text:
Colours in Coordination Compounds
I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion. It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$. When $\ce{H2O}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{15}$, though in the spectrochemical series, ammonia is far ahead of water (Source-Triump chemistry ,target publications).
So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as $\ce{Ni}$ is in +2 oxidation state.
The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region (PYQ- AIIMS 2005) the options are  $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(NH3)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ and    $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ <  $\ce{[Ni(NH3)6]^2+}$ .
I used $E=hc/\lambda$. So, E is inversely proportion to wavelength. Therefore the ideal answer should be $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ <  $\ce{[Ni(NH3)6]^2+}$ . As $\ce{[Ni(H2O)6]^2+}$ has a better stability constant, more energy is required to break it. Hence a shorter wavelength than $\ce{[Ni(NH3)6]^2+}$ , whose stability constant is less. This less energy is required, hence a longer wavelength. But the correct answer is $\ce{[Ni(NO2)6]^2+}$ < $\ce{[Ni(NH3)6]^2+}$ < $\ce{[Ni(H2O)6]^2+}$ according to the AIIMS . Please explain.

Hello, can you please break up your text into sentences and paragraphs? You can also use MathJax to typeset the chemicals nicely: type  e.g. $\ce{[Ni(H2O)6]^2+}$ to get $\ce{[Ni(H2O)6]^2+}$.

please just help me with one thing which compound will have a better stability constant [Ni(H20)6]^2+ or  [Ni(NH3)6]^2+

The absorption wavelength has nothing to do with the stability constant. The absorption of a photon doesn't break any bonds. It only depends on the d-d* energy gap (or $\rm t_{2g}$ and $\rm e_g$, to use their proper names).

ill just ask you a question and then you can apply the logic for yourself , match the following for me, will you ? a) [M(Cl)6]^3-  b) [M(H20]6]^3+  c) [M(CN)6] , your options for absorbed colors are orange,green and red . Just match the three complex with their corresponding colors for me

I’m sorry, but it’s not really my job to answer questions on demand. Perhaps someone else who sees this may do it. Also, there’s not much point in me giving you the answer straight up: teach a man to fish, and so on. Finally, I suggest also posting it as a different question; don’t use the comment section to ask a different question from your original one.

nah !! just wanted to you to realize the logic behind my original question and what method did i approached to get the answer of it ... nevertheless thanks for noticing ! shakabrah

@SiddharthBisht, See this if this helps https://chem.libretexts.org/Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Readings/Purgatory/Chapter_3%3A_Introduction_to_Advanced_Bonding_Theories/3.2_Ligand_Field_Theory/B._Ligand_Field_Theory/B.2._Spectrochemical_Series

Don't mix stability constants with spectrochemical series. Stability constants is a thermodynamic concept. Spectrochemical series is an optical issue. It will be better if you write the exact question you have and then provide your own explanation.

@M. Farooq here in this case the stability constants show us how strong the ligands are which is ultimately going to decide the CFSE of the complex (CFSE is inversely proportional to wavelegth according to E = hc/lambda).For example the stability constants of the complex formed by the metal ion (M+2) with NH3, CN^-,H20 and en are of the order 10^11, 10^27, 10^15 and 10^8 respectively. This depicts that CN^-1 is the strongest ligand

I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion.

(I assume that SFL = strong-field ligand and WFL = weak field ligand.)
No, this is not correct. Ammonia is neither a true strong-field ligand nor a true weak-field ligand. Instead, it can form high-spin and low-spin complexes depending on the properties of the metal. Thus, it is best to term it a medium-field ligand.
It is also not true that all metal(III) ammine complexes be low-spin while all metal(II) ammine complexes be high-spin. However, among the many different factors that influence the spin state of a complex, it is true that a higher oxidation state tends to favour (note the wording) a low-spin configuration.

It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$.

It would be very weird if the stability constants of all $\ce{[M(NH3)6]^2+}$ complexes have the same value. Think about it: the metals are different while everything else is the same. There has to be variation. The same is true for the aquacomplexes.

So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as Ni is in the $\mathrm{+II}$ oxidation state.

You can’t say much about the stability of the two a priori. What you can say is that both of these complexes should be relatively inert (although not as inert as, say, $\ce{[Co(NH3)6]^3+}$) from a simple CFSE estimation.

The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region

This has absolutely nothing to do with the stability of the complexes or their stability constant; only with their electronic structure and the energy difference between the orbital an electron is excited from and the orbital an electron is excited into. Furthermore, there has to be an error in $\ce{[Ni(NO2)6]^2+}$ – I find it hard to see a stable complex involving six stable radicals around it.
As for the order the question asks for, this is where the spectrochemical series actually comes in and in handy. Determine which orbitals are involved, what effect a greater field split has on these, find the ligands on the spectrochemical series, determine their relative field splits and translate that all into an photon wavelength. You will soon notice that the wavelength of the hexaaquanickel(II) complex must be longer than the wavelength of hexaamminenickel(II) which gives you your answer by inferring.

I am not being rude, but honestly I think you should review the meaning of the spectrochemical series and how the transition metal complexes are colored.
A ligand being strong or weak according to the spectrochemical series does not correlate to the complex being stable or not, but its ability to split the d orbitals of the metal. In the spectrochemical series, the order is $\ce{NO2- > NH3 > H2O}$. The stronger the ligand, the larger the energy difference between the d and d* orbitals.
I give you an example: the formation constant of $\ce{[Ag(NH3)2]+}$ is $\mathrm{1.6 \times 10^7}$, whereas that of the $\ce{[AgI2]-}$ is $\mathrm{10^{11}}$, although the iodide ligand is much weaker than the ammonia! To tell a complex is stable or not, it requires a very high amount of things to consider, from the HSAB theory, steric hindrance, stereochemistry, the 18-electron rule, the stability of the oxidation state of the metal, the willingness of the ligand to donate the electrons and sometimes redox properties have to be considered as well.
When a photon of UV or Vis spectrum is absorbed, it does not break anything. Breaking a bond is the job of X-rays, gamma rays and cosmic rays, simply because the UV/Vis photons do not have enough energy to break bonds. Instead the UV/Vis photons transfer their energy to an electron in the d orbital of the metal complex, and "promote" it to the d* orbital. As I have mentioned, the two levels have an energy difference to overcome, and to promote the electron, the photon needs to have its energy to be at least equal to the energy difference between the two orbitals. As you mentioned, energy is inversely proportional to the wavelength, and the energy differences of $\ce{NO2-, NH3}$ and $\ce{H2O}$ are of the order $\ce{[Ni(NO2)6]^{4-} > [Ni(NH3)6]^{2+} > [Ni(H2O)6]^{2+}}$, the wavelength order is exactly opposite. The answer of AIIMS is correct

Simply because the UV/Vis photons do not have enough energy to break bonds – radical chlorination would like a word. But not in sunlight, where it is too busy forming radicals.

What do you mean by d and d* orbitals? That sounds ambiguous


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Are molecular properties robust towards manipulations?
Machine learning algorithms analyze molecules by looking at their molecular representations (e. g. SMILES and graph) and turning them into feature maps that help the algorithm to distinguish them. In the case of graph representations, algorithm tends to generate similar feature maps for the molecules that have similar topologies.
I was wondering if it is a correct assumption, generally speaking, that molecules with similar molecular graphs tend to have similar properties. That is, if we slightly modify one molecule, the output is expected to be close to the original molecule.

EDITS:
It seems this assumption is the base for Chemical Similarity. It goes on to say

[...] descriptive network properties and graph theory can be applied
to analyze large chemical space

At the end of the day, I think it is true that it all depends on what type of properties I'm dealing with. But my understanding is that this chemical similarity provides a base for more complex problems to build upon. For instance, if I want to modify a particular property, I would just add that into my initial analysis of the molecule.

Depends on the feature, I guess. A small modification will not change molecular mass by much, for example; on the other hand, it may turn an aldehyde to an acid.

It would be good to have an idea of what "slightly" could signify in this context? For instance are Benzene and Dewar Benzene only slightly different? If so one is a stable molecule known for centuries, the other wasn't synthesised until 1962 and has half life of around 2 days.

Adding a substituent, is it a small change ?

