text:
Does boron form compounds without covalent bonds?
I have read that boron, due to the very high sum of its first three ionization energies, it is not able to form its +3 ions, and thus it generally forms only covalent compounds. But in a popular Indian engineering entrance exam: JEE MAINS-2014 19th April paper, this question was asked:

Which of these statements is not true ?

In aqueous solution, the $\ce{Tl+}$ ion is much more stable than Tl (III)

$\ce{LiAlH4}$ is a versatile reducing agent in organic synthesis.

$\ce{NO+}$ is not isoelectronic with $\ce{O2}$

$\ce{B}$ is always covalent in its compounds.

The answer given was 4.
I get that there might be an exception but I'm not able to find it anywhere. If there exists an ionic compound of boron, could someone please mention that to me?

Ever heard of MgB2?

@IvanNeretin  The difference in Pauling electronegativities between Mg and B is only 0.7 (for comparison, that's less than the 0.9 between C and O), suggesting that bonding between Mg and B would be polar covalent, rather than ionic, in character.

Electronegativity is hardly a precise measure of anything. MgB2 behaves like an ionic compound.

@IvanNeretin   The bonding in MgB2 seems complex.  I looked at a few articles, and it appears that it consists of graphene-like covalently bonded sheets of boron itercalated with Mg atoms, where the Mg acts metallically, i.e., that it consists of a combination of covalent and metallic bonds:  https://www.mdpi.com/2410-3896/4/2/37  (the borons are the small blue spheres, and the blue rods are covalent boron-boron bonds).

Covalent and ionic , I'd say. Well, in any case we both agree that it is not completely covalent, which justifies the claim.

Nope, it doesn't justify the claim.  These articles are saying the borons are bonded covalently, and the magesiums are interacting metallically.  For there to be ionic bonds between the borons and the magnesiums, there would need to be sufficent charge transfer between the borons and the magnesiums for their interaction to be considered ionic, and these articles don't indicate that.  I.e., the picture they're painting doesn't seem to be  negative borons bonded to positive magnesiums, but rather neutral covalently-bonded borons sitting in a sea of positive magnesium ions and negative electrons.

@theorist so are there any other 'non-controversial' examples of ionic/ coordinate bonds formed by Boron that justify the answer that I have stated in my post :| ?

I have no idea.   I was just commenting on the idea that MgB2 is ionic, which seemed surprising to me. Furthermore, I'm not saying that MgB2 definitely isn't ionic.  I'm saying the evidence I've found thus far in the literature seems to indicate it's not.  But perhaps a more thorough search would uncover papers showing that the atoms in the boron sheets are negatively charged. Or perhaps, because MgB2's nature is not yet well-understood, there is not a consensus on this in the literature, and so some think those sheets are negatively charged and some don't.

@IvanNeretin thanks for the information though! now I at least know that there are indeed some exceptions to the fact that Boron 'generally' forms covalent bonds and thus the 4th option was the right answer for the question :)

According to https://assets.researchsquare.com/files/rs-109340/v1_stamped.pdf, FeB2 has partial ionic character:  "It was established that in boride-rich compounds such as FeB2, Fe as cation holds a partial positive charge, and B as anion carries a partial negative charge.".  But this doesn't definitely mean it does; it's just one paper.     Ideally, an expert in inorganic chemistry (which I'm not) will, at some point, weigh in on your question.  I'll be interested to hear what s/he says.

See: https://chemistry.stackexchange.com/questions/100419/lewis-structure-for-magnesium-diboride/100420?r=SearchResults#100420

Boron can form ions but there is some fine print.  You won't get monatomic cations like the metals below it.  Instead, ionic boron structures are formed from clusters where the ionic bonding is driven by the molecular orbital structures in these clusters, not by electronegativity (cf. This answer).
Such clusters are internally held together by covalent bonds between the boron atoms, so in this sense boron is still forming covalent bonds.  The ionic bonds would be with atoms of other elements outside the boron cluster.  Since the valence shells of a neutral boron atom are less than half filled the clusters will likely have low-energy, bonding orbitals that require electrons from outside atoms.  Thus the boron clusters will be anionic and the ionic bonds will be most likely formed with electropositive metals.  As suggested in the comments, magnesium diboride, $\ce{MgB2}$, is one of the most widely studied compounds containing such boron clusters. It has drawn much research interest because of its relatively high critical temperature (39 K) for superconductivity, which may be related to the impact of ionic magnesium-boron bonding on the eletronic interactions that lead to superconduction.
Magnesium diboride has a layered structure in which magnesium layers alternate with boron layers.  The latter are covalently bonded into a hexagonal honeycomb, resembling a carbon layer in graphite.  However, in the boron layers each atom supplies only three electrons per atom instead of four, so the layers may act as electron-accepting structures to form macro-anion having the formula $\ce{B^-}$.  An ionic model for the diboride would then have the empirical formula $\ce{Mg^{2+}(B^-)2}$.  Here I discuss two references I have examined, in which the bonding is examined and the results may be compared with this model.
De la Mora et al. [1] compare magnesium diboride with other $\ce{MeB2}$ diborides using early transition metals and aluminum (the latter might also be regarded as having early-transition-metal character, as there is no $d$ block separating this element from magnesium).  They find that while all the diborides have significant ionic character, this ionicity is enhanced in the magnesium compound.  Thereby the magnesium compound has increased electrical-conduction anisotropy as the valence electrons are strongly localized towards the boron layers.  Zirconium diboride, with less ionic bonding and less electron localization, is also superconducting, but its critical temperature according to this reference is only 5.5 K versus 39 K for the magnesium compound.  The authors also suggest that an isoelectronic, even more strongly ionic $\ce{Li(BC)}$ compound may offer an even further enhancement in superconductivity.
Nishibori et al. [2] found that at room temperature, magnesium is essentially fully ionized to $\ce{Mg^{2+}}$ while the boron remains neutral; the negative charge is associated with the interstitial regions as if to constitute metallic bonds.  This still represents two-thirds of the theoretical charge separation for an ionic model and in that sense, the bonding between magnesium and boron may be deemed predominantly ionic.  At 15 K the electrons become more localized so the boron now has a significant negative charge and the percentage of theoretical charge separation exceeds 80%.
Thus both references agree that in magnesium diboride, the combination of an electropositive electron source with a favorable molecular structure for electron acceptance leads to strongly ionic bonding between the magnesium and the boron.  This applies especially in the low-temperature superconducting state, even as the boron-boron bonding within the boron layers themselves remains covalent.
References
1.
Pablo de la Mora, Miguel Castro and Gustavo Tavizonb,
"Comparative study of the electronic structure of alkaline-earth  borides (MeB2; Me=Mg, Al, Zr, Nb, and Ta) and their normal-state conductivity", Journal of Solid State Chemistry 169 (2002) 168–175, https://doi.org/10.1016/S0022-4596(02)00045-2.
2.
Eiji Nishibori, Masaki Takata, Makoto Sakata, Hiroshi Tanaka, Takahiro Muranaka and Jun Akimitsu, "Bonding Nature in MgB2", Journal of the Physical Society of Japan 70:8 (2001), 2252-2254,
https://doi.org/10.1143/JPSJ.70.2252.


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Calculating the percentage of (+)-2- butanol in the sample

Specific rotation of (+)-2-butanol is 12.5$^{\circ}$ . A sample of 2-butanol containing both the
enantiomers was found to have specific rotation value of –2.8$^{\circ}$ . What is the percentage of (+)-2- butanol in the sample?

I attempted it by letting the percentage of (+)-2-butanol be $x \%$ and therefore percentage of enantiomers in the mixture would be $2x \%$ due to equal contribution by (-)-2-butanol. Since the net specific rotation of the mixture is given to be negative, we know that percentage of excess of (-)-2-butanol would be $(100-2x)\%$. It is also given that specific rotation of pure (+)-2-butanol is 12.5$^{\circ}$, but I am unable to understand how is that useful here.
In it's solution, they have calculated the percentage of (-)-2-butanol as $\dfrac{2.8}{12.5} \times 100=22.4 \%$ and thus percentage of (+)-2-butanol came out to be $\dfrac{100-22.4}{2}=38.8\%$.
I didn't understand this last step, why is the division of 2.8 and 12.5 giving the percentage of (-)-2-butanol?

A different, more logical solution gives the correct answer as well.
Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $x$ and the mole fraction of the (-) isomer is $1-x$, then
$$-2.8 = 12.5x -12.5(1-x)$$
Solving this simple linear equation gives $x = 0.388$, which when converted to mole percentage, is $38.8\%$.

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What is meant by "basis" in "... its basis in particles leads to a material comprising crystallites or grains tightly bound together, ..."?
I am currently studying Ultrasonic transducers – Materials and design for sensors, actuators and medical applications by Kentaro Nakamura. In chapter 1.2.1 Ferroelectricity, the author says the following:

$\ce{BaTiO_3}$ is realised as a ceramic, originating as a mouldable paste, which is dried and sintered to form the final material. The properties of the paste result from the mixing of particles with a liquid and its basis in particles leads to a material comprising crystallites or grains tightly bound together, with properties differing between the individual grains and the collection of grains in the bulk ceramic because of the effects of the interfaces between the grains. Following the principle of minimisation of energy, and taking into account the resulting internal dynamics of the relative positions of the ions, $\ce{BaTiO_3}$ divides itself into multiple domains within each grain, each domain having a common polarisation, but with the polarisation differing in direction from that of neighbouring domains by either $90^\circ$ or $180^\circ$.

My concern is with this part:

The properties of the paste result from the mixing of particles with a liquid and its basis in particles leads to a material comprising crystallites or grains tightly bound together, with properties differing between the individual grains and the collection of grains in the bulk ceramic because of the effects of the interfaces between the grains.

What is meant by "basis" in "... its basis in particles leads to a material comprising crystallites or grains tightly bound together, ..."? I don't understand what this means. Is this possibly a typo or incorrectly used term by a non-native English speaker? If so, then does anyone know what they meant (or what it should say for this described process)?

I would suggest that "basis" can be here replaced by "content"

@Maurice In that case, what would "The properties of the paste result from the mixing of particles with a liquid and its content in particles leads to a material comprising crystallites or grains tightly bound together, ..." mean? I still don't understand what "content in particles" would mean.

It may mean that mixing a liquid with a powder makes a paste which is heterogeneous on the microscopic level. But this mixture may appear homogeneous at the observer. If you imagine the mud made by sand deposited in the bottom of a lake, you may have an idea of this substance. Mud is made of sand grains with water between them. The liquid may be considered as having a ¨basis¨ or a "content" of sand

@Maurice Hmm, I see what you mean. But, in the mud example, the water is not in the sand particles, it's between them, right? "content in particles" sounds weird. But your explanation makes sense, so perhaps that's what the author meant.

Of course. The water is not in the sand but between the grains.

