself.
Also, a cited half-cell reaction in the Daniell cell is:
$$\ce{Cu(II)L + 2e− → Cu(s) }$$
Or, can it proceed, at least partially also, in steps dependent on solubility provided by the ligand:
$$\ce{Cu(II)L + e− → Cu(I)L }$$
$$\ce{Cu(I)L + e− → Cu }$$
So, again another possible example of the role of the select ligand in facilitating the cuprous (as an intermediate) concentration.
As such, my supposition is that the presence/stability of the soluble ligand impacts copper ion(s) concentration thereby impacting the Nernst equation.
[EDIT] Based on further research, my speculated explanation is likely too simple. Here is a more technical source on an advanced Daniell cell design. In recharging, note the mention of problematic Cu2O formation which is rectified, in part, by an increased Cu(NO3)2 electrolyte concentration.


text:
Number of orbitals Lithium
I have a very rudimentary question on orbitals (I have basic chemistry knowledge, using for a comp chem project)
Lithium, to my understanding, has three electrons allocated to the 1s and 2s orbitals. 2s is not full. How many orbitals are in lithium then? My comp chem package returns 5, which I am assuming has to do with unfilled orbitals in the second shell. Can someone help explain this?
Also, if lithium has one unfilled orbital, it has spin of 1? and charge of 1?
EDIT: A followup question is how many unoccupied orbitals are in LiH? My guess is that it's still the 3 unoccupied orbitals from Li. Is this true?
Thank you for the help!

This is really an ambiguous question. If you look at the atomic emission spectra of lithium a slew of additional orbitals will be found (theoretically infinite). There is no definite number of orbitals unless the question is restricted such as: How many occupied orbitals are there in a free atom of lithium? // What was the exact wording of the question?

Basically I am looking at a calculation of the ground state energy of LiH. I would like to construct the hamiltonian describing this molecule. One step of doing so is to retrieve the number of orbitals in this molecule. PYSCF tells me it is 6, and I am trying to understand how it arrived at this conclusion.
Furthemore, the example online reduces the calculation by freezing the core and removing the orbitals indexed at 4 and 5. I assume this corresponds to two of the unoccupied orbitals of Li. Based on my basic knowledge, I think there is another unoccupied orbital in Li. I want to confirm.

If an orbital has no electrons in it, does it still exist?

@Zhe Orbitals do not exist but in our atomic models.

@Poutnik I was thinking more metaphysically. Does $f(x)$ exist?

@Zhe I guess there is more context-dependent definitions of existence, from very material to very abstract one. Being a member of a set may be one of them.

Just to be sure: Note that orbitals(*) themselves have neither spin, neither charge. They are not real objects, but theoretical constructs of quantum atom models, that fit well the observations.
Every atom has theoretically infinite number of orbitals ( 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc), all but some unoccupied. For most scenarious, it is practical to consider just few highest occupied and few lowest unoccupied orbitals (like 1s, 2s, 2p, 3s, 3p for Li), what may be the reason your chem package provides just few.
Electrons are fermions, so the unpaired $\mathrm{2s}$ electron has the spin 1/2. The spin of a lithium nucleus depends on the isotope. $^6\ce{Li}$ has the spin 1(**), $^7\ce{Li}$ has spin 1/2. The net charge of lithium atom is zero, the charge of the unpaired 2s electron is obviously $-\mathrm{e}$.
Similarly, $\ce{LiH}$ has unlimited number of atomic or molecular orbitals ( depending on the quantum model ), as they are features of quantum models, not real objects. And again, only highest "valence" occupied and few lowest unoccupied ones are worthy to consider. (For $\ce{LiH}$, s and p for atomic ones, $\sigma, \pi $ for molecular ones. )

The orbitals with very high principal quantum number $n$ converge for $\mathrm{n} \rightarrow \infty$ to energetic continuum with very low ionization energy. If occupied by careful excitation, they have very interesting behaviour like effectively disappearing quantization. Such electrons behave like orbitting classically a point-like central charge. See Rydberg atoms for more. They have also ridiculous values of atomic radius, which increases with $n^2$. E.g. $\ce{H}$ with $n=137$ has radius $\pu{1 \mu m}$ and $\ce{K}$ with $n=600$ would have reportedly size $\pu{0.1 mm}$.

(*) Note that the term orbital has 3 major, related but distinguished meanings:

orbital(1) - a wave function as the particular solution of the Schroedinger's quantum wave equation

orbital(2) - a quantum state of an electron(energy, orbital angular momentum, spin angular momentum), belonging to orbital(1)

orbital(3) - a 3D region (or 3D surface/shape), describing the region of significant probability of electron occurance (or its border threshold value), belonging to orbital(1,2)

(**) The only other stable nuclei with spin 1 are $^2\mathrm{H}$ and $^{14}\mathrm{N}$, as odd-odd nuclei are not generally stable, undergoing beta decay to even-even nuclei. The eventual other stable odd-odd nucleus candidate $\ce{^{10}_{5}B}$ does not count, as its spin is 3, not 1.