Another example is benzene vs toluene. Substituting an H with a CH3 might be a change for a graph or symmetry, but surely is not for a graph or symmetry as seen by pi electrons. Or one is a powerful carcinogen, the other seems to be safe. This can be and is done having the features in mind; topology and symmetry lead to rules, even. But the graph must be have something in common. Otherwise all biatomic molecules would have the almost the same properties, while in reality the only thing they share is x-x

When neither the algorithm nor the user have the slightest idea what they're doing ... ;-)

Even minor structural rearrangements can have big effects, eg n-butanol has a melting point of approx. negative 90 C, while its isomer t-butanol has an mp of positive 25 C. Both are aliphatic alcohols of formula $\ce{C4H10O}$.

@Karl When people have nothing to offer but still comment . . .

Neither classical chemometrics nor ML does require perfect connection between the similar molecules. Finding good descriptors is the job itself. If you find good ones, then you have found good ones...

Can you clarify the question? It is not clear what are asking or missing. The assumption that structure is somehow relevant to properties is about the same as saying chemical properties are related to chemical structure somehow, most of the time. I.e. that Chemistry kind of works. Since you are using these terms rather loosely, my statement seems more strict. You yourself acknowledge that the given assumption tends to work. What is the question?

@Greg While changing one atom or one bond in a molecule can drastically change the properties of a molecule, machine learning algorithms represent them with similar representations. Essentially, if we manipulate a molecule just a little bit, the ML representation would also change a little bit. So is this justified? Are molecular properties robust towards manipulations? It seems that the concept of chemical similarity is built upon robustness towards manipulations. But how can I justify drastic changes that we may see by just changing one atom or one bond or changing order of a bond.

@Blade It depends on the descriptor and metrics what is a small change. If you intentionally choose bad metrics, then of course they will not work. Chemometrics has collected literally hundreds of similarity measures that can work in a given context, the devil is in the details. I only can repeat: ML algorithms are universal approximators, ergo all you are asking here is that "is there any relationship between structure and property?", in short: "does chemistry make any sense?".

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Is 2-methyl propyl and 2-methyl propane the same?
I'm very sorry if this is a very simple question but it has been bugging me for a while Google searches for both 2-methyl propyl and 2-methyl propane give me the same structure .
My Intuition says propyl is the write one since it looses a hydrogen ,is there a very terrible flaw in my reasoning ?

2-Methyl propyl is the equivalent of 2-adjective-adjective looking for a noun to modify. 2-methylpropane is = to 2-adjective-noun, which is grammatically correct.

No they are not the same. 2-methyl propyl is a substituent group. 2-methyl propane is a molecule.
2-methylpropyl methanoate would be an example of a molecule with the methylpropyl substituent. Also known as isobutyl formate.

It would probably a good idea to show the difference; the question itself is not a stellar fit for our site.

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Does spectrochemical series trend depend on the metal center?
Can the trend of spectrochemical series vary/fluctuate according to the central metal atom (CMA) that is being used?
For example, for the CMA $\ce{Ni^2+}$ the ligand $\ce{H2O}$ will have more CFSE than $\ce{NH3}$ because $\ce{NH3}$ acts as a weak field ligand (WFL) for $\ce{M^2+},$ even worse then $\ce{H2O},$ though $\ce{NH3}$ is placed higher then $\ce{H2O}$ in the spectrochemical series.

The spectrochemical series of ligands, our instructor clarified, gives only a rough idea of the relative strengths of ligands, and may vary for different Central Metal Atoms (CMAs).
An important example that I have encountered is of $\ce{[Co(H2O)6]^+^3}$, wherein $\ce{H2O}$ acts as a Strong Field Ligand (SFL), despite its lower position in the series.
Also, JD Lee mentions in his Concise Inorganic Chemistry (5th ed., Wiley publications) that there is an increase in CFSE as we go down the group in d-block, and several instances may be found where the ligands designated as WFL begin to act as SFL for 4-d or 5-d members.
Overall, I believe there are certain deviations from the series, but it generally works well.


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IUPAC naming of a haloalkane
The IUPAC name of $\ce{(CH3)3C-CH2-CHI-CH3}$ should be 4-iodo-2,2-dimethylpentane as we consider iodine as a substituent, and we should start numbering from that carbon which is closest to the carbon with maximum number of substituents.
However, PubChem and Mol-Instincts state the IUPAC name of this compound as 2-iodo-4,4-dimethylpentane.
Why not start numbering from the left as there are two methyl group as substituents on the second carbon, and then count iodo on the fourth carbon as it has only one substituent in the form of iodine?

Neither PubChem nor Mol-Instincts is an authoritative source for chemical nomenclature. PubChem is at least honest with you and informs that the name is auto-generated ("Computed by LexiChem 2.6.6") and suggest the correct alternative in the synonyms list, 4-iodo-2,2-dimethylpentane (which you rightfully suggest as well). To answer the question adequately, one must look up for the applicable rules in the IUPAC Blue Book.

@andselisk It's hard to believe that "2-iodo-4,4-dimethylpentane" is really auto-generated. Such a simple case (only an alkane without any substituents would be more simple) should be correct with any automatic algorithm.

@Loong True. Maybe their algorithm also relies on how frequent a certain name is being used. For some reason there are plenty of patents and publications (https://patents.google.com/?q=SMILES=CC(I)CC(C)(C)C&scholar) going for 2-iodo-4,4-dimethylpentane.


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Cleaning with vinegar and sodium bicarbonate
Mixing sodium bicarbonate with vinegar (and sometimes adding dishwasher detergent) is quite a common tip on the internet. But I have never understood the reason.
I saw there was a question about using sodium bicarbonate and vinegar here to unclog drains, where the answer referred to the production of $\ce{CO2}$ being able to push whatever gunk blocking the drain.
However, for cleaning, the reaction is done when you use the mixture. Is there something I am missing? I am not a chemist, but my intuition is that using either bicarbonate or vinegar + dishwasher detergent alone would be more effective? Can somebody explain if the cleaning tip actually makes sense?

Mixing vinegar and bicarbonate is a non-sense. Bicarbonate destroys the acetic acid contained in the vinegar. The only advantage is that this mixture produces a huge amount of foam, as if this foam would unclog the drain. It does not. Adding detergent would still increase the volume of the foam. Only vinegar may unclog a drain. Either bicarbonate or vinegar + dishwasher is more effective

The advantage is that you're adding an abrasive to a weak acid, which neither mars most surfaces nor affects their color fastness when used as a cleaning solution.

Your intuition is on point. Pretty much all the "lifehacks" praising cleaning abilities of vinegar and sodium bicarbonate are ineffective. The reaction appears vigorous due to the formation of carbon dioxide gas, but what is left is a diluted solution of sodium acetate:
$$\ce{CH3COOH(aq) + NaHCO3(aq) -> NaCH3COO(aq) + H2O(l) + CO2(g)}$$
Sodium acetate doesn't have any pronounced cleaning abilities. It doesn't act like a surfactant, doesn't drastically increase solubility of the impurities, or participate in complexation helping to dissolve metal-containing compounds.
As for unclogging abilities—yes, this is something that my grandmother used to use, but these days there are much better alternatives.
The idea was to pour some hot water into the system and then put a couple of spoons of baking soda into the drain creating warm concentrated solution in the siphon.
This solution would also be mildly basic due to hydrolysis and help to dissolve some fat and thin hair:
$$\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH-(aq)}$$
After some time (approx. 30 min) it would make sense to add some vinegar hoping the reaction would cause mixing up the constituents of the clog and excessive pressure would help to push the waste further down the drain.

Better is using washing soda alone first ( instead of baking soda ), pouredby boiling water and kept there overnight. Next day vinegar may be used to make some bubbles desintegrating the clogging.
Washing soda has cleaning ability by itself and has better solubility than baking soda. Adding vinegar later is not to produce some extra powerful chemical cleaner, but it's effect is mechanical. It is producing CO2 bubbles and pressure to tear and push clogging, hopefully weakened by prior soda action.
Detergents then make the  clog desintegration easier.
An even stronger cleaner is sodium hydroxide, that is often used as well. But is cannot be used alone, without soda, to produce CO2 with vinegar.
Note that majority of clogging is caused by protein a/o fat rich materials, like hairs, that are chemically dissolved in alkaline solutions.

IANAC, but I find sodium percarbonate (e.g. oxyclean laundry powder) + boiling water to usually do well at unblocking drains (3 long-haired humans in my house).  Pack the dry drain with a cup or so of Na2H3CO, then pour a liter or more of boiling water over it.  I think the heat of the water amplifies the alkaline properties of the Na2H3CO, and the water causes it to release H2O2 which does some good oxidising.  It all washes away cleanly with no need for vinegar.  Na2H3CO works for 90% of blocked toilets too if given some time to react.