The author is attempting to indicate that the "paste" is a suspension of particles that aggregate as it dries (ie the liquid portion evaporates), in contrast to a solution of a dissolved solid that recrystallizes as the solvent evaporates. Because the particles never dissolve, there is no possibility for a single crystal to form that is any larger than an already existing particle. Therefore, the resulting solid is necessarily composed of many small crystals/particles with boundary zones between them, rather than having a large scale uniform structure.

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Mechanism for Lucas Test
In the mechanism for Lucas' Test for Alcohols, I am reading two different types. One says the first step is the abstraction of a proton from HCl, but on other references, I also see that the first step is the formation of a zinc chloride complex. Anyone can clear this out for me? Thanks

The first step is the formation of the complex $\ce{ZnCl4^{2-}}$. Then the alcohol reacts with this complex and makes a new intermediate complex having two new bonds, one Zn-OH, and one Zn-R. Later on, Zn would go back to a tetra-coordinate complex, by keeping the OH group, then loosing R and one Cl atom, which produces R-Cl.

See this question and the question linked in the comments.


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Change of entropy of a thermal reservoir
How to calculate/derive the change in entropy of a heat reservoir as function of its internal energy? If the constant temperature of heat reservoir is T, and
$$ S(U_0)=S_0,$$
what is $$ S(U)? $$

For a reservoir of finite thermal inertial, the change in entropy is $$MC\ln{(T_f/T_i)}=MC\ln{\left[\frac{(T_i+\frac{Q}{MC})}{T_i}\right]}$$
Here's a hint:  What is the limit of this as the thermal inertial of the reservoir MC becomes infinite?

Thanks! About this heat Q, is it equal to the change in internal energy? Can the volume of heat reservoir change and thus can it do work?

It's reasonable to assume that the reservoir is an incompressible liquid.

Or we could assume that the reservoir is at constant pressure, in which case $Q=MC_p\Delta T=\Delta H$.  Either way one arrives at the same result.


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Is it possible to have three stable liquid phases in a binary mixture?
With this system, it would be possible to have three liquid phases and two liquid-liquid coexistence regions at various compositions. Similar behaviour is common for solids such as alloys, sometimes with even more stable phases.
I have been searching for a liquid mixture with this behaviour, but haven't found any. Are there any liquid binary mixtures which exhibit this behaviour, or is there any physical reason why this isn't possible? Does it have something to to with the higher disorder liquids have compared to solids?

There was a similar question not long ago. I commented that triple points are known in liquid crystals. I have never seen a liquid triple point on a binary elemental phase diagram.

Ah, your similar question was over on Physics... So, how do you feel about liquid crystals vs binary liquids?

I have been looking for binary liquids specifically, but thanks for mentioning liquid crystals. It kind of makes sense that they appear in liquid crystals, since they are more ordered than liquids. Solid solutions like alloys can show even more complex behaviour, and it seems like it might have something to do with the higher order they have. The thing I'm interested about is whether there is some deep physical reason binary liquids don't exhibit this behaviour, or it is just very unlikely and doesn't appear in any mixtures researched so far.

I currently have (quickly checking) some 2270 papers on my hard drive on binary or ternary phase diagrams. Most are elemental diagrams, some small fraction are various salts (nitrates, sulfates, etc.) or oxides. I have never seen a diagram with a liquid triple point. Miscibility gaps are modeled by either a simple positive enthalpy of mixing or an 'associate' suggesting some preferred clustering at a specific ratio.


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Finding the structure based on spectra
I had to identify the structure of a drug based off the spectra I had. I suck at this. I gave it my best go and got some ugly structure, only to check it using a database and find out I was wrong. I gave up and searched for the chemical structure I was given, only to find out that the drug was diazepam and I was on the right track (my aromatic substitution was correct, and I got the correct degree of hydrogen defeciency, but I didn't know how to tie everything together and I was stuck with one more quaternary carbon that I didn't know where to place). I feel like the spectra I was given are hard to read (specifically the HSQC, HMBC, and COSY); do you all feel the same? I really want to know how to arrive at the answer.
The images are the spectra I was given. I was also given mass and IR strectra, but I cannot post too many pictures with my reputation. If you all need better resolution images, let me know. The molecular formula is C16H13ClN2O, and the groups I found with the IR were NH, C=O, and benzene. I am kind of afraid to ask the instructor at this point because I already found the answer. It wouldn't hurt to ask him anyway, would it? I just don't want to lose points.
EDIT: I took the original images down because I did not want the instructor to find it, if he hadn't already.

Additionally, I would love any tips to help me get the structure correct the first time around.

The most practical approach is to identify the structure of an organic compound by combination of spectroscopic techniques.  This is why you were provided with UV-Vis, IR, NMR and MS data, because it equally reflects the eventual approach in the lab after product isolation and purification, too.  Hopefully you have sufficient background what theses techniques assess (if not, consider consulting e.g., the spectroscopy section here on ChemSE, or on Chemistry LibreTexts).
Then, start training with simpler molecular structures in first place.  You may use, for example, Field's training booklet from the shelves (either print or/and electronic resource) of your school's library.  The obstacle here: for the majority of the sets, there is no answer in the book (the key is available only to instructors).  You may however list structures likely to be in agreement with the data per set with public references such as the SDBS spectral database.  You equally may train yourself picking an organic compound from Wikipedia and list the features a theoretical UV-Vis IR or MS or NMR spectrum would contain and again compare, e.g. with SDBS.
Yes, there are programs to predict the position and shape of NMR spectra, and you equally may find diazapam with a search via Hill formula on nmrshiftdb2.org, too.  They may help you to internalize how structural features (e.g., CH vs CH2) show in the APT spectrum because you may select either of the type of nuclei to be displayed, or to highlight their attribution of a calculated position of a signal to an atom in a given structure, or their indication of a cross peak e.g. in the HSQC.  Aside from an exam where these programs might not available to you, don't rely too much on them.  E.g., the position of signals in the $\ce{^1H}$-NMR spectrum depends at least on the concentration of your analyte, the choice of solvent (e.g. chloroform or DMSO) may influence if exchangeable protons on alcohols and amines yield a broad signal, or split into a multiplet (if there are coupling nuclei).
The only way to get better in this is training.  If the structure suggested is not in agreement with all of the spectral observations, the assignment is erroneous and one has to start again. It may be demanding, but don't give up.

As a starter:

N-NMR:  there are two types of N with different chemical environment
H-NMR:  there are two sets of alkyl-H; by integration possibly one $\ce{CH3}$ and one $\ce{CH2}$ or integer multiples of both.
APT-NMR:  again indication for a methyl, and a methylene group; count of CH: ... which are in the region expected for aromatic CH, count of C: ... etc.


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Role of catalyst in elementary reactions towards rate law
We do know that the concentration of a catalyst might affect the rate of reaction. But what if the catalyst is included in the rate law? For example,
$$\ce{O3 + M <=>[$k_1$][$k_{-1}$] O2 + O + M},\tag{1}$$
where $\ce{M}$ is the catalyst, when I am writing the rate change of $\ce{O},$ do I write
$$\frac{\mathrm d[\ce{O2}]}{\mathrm dt} = k_1 [\ce{O3}][\ce{M}] - k_{-1}[\ce{O2][O][M}],\tag{2}$$
or
$$\frac{\mathrm d[\ce{O2}]}{\mathrm dt} = k_1 [\ce{O3}] - k_{-1}[\ce{O2][O}]?\tag{3}$$
Which is correct?

can the downvoter explain the downvote?

The catalyst uses a different reaction path with the zero net catalyst consumption, compared to not catalyzed reaction.
E.g. instead of $\ce{A + B <=> C}$, there is $\ce{A + M <=> AM}$, $\ce{AM + B <=> M + C}$.
If there is a homogenous catalyst, then its concentration in rate equations is processed by the same way as the reactants or products. But the different reaction path, compared to non catalyzed reaction, remains.
So it would not be $$\ce{O3 + M <=> O2 + O + M}$$
but rather e.g. :
$$\ce{O3 + M <=> O2 + MO}$$
and
$$\ce{MO <=> O + M}$$

OR:
$$\ce{O3 + M <=> O + MO2}$$
and
$$\ce{MO2 <=> O2 + M}$$
So you need to solve the kinetics of the system of linked reactions, considering catalyst form concentrations in steady states.

If it is a heterogenous contact catalysis, then the amount and activity of catalyst is implicitly involved in the catalyst-dependent kinetic constant:
$$\frac{\mathrm d[\ce{O2}]}{\mathrm{dt}} = k_{1,\mathrm{M}} [\ce{O3}] - k_{-1,\mathrm{M}}[\ce{[O2][O]}$$
where the ratio $\frac{k_{1,\mathrm{M}}}{k_{-1,\mathrm{M}}}$ remains constant, independent of the catalyst activity.


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What is a common cause of peak tailing?
What is a common cause of peak tailing?
a. The distribution constant decreases as analyte concentration increases.
b. The distribution constant increases as analyte concentration increases.
c. The plate height increases as analyte concentration increases.
d. The plate height decreases as analyte concentration increases.

Simple answer to your question should be - None. Whoever designed this question for students does not understand the basic principles of analytical chromatography.
Under low concentration of analyte, which is the usual case, peak tailing has kinetic origins, or it could be from secondary interactions or due to the structure of the packed bed. All peaks in real chromatography tail to some extent i.e., none of them are 100% perfect Gaussians.
In rare cases, you can inject a lot of sample on an analytical column. There peak tailing  originates due to non-linear isotherms. This is what the test question is about. You have to read this section on isotherms Distribution Isotherms in Chromatography to choose the right answer.
There is another form of chromatography, called preparative mode, where you really inject a lot and this is where isotherms come in.


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How to read mutual solubility curves of salts with a common ion?
Came across a few graphs that represent the mutual solubility of two compounds in a solvent, for e.g., solubility of $\ce{KCl}$, and $\ce{NaCl}$ in $\ce{H2O}$.
On these particular graphs, I can see $x$-axis as $\pu{g}/\pu{100 mL}$ of $\ce{KCl}$, on $y$-axis as $\pu{g}/\pu{100 mL}$ of $\ce{NaCl}$. And the graph has multiple lines each representing a curve at a specific temperature.
I just wanted to understand how to read these type of graphs:

To me the graphs represent a point wise mechanism. For example, if I pick a spot on the $x$-axis, with $\pu{50 g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$, does that mean if I dissolve $\pu{50g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$,then I will not be able to dissolve $\ce{NaCl}$ in it?
Or does it mean if I have a starting condition where I have $\pu{50g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$, and I start adding $\ce{NaCl}$, then $\ce{KCl}$ will precipitate out at a slow rate till I reach point A at a very rapid rate?
The graph doesn't appear complete, or is at least missing additional info to make sense of it. I want to know what that info is.