This is very helpful! Thank you for explaining!


text:
How does chromium behave in a carbonate solution?
I'm wondering how chromium (as a metal or as a salt) behaves in a concentrated carbonate solution. Will it form any, one or several carbonate specie(s)? Will the metal solubilize? How does it affect chromium salts and their behavior? Does it matter which carbonate salt the solution is made from (Ca/Mg/Na/K/Mn/Fe/Zn/Cu carbonate)?
The question is a bit broad but it only reflects my ignorance of its behavior. I'd welcome more understanding of any of the aforementioned aspects of its behavior.
In some ways I think my question amounts to finding out how it will behave at high pH but I'd rather be asking specifically in case I might be missing some aspect of what is going on.
Mo is a transition metal and seems to form an insoluble carbonate. But I haven't been able to find evidence showing chromium (and Tungsten for that matter) do so as well (and I don't know if extrapolations can be that easily made amongst the behaviour of elements of the same group), nor have I find any numerical data as to their solubility.

Solids precipitating from heavy metal salt solutions in alkaline environment, with or without carbonates, is always tough subject to evaluate. I expect a mixture of hydroxide, hydrated oxide and basic carbonate, all in not well defined composition.

Thanks. So there not a simple answer? But chromium carbonate does (in part) form? If so, would you know the solubility (in g/100ml) of chromium carbonate(s)?

The solubility product of  Cr(OH)3   Ksp = 6.3 x 10-31. Solid chromium does not dissolve, but may react, even if not in these conditions.

So at e.g. pH=11, [Cr^3+]=6.3x10^-22 mol/L, so literally few ions per litre. This does not obviously address colloid particles.

What you get will also depend on the oxidation state. I would expect Cr2+ and Cr3+ to behave somewhat differently

Chromium metal is unaffected. In simple terms that is why it is added to make alloys corrosion resistant .  Alloys like Vitallium ( Cr + Co )  are immune to almost  any solution.

Chromium(III) ions are like Fe(III) and Al(III) ions. Exactly like aluminum and ferric carbonates, chromium(III) carbonate does not exist in aqueous solution. In principle it could be obtained according to $$\ce{2 Cr^{3+} + 3 CO3^{2-} -> Cr2(CO3)3}$$ But like with the aluminum and ferric carbonates, the chromium carbonate is immediately destroyed according to : $$\ce{Cr2(CO3)3 + 3 H2O -> 3 CO2  + 2 Cr(OH)3}$$ As a consequence, chromium carbonate does not exist in aqueous solutions and, if carbonate ions are added to the solution, the ions of Cr(III) are transformed into chromium hydroxyde $\ce{Cr(OH)3}$ precipitate. And some bubbles of $\ce{CO2}$ may appear in the solution.
Furthermore, like $\ce{Al(OH)3}$ and $\ce{Fe(OH)3}$, $\ce{Cr(OH)3}$ cannot be redissolved in concentrated solutions of carbonates. A pH higher than $12$ is needed to redissolve $\ce{Cr(OH)3}$ according to : $$\ce{Cr(OH)3 +  OH^-  ->  Cr(OH)4^-}$$ And such a high pH value cannot be obtained with concentrated solutions of carbonates. , ,

$\ce{CO2}$ would not be generated in carbonate presence. It would be rather $\ce{Cr(H2O)6]^3+ + n CO3^2- ->  [Cr(H2O)(6-n)(OH)n]^{3-n} + n HCO3-}$, particularly $\ce{Cr(H2O)6]^3+ + 3 CO3^2- ->  [Cr(H2O)3(OH)3] + 3 HCO3-}$.

Thanks for your answer. If I understand well throwing in either Cr/Al/Fe2(CO3)3 in water would fizz and precipitate 100% of it as the corresponding hydroxide?

What about the other oxidation states of Cr/Al/Fe; and about the respective metals, do they form stable carbonates or also turn and precipitate into corresponding hydroxides?

And what about Mo and W. Do they behave entirely differently even though they are in the same group as Cr?

@Poutnik. CO2 will probably not be evolved, as it will be transformed into hydrogenocarbonate ions. And I have not mentioned the H2O molecules acting as ligand around Cr, because of the problem of the two series of chromium(III) salts, green and violet.

Rather, carbonate would stop at hydrogencarbonmate stage and would not go further. Sure, variability of Cr coordination chemistry complicates it. Equetions could be written without coordinated H2O as well.

FeCO3 is the well known mineral siderite.