What exactly is "washing soda"?

@LawnmowerMan Is not faster to google "washing soda" ? It is a common household name for $\ce{Na2CO3 . 10 H2O}$, being part of washing powder formulas and, mainly in past, used even separately to prevent coagulation of insoluble calcium and magnesium salts of fatty acids. Baking soda is $\ce{NaHCO3}$

@DigitalTrauma You must mean $\ce{Na2H3CO6}$, more properly written as $\ce{2 Na2CO3 . 3 H2O2}$

@Poutnik yes you’re right


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Is it possible for graphite which has shed off metal rods during water electrolysis to bond with the hydrogen being produced to form hydrocarbons?
I recently conducted a simple experiment for school involving the electrolysis of water.
I changed the concentrations of an ionic compound to determine its effect on the rate at which the products of the lysis of water are formed. I used graphite rods for the electrodes. These shed quite heavily.
I noted this down in my qualitative data, but I was wondering whether this excess graphite may have bonded with some of the hydrogen produced at the cathode to form hydrocarbons?

Probably not. It's rather weird that was any damage to electrodes, though. I think yo should provide more details.

So the graphite is coming from graphite rods. I have used graphite rods for electrolysis and found they shed. I assumed it was a mechanical affect of oxygen bubbles forming below the surface. I switched to platinum .


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Concentration of sulfur in turbid solution
I recently participated in an experiment where I need to calculated the reaction orders and the reaction constant of the reaction:
$$\ce{8 Na2S2O3(aq) + 16 HCl(aq) -> S8(s) + 8 SO2(g) + 16 NaCl(aq) + 8 H2O(l)}$$
by measuring time until the solution becomes too turbid for light to travel through the fluid. Calculating the orders of reaction was possible using proportions of the known starting concentrations and time.
In order to calculate the reaction constant, I need to know a specific change in the concentration of the sulfur (I assumed that this is the element that makes the solution turbid), but I couldn't find any information on the internet. If there is no exact answer, what would be an acceptable estimate?
My data is: measured time until turbidity as a function of changing initial thiosulfat concntration (while initial HCl con. remains constant), and data with reversed roles for HCl and thiosulfat. Need to find the constant of reaction k.

$\ce{S8}$ is very insoluble in aqueous solution. So assume that there is a constant particle size as more of the  $\ce{S8}$ is created. Does this help?

You cannot "find" concentration of sulfur directly. You have to use turbidity as proportional to concentration of sulfur.

Im not sure what do you mean... I tried alot but there is no way to escape this mathematically. The orders of reaction can be calculated only using initial concentration proportions but its not enough for calculate k (reaction constant), the lab guide says nothing on the matter

I really think you do not need sulfur conc. You need to plot the initial sodium thiosulfate concentration on x-axis and 1/time on the y-axis. This is why nobody is talking about sulfur concentration.

I doesn't make sense to me... the time is measured until the water is turbid (I draw X on paper and wait until I can't see it), the t is not directly related to the initial thiosulfate concentration.

Add details what you have exactly done in detail in the original post. Nobody can guess what you were doing.

Sorry, added now

Okay, so you have several initial concentrations of thiosulfate and you must have notice the time of turbidity. You do not need sulfur concentration. See this one
Kinetics of Thiosulfate in Acid www.science.purdue.edu › science-express › labs › labs. Google this title, the first result should solve some of your problems. It is a word file.


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Problem in the calculation of ionic strength of a solution
Recently I've been trying to answer the question 10-13 (c) of 9th edition of  Fundamentals of Analytical Chemistry of Skoog et al. Comparing my answers with the student manual , I realized that my numbers didn't agree to the amount of $\mu$ of solution calculated by the book.
Here's the full question:

Use activities to calculate the molar solubility of $Zn(OH)_2$ in the solution that results when you mix 40.0 mL of 0.250M $KOH$ with 60.0 mL of 0.0250M $ZnCl_2$.
(Available information : $K_{sp}$ $Zn(OH)_2$ = $ 3\times 10^{-16}$ , $\alpha_{Zn^{+2}}$ = 0.6 nm , $\alpha_{OH^{-}}$=0.35 nm , $-\log\gamma_x=\frac{0.51{Z_x}^2 \sqrt{\mu}}{1+3.3\alpha_x \sqrt{\mu}}$)

I'm not looking for the full answer. I'm okay with the whole provided answer except with $\mu$.  To be brief the final solution should contain $[K^+]=0.1M$ , $[OH^-]=0.07M $ , $[Cl^-]=0.03M $ , $[Zn^{+2}] \approx 0 M$.
Then
the manual says : $\mu = {1\over 2 }{(0.1 \times 1^2 +0.07 \times 1^2 + (2) \times 0.03 \times 1^2) }$ which means $\mu = 0.115$  . From that, the solubility is found to be 2.8 $\times$ $10^{-13}$ M
From my point of view:
$\mu={1\over 2 }{(0.1 \times 1^2 +0.07 \times 1^2 + 0.03 \times 1^2) }$
Therefore $\mu=0.1M$ and finally Solubility = 2.6302 $\times$ $10^{-13}$M.
I can not understand why we have to multiply the corresponding portion of $Cl^-$ by two (In the parenthesis). Is the book making an error or am I missing something ?
According to the ending pages of the book the answer is 2.8 $\times$ $10^{-13}$ M and not 2.6302 $\times$ $10^{-13}$M. Please be aware that the differences do not result from approximations.

Every dissociating mole of ZnCl2 generates 2 mole of Cl- ?

You're right, the book is wrong. However you have far too many significant figures in your final answer. The Ksp only has one significant figure. // Both you and the book show to few significant figures in the ionic strength calculation.

The problem is worked correctly in the errata for the book on page 5 of the pdf.  https://www.cengage.com/resource_uploads/downloads/0495558281_519368.pdf

@MaxW thank you very much for that document. I wasn't aware of that. It helped me a lot. :)


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Stability and Rate of Formation of Carbocations
Consider the dehydration reaction of the following compounds with $\ce{H2SO4}$:

(a) Tert-butyl alcohol
(b) 3-Methylbut-2-en-1-ol

Consider the protonation of the $\ce{OH}$ group followed by the removal of a water molecule. In (a), we get a tertiary carbocation whereas in (b) we get a primary allylic carbocation. As far as I know, a tertiary carbocation is more stable than a primary allylic carbocation. Therefore I expect (a) to be dehydrated at a faster rate.
But if I consider a resonating structure of the carbocation formed in (b), I get a tertiary allylic carbocation which is more stable than the tertiary carbocation.
My main question is when comparing the rate of reaction of 2 different compounds, do we rank them based on the stability of the initial carbocation formed, or based on the most stable carbocation formed by these compounds after resonance?
I believe that the compound having the tertiary carbocation would be formed at a faster rate as the initially formed intermediate is more stable.

Could anyone please shed some light on this?

Rank it based on the stability of carbocation formed before rearrangement.

I think you're missing a locant for the alcohol.

I changed it. Could you please check now


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Do electrons of opposite spin attract each other?
Electrons of opposite spin create magnetic fields that are opposite to each other. So, is it possible that, 'such electron pairs attract each other' or ' they distract each other less than electron of same spin'? Thanks.

Your title seems to contradict your question. Are you talking of the two electrons (presumably attracting each other), or of the two electron pairs? Make up your mind.

I would say two electrons in the same orbital avoid each other, rather.

Quite the opposite; electrons with the same spin repel each other less. This is a feature commonly known as exchange energy, and is discussed in detail at https://chemistry.stackexchange.com/questions/58625/what-is-the-physical-basis-for-hunds-first-rule, although the full explanation depends on quantum mechanics. You will certainly come across the notion of exchange energy when you do electronic configurations, although understanding the QM may have to wait until a bit later.

Electrons can sort of attract. Let's review magnetic dipoles. First, suppose we have a current loop $I$ rendered in blue at radius $a$ around the $z$-axis in the $xy$-plane and we want to compute the magnetic field at some point on the $z$-axis marked in red. See Figure $1$.