My interpretation is:
I assume both cross solubilities are related to the volume of water used for the initial solution, not of the initial solution volume, i.e. both in $\pu{g}/\pu{100 mL}$ of water.
If you start at $x$-axis with the $\ce{KCl}$ solution, you can dissolve $\ce{NaCl}$, climbing upwards until you reach the curve for the given temperature. If you cross it, some $\ce{NaCl}$ will not dissolve, and some $\ce{KCl}$ may precipitate, because of the common ion effect.
The same of the initial $\ce{NaCl}$ solution, starting on $y$-axis, and dissolving $\ce{KCl}$, going to the right, again until you reach or cross the curve.
$\ce{KCl}$ solubility is lower, if some extra $\ce{Cl-}$ ions from $\ce{NaCl}$ are present, because of the solubility product $K_{\mathrm{sp}, \ce{KCl}} = a_\ce{K+} \cdot a_\ce{Cl-}$.
Similarly for $\ce{NaCl}$,  $K_{\mathrm{sp},\ce{NaCl}} = a_\ce{Na+} \cdot a_\ce{Cl-}$.
Problem is difficulty with both measuring and predicting ion activities or activity coefficients for concentrated ionic solution. Replacing activities with concentrations could be done only with lot of grains of salt and both eyes closed tightly.
I do not have clear interpretation of the about horizontal edge in the upper chart part. It looks like a kind of "precipitation eutecticum", an analogue to a binary mixture phase chart.

Feedback to comments:
If the system is at the cross-saturation curve, and there is excess of one solid salt, then due dynamic nature of processes, this salt dissolves and the other precipitates. The system is moving slightly along the curve, until for each of salts the rates of dissolving and precipitating are equal. This process is slow.
If a cross-saturated solution cools down, both salts precipitate, according to the ratio of kinetic rates of both processes, until the curve for new temperature is reached.
The ratio can be only very roughly estimated from

the chart position
the direction of the normal vector of the curve
the different ion mobility.

" both in g/100 mL of water". Yes this is my intrp as well, its about x,y gms of Kcl/Nacl in 100ml water.
Okay so, if I am at 80c, with an initial KCl soln of 40gm/100ml, then I can dissolve upto a little less than 15gm of NaCl before, something will happen(ie. Some ppt of KCl, or NaCl).
And I guess just for completeness , if I start with 20G NaCl in 100ml water at 80C, then I can dissolve ~35gms of KCl.
Is that right ?
Also, any idea how this graph can be used to predict what ppts out when temp drops say from 80C to 75.

See the answer update.


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Is there any relation between resonance and smell of a hydrocarbon?
I came across several examples (such as many benzene derivatives) of hydrocarbons having resonance that also seem to have a sharp odour. So, does resonance have an effect on smell?

Resonance is not a physical phenomenon, it is just a mental picture.

Aromatic compounds were called aromatic because they have aroma.

Carcinogenic aroma :-( Salute to the good old organic chemists of the 18th century who risked their lives out of curiosity driven research.

Then again, many non-aromatic hydrocarbons have aroma as well.

I wish I could convey the spirit of Professor Louis Fieser on the development of organic chemistry. I couldn't find it in his textbooks, but in class, it went something like this:
Coal distillation led to a large number of different unsaturated compounds, most of which had strong odors, so they were classified as "aromatic". When more quantitative measurements (heats of combustion and heats of hydrogenation) were available, double bonds were investigated, and some compounds were found to be unusually stable. A single $C-C$ bond should have a certain energy, and so would a $C-H$ bond, and so also should a $C=C$ double bond, but some compounds gave surprising results.
Some molecules (like benzene) could be written with two positions for the double bonds - they could "resonate" between the two arrangements. Most of the compounds that had this sort of possibility were odorous (aromatic), so any compound with this sort of resonance possibility was called "aromatic", even if it had no appreciable odor. And they all had some sort of extra stability ("resonance energy") due to the double bond shifting.
Now while "resonance energy" ought to have some quantitative significance, it had some considerable variability, depending on what values you used to calculate the expected value for heat of hydrogenation or combustion. Another term popped into use: "aromaticity", which was never quantitatively defined, but was really a measure of the surprise at the difference between the calculated value and the observed value of the heats.
But I expect that odor is not a result of extra stability, but more likely a result of some reactivity of the multiple double bond system. And volatility.

There is, but it is somewhat accidental
Resonance is an odd, ill-defined concept to start with. It was an early attempt to explain the structure and properties of molecules like benzene (which doesn't behave as the entirely hypothetical molecule cyclohexatriene).
The link arises because of the constituents of crude oil. In the early days of chemistry, when experts were trying to understand all the constituents of crude oil they discovered that some distillation fractions contained things with strong, distinct smells (also, as James Gaidis notes, this is true for many products from the distillation of coal). They called them aromatic hydrocarbons. As a better understanding of their structures emerged, it turned out that many of these aromatics contained benzene and its derivatives (toluene, xylene etc.) These were the source of the "aromatic" odours. But their structure posed the above problem noted above: they were clearly unsaturated (ie contained carbon-carbon double bonds) but had distinct properties from simpler alkenes. This led to the idea of resonance (multiple different structures exist in "resonance" where the double bonds move around). Today we would usually talk about delocalised pi-orbitals instead of resonance.
So the group of compounds that led to the idea of resonance happened to be particularly smelly (which is why they were called aromatics). In this sense resonance is related to smell. Many volatile benzene derivatives and related compounds happen to have a strong smell. But, more generally, not every molecule with delocalised bonds (or resonance) will be smelly, though it seems very common in simple hydrocarbons.


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How to calculate ∆rGº from entropy and ∆fGº in different temperatures?
"The total oxidation of glucose occurs according to the following chemical equation:
C6H12O6 (s) + 6O2 (g) -> 6CO2 (g) + 6H2O (l)
The following table gives us the free energies of standard formation and the standard molar entropies of
compounds involved in the previous reaction.

Compound
∆fGº(298 K) kJ/mol
Smº (298 K) J/K/mol

C6H12O6
-917,2
212,10

O2
0
205,14

CO2
-394,36
213,14

H20
-273,13
69,91

Based on the previous data, determine the ∆rGº
for glucose oxidation at 308 K"
It's a question of a doctoral selection process and I want to figure out if it is formulated wrong or if there is a way to an answer.

Yes, it is answerable.  From the individual free energies and entropies of formation, you can get delta_G and delta_S for the overall reaction at 298K, from which you can then get delta_H (from delta_G=delta_H – T delta_S).   delta_H in turn gives you the temperature-dependence of the reaction,  from which you can get delta_G at 308K.  Now that you have a roadmap, try working through as much as possible on your own.

Should not it be rather $\Delta G^{\circ}_f$ and $\Delta G^{\circ}_r$ ?

For a small temperature change like that, you can make use of dG=-SdT.  Determine $\Delta rG^0$ and $\Delta r S^0$ at 298, multiply the latter by 10 , and subtract it from the former to get $\Delta rG^0$ at 308K.


text:
Energy bands in Liquid

In solid, energy bands means the 1s orbital with spin up and down become 3 states when there are 3 atoms.
But then. If you look at hydrogen atoms, the allowed bound states have energies equal to En=Ry/(n2) and n is the main quantum number ranging from 1 to infinity. In other words, there is an infinite number of available electronic states in a single hydrogen atom.
When we talk about energy bands. Which of the above is usually referred to? The few states orbitals or the almost infinite allowed bound states?
For the latter, how do you characterize the energy bands in liquid like
water? What kind of bands are formed in such cases?

Liquids, and especially water, are a bad example for discussing band structure of semiconductors.

No i chose semiconductor bec i cant find the energy band tag. But my question is about energy bands of liquid. What tag must i use? I just changed the tag to liquid

When we talk about energy bands, we are talking about mixing of orbitals on different atoms (AOs) with similar energies. First, assume simple models. In the solid, the structure is very regular and all atoms have AOs with identical matching energies. This makes the number of orbitals available for mixing in solids very large, and mixing results in bands. In a liquid, on the other hand, structure is highly irregular. Orbitals on different atoms often have energy or overlap mismatches, but occassionally match and mix, and this results in much local mixing (say due to hydrogen bonding) but not extended delocalized orbitals. Because of energy mismatch you can also end up with bands in liquids but their origin (in this model) is different than in solids.
The difference between the origin of bands in liquids and solids is similar to the different cause of homogeneous and inhomogeneous broadening. In both cases you get bands (a distribution of energies), but in one case broadening involves delocalization over resonantly matched orbitals on different atoms/molecules and in the latter it involves variations in the local environment due to randomly fluctuating perturbations caused by neighboring molecules  ("the solvent"). Extensive orbital mixing and Pauli's exclusion principle results in homogeneous broadening in the first case. Motional jostling in the case of a liquid results in inhomogeneous broadening.
Energy levels of individual molecules in liquid water are "perturbed" by neighboring molecules. These perturbations are small compared to the ionization energy of the valence electrons. The stability conferred by an H-bond in water is in the order of 20 kJ/mol. Compare that to the ionization energy, ~1200 kJ/mol. Therefore electrons remain bound to local waters even as their energies are slightly perturbed by interactions with other molecules. H-bonds do contain covalent character, that is, involve sharing of electrons between two molecules (specifically between H-bond donor and acceptor atoms). However this does not imply electrons in liquid water are extensively delocalized beyond local bonds, certainly not on the spatial scale possible in solids.
As a note of interest ice under pressure exhibits metallic phases.
Note perturbations due to the solvent are evident in the difference between vibrational spectra obtained in different phases.

What if there is no solvent. Let's just take the case of pure H2O. We mostly have hydrogen bonding, how do the electrons behave in water? That's what I wanted to know in the original question.

I included some additional comments. In pure H2O the solvent can be regarded as waters surrounding one molecule being inspected.

I think this makes it seem that there are very different reasons for the electronic structure of solids and liquids - there aren't, within an orbital view of the world they both exhibit bands ultimately due to same thing, and those bands can be found by exactly the same method - solving an appropriate approximation to the Schrodinger equation. Yes the topological disorder does result in differences in the solutions in certain cases, for instance Anderson localisation https://en.wikipedia.org/wiki/Anderson_localization, but the effective Hamiltonian is the same.

@IanBush I am attempting to convey the difference between a regular extended bonding network in a solid involving relatively stable bonding as opposed to continuously rearranging bonds in a liquid. Bands in a solid are basically regarded as due to highly delocalized orbitals rather than localized covalent bonds. In a liquid the energy is altered not only by bonding but also due to perturbations by other (not necessarily bonding) molecules which do not share orbitals.

My understanding is also that a high degree of translational symmetry in solids allows a different treatment than liquids, which is why the latter can be more difficult to model (sample).

Must ... Not ... Go ... Too ... Deep ... My doctoral thesis was on electron localisation in materials with liquid like disorder. And yes, the translational symmetry can be used in periodic systems, but the theory is really no different from using symmetry adapted orbitals to block diagonalise the Hamiltonian, which is standard practice in quantum chemistry - it is just that space groups are being used rather than point groups. So from this point of view and within the BO approximation a liquid is simply an example of a solid with C1 symmetry.