@ Hans. 1) It is better to say that Cr, Fe and Al carbonates do not exist. They will not fizz in water : they don't exist, even in the dry state. 2) Fe(II) makes a carbonate FeCO3 that is easily oxidized in humid air to from Fe(III) hydroxide.  3) I don't know the situation of Molybdenum and Tungsten. 4) Cr(II) hydroxyde is unstable. It absorbs the tiniest traces of O2 to produce Cr(OH)3.  And if dry it get decomposed according to the extremely strange equation  : $\ce{2 Cr(OH)2 -> Cr2O3 + H2O + H2}$


text:
HDPE vs PP for storing pure aldehydes
I know both HDPE and PP have at least some degree of resistance to aldehydes, esters and aliphatic hydrocarbons.
I am concerned however about minute contamination over time of pure samples and so I wonder how PP and HDPE compare to each other as to their resistance to aldehydes (mainly) as well as to the other two groups.

I am just guessing the isotactic PP should be somewhat more resistent to contamination penetration due tougher structure. But the surface adsortion though, it may be similar for both. The short-time contamination leading to adsorption may be the major problem, depending on the case.

General idea is storing organics ( especially nonpolar ) in glass and storing trace-contamination sensitive inorganics ( very dilute solutions for AAS) or glass attacking inorganics ( HF ) in HDPE/PP plastic. But you know that.

Plasticizers can migrate. Not all PP and HDPE are plasticizer-free.

DrMoishe Pippik: How does one know which plasticizers are used in a given PP/HDPE item? Is there any standard marking on this like an adjunct to the plastic type logo? I noticed sometimes there's a number inside the logo.


text:
Oxygen depriving gas/compound
A scene from the movie TENET made me think of a question. In the scene a gas is released into a room that is filled with precious artwork. The gas is used to rid of all oxygen in the room so fire cannot enter the facility. What kind of gas would be used to "gobble up" oxygen? Assuming there is no ventilation system to push the oxygen out, is there a gas/compound that can rapidly absorb or destroy oxygen atoms? Thank you!

Such reactions that remove oxygen do so with the release of a lot of energy, usually as heat. This does artworks no good at all.

yea, you can purge the room with $N_{2}$ or some other inert gas. Maybe the addition of an oxygen scrubber will reduce $O_{2}$ levels.

Not a gas, but a solid can do that. apparently: https://phys.org/news/2014-09-material-oxygen-air.html . Leaves me a bit sceptical though. Why isn't it a world-renowned success story, if it really works?

The easiest, quickiest, cheapest and most efficient thing to do is to purge the room with nitrogen $\ce{N_2}$

@Maurice, it sounds like this guy wants something that reacts with oxygen, not that chases it away mechanically.

EDIT: yes I am looking for a chemical that reacts with oxygen.

Well, one is the powder I referred to above, if we believe it. Then, I just looked up my old Vogel's Practical Organic Chemistry. Apparently alkaline solutions of pyrogallol, and Fieser's solution, can be used to scrub small quantities of oxygen from a gas stream. For your case, removing the oxygen in a whole room, especially by another gas, I think we are in the realm of science fiction. Otherwise it would be the perfect crime :).

In fact hydrogen can react with oxygen, so yes, you could inject some hydrogen in the room and then ignite it. Not sure it would be a very inconspicuous process, though, as pointed out by Waylander :)

Why is displacing the oxygen not considered? Put in dense halon at bottom, N2/O2 comes out the top. This is, indeed, the way server-room fire extinguishers operate.

@DrMoishePippik : why? Original post: "Assuming there is no ventilation system to push the oxygen out".

@DrMoishePippik - Halon does more than displace oxygen. It is quite good at gobbling up reactive species and shutting down reaction paths. Done properly, a human should be able to breath in a room after a halon release, if sized properly.

Actually, I was at a site where they accidentally "tested" the halon server-room extinguishers while the glassed-in room was occupied. Though the over-pressure rather dramatically blew out the heavy windows, no one was badly injured (or suffocated).

I’m voting to close this question because we really shouldn’t be entertaining a question based on such a wrong assumption.

Actually there are commercial fire extinguishers that work by displacing the oxygen in a room. I forget the name but the safety experts put spheres containing an inert gas ( not a noble gas ,not CO2) to auto automatically be released in event of fire . Hydrogen was used in these lab rooms , but I don't know if that is why these extinguishes were chosen.