The source point vector is
$$\vec{r}_s=\langle a\cos\phi,a\sin\phi,0\rangle$$
So the element of arc length along the current path is
$$d\vec\ell=d\vec{r}_s=\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi$$
The field point vector is
$$\vec{r}_f=\langle0,0,z\rangle$$
So the vector from the source point to the field point is
$$\vec{r}=\vec{r}_f-\vec{r}_s=\langle-a\cos\phi,-a\sin\phi,z\rangle$$
And its magnitude
$$r=\left\lVert\vec{r}\right\rVert=\sqrt{a^2+z^2}$$
And a unit vector in its direction
$$\hat{r}=\frac{\vec{r}}{r}=\frac{\langle-a\cos\phi,-a\sin\phi,z\rangle}{\sqrt{a^2+z^2}}$$
So we can get the magnetic field at $\vec{r}_f$ via the Biot-Savart law
$$\begin{align}\vec B&=\frac{\mu_0}{4\pi}\int\frac{Id\vec\ell\times\hat{r}}{r^2}\\
&=\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\langle-a\cos\phi,-a\sin\phi,z\rangle}{(a^2+z^2)^{3/2}}\\
&=\frac{\mu_0I}{4\pi(a^2+z^2)^{3/2}}\int_0^{2\pi}\langle az\cos\phi,az\sin\phi,a^2\rangle\,d\phi\\
&=\frac{\mu_0}{4\pi}\frac{2\pi a^2I}{(a^2+z^2)^{3/2}}\hat k\approx\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k\end{align}$$
Where the magnetic dipole moment of the loop is $m=\pi a^2I$ and we are looking at the far field of the dipole so $a^2+z^2\approx r^2$. The magnetic scalar potential of a dipole, valid away from its source current, is
$$\Phi_m=\frac{\mu_0}{4\pi}\frac{mz}{r^3}$$
so
$$\begin{align}\vec B&=-\vec{\nabla}\Phi_m=-\frac{\mu_0m}{4\pi}\left\langle-\frac32\frac{2xz}{r^5},-\frac32\frac{2yz}{r^5},-\frac32\frac{2z^2}{r^5}+\frac{r^2}{r^5}\right\rangle\\
&=\frac{\mu_0m}{4\pi r^5}\langle3xz,3yz,3z^2-r^2\rangle\end{align}$$
Along the $z$-axis, $x=y=0$, $r=z$ so
$$\vec B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k$$
So the scalar potential checks. Along the $x$-axis, $y=z=0$, $r=x$ and
$$\vec B=-\frac{\mu_0m}{4\pi x^3}\hat k$$
so
$$\frac{\partial B_z}{\partial x}=\frac{3\mu_0m}{4\pi r^4}$$
This is the same as for a scalar potential of
$$\Phi_m=-\frac{\partial B_z}{\partial x}\,xz$$
because then
$$\vec B=-\vec{\nabla}\Phi_m=\frac{\partial B_z}{\partial x}\langle z,0,x\rangle=\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle$$
For the field at our loop due to an identical loop at $\langle-r,0,0\rangle$. So then the force on our loop is
$$\begin{align}\vec F&=\int Id\vec\ell\times\vec B\\
&=\int_0^{2\pi}I\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle\\
&=\frac{3\mu_0mI}{4\pi r^4}\int_0^{2\pi}\langle a^2\cos^2\phi,a^2\sin\phi\cos\phi,0\rangle\,d\phi\\
&=\frac{3\mu_0mI\pi a^2}{4\pi r^4}\hat i=\frac{3\mu_0m^2}{4\pi r^4}\hat i\end{align}$$
So in this configuration the force is repulsive if the dipoles are aligned but attractive if pointed opposite one another. You can see this if you have a neocube where oppositely aligned strings of magnet spheres attract at the equators.
For electrons things are more complicated. If we have little bits of charge $\Delta q$ and mass $\Delta m$ orbiting around an axis at distance $a$ and angular frequency $\omega$ then the charge passing a point in the orbit per unit time is $I=f\Delta q=\frac{\omega\Delta q}{2\pi}$ so the magnetic dipole moment is $m=I\pi a^2=\frac12\omega a^2\Delta q$. The angular momentum is $L=rp=a\Delta ma\omega$ so we would get
$$\frac mL=\frac12\frac{\Delta q}{\Delta m}$$
But life isn't so simple for a couple of reasons: the ratio is actually a factor of $g_e\approx2.002$. Also there is a question of whether we want to consider the total angular momentum of the electrons or the component at right angles to the vector from one to another. We choose the latter and this component is $L_z=\frac12\hbar$ so we get
$$m=\frac{g_eq_e\hbar}{4m_e}$$
So the magnetic force is
$$F=\frac{3\mu_0}{4\pi r^4}\left(\frac{g_eq_e\hbar}{4m_e}\right)^2$$
Comparing with the electrostatic force,
$$F=\frac{q_e^2}{4\pi\epsilon_0r^2}=\frac{\mu_0c^2q_e^2}{4\pi r^2}$$
We see that the magnetostatic force is greater when
$$r\lt\frac{\sqrt3g_e\hbar}{4m_ec}=\frac{\sqrt3(2.002)(1.055\times10^{-34}\text{J}\cdot\text{s})}{4(9.11\times10^{-31}\text{kg})(2.998\times10^8\text{m}/\text{s})}=3.35\times10^{-13}\text{m}$$
This is so close that it kind of negates the whole calculation though: the electrons are going to be moving pretty fast so there will be a magnetic force due to their relative motion and at these distances you really need quantum electrodynamics to analyze the problem. Still, it gives you an idea of how close you have to get so that the net force has at least a chance of being attractive.


text:
Thinking of reducing aluminum hydroxide with magnesium
Is it possible to reduce aluminum hydroxide to aluminum metal by using a sacrificial magnesium electrode in a single displacement reaction?
I checked my revision of redox charts and it looks like it should work.
For more detail, consider: $\ce{Al}$ is very fond of bonding with oxygen. It's such a strong reaction that the leading industrial process to refine it involves adding $\ce{Na3AlF6}$ (cryolite) to lower its melting point, and using sacrificial carbon electrodes, generating large amounts of carbon dioxide.
I was curious as to whether anyone has tried using magnesium metal instead, since $\ce{Mg}$ is easier to obtain electrochemically, Together, they form a eutectic that melts over 200 °C lower than either metal alone, and $\ce{Mg}$ seems to support reacting more favorably to oxygen and hydroxide in comparison to $\ce{Al}.$
So, what am I missing? Does it work?

The reaction $$\ce{Al2O3 + 3 Mg -> 3 MgO + 2 Al}$$ is slightly exothermic, with a $\Delta H = \pu{-130 kJ mol^-1}.$ It is also exoergic at all reasonable temperatures, as $\Delta S $ is very small $(\pu{-1.5 J mol^-1}).$ So, the reaction is in principle feasible. The trouble is that $\ce{Al2O3}$ and $\ce{Mg}$ are solid at room temperature. And the reactions between solids are never efficient, if no gas is produced. They only happen at the points of contact of the solids. So, the yield is poor.
The only way to improve the contacts between reagents is to use liquid reagents. Here the first substance to melt when heated is magnesium metal, which melts at $\pu{654 °C}.$ So, the reaction should be pretty efficient above $\pu{654 °C}.$ But magnesium takes fire in air much before $\pu{600 °C},$ and it reacts violently with $\ce{O2}$ and with $\ce{N2}.$ So, this reaction should be carried out in argon gas or in a vacuum. This prevents this process to become commercial.

Your last sentence suggests that a process of melting magnesium in Ar (or vacuum) would prevent the process from commercial use. Do you have info on how it's worse than current practice?

The use of important volumes of pure argon in metallurgy must be rather expensive, just for producing a common metal like aluminum. I don't have any reference to back my purposes : it's pure logic.

Argon is a byproduct of air separation so it is unlikely to be relatively expensive.  In the production of anhydrous ammonia the process gas is recycled except a percentage is discharged to the atmosphere because of argon concentrated in it ; as I recall ,discharge has 10 % argon . If argon was very valuable I expect it would be collected from the discharge gas.