Hmmm, but what's relevant to the question is not necessarily that you can write Hamiltonians with the same elements (as you might for solid ice at low and high pressures), but rather the differences between the systems, and the regularity of solids is certainly instrumental in determining the nature of the solutions, not just the sparsity of your matrix.

@IanBush I would be interested to read your interpretation of an appropriate answer. Is a liquid to be thought of as a defective solid? How do you address temporal variations in a liquid which continuously disrupt bonding networks?

OK - will do, prob at the weekend now. But note in the same AO basis the Hamiltonian matrix for a solid and a liquid at the same density will have approximately the same sparsity, Translational symmetry means that there exists a basis within which the Hamiltonian becomes block diagonal, and that is related to the original basis by a unitary transformation. And as only the solid has translational symmetry this new basis only exists in that case. In any other basis, almost certainly including an AO based one, the matrices will look very similar.

@buck thorn u only answered the last paragraph. I want to know the rest.Quoting agen "In solid, energy bands means the 1s orbital with spin up and down become 3 states when there are 3 atoms. But then. If you look at hydrogen atoms, the allowed bound states have energies equal to En=Ry/(n2) and n is the main quantum number ranging from 1 to infinity. In other words, there is an infinite number of available electronic states in a single hydrogen atom. When we talk about energy bands. Which of the above is usually referred to? The few states orbitals or the almost infinite allowed bound states?"

@Jtl I tried to address this. The origin of bands is mixing of orbitals of same (or similar) energy on different atoms. By bands one does not mean the range of excited orbitals in one atom such as H.


text:
energy difference vs absolute energy
When dealing with determining the ground state energy of a system, people often say that it is better to know the energy difference rather than the absolute minimum energy. I understand that the absolute minimum energy depend on the model so the more complex your model, the lower or closer the energy you found to the true ground state energy.
But what do people refer to when they say energy difference here? Different between what energies exactly?

They are referring to the energy difference between electronic or vibrational states or between rotational states.  Energy in QM calculations must always be in reference to something.  Usually it is specified as the minimum of the PES for the lowest electronic state.  As you mentioned, absolute energies will depend on the method and choice of expansion, basis, etc.

Absolute energies returned from ab initio software is usually with reference to infinite separation of all particles which is usually hundreds of hartree away from the minimum energy of the PES so that it is a useless reference point.

The energy difference can also be from the bottom of the potential well, but being trapped at the bottom of the well is not an accessible state.

At the end is due to the fact that nothing has absolute zero energy, just a minimum. As far you are able to get or locate values versus the same reference, than you can compare them.

@Bertram Regard to your first comment: Let's say I use two different methods to calculate the electronic ground state energies. Then is it right for me to say that the energy differences here refer to the differences in energy I get from different geometry configurations? For example, I calculate it the ground state energy at bond distance of 1,2, and 3 Angstrom. I found that I get min energy occurred at 2 Angstrom on both method. Now, I will take the difference between the energies between 1 and 2 Angstrom and 3 and 2 Angstrom on both methods and compare?

@Alchimista am just trying to understand how can I calculate this energy difference that people tend to refer to. I understand that if you give me a Hamiltonian, I can discretize using some basis, then solve for the lowest eigenvalue. What value can I take as a reference point for me to calculate what people often refer to as energy difference? Thank you for your response!

@user102247 Yes the process you described would be reasonable.  As long as you are looking at the energy differences at different geometries within the same method.  And as you might expect there will be differences in the results depending on what method you are using.

@user102247 if you look for a technical answer be clearer in your question. The point is that you get different values, the difference is naturally there. It can be that whatever method is more or less accurate depending on the absolute energy, but this is already too technical. Perhaps I don't get your question. Conceptually, is like to graphically split AO in MO. The difference between the levels is in there.

Perhaps I got it. No one would ever state an energy difference for just the ground of a species being calculated. If done, it will be implicit there is a common reference for all cases treated the same way. This doesn't happen in quantum calculations only. Think about thermochemistry or thermodynamics. You can sort a table of H of formation for compounds. But you must have defined standard conditions, as well as what "formation" stands for.

To clarify, the meaning of "better to know the energy difference..." is more understandably "keep in mind that the energy of the minimum is rather a difference in energy versus some kind of reference". There is no difference to be calculated, it is just a statement that the value is not absolute.


text:
How solubility in alcohol imply covalent nature

lithium chloride is considered to be an ionic compound,but it is soluble in alcohol,which suggests that it also posses a small amount of covalent character.

This is an extract from my book ,when I looked it up it said ionic bonds are polar so they are soluble in alcohol.
So how does lithium chloride's solubility in alcohol prove it has small ammount of covalent character?Is my book wrong ?

Ever saw chalk? It is ionic. Ever saw sugar? It is covalent. Ever saw a book full of unjustified generalizations?

A hint can be given, that CaCl2 crystalizes with 6 H2O, or 4 CH3OH, or 3 CH3CH2OH, therefore anhydrous CaCl2 cannot be used for alkohol drying.

See answers here:  https://chemistry.stackexchange.com/questions/115112/why-do-we-use-licl-as-a-source-of-chloride-ions-as-nucleophile-but-not-other-alk/115149?r=SearchResults#115149

Also consider that ionic bonds break apart leading to solutions of ions. In a way there is a difference while dissolving, say, NaCl or sugar. The former is an ionic solid and goes into ions, while the second one retains its molecular structure.

Your title doesn't really match the actual question. This will score badly here so you should fix it.


text:
What is the purpose of balancing a redox half equation using H+ and H2O and electrons?
Is it possible to say that:

H+ is used due to the provision of an acidic medium?

H2O used as it is an aqueous medium and products have oxygen, so logically oxygen must come from H2O?

Electrons added to conserve charge?

You must have heard about the fundamental conservation laws od nature. The consequence is, if the equation is not balanced, it is wrong. Note that $\ce{H+}$ is just a convenient simplification of $\ce{H3O+ . n H2O}$

that doesn’t explain why it must be added using those... i can easily add oxygen gas to balance...

I was addressing the title. The title is in agreement with the body, isn't it ? BTW, try to reformulate the body, it is not very easy to understand, what you mean.

You can add anything, but it may not follow the reality. Equations describe what reacts and what are the products.

It is not about what is used, but what mutually interact and react. E.g. $$\ce{2 H3O+ + 2 e- -> 2 H2O + H2(g)}$$

how is this rule even generalised? I mean the fact that H+ and H2O is always used, you’re saying that every redox half equation has it? sorry i’m very confused by this... i always thought there was a purpose to it (like the one i highlighted in the qn since that’ll make sense)

It is not tule, it is following the reality. And no, I am not saying H+ and H2O is always used. Where are they in Cu2+ + 2 e- -> Cu ? The best would be the illustration on particular full reaction and its separation to half reactions in the question.

What about balancing redox reactions in aqueous base? No protons available there.

One reason to track solvated hydrogen (or hydroxide) ions and water in a redox reaction is that the hydrogen ions often get transferred too, and you need to know how that takes place to render the products.  We should expect hydrogen-ion transfers as part of the process in a protocol solvent such as water or a concentrated protic acid, as the electrons are charged and some compensating charge-transfer process must take place to keep the charges in balance.
Copper in concentrated nitric acid, where nitrogen is reduced (primarily) to the +4 oxidation state, is a well-worn example.  The oxidation half-reaction is, of course, copper being oxidized to $\ce{Cu(II)}$.  So we look at the reduction half-reaction.  In its "raw" form we would have
$\ce{HNO_3 -> NO2}$
We balance the elements other than hydrogen and oxygen, which is already done, and then compare the combined oxidation states of those elements in the reactants and products to find that the nitric acid must have taken one elecron:
$\ce{HNO_3 + e^- -> NO2}$
Now the charges are not balanced, and so we invoke that compensating hydrogen ion transfer.  In acid this is set up by adding (solvated) hydrogen ions to balance the charges:
$\ce{HNO_3 + \color{blue}{H^+} + e^- -> NO2}$
Water is used to balance the hydrogen which, if we did things correctly, balances the oxygen as well giving our familiar result
$\ce{HNO_3 + H^+ + e^- -> NO2 +H2O}$
and upon combining with the copper oxidation:
$\ce{Cu + 2 HNO_3 + 2 H^+ -> Cu^{2+} + 2 NO2 + 2 H2O}$
in which we see that in the final reaction, hydrogen ions are indeed transferred.  So we needed to account for them.


text:
Help with Forster distance conversion for FRET
For my project on Fluorescence, I'm trying to simplify the equation for the Forster distance (I obtained this from the Lakowicz book). To keep track, the units are in square brackets.
$$R_0 = \left(\frac{9000\kappa^2\Phi_D\ln(10)}{128\pi^5N_An^4}J(\lambda)\right)^{1/6} \approx 0.211 (\kappa^2n^{-4}\Phi_DJ(\lambda))^{1/6}\, [Å]$$
I'm using $J(\lambda)$ in units of $M^{-1}cm^{-1}nm^4$ but for some reason I can't seem to obtain the $0.211$ no matter how I do the conversion. I always seem to be off by a factor of $10^{-3}$, which I can't seem to locate.
For reference, $J(\lambda)$ is the overlap integral:
$$J(\lambda) = \frac{\int_0^{\infty} F_D(\lambda)\varepsilon_A(\lambda)\lambda^4 d\lambda}{\int_0^{\infty} F_D(\lambda) d\lambda}$$
For example, I simplified $R^6_0$, attempting to convert to Angstrom:
$$8.79\times10^{-25}\kappa^2\Phi_Dn^{-4} J(\lambda) [mol\,M^{-1}\,cm^{-1}\,nm^4]$$
$$=8.79\times10^{-25}\kappa^2\Phi_Dn^{-4} J(\lambda) [0.1m^210^{-36}m^4]$$
$$=R^6_0 = 8.79\times10^{-62}\kappa^2\Phi_Dn^{-4}J(\lambda) [m^6] = 8.79\times10^{-2} \kappa^2\Phi_Dn^{-4}J(\lambda)\,[Å]^6$$
Which, upon taking the 6th-root, I don't get the number $0.211$. If the power is $10^{-5}$ I do get the number.
What am I missing?
Any help appreciated. I'm not sure if I'm overlooking something painfully obvious, but I can't see it.
Cheers!

Might be of use - in this publication, there is a 9 where you have a 9000. Haven't worked through the rest and not sure if your expression is equivalent to their eq (2), p. 5, but might be a clue?

I did indeed wonder about that. I'm not 100% sure where the prefactor is derived, and most sources (including the lakowicz) appear to use 9000, and then have the 0.211 as well.