Interestingly, no exotic compound is required to consume the oxygen in a room, or, for example, a large sewer pipe.
In fact, an interesting report of workers who died when drainage occurred in a sewer pipe, resulting in the exposure of fresh iron metal surfaces, likely in the presence of acidic conditions and a complexing agent. A reaction pursed resulting in a significant consumption of the available oxygen.
Here is the underlying electrochemical reaction:
$\ce{ Fe (s) + O2 (g) + 2 H+ (aq) -> Fe(II) (aq) + 2 OH- (aq) }$
Source: See http://corrosion-doctors.org/Experiments/rust-chemistry.htm
Further, a corresponding reaction based on ferrous converting into ferric.
In the current context of the question, spraying a very fine mist of aqueous nano-iron particles, freshly mixed with an acid source, say CO2 or citric acid, may have a similar effect.
[EDIT] Here is a citation from the CDC on the alluded to sewers workers deaths, to quote:

On August 19, 1988, a 31-year-old male assistant construction supervisor (victim) entered an oxygen-deficient manhole to close a valve and collapsed at the bottom. In a rescue attempt a labor foreman (male, age 34) and the victim’s supervisor (male, age 36) also entered the manhole and also collapsed. All three workers were pronounced dead at the scene by the county coroner.

Per another related report, to quote:

In underground sewer holes, high levels of methane gas and depletion of oxygen are common, Sherman said.

Now, some may feel confident in dismissing electrochemical based oxygen depletion, but please note it is actually killing essential workers.
[EDIT 2] Did find an explicit report citing the action of rusting leading to oxygen consumption and a fatality in this source addressing confined space accidents. To quote:

The interior of the compartment was rusty, Kingsley noted, and there was standing water inside. "The chemical process of oxidation or rusting uses up oxygen, thereby creating an oxygen-poor atmosphere," she wrote. According to Workers' Compensation Board findings included in the report, the air in the Apex No. 1's other eight compartments contained 20.9-per-cent oxygen, which is in the normal range for the Earth's atmosphere. The sealed compartment's air was between 15.7 and 16.4 per cent oxygen -- low enough that it can't support human life for long.

Please cite why the documented reaction is not applicable here. Thanks. By the way, do not place your head in a vessel with such a wet iron metal mix to prove your point, as I am sure you will suffer oxygen deprivation.

In the case you cited the speculation  was that decomposition of organic matter (ie biological activity) was responsible for oxygen depletion. No where is it suggested that exposed iron was a factor.

Andrew: Yes, you are correct, but only in the references I was able to find recently with respect to the fact that Fe exposure is not explicitly mentioned. However, my original source did mention something about drainage and freshly exposed Iron surfaces which I immediately linked back to my claimed key electrochemical reaction. More recent sources above only affirm O2 consumption but not how.

Andrew: Was able to find a source confirming my essentially rusting-based electrochemical reaction. See [EDIT 2].


text:
Are all gases equal in their ability to carry with them lower volatility matter at a given temperature?
As we all know water is a great solvent in it's liquid form relative to other substances. I wonder however how it performs as a gas relative to other stuff.
Does anyone know how well steam carries compounds with it compared to other gases at the same temperature and pressure?

Arguably, no gas is a good solvent.

Or they are all perfect, depending on what on Earth you mean!

Supercritical water dissolves for a lot of things you usually think are perfectly insoluble, like many inorganic solids.

Are all gases equal in their ability to carry with them lower volatility matter at a given temperature?

@Hans, think about gas chromatography, if you are familiar with it. Does the retention time change if you change gases, say, nitrogen to helium or hydrogen? No, it does not.

It depends on how a solvent is defined. Regardless of semantics, supercritical steam 217.75 atm, 373.946 °C is a very good solvent. It will dissolve fats, hydrocarbons or plastics. It also becomes very corrosive to metals.

Thing is, you're talking about liquid-like supercritical H2O, which I wouldn't call steam.

Supercritical "fluid" is neither a typical gas nor liquid. It does not matter whether you accept the term supercritical steam or not, I would rather worry if the scientific community accepts it or not. "Supercritical steam" brings up at least 9700 results in Google Scholar.

Perhaps read my comment again and check out on supercritical fluids, before responding...

Mithoron, Please bring some positivity in your tone. Instead of criticizing answers mainly on semantics or trivial issues and downvoting almost every other question, why don't you write a separate answer if you have a better explanation. Perhaps in a better way, instead of cursing the darkness, why not light a lamp? SE Chem is a volunteering Q&A site. Make this a productive place. BTW, what is your area of specialization?

https://chemistry.stackexchange.com/questions/22185/does-the-density-of-a-liquid-and-gas-phase-of-a-substancenitrogen-or-water-con https://chemistry.stackexchange.com/questions/122239/is-oxygen-above-the-critical-point-always-supercritical-fluid-would-it-still-ap


text:
Thermodynamics- How are various property tables made?
I've always been curious about how different thermodynamic property tables are made. It's really interesting to me. I want to know how different things such as critical properties and variations of properties of different substances are determined. I'm not familiar with statistical physics, but any derivations using it would be great for future reference.