Magnesium was melted in air in about 1958 , a charcoal blanket was used and a hood to reduce air circulation. College metallurgy class tour to Chicago Magnesium Foundry , don't remember much except a brilliant flame when the charcoal was pushed back in a small spot.. I was much younger then.


text:
What is the essence of chemistry?
I am an engineering student who loves physics but didn't quite enjoy chemistry.
a. When I was in high school chemistry to me was something not as inspiring as physics but in which I couldn't solve problems well. Things looked no more than the arithmetics, but I got the answers wrong.
b. Reading university chemistry textbooks (e.g. Oxtoby, Principles of Modern Chemistry 7E) I thought it was more of the physics behind the chemical objects and their reactions. Like quantum mechanics, electrodynamics, thermodynamics, statistical mechanics, ... they are just called differently - quantum chemistry, thermochemistry, chemical kinetics.
c. So I question the significance of existence of chemistry. No offence, it's a pure curiosity. I now believe that all fields of study must be respected, and I want to take a step to do so by understanding the true meaning of chemistry.
Q1. Why isn't chemistry as science-looking as other natural sciences?
Q2. Why isn't chemistry a mere branch of physics? (Why is it distinguishable?)
Q3. So what do chemists research?
Q4. Please give me any important insight about chemistry as a discipline.
Nowadays, I am very interested in chemistry, and want to learn it again at a broader viewpoint, with the proper philosophy of appreciating it. Could any chemist help.
Edited 1. I guess many people dislike and downvoted this post simply because I proposed a view (kind of) against chemistry. I will not delete the post - it is definitely worth leaving it as it is, because there is a nice answer and I consider that can help many other students ovecome difficulties of studying chemistry.
Edited 2 (My answer to this question).
Now I fully understood the need for chemistry. Sadly I couldn't post my own answer as the question was closed, so I am simply adding it here. Studying some more quantum mechanics, I realised that it provides a good basis for the theoretical understanding of facts through idealised models, but perhaps the only way it can directly contribute to the real world is through experiments and useful numerical values. That's the experimental physics aspect of chemistry that makes it meaningful.
Another important fact is chemistry itself being a theoretical basis for many other fields, such as molecular biology or chemical engineering. As an example, I would say that it is both practically and theoretically impossible and meaningful to think about the wave functions every single time we consider chemical processes. Various numerical measures in chemistry, such as the equilibrium constant, allow us to focus more on the essence of what's happening in other natural phenomena.
The final thing I want to emphasise is that contemporarily the boundaries between all of the natural sciences and even engineering have become much less explicit and most research is being conducted under interdisciplinary circumstances. I realised this after learning some proper physics and general chemistry. And general chemistry is like a collection of 100-year-old theories. The interconnections between various fields of studies in the modern sciences are so complex that I cannot, or anybody else would also be unable to, express.

Everything that happens in your body is the result of a chemical process, why wouldn't you want to study that?

Chemistry is a branch of physics, albeit a huge one.

@Waylander Hmm I realised that the title was inappropriate, so I now changed.

The division of sciences is historical and in most real problems requires the utilization of more than one. You can cut them however you want. It is not much to do with the question if you enjoy them or not.

See https://en.m.wikipedia.org/wiki/Comparison_of_chemistry_and_physics

If you are an organic chemist, there is plenty of essence!

I've started an answer going through your points but I've realised I was in all cases repeating the same two concepts. Chemistry is a variegated discipline with a lot of physics involved. Even taking apart the physics of chemistry, the latter remains a discipline itself. This is because you can't reduce the enormous number of chemical problems and applications to just physics. Note that this happens without considering branch of engineering type, for which the situation is then similar in other sciences/discipline. Consider that chemical engineering as a speciality usually requires....

... a completely different background. Q4 is basically an answer,as for you might question how much of science is in organic chemistry at the bench, but surely it is a discipline, and is part of a bigger one which is grounded on the physics of what we call chemical species.

No one has yet mentioned that Chemistry is the uniquely the study of the properties of molecules and synthesis of molecules of all sorts from H2 to DNA, of necessity this, of course, pulls in lots of other disciplines as pointed out in answers and comments. No science stands on its own, all need to have inroads into others, otherwise, as it were, the wheel keeps on getting reinvented.

Your post was closed because answers are bound to be opinion based, particularly as regards Q1. Also, it is unfocused and unclear.  What is " science-looking"`? What kind of insight are you looking for? I would remove at least Q1 and Q4. Regarding Q2 you may want to ask this at science history SE. There is a long and intimate relationship between physics and chemistry since the dawn of science and the border between the two is murky. Finally, there are many branches of chemistry (related to your Q3). See the Wikipedia link provided by Karsten Theis in comment above.

Edits are welcome if you incorporate them as part of the question (to motivate the question). Please edit your post to this end. You should not answer your own question as an edit of the original post.

Interesting question but keep in mind that it is normal to dislike a subject. You don't have to like chemistry. Many people loathe physics and mathematics. The world is not affected. I am sure your experience is limited to general chemistry and Oxtoby. In statistical terms, one cannot and should not trust a single sample (=experience). However, some of your complaints and thoughts are worth commenting.
a) First of all beware that real life is greyish without clear cut boundaries and modern science is more so. One cannot separate the knowledge of physics and mathematics from chemistry. As someone jokingly said, "Chemistry is the physics of the outermost electron." So a "chemist" must also study the traditional topics studied by physics students such as quantum mechanics, thermodynamics, and statistical mechanics etc. Since they wish to study the applications of thermodynamics, quantum mechanics, kinetics it is fair to call that thermochemistry (=thermodynamics applied to chemical problems not engines), quantum chemistry (quantum mechanics applied to chemical problems, molecular structure), chemical kinetics (study of the time behaviour of chemical reactions, rate laws, differential equations).
b) Using the same fallacious argument one may argue that engineering is an application of mathematics or it is a branch of applied mathematics. It does not have that science-looking characteristics. Why do engineering courses have differential equations? Matrices, Vectors? That is pure mathematics, Why do you need to learn programming, that is computer science, why do they need to learn English, this belongs to linguistics, why do you need to learn drawing, that is the artist's job. Again there are no clear cut boundaries in modern science and engineering.
(c) Regarding the "existence" of chemistry: This is a major misconception and it means you have not looked around carefully. Most of the modern and indispensable objects owe their existence due to the knowledge generated by hundreds of thousands of dedicated chemists in 300-400 years around the world. I am not counting the contributions of medieval scientists. Just look at your own laptop, how many plastic/polymeric and metallic components do you have? How did it become possible to make pure silicon chips? Organic LEDs? What about medicines in your cabinet, printer inks, what color of clothes are you wearing? How many lives the knowledge of chemistry and biology has saved? ALL modern medicines owe their existence due to the knowledge of chemistry and chemical synthesis including this corona vaccine.
Most of the early and important knowledge about atoms and elements came from people who were interested in chemistry.
Coming to the unfortunate part of Chemical Education. Yes, the early undergraduate chemical education is indeed in shambles whether it is a rich country or an economically deprived country. General chemistry teaching should be abolished from the first year courses because it is a mere collection of random facts. Gillespie, the man behind VSEPR model and a well respect famous Canadian chemist had written an interesting article 50 years ago. Chemistry-Fact or Fiction. He complained that chemistry does not attract the brightest students because the way early chemistry courses are taught. He emphasizes the same thing that there the emphasis is more on memorization of facts. Basically he points out a lot a flaws in the current chemical educational system He is making a general remark. But any serious chemistry teacher should read this article for the sake of inner reflection. It does not by default mean that all engineers are bright by default. There is no shortage of incompetent engineers in the world, just as there is no shortage of gifted and bright chemists.

Wow, what a gorgeous and clear answer. You summarise everything. Thank you very much

If I want to learn general chemistry properly what do you think is the best way?

I would say buy Linus Pauling's General Chemistry for bed-side reading. I dislike these modern gen chem. books.

Why isn't chemistry as science-looking as other natural sciences?

For some reason chemistry is often taught badly (Khan Academy is probably the best place to start if you want to study the subject on your own). And there is less interesting information about chemistry on the internet relative to other disciplines. It's too far from the reality of a typical non-scientific person (relative to biology) and it's not as "dreamy", reality-quenstioning as let's say astrophysics/QM/etc.

Why isn't chemistry a mere branch of physics? (Why is it distinguishable?)

If you go deep enough you'll need lot's of physics. But:

At a more practical level there's a huge jump in abstractions. And these high-level abstractions allow drawing useful conclusions and do useful work.
Physics is too low-level and you'd need immense computing powers or different calculation approaches to calculate something practical at high  level.

So what do chemists research?