@FrankieS.Palmer, Your numbers seem to be right, a back of the envelope calculation also results in 0.0879, which is same as yours. However, other publications also use 9000. It has have a power 10^-5 in order to get 0.211. It means there is an extra hidden factor of 1000 in the numerator.

@M.Farooq Yeah, as mentioned before by Todd, if the numerator is simply a 9 then I get the right result- the issue is now more a case of why that's the case- as several sources (the Lakowicz book included) appear to give the 9000 as I have, but with the correct 0.211 result, too.

It has to be 9000, I checked several references, Lakowicz will not hopefully do this mistake of writing 9000 for 9. We are missing a 1000 factor somewhere.

It is probably in how the extinction coefficient is defined; usually as $dm^3/mol/cm$  as we use concentration in $mol/dm^3$ but may defined be as $m^3/mol/cm$ in some texts.

But the author, Lakowicz, after already clarified the units of the overlap integral.

what I meant was that as $\epsilon$ is in $dm^3/mol/cm$ but we work in SI then there is a factor of $10^{-5}$ in J .

Afaik, $M^{-1}cm^{-1} = 0.1m^2mol^{-1}$, which I've used

A relevant discussion of the 9000-form vs 9-form can be found in section 3.4 of Förster Resonance Energy Transfer-From theory to application edited by Medintz and Hildebrandt and discussion in Braslavsky, et. al., 2008.
I think the confusion might be stemming from how Förster used $N'$ to represent the number of particles per mole instead of $N_{A}$.
Both ($N'=\pu{6.02E20mmol^{-1}}$) and ($N_{A}=\pu{6.02E23mol^{-1}}$) represent equivalent numbers of particles per mole. But, it is improper to use $N_{A}$ with the 9000-form.
For example the first term should be $\frac{9ln(10)}{128\pi^{5}(6.02*10^{23})}$ which is $\pu{8.79E-22}$.
I think you may be inadvertently using $\frac{9000ln(10)}{128\pi^{5}(6.02*10^{23})}$ which is why you are getting $\pu{8.79E-25}$.
Otherwise I think you would need to correct your units by using:
$$\pu{8.79E-25}\kappa^{2}\Phi_{D}n^{-4}J(\lambda)[mol^{-3}M^{-1}cm^{-1}nm^{4}]$$

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

Thanks for the insight! I used the 9000 form due to the fact that that was what was given in my primary source (Principles of Fluorescence Spectroscopy, 3rd Ed. by J.R. Jakowicz) - In section 13.2 the formula is shown (equation 13.4), and in the text it's indicated that $N$ is just "Avogadro's Number"... Why would he have used the $\text{mmol}^{-1}$ version over the normal one?

To be honest I'm not sure, I'll have to go back and re-read Lakowicz.


text:
What is the geometry of a tri-substituted methyl free radical?
My teacher explained that free radicals could have a geometry that is planar or pyramidal(tending to tetrahedral) based on whether or not it is substituted by lone-pair containing atoms.
For example,
$\ce{(CH3)_3C}$ would exhibit sp2 hybrid geometry of planar.
But when tri-substituted with Flourine/Chlorine, it will exhibit a geometry that is tending towards tetrahedral(pyramidal) because the free radical electron in the P-orbital repels the lone-pairs on the Halogen.
It does not seem convincing because, the P-orbital is above and below the plane, then why does the electron in P-orbital repel the groups into a geometry tending towards tetrahedral?
That's like as if the electron was present only on the top and was repelling the groups down.
Here is a picture that might help visualise the p-orbital above and below the plane: https://www.google.com/imgres?imgurl=https://chemistryonline.guru/wp-content/uploads/2016/09/carbonium-1280x720.jpg&imgrefurl=https://chemistryonline.guru/carbonium-ion/&tbnid=1bNjv-81SP8QxM&vet=1&docid=MPke176Wg2AS2M&w=1280&h=720&source=sh/x/im

https://images.app.goo.gl/q8mvwqGhvbyYRyVX6

Yes, I agree that the attached groups due to the sp2 hybridisation, are aligned into a planar geometry, but the vacant p-orbital must remain perpendicular to the plane of the sp2 orbitals. So, that means the p-orbital being dumbbell shaped, should be above and below the plane of the sp2 bonds.

Yes, that is just a representation I could find off the internet. The point is, when it is a FREE RADICAL (having a single electron in P-orbital) it is going to repel any bonded lone-pair containing atoms like for example halogens.

I even took it as we were discussing of carbocations. I delete all the above :) Do the same otherwise is a mess. You'll get the answer....

.... from here https://chemistry.stackexchange.com/questions/49754/what-is-the-hybridisation-of-trifluoromethyl-free-radical

And as suggested in a comment therein, a look at https://en.m.wikipedia.org/wiki/Bent%27s_rule

Please don't link through Google tracking algorithms to images. If possible and allowed, please include them directly in your post. Otherwise try a permanent URL.

I return here specifically for you. While the explanation given by your teacher might be not accurate (as for the material posted below) however it is not inaccurate because the reasoning leading to your doubt. It doesn't matter that the e is repelling equally from above and belove. It matters that with a more pyramidal or sp3 like geometry all repulsions are minimised.

As already pointed out by Martin to not use Google links, here is the relevant meta post to give you the context: https://chemistry.meta.stackexchange.com/questions/4830/can-outdated-versions-of-answers-be-removed/4832#4832

A priori, I would expect substituted methyl radicals to adopt a planar structure due to Bent’s rule.
However, I consider it possible for methyl radicals with electron-withdrawing substituents to adopt a trigonal-pyramidal structure too. In this case, Bent’s rule would imply that the electron-withdrawing substituents are ‘more electron-withdrawing’ than the radical is electron-poor. In such a case, it might be beneficial to distort the planar structure, causing rehybridisation and thus resulting in $\mathrm{sp}^n$ orbitals across the board so that all orbitals together add up to $\mathrm{sp^3}$. This structure may not be perfectly pseudotetrahedral but close enough to have a significant s-contribution to the singly-occupied orbital.
Note that $\mathrm{sp^3}$-type radicals are not inherently unstable (unlike $\mathrm{sp^3}$-type cations) as can be seen in bicyclo[2.2.2]octa-1-yl radicals (or bridgehead radicals in general). This points to $\mathrm{sp^3}$ radicals being accessible and thus the finer electronic environments mattering.

Addendum: Ron has shared links to two other questions he answered (1, 2) which argue the case that methyl radicals would be (slightly) pyramidal rather than planar in their ground state. So my a priori expectation might be wrong. I believe it might be very hard to prove the structure conclusively, as it is expected for a pyramidal radical to rapidly inverse as nitrogen does in ammonia and its derivatives so that any non-short timescale measurements would result in an averaged structure (which would be planar).

Hi Jan...I find it difficult to believe that nature would build a molecule where only the 3 two-electron $\ce{C-H}$ bonds are stabilized by mixing in s-character, while the orbital containing a single electron is not stabilized by mixing in at least a sliver of s-character.  I've suggested previously (to some objection) that the methyl radical is likely slightly pyramidal due to such s-character mixing.  See What is the geometry of an alkyl radical? and

Geometries of methyl and silyl radicals.

@ron That is a very good point and to be perfectly honest I hadn’t yet fully thought it through all the way.


text:
Finding excited stage of electron from its potential energy

The potential energy of an electron in the hydrogen atom is $\pu{-6.8 eV}.$ Indicate the excited stage in which electron is present.

Total energy would be equal to $\pu{-3.4 eV}.$
I used the formula
$$-13.6\left(\frac{1}{1^2} - \frac{1}{n^2}\right) = \pu{-3.4 eV};$$
$$n^2 - 1/ n^2 = 4.$$
I got answer as $3n^2 = 4,$ which is obviously not correct. Where am I going wrong? I need help either with the formula or calculation I did wrong.

Replace n by infinity and right side by 0 eV for a free electron and you will see the error. Or, use n=1 and right side -13.6 eV.  You would probably avoid the error,if you used algebraic symbolic equation first, as error in thinking would be easier to spot. In fact, if you spent the time for writing the question by additional thinking, you would have the answer much sooner.

Why is n = infinity. If n is equal to infinity , then we don’t have any variable . Do you mean to say that we don’t know whether electron is in first orbit or 2nd orbit ? I didn’t understand why n = infinity@Poutnik

If n is from 1 to 2 , it is 1st excited change.Why does n has to be = infinity ?

n=infinity to make your error more obvious. :-)

Have I done calculation mistake

Big one. I intentionally do not say which one, as you would learn more if you discover it yourself

Let us continue this discussion in chat.

The potential energy ($U$) of an electron in nth orbit of hydrogen atom is given below :     $$ U = \frac{-27.2}{n^2} \space (in \space eV)$$ $$ So, \space -6.8 = \frac{-27.2}{n^2} $$ $$ \implies n=2 $$ As the electron is in second orbit it is said to be in its first  excited state.

Note : The formula that you used is the energy of emitted photon during transition but not the potential energy.


text:
Difference in collision cross-section obtained by measurement method
I have a simple question of Physical Chemistry, but I don't know the answer.
We can find collision cross-section(σ) through experiments in three ways.
first, diffusion flux experiment. diffusion coefficient D = (1/3)x(vλ) = (1/3)x(8RT/πM)x(RT/PNσ). So, σ can be found by measuring the diffusion flux.
Second, Thermal conduction experiment. κ = (1/3)x(C/N)x(8RT/πM)x(RT/PNσ). So, σ can be found by measuring the thermal conductivity.
Third, viscosity experiment. μ = (1/3)x(m)x(8RT/πM)x(RT/PNσ). So, σ can be found by measuring the viscosity.
This is the question. I do not know why the sigma obtained through each experiment is different. The sigma obtained by measuring the diffusion flux is different from the sigma obtained by measuring the viscosity. The same is true of thermal conductivity. Why is this happening?

For eventual writing and formatting of chemical/mathematical formulas or equations, see  MathJax usage guide   It is not mandatory, but highly recommended and it gives huge advantage for writers and readers.

Before starting to think about this question, why do you write RT/PN and not V = Volume of one molecule ? And can you simply give us your calculated value for sigma with these three approaches ? Are they different by a factory greater than 1000 ?

In which sense they differ? Because each methods can have errors and especially under or overestimate a quantity. This can be dramatic if any approximation is required to theoretically attain the formulae you've list (which I think is the case, by feeling).


text:
How is the UV-visible spectra of carbonatotetraamminecobalt(III) nitrate interpreted?
In a laboratory practice I have done a characterization of the $\ce{[Co(NH3)4CO3]NO3}$ and obtained the following spectra:

I don't know if the band between 300 nm and 400 nm can be due to the effect of the counteranion or if it is a d–d transition.

This UV spectrum is useless. Please see the y-axis. Where is there absorbance? An absorbance of 2 means almost 100% of the light has been absorbed. Your sample is too concentrated. Do the experiment again and dilute it till your absorbance is around 1. Collect two spectra, in the UV range (220 to 400 nm) and one in the Vis range (400 to 600 nm) with appropriate heights.