How would you react if I was asking you : I would like to learn the History of England, or  the geometry, or the laws of my country. What should I do ?

@Maurice Recommend some books.

Books about what ? Are you looking for books about (I copy your second line)  "different things such as properties and variations of properties of different systems" ? Well ! What does it mean, apart from ... all our scientific knowledges ?

@Maurice I meant different substances.

Today the number of known substances is greater than 10 millions, and is increasing every day.

Property tables are made by a bunch of lab slaves repeating the same experiments with different compounds and tabulating the results. Sometimes, the experiment is essentially decentralised because any individual who synthesises a compound will routinely measure a property which then just needs to be aggregated by CAS or similar databases.


text:
Change in surface tension of soapy water as oil is added
I am wondering about how the surface tension of clean soapy water changes as oil is added to it?  Is the change gradual or is it abrupt as the amount of oil added exceeds the amount of soap available to surround the oil.  So if the surface tension of clean soapy water is 25 dynes/cm, then as I add oil, will that surface tension go up or will it stay essentially the same until the amount of oil added exceeds the amount of soap needed.  Or perhaps something else happens??

Oil will reduce the surface tension of water, as will soap. However, depending on the soap used, the concentration of soap, and it’s ability to form micelles, adding oil might pull surfactant molecules away from the surface and damper the surface tension lower effect of the soap by forming micelles. However, adding to much oil might cause the micelles to burst.


text:
When does replacing a double bond with a lone pair result in a valid resonance structure?
Take molecules A and B below. In A, we can move the electrons from the green pi bond into a lone pair on oxygen, then move the blue lone pair of electrons from the other oxygen to form a new π bond, and this is a valid resonance structure. But in B, I was told that if we move the electrons in the green π bond into a lone pair on carbon, and then move the blue lone pair of electrons from the nitrogen to form a new π bond, the result is not a valid resonance structure.
Why is this?
The only explanation I heard is that in B, the blue electrons on nitrogen are in $\mathrm{sp^2}$, while a π bond is $\mathrm p$, and since $\mathrm{sp^2}$ and $\mathrm{p}$ are orthogonal, electrons cannot delocalize between them. But that doesn't make sense, since the same thing is true in A! In A, the blue electrons are in $\mathrm{sp^3}$, and to form a π bond they will have to move to a $\mathrm{p}$ orbital, but doing this does give a valid resonance structure, so here they can delocalize between $\mathrm p$ and $\mathrm{sp^3}$. So this does not explain why we can't do the same thing for B.

But that doesn't make sense, since the same thing is true in A!

No, it isn’t; and identifying which of the two cases is true is precisely the key point to understanding when resonance can happen and when it cannot.
First and foremost: we usually draw in two dimensions to represent three-dimensional molecules. Obviously, this is inadequate and often we need some way of determining the three-dimensional orientation of the two-dimensional images we draw. This is especially problematic with lone pairs which we typically draw in four corners (like a square) but which can be in any of many different orientations.
A lone pair can be in a pure s, a pure p or a hybrid orbital. A pure s orbital cannot participate in resonance with other π-bonds. A pure p orbital can, if it has the correct orientation and a hybrid orbital can if it has both the correct orientation and a sufficiently good hybridisation. In most cases, it is only the orientation that truly matters.
To determine the type of orbital a lone pair is in, one needs to look at the other bonds (and lone pairs, if applicable) that this atom has formed with other atoms. For pyridine, this is relatively simple:

there are two $\ce{C-N}$ σ bonds to neighbouring atoms of the six-membered ring; and
there is one π bond which is part of the ring’s aromatic system.