An example of a task for cheminformatics (that I know of): reactivity prediction. You don't necessarily know if molecule A will react with molecule B (and how well) until you try it. And it would be beneficial to predict this better - so that bench chemists conduct fewer reactions to find the most efficient conditions (in terms of time/chemicals/environment) for synthesis of some molecule.
A more hands-on chemical example is drug discovery in pharma. You need to find a molecule which has therapeutic effect and doesn't kill you. For that chemists have to try to synthesise tens of thousands of drug-like molecules (they have to choose which ones to synthesise because there are millions or even billions of possible molecules to choose from), find the right conditions for a reaction to succeed, test whether the candidate has the right effect on a protein, has right chemical properties, whether it's not toxic, etc, etc.
This requires minituarization (running reactions at nano/micro scale) and automation to be done quickly & effectively - which is a big task on its own.

give me any important insight about chemistry as a discipline

I probably don't know such facts about the discipline itself, but just to interest you more in the subject.. If you look at periodic table (this one is great) - elements to the left of it will be less electronegative and thus will easily let electrons go and become positively charged. Elements on the right want electrons more and thus will try to take electrons away from other atoms/molecules and become negatively charged. In your body (it's mostly water - the solvent matter for these things) you have plenty of positive (Na+, K+, Ca++) and negative (Cl-) ions. These ions (and the fact that they are ions and not neutral particles is important) participate in physiological processes like muscle contraction, firing of nerve impulses.
PS: I'm not a chemist (though I work closely with chemists), take everything that I say with a boulder of salt.


text:
A question about rotational symmetry of organic molecules and in particular, the cyclic Isomers of C3H4X2
I was thinking about an organic chemistry problem a few days ago but reached a dead end. There were 2 parts to this question.
The first was thinking about the isomers of 'C3H4X2' [where X could be anything like a halogen or a deuterium etc].
I came up with the following 7 isomers:

(I've really tried to make it as clear as possible! Hopefully it has worked!)
...but a website I found stated that there were only 4:
A -
B -
C -
D -
;with their 'D' representing a deuterium.
So this is the first aspect of my quandary - what am I doing wrong when I draw these isomers!
The second aspect to this question was about considering the symmetry of these organic molecules. In particular, considering firstly 'plane symmetry' [which I understand] and 'rotational symmetry' [which I thought I understand but clearly I don't!]
The same website says that molecule A (from above) plane symmetry and rotational symmetry, which is fine.
But it also says that molecule C has plane symmetry but not rotational. Why is this? Surely if I rotate this molecule 60 degrees I get the same shape?
And then finally, they say that molecules B and D have rotational symmetry but not plane symmetry which I think I understand but would appreciate any further comments if anyone had them.
Ultimately, I would be very grateful if someone could point me towards something which could help me understand rotational symmetry in organic molecules and in particular why these examples given above have their particular symmetries! Thank you very much, everyone!
Best wishes,

Your molecules are all $\ce{C3H4X2}$ and not $\ce{C4H3X2}$ !!! What are you looking for ? Anyway, 1) and 2) are impossible. 4) and 6) are identical,

Sorry, @Maurice - typo!! I am meant to be looking for C3H4X2 not C4H3X2...why are 1) and 2) impossible? And yes, I do see that 4 and 6 are identical now - thanks!...what about the rest?

and 2) are impossible, because the four bonds C-C or C-X must define a tetrahedron centered on the C atom. One X must be in front of the triangle plane. The second X cannot be on the same side of this plane. It must be on the other side of this plane, backwards.

@Maurice - Oh right, I see! Now the question is about the symmetry...

A has a plane and an axis of symmetry . B and D have an axis of symmetry defined by the bisecting line coming from the third Carbon atom. C has a vertical plane of symmetry including the bisecting line coming from the third Carbon atom.

But @Maurice - why does C not not have an axis of symmetry? Surely if C is rotated by 60 degrees, or a multiple of 60, it will give the same molecule? Please could you clarify?

A rotation of 60° or 120° does not reproduce the molecule. The substituants are transformed in q different configuration.

@Maurice or anybody else - I have tried drawing the molecules on here https://molview.org/?smiles=C(Cl)1C(Cl)C1.[H].C1C(Cl)(Cl)C1     - but I still don't understand what I'm looking for when I want to find an axis of symmetry. Can anyone still help by any chance?

Sorry. I am not able to open your reference, without paying. So ...

If the website diagram didn't clarify the situation immediately, the issue is not to answer questions about 1 to 7, but to understand why the question is wrong.
My organic chemistry course had us playing around with molecular models - 3D representations of reality, one type with a metal framework representing the multivalent nucleus and plastic tubes representing bond lengths, the other type with wooden balls and springs. They are a far cry from reality - but molecular models give an added dimension of understanding.
Lacking such 3D models, and lacking an answer sheet, one possibility would be to go to 3D images on the web: they are actually in 2 dimensions, but with perspective in the drawing, so it could be sort of 2.5 D, like this:

Only after knowing what is represented by the dotted lines and tilted bond can you understand what you are drawing. The proposed differences between structures 1, 2 and 3 indicate an unfamiliarity with what the bond representations actually mean. One look at the picture above should clarify everything. Perhaps one more comment should be added: the bonds between a carbon and its two substituents are never coplanar with the 3-carbon ring. Interestingly, I did not find any other similar images (of dichlorocylopropane). Everything was of the straight line or dashed line/tilted bond type. This just assumes that the lines/dashes have an obvious meaning. Sometimes a good picture is worth a thousand words. (This answer is worth 1254 words, if you include 1000 words for the picture.)

Thanks @James Gaidis - I get it now! 1254 words well spent!

would you be able to help me with the second part of the question now? The one about the symmetry?

Symmetry operations are ways to move the molecule and compare the new position with the original. Plane symmetry of C is shown by putting a vertical mirror thru the altitude of the triangle containing the CH2 carbon and the midpoint of the CDH--CDH bond. The mirror makes the molecule look the same - unshifted. C cannot be rotated less than 360 degrees without being able to notice it.  A link to rotational symmetry: https://www.mathsisfun.com/geometry/symmetry-rotational.html . Note: math is fun - till they make it complicated!

@JamesGaidis You should look at https://pubchem.ncbi.nlm.nih.gov/compound/74970#section=3D-Conformer

In your drawings you are correct that 5 and 7 are enantiomers.  But 1 and 2 are inconsistent with tetrahedral bonding at the carbon atom (when two bonds are in the plane of the paper the others must be one above and one below), and 4 and 6 are just rotations of each other (around a twofold axis in the plane of the paper, not the perpendicular threefold one).  Therefore the only actual and distinct isomers are 3, either 4 or 6, and the 5/7 enantiomeric pair.

For the molecule to have symmetry, the atoms need to match after the symmetry operation.
For A, there are 2 planes of symmetry (the plane of the 3 carbons and the plane through the 2 hetero atoms and the attached carbon) and 1 axis of rotation symmetry (the intersection of the 2 planes).
With B (and D), there are no planes since the atoms cannot be reflected through a plane and have the same element. It has axis of rotation through the carbon without the hetero atom and the middle of the bond in the ring.
For C, there is neither a plane of symmetry nor an axis of symmetry.


text:
Particules speed in ideal gasses
According to Graham's law, we can calculate de average speed of gasses by using the formula
$$v^2=\frac{3RT}{M}$$
I tried to use this simulation and got these data by keeping same particles amount and increasing the temperature(average speed(m/s) - temperatura(°C)):

T in C
V  in  m/s

27
454

50
480

90
510

130
530

225
590

350
660

425
710

500
740

But, shouldn't be the speed ($v$) a square function of the temperature? ($T=\frac{Mv^2}{3R}$ ) What am I doing wrong?

The shoe is on the other foot (the square is on the other quantity), and you should use the Kelvin scale. Using the Celsius scale, RT will not be proportional to the kinetic energy of particles (just try below freezing to convince yourself).

You are confused about the equations. Your first equation is a function of temperature and the second is a function of v squared.

T in K
$V^2$  in  $\pu{m2 s-2}$

300
206116

323
230400

363
260100

403
280900

498
348100

623
435600

698
504100

773
547600

These are proportional.


text:
Does DIBAL-H reduce alcohol?
Does DIBAL-H reduce alcohol? I was reading an article by Rentsch and Kalesse [1] and I came across this reaction, but could not figure out how the final compound is formed:

Reference

Rentsch, A.; Kalesse, M. The Total Synthesis of Corallopyronin A and Myxopyronin B. Angew. Chem. Int. Ed. 2012, 51 (45), 11381–11384. DOI: 10.1002/anie.201206560.