@M.Farooq.  The spectrum is surely not useless for all wavelengths greater than 300nm? (I have used spectrophotometers that obey the Beer-Lambert law up to absorbances of 2 (at 340nm) and those that markedly deviate from linearity at an absorbance of 1 (in the low uv region, for example))

@M.Farooq Ok, I didn't know that. But anyways, is it possible to know the type of band placed between 300 and 400 nm?

You will have to do little bit more controls. Are you using a quartz cuvet in the UV range? If not do not collect the UV data. If you suspect it is the counter ions, you will have to think about the color of the counter ions. Is carbonate ion coloress? Is nitrate ion colorless? And finally how pure is your preparation? What is the color of your complex. Do the Vis bands qualitatively match with the color of the complex? For example KMnO4 is purple, it absorbs in the green region.

@tomd, Yes, after 300 nm it is fine, but students should be aware of what they are collecting and making sure the spectrum should like a spectrum.

You could check the 300-400 nm band with soln of , say sodium nitrate, to see if its the counter ion. Check that you are using quartz cuvettes, otherwise uv absorption may be due to cuvette (or solvent), as a 'rule of thumb' ignore any absorption above approx 2.5 at any wavelength and below 220 nm anyway, most spectrophotometers do not work well there.


text:
Why are we finding the integrated rate law by starting with r=k[A]?
When deriving the integrated rate equations, we start off with r =k[A] for example, and then we do the whole integration thing to find an equation of [A] in terms of time. But why are we ingnoring any other reactants here? Wouldnt the r=k[A][B] if B was was another reactant?

You'd just get a different integrated rate law. Each version of the integrated rate law corresponds to one version of the (non-integrated / differentiated) rate law. The ones with different reactants (like your suggestion of $k[\ce{A}][\ce{B}]$) can get quite tricky to solve, though, which is why those are rarely touched on.

Note that  r =k[A] and r=k[A][B] assume elemental reactions.


text:
Why does a spur of ice sometimes appear when freezing water in my freezer
From time to time, a long spur of ice grows on one “cube” of my ice cube tray.
It’s quite slender, about 3mm or 1/8 inch at the base tapering to about 1mm at the tip, but it’s about 30mm or over 1 inch long!

The angle of the spur seems to always be the same - about 45 degrees.
What going on here? It seems to defy gravity!
Probably relevant: I seems to happen when the ice tray is placed near the chilled air outlet of the freezer.
Plain tap water.

Check here: https://physics.stackexchange.com/q/70974. They form on lakes as well.

Wiki article as well: https://en.wikipedia.org/wiki/Ice_spike. Loved the bit about the guy hiking across frozen Lake Erie!

It does not defy gravity for sure. It is not the ice that protrudes out the tray, but more water freezing on it from around/above. It is a region of higher potential, like in snow flakes or in lightning rod. For that region to begin, then refer to thread linked in the comment above.

Is the angle of the spur such that the chilled air blows it in that direction, or is the spur growing into the chilled air source?

@james the spur is angled towards the cold air flow, more or less pointing straight at it in 3D space.


text:
Why does a proton react with water to form hydronium ion and why does that ion has an overall positive charge?
Take this is an example:
$$\ce{HCl(aq) -> H^+(aq) + Cl^-(aq)}$$
and
$$\ce{HCl(aq) + H2O -> H3O+ + Cl^-(aq)}$$
Clearly the proton latches on to the water molecule but why? Also, why does the hydronium ion has an positive charge? Did the proton settle in the nucleus of the some atom?
What exactly is happening here?

About the charge: 0 + (+1) = +1, i.e. water is uncharged, the hydrogen ion is charged, so the hydronium ion as the product of the two will be charged as well. It does not matter where exactly the protons and electrons are, the net charge will always be +1.

You may found many interesting info and some references in Wikipedia - hydronium page. See also that a hydronium ion is the most simple particular case of more general oxonium ions

The evidence of a positively charged hydronium ion or a simple hydrogen ion comes indirectly from electrolysis experiments. This was demonstrated by Arrhenius more than hundred years ago. When he proposed these ideas during his PhD work, his mentors did not like these ideas which are bothering you now.
If we take ultrapure pure water and dip two electrodes connected to a negative and positive terminals of a battery, no current passes through it. It means there are not many ions in water.
The moment you add HCl or any strong acid, water becomes a very strong conductor of electricity. As you might know in solution, electricity is carried by charged particles. This proves that there must be ions in an acidic solution. How do you we know hydrogen ion carries a positive charge? The simple demonstration is that hydrogen ion is attracted to the negative electrode- because hydrogen gas always appeared at the negative electrode.

Clearly the proton latches on to the water molecule but why? Also, why does the hydronium ion has an +ve charge? Did the proton settle in the nucleus of the some atom?

Well, this picture is still not very understood as far as I remember. The symbolism $\ce{H+}$ or $\ce{H3O+}$ is a very simple picture. Probably it is a solvated proton $\ce{H3O+.(H2O)n}$ but people have demonstrated spectroscopically that it exists. I do not know the fundamental and mechanistic reason as to why the HCl molecule when dissolved in water, it would protonate water. Yes, people can put forward thermodynamic reasons but thermo never talks about the mechanistic why's and how's.

But what seems to be missing in your opinion? Besides that we can write down H3O+ (even H+, if we keep it mind that is H+aq, where aq is specially strong) in places of the various hydrated species (3 water, 9 or whatever) what is missing from a mechanicist viewpoint, at list at textbook level? Does not Arrhenius acid base theory suffices? Thermodynamics only guarantees a net energy gain. (I am referring especially to your passage "I do not know the fundamental reason....".

I wanted to say Lewis instead of Arrhenius, being the first one a mere requisite to the formation of hydronium(s) ions.

Let me explain it in a less complicated and direct way.
H+ 'reacts' with H2O to form the hydronium because the oxygen in the water molecule has a lone pair of electrons that attracts the H+ ion.
Why does it have an overall positive charge? It is something you will be comfortable seeing in chemistry.
Right now you seem to be just starting to question why things are the way they are in Chemistry. You will gradually get used to the way things are supposed to be.
The reason for why hydronium has a positive charge is similar to why ammonium has a positive charge.
When atoms with lone pair, like oxygen in hydronium or nitrogen in ammonium form a coordinate bond, the atom having donated its lone pair gets a positive charge on it instead of the H+

It can be also said a proton is an extremely strong Lewis acid and reacts  with H2O as a Lewis base, attaching itself to the free electron pair of H2O.

And thermodynamics turns useful, otherwise one can argue why HCl dissociate in water if charges are inherently less stable than HCl itself. But perhaps it is a trivial remark.

"hydronium because the oxygen in the water molecule has a lone pair of electrons that attracts the H+ ion" This does not make sense at all.


text:
Electrolysis with two similar metals
I've researching electrolysis recently, and all of the examples I've seen use two different metals as the cathode and anode.
In my attempts to make a hydrogen producing cell (table salt used as the electrolyte in water) I've used two steel nails as both my cathode and anode. When I power it up, one side bubbles and the other stays normal. The side where the bubbling occurred, corrodes and makes the water murky with what I presume to be rust. The other side does nothing, not even get electroplated with the other side's metal.
My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

Can there be a word in which the same letter occurs twice? Yes. Why? Because (1) we can't possibly think of a valid reason that would prevent it and (2) we have a working example. The same applies to your case.

You cannot electroplate anything from a solution that hardly contains the respective ion, and I'm pretty sure you cannot electroplate with iron at all. You need a positive standard potential, otherwise you always get hydrogen.

My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

The battery or the power source does not care about the electrode material. Its job is to pump the electrons in the circuit. Now the problems originate from the chemistry rather. By definition, if a direct current passes through an electrolyte, electrolysis (=chemical decomposition) has to occur. Something has to reduce at the cathode and something has to oxidize at the anode. Here, we enter the competition, as shown below in the case of your simple aqueous NaCl set up.
At the anode (the electrode connected to the positive terminal of the DC supply), you have to think...
(a) Will the electrode oxidize itself? Possible
(b) Will the solvent i.e., water oxidize to oxygen? Possible
(c) Will the chloride ion oxidize to chlorine? Possible
(d) Na+ ions, they should be repelled
(e) Can the electrode react with the electrolytic product? Very much possible
Similarly at the cathode (the electrode connected to the negative terminal of the DC supply), think about ...
(a) Can the electrode reduce itself? If it is a metal, it cannot reduce further.
(b) Will the solvent i.e., water reduce to hydrogen? Possible
(c) Can the Na+ ions be reduced to metallic sodium? Possible, but not in plain water.
(d) Cl- should be pushed away.
(e) Can the electrode react with the electrolytic products? Very much possible
So you can see here, that depending on the electrode material, yes it can be oxidized if used an anode. It can also be oxidized indirectly, if the evolving chlorine begins to attack the metal, which happened in your case.
In any case, electrolytic cell does not care about the electrodes, the electrodes can certainly be the same but they have to be inert. People always choose inert electrodes so that the metallic cathode or anode do not interfere chemically during the electrolysis. Platinum is a good choice but it is too expensive for home experiments. Carbon/graphite electrode is another choice. As you have seen, stainless steel is not a good choice as an anode. It is readily oxidized by chlorine.

If $\ce{NaCl}$ is electrolyzed between two iron electrodes, the iron is not modified at the cathode. The cathodic reaction is : $$\ce{2 H2O + 2 e-  ->  H2 + 2OH-}$$ So gaseous $\ce{H2}$ is produced in bubbles, and the solution becomes basic, because of the ions $\ce{OH-}$. The nature of the metal at the cathode does not matter. The reaction at the anode will be :  $$\ce{Fe  ->  Fe^{2+} + 2 e-}$$ But after a while the ions produced at both electrodes interact and produce a dark precipitate $\ce{Fe(OH)2}$ , $$\ce{Fe^{2+} + 2 OH^- -> Fe(OH)2}$$ which is soon oxidized by air into a brown precipitate $\ce{Fe(OH)3}$ that looks like rust. $$\ce{4Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$


text:
Order of −I effect among ethenyl and 1-propenyl groups
We know that both ethenyl $\ce{-CH=CH}_2$ and 1-propenyl $\ce{-CH=CH-CH3}$ are electron-withdrawing groups when attached to hydrocarbon chain. Which one has greater electron withdrawing power?
The $\ce{-CH3}$ is electron-donating, hence allyl should have lesser electron withdrawing effect. Is this correct?

I would say yes, the -CH3 group adds a little +I effect, making the -CH=CH-CH3 a little less electron withdrawing compared to the vinyl.


text:
Is it safe to mix methylated spirits with isopropyl alcohol?
I have about 200ml of methylated spirits and 500ml of 100% isopropyl alcohol that I want to mix together in a plastic drink bottle.  I will use it for fuel in an alcohol burning stove while camping.  I can't seem to find any other information about mixing them so...