The π bond in a non-metal must derive from a p-type orbital. That locks one of our orbitals in as p. The remaining two depend on the $\ce{C-N-C}$ bond angle which we can approximate as $120^\circ$ which in turn leads to the necessary conclusions that these two orbitals must be $\mathrm{sp^2}$ orbitals. If we have two $\mathrm{sp^2}$ orbitals, there must be a third one (the lone pair) and all three must be in the same plane. Furthermore, these three $\mathrm{sp^2}$ orbitals must be angled at $90^\circ$ to the p-type orbital participating in the π bonds. This last tidbit is the key: As these two orbitals ($\mathrm{sp^2}$ and p) cannot be made parallel, they cannot be part of the same π system and thus they cannot be flipped in resonance structures.
Acetate’s $\ce{C-O-}$ oxygen is very different. Before assuming anything about potential resonance or not, the only orbital that has to point somewhere is the $\ce{C-O}$ bond. This bond could be realised by either a pure p orbital or any $\mathrm{sp}^n$ hybrid orbital. Thereafter, we have no bond angles to work with so no information to lock in any of the other orbitals. What the set of orbitals with the lowest energy will be depends on the molecule’s overall symmetry and various considerations that are impossible to reproduce with pen and paper in an exam. Thus, the best course of action is to assume a completely unhybridised oxygen atom. (This assumption turns out to not be entirely correct if one runs the calculation, but it certainly suffices for our considerations!)
As we have only pure atomic orbitals, we have two orthogonal p-type orbitals which in turn are orthogonal to the $\ce{C-O}$ bond. Critically, due to the constraints on the carbon atom connecting the two oxygens (it forms a π bond and three σ bonds; thus it must be $\mathrm{sp^2+p}$), the $\ce{C=O}$ π bond will itself be orthogonal to the $\ce{C-O-}$ σ bond (as all of carbon’s σ bonds must be in the same plane) and assuming Euclidian geometry this means one of the lone pairs on oxygen must be parallel to the $\ce{C=O}$ π bond.
Long consideration, short result: we have a p-type lone pair on oxygen that is oriented in parallel to the $\ce{C=O}$ π bond; these can overlap in resonance and that is why you can shift do electron-pair flipping with that lone pair.
As an exercise, you might want to analyse the situation in furane ($\ce{C4H4O}$), a five-membered ring made up of oxygen and four $\ce{C-H}$ groups.

In figure B, nitrogen is $\mathrm{sp^{2}}$ hybridized which means that the lone pair sticking out from the ring is contained in a $\mathrm p$ orbital. This $\mathrm p$ orbital is geometrically restricted from interacting with the $\mathrm p$ orbital of the adjacent carbon. The $\mathrm p$ orbital containing the lone pair on nitrogen is in a different plane then the $\mathrm p$ orbitals of the adjacent carbon. Because of this, they cannot interact and therefore, no π-bonds are formed.

eletronic configuration is meant to be written as $\mathrm{sp^x}$ and not $\mathrm{sp_x}$, also they have to be upright.. you can use $\mathrm{}$ or $\ce{}$ for this.

@Safdar much obliged

Generally speaking, you can convert a $\pi$ bobd into an electron pair under either of two circumstances:

You can redraw the bond elsewhere, usually in an adjacent linkage, so you are really moving the $\pi$ bond rather than eliminating it.

The remaining $\pi$ bonding becomes more strongly stabilized, most commonly by forming an aromatic ring.

Your carboxylate ion is a good example of possibility (1). Another case involves enolate ions, which figure into reactions of carbonyl compounds with bases:
$\ce{R2C(CR3)\overset{\overset{-}{..}}{O}<->R2\overset{\overset{-}{..}}{C}-(CR3)C=O}$
Equally important is a variation presented by the allyl cation, where the $\pi$ bond is traded with an empty orbital. Allyl cations are prominently featured in reaction mechanisms that involve carbocations because of their resonance stabilization:
$\ce{R2\overset{+}{C}-CR=CR2<->R2C=CR-\overset{+}{C}R2}$
The second possibility typically involves the remaining $\pi$ bonds becoming more stabilized through firmation of am)n aromatic ring. Cyclopropenones offer an example here (drawing taken from this answer):


text:
How would one determine the temperature needed such that hydrogen peroxide will start to decompose?
Hydrogen peroxide decomposes at room temperature and so I was wondering what temperature is sufficient for this decomposition to occur.
Can someone explain how we can determine this? (theoretically or experimentally - but to me it seems like it would be easier to determine experimentally)
I don't know if there is a technical word for this - maybe activation temperature?
I also googled the activation energy for the decomposition of hydrogen peroxide (without a catalyst) and it is around 75 kJ/mol which seems really high since it decomposes even at room temperature.

Ther Is no temperature above which it decomposes and below it does not. It is rather about an arbitrary choice of acceptable decomposition rate.

@Poutnik So H2O2 will decompose at any temperature greater than absolute zero, just very slowly at lower temperatures?

The question can be reformulated to:
"At what T is the equation $\exp{\left(-\frac{E}{RT}\right)}=0$ true ?"
There is no temperature above which $\ce{H2O2}$ decomposes and below which it does not. It is rather about an arbitrary choice of acceptable decomposition rate.
It is similar to solvent evaporation. There is no temperature above which solvent evaporates and below which solvent does not evaporate. That includes frozen solvent.
At very low temperatures, $\ce{H2O2}$ is kinetically very stable, but thermodynamically still unstable. The kinetic stability can be so high you would not notice the decomposition.
E.g. Let assume the supposed the shelf life of a substance is 2 years. If T is temperature at/below which there is insignificant decomposition (product purpose dependent) after 2 years, there can be arbitrarily stated it is stable at/below T.