Dibal-H is reducing the ester to the allylic alcohol which is oxidised to the aldehyde by Manganese Dioxide.

The highlighting in your image seems to suggest you are stuck on a different step (c) than the DIBAl-H reduction (d). Could you clarify which step you are stuck on: (c) or (d)? If you understand (a) and (b), what do you think the structure is after (b)?

i understand step (a) follows The Mitsunobu reaction, and steps (d) and (e) gives the aldehyde moiety in the final product, im stuck on how the C=C is formed in steps (b) and (c), the article says sulfer is oxidized to sulfone.Then, KHMDS deprotonate the carbon next to the sulfur and reacted with acetaldehyde

thank you, i guess im not understanding how the C=C is formed then and what would be the structure of step (b) and (c)

For the early steps, look at the Julia Olefination reaction: https://www.organic-chemistry.org/namedreactions/modified-julia-kocienski-olefination.shtm


text:
What favors the active transport in a membrane?
I was reading about active transport in membranes where ATP is used. ATP "reacts" with the protein pump and converts into ADP and also make a conformational change to the pump. Now this conformational change can transport molecules from low concentrations to high concentrations areas i.e. opposite in their concentration gradient. My problem is that for this to happen we must couple the hydrolysis of ATP with the "reaction" of transport along the membrane.
But I can't find it in textbooks and also I can't imagine if such coupling makes sense. In many textbooks it is stated that the energy that released due to the hydrolysis fo ATP is used to drive the pumping. I don't know if it is possible to couple two reactions with no common intermediates. Also in the pictures that show the active transport like the one in the left side the molecules can't go through the lipids.

So the molecules just move from low to high but the moving from high to low (diffusion) is blocked because the molecules can't move through the lipids. In other words it isn't like molecules make something that it would be otherwise violated because at first the motion from one side to the other is blocked. I mean even if we have no ATP the molecules won't flow from high to low concentration. Then what is the point of "active" transport? For a moment I thought to describe the above scheme for active transport as the following reaction: $$\ce{ATP + H2O + M_{in}} \rightleftharpoons \ce{ADP + P_i + M_{out}} $$
Which can be thought as the sum of:
$$\ce{ATP + H2O } \rightleftharpoons \ce{ADP + P_i}$$
and
$$\ce{M_{in}} \rightleftharpoons \ce{M_{out}}$$
But as I stated I haven't find a coupling reaction with no common intermediates and also it doesn't make sense to write the second equilibrium as the molecules can't go through the lipids. Also from the second reaction we should expect the concentrations on both sides to be equal (because the equilibrium constant must be equal to unity) no matter the concentration of ATP.
So can the first equilibrium alone describe the "active" transport? I mean if this reaction is exergonic then indeed by varying the amount of ATP we can alter the concentration of M outside of the membrane due to the Le Chatelier principle.

The coupling is achieved physically by linking the hydrolysis of ATP to a change in the protein structure that pushes the carried metabolite from one side of the membrane to another. Most biochem texts will have good diagrams of this process.


text:
Dynamic viscosity of gas mixture (appropriate estimation)
What estimation of the dynamic viscosity $\eta = f(T)$ $([\eta] = \pu{Pa·s})$ of the gas mixture components would you recommend during adsorption at a low temperature, approx. $\pu{293 K}$ to $\pu{333 K},$ and a relatively low pressure?
Gas mixture contains $\ce{N2}$ (the major component), $\ce{CO2}$ and $\ce{H2O}$ in the form of water vapor.
I assume the gas mixture to be ideal.
About a year ago, I use to calculate the dynamic viscosity using Lennard–Jones coefficients for a slightly different conditions if I recall correctly. However, I cannot find my calculations, hence I wanted to check whether using these coefficients would be relevant for such case.

What is "gas separation" ?

@Karl I referred to adsorption - corrected.

Ah, OK. Still: I could make a lot of assumptions about the process you have in your mind, and then the question would make sense. But I'd rather you told us.

See Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 1 for mixing rules to get viscosity.

@ChetMiller Thanks, I think I got it by combining either Sutherland Equation with Herning-Zipperer Eq. or probably more precise Chapman-Enskog theory with Herning-Zipperer Eq. If I can ask you one more question... I can apply the Chapman-Enskog theory to obtain values of gas components/mixture conductivity $\lambda_k$ and binary diffusion $D_{AB}$. What would you recommend to use for molecular diffusion $D_m$, between two gases, which I need to obtain axial dispersion coefficient $D_{ax}$, which lumps all the negative effects on separation, thanks. (I'd neglect $H_2O$ in the first attempt.)

I believe that is also covered in Bird, et al, in a later chapter.

@ChetMiller Found it in the chapter 17 :), thanks.


text:
In the pH system of acids and bases, are there 10^-7 loose ions per liter (at pH of 7) or per mole of solvent? (Or per mole of total substance?)
Most places tell me that the number on the pH scale tells you the number of dissociated ions, in moles, per liter of substance.    But, Wikipedia's article pH says this....:

... For a commonplace example based on the facts that the masses of a mole of water, a mole of hydrogen ions, and a mole of hydroxide ions are respectively 18 g, 1 g, and 17 g, a quantity of $10^7$ moles of pure (pH 7) water, or 180 tonnes ($18\times10^7$ g), contains close to 1 g of dissociated hydrogen ions (or rather 19 g of H3O+ hydronium ions) and 17 g of hydroxide ions.

But, if there are 10^-7 moles of hydrogen cations (protons) per LITER, and one mole of protons is approximately one gram (as Wiki's example says), then there is one mole (one gram) of protons per 10^7 liters of water, right?  One gram of hydrogen cations per 10,000,000 liters?
As opposed to what Wiki says, claiming one gram of loose hydrogen ions per 180 tonnes of water?
(One liter of water weighs about one kilogram, and there are a thousand kilograms in a tonne, correct?)
EDIT: Wiki's section on pH and all the incorrect numbers have been corrected.

Wikipedia certainly has errors, but typically they catch the most glaring ones.  And saying that 18 g/mol x 107 mols = 180 tonnes is a pretty glaring error, so I'm suprised that would be in Wikipedia. Thus please provide a link to the Wikipedia reference so we can see what is going on.

But if the Wikipedia page is saying pH 7=> 10^-7 moles H+/(mole H2O), that would be wrong, since pH 7 => 10^-7 moles H+/(L H2O) =10^-7 moles H+/(55.6 L H2O), b/c 1 L H2O = 1000g/(18 g/mol) = 55.6 mol H2O

Think this through:  1. How can there be a fraction of an ion? What is 1/2 ion, or 1/10,000,000  [10^-7] ion? 2.Start with Avogadro's Number: There are ~6*10^23 molecules/mole.

I added the source of the quote to the question. I also did multiple edits to correct the formatting of the exponents in the text.

In the pH system of acids and bases, are there 10^-7 loose ions per
liter (at pH of 7) or per mole of solvent? (Or per mole of total
substance?)

In a rigorous sense none of the options is correct, because pH being approximated by molarity is an approximation itself. The problem lies in the measurement itself and pH should be
$$pH= -log (a_{H^+})$$
The pH probe "senses" the activity of hydrogen ions not the molarity. This is why the pH of 5 M HCl is not equal to $$pH= -log (5)$$
There are no free or loose hydrogen ions in water, like a bare proton, rather this proton is attached to a network of water molecules.
Wiki is talking about an approximation. 1 ton is equal 1000,000 grams, so 10$^7$ mole/g x 18 g $\ce{H2O}$ is indeed 180 t. This part is correct so far. Note the author must be following metric ton. The whole world does not follow the US ton.
1 liter of water weighs ~ 1000 grams. Then 1000 g water contains 10$^{-7}$ moles of $\ce{H+}$, and 18x10$^{7}$ g water should have 1.8 x 10$^{-2}$ mol $\ce{H+}$ or 0.018 g $\ce{H+}$. I don't get 1 g.

To settle the basic question, at 25 °C there are $10^{-7}$ moles of $\ce{H+}$  and $\ce{OH-}$ ions per liter of pure water. The value of the ionization constant of liquid water depends on the temperature of course.
The cited sentence jams so much together that the sentence is extremely poorly constructed.