Is it safe to do so?
Is a clear plastic drink bottle OK to keep them in? (worried about melting the bottle - adverse reaction)
If it is safe to mix them, will I end up with a more flammable mix?

Thanks for any help!

What is methylated spirit ? You may mean methylalkohol aka methanol.  I would avoid a plastic drink bottle to prevent some health endangering accidental drinking, if other persons get hand on it. Or mark it very obviously as not drinkable. If people can drink acids, they can drink this as well.

@Poutnik Methylated spirit is a common name (at least in some anglo saxon countries) for cheap solutions of ethanol. To avoid making this a cheap source of drinkable ethanol (which needs to be taxed and food-safe), it is usually denatured with methanol (making it toxic), coloured with a strong dye (to make it visually obvious it isn't drinkable) and has a strong bitter flavour (often bitrex) added to make sure you know immediately it should not be drunk.

No reason to mix it at first. This is pedantic, because I would mix them :) after being sure that a methylated spirit is indeed denatured ethyl alcohol. But normally flammables should be kept in smaller portion, if their use allows that. And this is your case.

@matt_black Well, it should come to my mind, but for some reasons it did not. I know methanol is used for ethanol denaturation for some fine technical applications, but I am rather used for quite nasty denaturating staff, used for combustion purposes, propably including the bitrex you have mentioned.

As both ethanol and isopropanol have similar boiling point, it would be compatible. Isopropanol would have somewhat better heating power, having bigger molecule and lower relative oxygen content.

A drinking bottle is not a good option, someone could mistake it for something to drink. Use an appropriate container.

I'm not at all sure why this has been closed, especially as "opinion based". It is a fairly simple question about chemical safety and not an issue about personal preferences or opinion.

@Martin-マーチン While you are broadly correct, especially if the bottle has an existing label for a soft drink, it is somewhat moot as methylated spirits has routinely been sold in many countries in the same sort of bottles though with appropriate labelling. Also the ingredients are designed to prevent accidental ingestion.

What are the two alcohols in at the moment? As for mixing, not an issue.

It is technically safe to burn the mixture. Be prepared fuel enriched by isopropylalkohol will need more air to burn and the flame will be more yellow and "sooty", possibly staining the camping gear bottom, because of the longer carbon atom chain. All depends, if the stove is burning pure vapors, or vapours premixed with air. The mentioned issue is not much of a problem for  the latter, if there is possible to manipulate fuel:air ratio.

Yes, as far chemical safety is the concern. No dangerous reactions nor dangerous phenomena will occur. The mixture, a solution to be more precise, will retain at least qualitatively the properties of the two components, eg will be a disinfectant, a flammable liquid, a solvent, and so on.
Flammable materials are better stored in separated containers in separated places.
For the case in question, going camping with two small bottles or one double sized bottle in the backpack won't make any real difference.


text:
Is it rigorous enough to change a DFT functional after geometry optimization to calculate excited state properties?
I have done the geometry optimization with the B3LYP functional, however, there is a need to calculate the excited state energy, so the CAM-B3LYP functional should be better suited for that. Do I need to reoptimize the geometry to use CAM-B3LYP or are these functionals simillar enough so no reoptimization is needed?

Assuming you have enough time, you could probably try it out anyway and see whether you get much of a difference. Otherwise, when it’s time for you to present your work, somebody will ask about it anyway...

@orthocresol, I have seen in a couple of articles, which model really similar molecules, that first they optimize with B3LYP and then calculate excited states with CAM-B3LYP or even other functionals and compare the results, so I was wondering, if it is really necessary.

What is sometimes done is use the geometry and thermochemical corrections (i.e. the results of the frequency calculation) of method X, but use the electronic energy (SCF energy) of another method Y. This is generally noted Y//X. Since the geometries and thermochemical corrections typically are not much influenced by the quality of the method (beyond a certain point), this scheme gives good result. The electronic energy can however benefit from a larger basis set or a better computational method (e.g. MP2 with DFT geometries).
I am assuming this is what you have seen in the literature, as it is relatively common, especially for large molecules like organometallic complexes. I am not too familiar with excited states, but the same thing likely applies.
However, you must always use the exact same method for the geometry optimisation and the frequency calculation. The frequency calculation uses the curvature of the potential energy surface (PES) to calculate approximate corrections. If you are not at an extremum (minimum for equilibrium structures or saddle point for transition state), the results are meaningless. Changing the method will move you off the extremum sufficiently and your frequency calculation will likely have several imaginary frequencies.

Thank you very much for the comprehensive answer. Yes, I am actually doing calculations for a large organometallic complex, however, I am planning to still to use the DFT method for calculations and just to change functional of it for energy calculations (and the same was done in the articles). Does the following still apply?

Yes, you can use the SCF energy of a different DFT functional as well

Excellent points! I'd like to add that in most cases, you should probably do a calibration of different functionals, see here. You might also find this question on Matter Modeling SE helpful. @aerospace

@Martin-マーチン, thank you for the recommendations!


text:
What is the difference in mosM and mOsm units, if there is any at all?
I'm reading Agre's paper on aquaporins [1] and he uses a unit called "mosM" throughout the paper, for example:

… ocytes were transferred from 200 mosM to 70 mosM modified Barth's buffer.

I just want to understand what this unit means, is it the same as mOsm? I know this might be an very stupid question but hey, I'm only a undergrad.
Reference

Preston, G. M.; Carroll, T. P.; Guggino, W. B.; Agre, P. Appearance of Water Channels in Xenopus Oocytes Expressing Red Cell CHIP28 Protein. Science 1992, 256 (5055), 385–387. DOI: 10.1126/science.256.5055.385.

Remember that "paper never refused ink". Having a wrong notation in Science magazine does not endorse its correctness.

@M.Farooq It's not just Science, Nature uses this symbol too: https://doi.org/10.1038/46045.

Judging from the following table consisting of compiled notations presented in offline edition of The ACS Style Guide and AMA manual of style, section 13.12 Units of Measure:
$$
\begin{array}{ll}
\hline
\text{Unit of measure} & \text{Symbol} & \text{Ref.} \\
\hline
\text{osmolar} & \text{osm (also osM, Osm)} & \text{[1, p. 191]} \\
\text{osmole} & \text{osm} & \text{[2, p. 635]} \\
\text{microosmole} & \text{µOsm} & \text{[2, p. 634]} \\
\text{milliosmole} & \text{mOsm} & \text{[2, p. 634]} \\
\hline
\end{array}
$$
symbols $\pu{Osm}$ and $\pu{osm}$ refer to the amount-of-substance of an osmotically active entity, and symbol $\pu{osM}$ refers to osmolar concentration suggesting the following relation:
$$\pu{1 osm L^-1} = \pu{1 Osm L^-1} = \pu{1 osM}$$
The letter “o” in $\pu{Osm}$ appears to be capitalized for the sake of better readability when used in conjunction with metric prefixes, and the letter “m” in in $\pu{osM}$ is likely capitalized to resemble analogy with the unit symbol for molarity (small capital letter “m”: ᴍ).
References

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.
AMA Manual of Style: A Guide for Authors and Editors, 11th ed.; American Medical Association, Ed.; Oxford University Press: New York, 2020. ISBN  978-0-19-024656-3.

Does the ACS Guide say anything about the osmolality? IUPAC still has this term with a different notation.

Thanks for the answer, please correct me if I'm wrong, it is the same as in 1mol/L = 1M, hence 1osmol/L=1osM?

@M.Farooq No, the third edition of the ACS Style Guide doesn't mention osmolality. Good point regarding IUPAC though.

@RafaelFranco Yes, this would be my interpretation. Minor correction: don't forget to put a space between the numerical value and the unit symbol.

I think the clinical chemistry notation has not been standardized but certainly it has been inspired by chemists.
The capital M in mosM emphasizes that it is milliosMolar. Chemists use capital M to denote molarity.
On the other hand, mosm indicates milliosmolal. Chemists use small m to symbolize molality.
This notation is not universal.
From this slightly dated article on Osmolality, which is behind pay-wall
Osmolality, Harry F. Weisberg Laboratory Medicine, Volume 12, Issue 2, 1 February 1981, Pages 81–85.

Terminology and Units
In 1934, Gamble, et al.,11 introduced the term ”osmillimole” which has been successively transformed into mosM (milliosmolar), mosm (milliosmolal), the noncommittal mOsm, and the present-day mosmol (milliosmole). Raoult’s law states that the lowering of the vapor pressure (one of the four ”colligative”
properties of dilute solutions) is proportional to the mole-fraction (more easily calculated from molality) of solute present in the mixture.


text:
Conversions in amines
We have to find A, B, C, D, E, F, G, H.
I got C is acetophenone and D is benzoic acid and further I got what are  E,F,G,H
I am just a bit confused in A and B please any hints..
Edits : C is acetophenone and the reagent in first reaction where A and B are reactants is AlCl3 not H2O

There's not enough information to answer this unambiguously. A could be benzyl alcohol and B an oxidising agent but there are other possibilities.

Or -- as an alternative to the oxidation suggested by @Waylander -- a Grignard reaction of $\ce{PhMgBr}$ and, e.g., ethyl formate.

Sorry ,C will be acetophenone I think now ,if I am wrong do correct me

@Rover If the entry about «2,4-DNP» is about 2,4-dinitrophenylhydrazine, then this may react on either aldehyde, or cetone (reference). This then requires a cross check in a database (e.g., Reaxys) or/and table (e.g. CRC Handbook Of Tables For Organic Compound Identification) if your diagram is supported by experimental evidence.  We got colored hands preparing derivative hydrazones, picrates, etc.

There still is not enough information to identify A&B if C is acetophenone. The range of possible reactants just got wider.

So if the conditions of the first reaction are AlCl3 then we are looking at a Friedel-Crafts reaction with benzene and acetyl  hloride

As @Waylander mentioned in comments, more information is needed to guess A and B.
Since, in the question other products have not been mentioned, I am writing the answers for them with reasoning.
1) Since on reaction of D with soda lime gives benzene, which is a decarboxylating reaction, we guess that D should be benzoic acid or sodium benzoate as you have correctly guessed.
2) Now we see that C on reaction with $\ce{NaOH}+\ce{Br_2}$ gives benzoic acid, and as the mentioned reagents are reagents for iodoform reaction, we conclude that C must have been a methyl ketone and hence acetophenone which also gives 2,4-DNP test.
3) Now D i.e, benzoic acid on heating with $\ce{NH_3}$ gives benzamide which is a condensation reaction in which water condenses out. So, E is benzamide.
4) Benzamide on reaction with $\ce{NaOH}+\ce{Br_2}$ will give aniline, which is the Hoffmann Bromamide degradation reaction. So,F is aniline.
5) Aniline on reaction with $\ce{CHCl_3}+\ce{KOH}$ will give phenyl isocynaide which is the carbylamine reaction.
6) Ozonolysis of phenyl isocyanide gives phenyl isocyanate. So, H is phenyl isocyanate.