I was originally thinking about it with Gibbs energy of activation and setting that to 0 and solving for the temperature (which would be the temperature such that the reaction is barrierless). So something like this: $T = \frac{\Delta H}{\Delta S}$ in terms of activation. And then I was thinking this would give me the temperature when hydrogen peroxide will start decomposing without any additional energy needed (which is what we see when storing H2O2). This would in fact get me some temperature which I guess isnt what I think it is. Could you tell me where I am wrong in my reasoning?

From what you said I guess it is wrong to say that only a barrierless reaction results in decomposition without any extra energy. So if this is true, what does a barrierless reaction mean?

Well, the reaction is exothermic, so lowering the temperature shifts the equilibrium constant by the van't Hoff equation toward the decomposition.

Sorry I am just going to learn equilibrium stuff next in school so I dont know much about the equilibrium constant (or that equation), but are you saying that by lowering the temperature the reaction is more likely to proceed by decomposition than the reverse?

I was just wondering about what it means for Gibbs energy of activation to be 0 because I dont think I understood properly, I thought it meant that its 0 when the reaction no longer needs any energy input to proceed (activation energy is met) and so the temperature when $\Delta G = 0$, is the minimum temperature needed for the decomposition to occur but you were saying that H2O2 decomposes at any temperature, so I must have misunderstood something about the Gibbs energy of activation and what it means for a reaction to be "barrierless".

Keep in mind $\Delta G(T)=\Delta H(T)-T \cdot \Delta S(T)$. Its eventual zero value would mean K for the equilibrium with the activated complex is 1. For H2O2 case and lowering temperature, both rate constants for peroxide creation and decomposure decrease, but the former faster, so equilibrium constant shifts in favour of decomposure.


text:
Where are the tetrahedral voids and octahedral voids located in an FCC?

A crystal is made up of atoms X, Y, and Z. Atoms X are in FCC packing, Y occupy all octahedral voids and Z occupies all the tetrahedral voids. If all the atoms along two body diagonal are removed, find formula of crystal.

Since there are 4 atoms in FCC, number of X=4. Tetrahedral voids occupy twice the number of atoms and octahedral voids occupy half of tetrahedral, so Y=4, and Z=8.
Now how I do proceed from here? I do not know where these voids are located.

Have you tried searching on google for the structure of an FCC and its voids?

@Safdar I have but various sources are having various answers so I couldn't confirm

@Safdar thankyou very much, this gives a perfect explanation. I was wondering if there is a post for the same concept done for CCP or HCP?

https://chemistry.stackexchange.com/questions/102248/number-of-octahedral-and-tetrahedral-voids-present-in-a-hcp-structure?rq=1

CCP and FCC are same..


text:
In what solvents does ionic association occur?
When a ionic compound like $\ce{NaCl}$ dissolves, it doesn't always full dissociate due to ionic pairing. This is measured by the van 't Hoff factor. Now, this is what I'm looking at for my chem experiment - how does ionic pairing affect freezing point depression. With changing radius the ionic pairing should change, and hence the van 't Hoff factor and freezing point depression should vary. Is this a valid hypothesis?
If so, my main issue lies with what solvents does ionic pairing occur in? Does it occur in all 3 of polar protic, polar aprotic and non polar solvents? I'm having trouble deciding which type of solvent to use. I was initially modelling water, but now I don't know how solvent type affects ionic association. Any help is appreciated. I'm not sure if ionic pairing occurs at all in all 3 types of solvents.

When you dissolve $\ce{NaCl}$ in water it almost completely dissociates into free mobile ions!

Have you seen https://pubs.acs.org/doi/10.1021/cr040087x?

All of them, maybe?

The solubility depends roughly inversely on dielectric constant of the solvent, approx values are water 78, acetonitrile 37, acetone 21 cyclohexane/benzene, hexane all about 2 to 3. So  generally soluble in high dielectric solvents and less so as this value decreases.  (The electric field of ion spreads out into surrounding solvent, when dielectric is low this filed goes along way so ions will remain in contact with one another as energy to separate is too large  vs energy to become separated & solubility is low)


text:
electron - nuclei recombination within stars
Nuclear fusion within stars combine atom nuclei, so the atoms are ionized and electrons run for free.
My question is, when or how these electrons recombine with the nuclei to form neutral atoms? inside the star? outside?

There are no neutral atoms in stars. It is too hot for them.

so do these neutral atoms form in the interstellar medium? in the planet formation?

Interstellar medium, I guess.