Were there any options in the question from where this has been asked? If there were, it would make it easy to guess A and B.


text:
When multiple sources give the content of an ingredient my an unqualified percentage, can I safely assume this is by weight? Or; it's not a wart
I am trying to make a simple scented shampoo and body soap for my own use and for Christmas gifts.
I statr with a base of 32 oz – (a bit under a kg) of a mild commercial castille soap (Dr. Bronner’s Peppermint) to which I have added modest amounts of essential oils ranging from 2 tablespoons (lavender) to a few ml (jasmine  and pine needle) down to a few drops (vetiver, bergamot). Because I have dandruf on my head and psoriasis on my elbows, and because I know both of the other people I am gifting use cosmetics containing exfoliants, I decided to add one percent of salicylic acid to the shampoo. Over-the-counter cosmetics use up to 2 percent, and prescription shampoos are available with up to six percent. This seemed adequately cautious for a first-time emdevor, and assuming thee percentage was by weight, I added a bit under10 gm of powdered salicylic acid to the almost a kg of shampoo.
Here is my question: at least half of the salicylic acid has not gone into solution after considerable agitation. This make me wonder if I am not adding too much salicylic acid because I am measuring the percentage incorrectly. Should I be measuring the shampoo by dry weight? Or reconstituting the salicylic acid somehow before weighing it? Or defining the percentage in molar terms? Or should I be doing something else to encourage solubility (boil it?) or to make it into a suspension or something?
I was hoping to try it out, but I thought I should try to establish that I am not using way too much acid.  A 24 percent solution is sold as wart remover.  I clearly don’t want to apply that on my head.

Yes, weight percents should be assumed. Salicylic acid solubility depends on what you are dissolve it in.

Solubility of salicylic acid is strongly dependent on the pH. In pure water, the solubility is only  $1.8$ g/L at $20$°C. If totally neutralized by sodium, the solubility is huge : $1110$ g of sodium salicylate can be dissolved in $1$ L water. In hard water, the solubility is increased a bit, as calcium salicylate solubility is $27$ g/L. In your case I would recommend to add an equal amount of soda ($\ce{Na2CO3}$). This will improve the solubility of your salicylic acid.

@Maurice OTOH, alkaline pH is not desirable for shampoo, so it should be neutralized by the shampoo components.

@Poutnik. pH 8 should be sufficient to dissolve the given amount of salicylic acid.

Thanks @Maurice and @Poutnik! More of it dissolved overnight. I'll try just a couple of grams of soda and see if that does the job.

@Maurice Worked like a charm. If you'd make your comment as an answer, I'd be happy to accept it as such.

If you appreciate a comment you can express it by clicking on the small pale triangle which is in the margin before the comment. This will produce a number like 1 just before the triangle.


text:
What is the implication of having the internal energy equal the Gibbs free energy?
I came up with a hypothetical equation of state, where one of the consequences is that $U=G$, where $U$ is the internal energy and $G$ is the Gibbs free energy. I wanted to ask the chemists, is there a fundamental problem with this? Are there any real systems where this could happen? Thank you for your time

I am curious about your equation.

$G=U+p \cdot V-T \cdot S$ by its definition and is also often called free enthalpy, as $G = H - T \cdot S$ and $H = U + p \cdot V$, where $H$ is enthalpy. So if $G=U$, then $S=p \cdot V/T$.
$pV/T$ is about constant at usual conditions for real gases, or even exactly constant for ideal gases ( $pV/T=nR$ ). It decreases with $T$ for condensed phases at constant pressure, or can dramatically grow with $T$ for condensed phases at constant volume.
Entropy $S$ increases with $T$ via $\mathrm{d}S=đQ/T=C(T) \cdot \mathrm{d}T/T$.
So $G=U$ could be true just accidentally for particular conditions, where both functions $S=f(T,p)$ and $pV/T = g(T,p)$ cross each other.
$G=U$ has no special meaning, but that the non free part of energy $T \cdot S$, blocked for work usage by the second law of thermodynamics, is accidentally equal to the $pV$ term.
$G=H$, or $A=U$  would be a different case, as it would mean a system must have zero entropy, what is not possible for $T \gt 0$.
( $A=U-TS$ is the Helmholtz (free) energy  )

Thank you! Can you reference any books that specifically state this somewhere?

Any textbooks with foundations of thermodynamics. See e.g. CH  SE : resources for learning chemistry. It may not be stated explicitly like the above, but is is the direct consequence of what is there. BTW, you did quite a poor job in a priori research before asking the question.

sorry, man. I'm just a lowly CS major.

NP.  Even humble CS majors are able to search for definitions of the basic TD quantities, if they operate with them.

I did know about the definitions you listed, however, the rest of the stuff you wrote wasn't immediately obvious to me. Sorry if I fucked up

No, it is OK, every wise person  is learning the whole life.

by the way, I'm guessing the quantity TS doesn't really have a name, or does it? I know that -PV is the Landau free energy for homogenous systems.

Let us continue this discussion in chat.


text:
When Iron reacts with any other species, how will we know that the reaction will be of Fe(II) or Fe(III) i.e. of ferrous or ferric?
Like when iron reacts with water there forms iron oxide and hydrogen but how'll we come to know which of three equation will be right equation for reaction?
$$\ce{Fe + H2O → FeO + H2}$$
$$\ce{2Fe + 3H2O → Fe2O3 + 3H2}$$
$$\ce{3Fe + 4H2O → Fe3O4 + 4H2}$$

There are no general rules ! Of course the oxidation of iron passes always  by the level Fe(II). But sometimes the oxidation goes immediately to the the level Fe(III). It depends on the nature of the oxidant. If $\ce{Fe}$ reacts with $\ce{HCl}$ it produced $\ce{FeCl2}$, as the $\ce{H2}$ produced would be able de reduce $\ce{FeCl3}$ to $\ce{FeCl2}$. If $\ce{Fe}$ reacts with $\ce{Cl2}$, it produces $\ce{FeCl3}$, as $\ce{Cl2}$ is able to oxidize $\ce{FeCl2}$ to $\ce{FeCl3}$

Note that there is an almost general rule, that oxides of metals with multiple (and close) oxidation states, like Fe, Mn, Co, Ni form frequently non-stoichiometric oxides, where atomic ratios are not a ratio of small integers. E.g. iron(II) oxide with  theoretical composion FeO has the real composition Fe(0.84)O to Fe(0.95)O.

It depends, in many cases, on how much of the reactant is present to force the reaction to a higher oxidation state. See https://en.wikipedia.org/wiki/Law_of_mass_action

The stability of the various iron oxides depends on the oxygen fugacity (or oxygen partial pressure, which is a good approximation to fugacity in near-atmospheric pressures).
$\ce{Fe^3+}$ is the stable form of iron in atmospheric oxygen partial pressures, this in such conditions $\ce{Fe2O3}$ (hematite) will always be the stable product. The lower oxidation state oxides (such as magnetite $\ce{Fe3O4}$ and wüstite ~$\ce{FeO}$) can only stably form in deep earth environments, or in special controlled atmosphere reaction chambers.
That said, in normal conditions it is even more likely for iron hydroxides to form hydrated compounds, especially when water is a reactant, or is present. In these cases, you get a zoo of various Fe-oxides-hydroxides, some of which are amorphous (not crystalline), and others are metastable. These compounds are generally referred to as "rust".

As Fe2O3 is reduced by H2 to Fe3O4, I bet rather Fe3O4 will be formed. Note that iron reacts with water mostly at conditions of hot fine iron particles passed by water vapour.  See also https://en.wikipedia.org/wiki/Iron%28II%2CIII%29_oxide


text:
Is the lattice point in ZnO crystal structure occupied by the molecule ZnO or the ion O^2-/Zn^2+?
I learnt that ZnO has a hexagonal crystal system. I was expecting that cations or anions would occupy the lattice points in hcp while voids would be occupied by anions or cations respectively. But while googling the ZnO structure, I found this structure below which was confusing because of my notion about lattice point occupied only by cation/anion while voids occupied by its countercharged part.

But after close inspection of this above structure, I thought maybe the lattice point is occupied by the molecule ZnO itself while voids are left unoccupied as opposed to the cation/anion occuping the lattice point and anion/cation occuping the voids. Because only then,this structure is possible. Am I right in thinking so?

It could be reformulated to the question Are there any non-equivalencies in Zn-O distances for a single chosen Zn atom, leading to a hypothesis the cause can be a ZnO molecule ? . The Zn in the centre of the blue tetrahedron seems to have all 4 Zn-O distances like the same, but it is just a visual guess. The same for the central O tetrahedron.

You could think of it either way, the crystal doesn't particularly care. One issue of talking about the 'voids' is which voids, since there are often different voids with different symmetries in a crystal. My preference would be that the unit Zn+O occupies a lattice site.

You are falling in the rather common mistake of confusing the crystal structure with the lattice. The lattice is a mathematical object (it does not exist as a real object), whereas the crystal structure is the convolution of the lattice with a basis (an atom or a collection of atoms).
In some cases atoms are distributed exactly at the same positions of the lattice points; for example Cu has a face centered cubic structure where copper atoms occupy the same positions of the face centered cubic lattice. In this case Cu at 0,0,0 position is your basis.
In your case neither the Zn atoms, nor O atoms occupy lattice positions. Your basis is contituted of Zn at 1/3,2/3,0 plus O at 1/3,2/3,z; 'attaching' this basis to each lattice point (taking into account the correct distances of both Zn and O atoms from each lattice point), you obtain the final structure.
In any case, the picture that you posted does not represent the real hexagonal structure; rather, I would that is the trigonal polymorph.


text:
Converting bromobenzene to 1‐bromo‐3‐methylbenzene

I started by taking $\ce{FeCl3}$ to reduce electron-giving capacity of $\ce{Br}.$ Then we need a methyl at meta, so I thought about Friedel–Crafts acylation, but then also methyl will come at para.

This is one of those pointless paper exercises that you would never do in the real world. I would not start from bromobenzene as $\ce{Br}$ directs ortho/para, so this is going to take several steps.
You could make the phenyl Grignard or phenyllithium from bromobenzene quenching with DMF followed by aqueous acid workup to give benzaldehyde. Brominate this with $\ce{BrCl}/\ce{AlCl3}$ described here to give 3-bromobenzaldehyde then do a Wolff–Kishner reduction $(\ce{H2NNH2}/\ce{KOH})$ on the aldehyde to give 3-bromotoluene. I would not risk a Clemmensen $(\ce{Zn}/\ce{Hg}$ & conc. $\ce{HCl})$ here as I think there is a danger of reducing the $\ce{Ar-Br}$ to $\ce{Ar-H}.$