Nuclear fusion in stars produces huge amounts of energy that raise the temperature of the star (e.g. our sun has a surface temperature of about 6000K). At such high temperatures, a lot of atoms possess enough energy to become ionized. However, what percentage of atoms at a particular temperature has what amount of energy is determined by the Boltzmann distribution. So, there will always be some atoms which are not completely ionized (i.e. maybe the electrons are excited from the ground energy level, but they have not completely 'left' the nucleus).
The upshot of all of this is that as the temperature rises, more and more atoms get ionized. Atoms with lower ionization energies like Na, Mg and other metals lose their electrons first, whereas atoms with high ionization energies like He hold onto their electrons until the temperature reaches ~30,000K. Elements like C, N, O, H come somewhere between them. Also, the Boltzmann distribution means that as the temperature rises, more number of a certain atom become ionized. (This page contains a nice diagram showing which ions form at which temperature)
Conversely, if you lower the temperature, you start to form more neutral atoms, and even molecules after a point. However, that happens only below ~3000K, where it starts to look less and less like a star, and more and more like a huge gas giant planet (see Brown dwarf).
As you lower the temperature even more, you can get big complicated molecules like the ones we have on Earth. All of this happens outside of stars, of course*. (*It requires elements like C,N,O,P all of which were synthesized in a pre-existing star, see how planets form )

As an aside, there is another way electron and nuclei can 'recombine' in stars. At the end of the life of a heavy star (> 8 times the mass of sun), the nuclear fusion (that provides an outwards radiation force) suddenly stops. The star then collapses inward on itself due to the heavy gravitational force. This force is so massive that it crushes the electrons and protons together to form neutrons and neutrinos and a huge amount of energy (a type of nuclear reaction). That type of rapid nuclear reaction causes the star to detonate in a supernova.


text:
Major product of Crossed-Cannizzaro Reaction between benzaldehyde and formaldehyde
I came across a question:

Predict the major product when conc. $\ce{NaOH}$ is added to a mixture of $\ce{PhCHO}$ and $\ce{HCHO}$.

According to me, as $\ce{HCHO}$ is less hindered and $\ce{>C=O}$ of $\ce{HCHO}$ is more reactive, so the major products must be $\ce{CH3ONa}$ and $\ce{HCOONa}$. The hydride transfer should be from one Formaldehyde molecule to the other.
However, the answer was given as $\ce{HCOONa}$ and $\ce{PhCH2ONa}$. Is there a mistake on my part?

Rate determining step of Cannizaro is the hydride transfer and not nucleophilic attack

Why would the hydride prefer transferring to the PHCHO?

Exactly...  Thats how I get HCOONa... But I am concerned about the Alcohol product... HCOONa matches with my answer... PhCH2ONa does not

So what happens to the benzaldehyde>

It might also form some PHCH2ONa... But the major product must be CH2ONa according to me... And that isn't the question what happens to benzaldehyde. What is the major product is what the question demands

See this earlier Q&A.

Ok but I still dont understand why the cross reaction is favored over one formaldehyde molecule reacting to the other

Effectiively there is no free formaldehyde to accept hydride.


text:
Tools like Litmus Paper
Litmus papers are used to test acidity and alkalinity of a fluid. Are there other paper-based tools to test for other stuff like dissolved oxygen, presence of toxicity, electric potential etc?

Hi, Welcome to Chemistry SE. This is good question, it reminded me of paper based battery indicator that worked just like mercury-type thermometer so when the battery potential is full the paper showed full range and if the battery is low then partial range. I can also imagine the plastic based coating on the mugs that turns colorless revealing a picture when hot fluid is poured, there are inexpensive methods to test for various properties.

Pregnancy test kits. Lead paint test kits.

@metron, yes i know temperature can be sensed by such materials but wasnt aware of paper-based battery indicator

@JonCuster what is lead paint test kit? is it also paper based?

Yes, available at your local hardware store. Moisten and wipe a painted object, turns color if lead is picked up from it.

@DrMoishePippik this is valid answer to my question. I am considering writing a paper on all the test papers, you read my mind :) Thanks.

@JhonnyS, I've enlarged the comment a bit and made it an answer.

There are paper thermometers (which actually employ a type of liquid crystals) that change colour and show temperature

@ShoubhikRMaiti well thats quite interesting actually its quite remarkable that we have all these tests available not in sophisticated devices but dispensable papers for easy home chemistry.

It's easy to find a slew of test papers and strips online. For example, a quick search of Amazon for "test papers" turned up chlorine test strips for pools and sanitation, glucose test strips for diabetes, phenylthiourea genetic-linked taste tests, UV-C dose test cards, lead acetate paper test for $\ce{H2S}$, WBC and nitrite test strips for UTI... You could write a paper on all the test papers available.
Increasingly, medicine is making use of paper tests because of ease of distribution, storage and use, particularly important in rural location and during emergencies.
BTW, you can make a test paper for polarity testing, and for detection of peroxides, by dipping filter paper into a suspension of starch and potassium iodide, $\ce{KI}$, rather than buy it. Dampen and use it as you would a potato to determine the polarity of an electric cell.