The verbatim quote in Wikipedia is plagiarized from an engineering book (or vice versa), called Engineering Chemistry by Nicky Rutledge, 2018. It has the same mistake, as in Wikipedia. Does 180 ton of water have 1 g $\ce{H+}$ ions? I get 0.018 g rather.

Yes, the article's math is right. Remember that a liter of water is about 55 molar.

Yes, but what is the omission here? 1 liter of water weighs ~ 1000 grams. Then 1000 g water contains 10$^{-7}$ moles of $\ce{H+}$, and 18x10$^{7}$ g water should have 1.8 x 10$^{-2}$ mol $\ce{H+}$ or 0.018 g $\ce{H+}$.

@M.Farooq - You are correct. I was multiplying 10^7 moles of H2O times 10^-7 molar for H+ to get 1 gram of H+ ions. But that is wrong. You need 10^7 liters of water to get 1 gram of H+ ions. So there is a fundamental mistake in the math in the sentence as you pointed out.

I was also wondering if both the book and Wiki can make such a mistake.

The 00:44, 26 May 2016 edit of the Wikipedia article does not have the exact sentence, but the next edit by Vaughan Pratt (talk | contribs) at 01:02, 26 May 2016 does. So it would seem that Rutledge plagiarized the Wikipedia article.

Someone should correct the Wikipedia section. The 180 ton part is fine. The rest is wrong.

The sentence is just ludicrous. It is supposed to be a "commonplace example." Who can really imagine 180 metric tons of water?

I corrected that Wiki section so that it does not confuse the rest and reach the books.


text:
Steel Wire Rusting & Glowing Orange
In my backyard I have a DIY wire trellis made in part by wrapping a steel wire around a water wellhead. I noticed out the window in the dark a tiny pinprick of orange / yellow light coming from the area, and upon closer inspection it originated from a point of contact between the trellis wire and the side of the well. The wire is 5-10 years old and rusted only around this point of contact. The glow disappeared when I disturbed the wire slightly. The outside temperature was 22°F.
Any idea what could be causing this phenomenon? I might be searching the wrong set of keywords, but I can't find anything online.
Upfront I'll say the closest I found is this question. My question could perhaps be more suitable as a comment on linked question, but

I'm new to the stackexchange and cannot write comments
I'm unsure if reviving a >5yr old question is the right thing

Have you searched the "flash rusting" concept?

I think it is a bit more complicated than just rusting. I'd suspect some sort of galvanic corrosion. I'm guessing that there must be some sort of mechanism that produces electroluminescence, like in a LED. It can't be getting hot enough to make light via blackbody radiation.

I collect rocks that fluoresce under UV, and so find the question interesting.  Thus I upvoted it.

The original 5 year old query talks about sparks. This could be an earthing (grounding in US) issue. A qualified electrician should these things in your area.

Something just to attempt... Could be that corrosion lead to some micro paths be formed and they behave as lamp filament just prior to burn? I am working with thin metals electrodes and you might burn them quite easily.

As @M.Farooq suggests, you might have a serious electrical hazard. Does the well have inside an electric pump? There could be considerable leakage from electric mains to well head, and someone touching it could be electrocuted.

Yes, my comment above also suggest hazard. I shout have stressed that point. It is likely there is an exposed circuit. The comment itself was just referring to the light mechanisms, basically the burning of micropaths due to their high (geometrical) resistance. The paths being built/exposed by the corrosion. @DrMoishe Pippik


text:
Which ligand names have to be parenthesized while naming the Mononuclear Coordination entities
While studying about the Nomenclature of Coordination Compounds I came to know that

While writing formulae of Mononuclear Coordination entities polyatomic ligands are enclosed in parenthesis and also their abbreviations.

But while naming the entities should I follow the same rule?
Like for $\ce{[Ni(H2O)2(en)2]SO4}$ should it be named as

Di(aqua)bis(ethylene-1,2-diamine)nickel(II) suphate

Or

Diaquabis(ethylene-1,2-diamine)nickel(II) sulphate

I Mean is there any rule that says you only need to close certain species of ligands like specifically bidentate (such as en) and polydented ligands (such as EDTA) and can keep the others as they are?
Thanks.

If you ask which ligand names have to be parenthesized then simply ask that.  Right now the title is meaningless.

Thanks for the help @Mithoron  I'm really confused about only the polyatomic ligands.

Greek prefixes are used to designate the number of each type of ligand in the complex ion, e.g. di-, tri- and tetra-. If the ligand already contains a Greek prefix (e.g. ethylenediamine) or if it is polydentate ligands (ie. can attach at more than one binding site) the prefixes bis-, tris-, tetrakis-, pentakis-, are used instead.
Bottom line - parenthesize only those ligands which already has a greek prefix in its name so that you can highlight the use of the alternative prefixes such as bis-,tris- etc.
Ps -Diaquabis(ethylene-1,2-diamine)nickel(II) sulphate is the correct naming but you can make it more appropriate by using ethane instead of ethylene


text:
Milk digestion procedure for AAS analysis
I am working on a project of analyzing heavy metals in milk samples using flame AAS. So far I have only analyzed cadmium and the results of spiked and unspiked have large differences. I was wondering if something went wrong with my digestion procedure. The procedure is listed as followed:

2g of milk sample, add 5 ml of 68% HNO3 (spiked solution spiked with 1ml 100ppm of Cd stock solution)
solution heated at 70 °C for 2 hours with hot plate until solution clears up
3ml of H2O2 added to solution and heated to effervescence and cool solution
filter solution through filter paper
dilute solution with minimum HNO3
dilute with 5% HNO3 in 100ml volumetric flask
filter with 0.7 nm syringe filter
analyze with AAS

after the digestion of HNO3 and H2O2, there are some floating bits which I am not sure what are the compositions, which cannot be digested or dissolved even when I add extra acid. I suspect that filtering with the filter paper may have affected the metal content to be lower than expected. The spiked sample also do not show consistent readings. Glasswares have all been soaked in 5% HNO3 overnight.
I want to know is there anywhere I can improve my digestion procedure? Any relevant literature I can review? Thanks.

Also mention other experimental variables i.e, wavelength, nature of flame (fuel plus oxidizer), burner length, calibration range, and where does filter paper enter the description when you are using syringe filter. What is the material of the syringe filter and last but not the least, what is meant by "non consistent readings". are they high or low?

Homework: How can you get along without the paper filter?

It also seems that filtration step is the culprit along with incomplete digestion of milk. Where did you get the temperature protocol?


text:
Why are S-thiocarbamates less toxic than carbamates?
According to Haley and Rhodes, neostigmine bromide (alternatively known as Prostigmine) has an LD50 in mice of around 0.165 mg/kg by IV injection. Pubchem claims that this is also the LD50 for neostigmine iodide. According to Pubchem, Ro-1837, the analog of neostigmine in which the ester oxygen is replaced with sulfur, having the chemical name (3-Mercaptophenyl) trimethylammonium iodide dimethylcarbamate, has an IV LD50 of 18 mg/kg.  Also according to page 209 of the book Chemical Warfare Agents and Related Chemical Problems, other S-thiocarbamates are likewise significantly less toxic than carbamates (ie. TL-1216 (subcutaneous LD50 of 0.17 mg/kg) vs TL-1239 (Subcutaneous LD50 somewhere over 80 mg/kg) ). Why is this so?
References:

Haley, T. J., & Rhodes, B. M. (1950). A note on the acute toxicity of neostigmine methyl bromide in the rat. Journal of the American Pharmaceutical Association (Scientific Ed.), 39(12), 701. doi:10.1002/jps.3030391218

https://apps.dtic.mil/dtic/tr/fulltext/u2/234270.pdf


text:
What happens when salt water is electrolyzed with a copper anode and a copper cathode?
I heard that copper oxide is formed on the cathode and copper chloride is formed on the anode, is this correct?
I don't have much knowledge of chemistry, so I can't find out much by searching on Google.

No copper oxide should be formed on the cathode. The only thing that should occur at the cathode is the formation of $\ce{H2}$ bubbles, plus some $\ce{OH^-}$ in solution. After a long time, the $\ce{OH^-}$ ions formed at the cathode should be mixed with $\ce{Cu^{2+}}$ formed at the anode, and a blue precipitate of $\ce{Cu(OH)2}$ should be the consequence, due to the reaction $\ce{Cu^{2+} + 2 OH^- -> Cu(OH)2}$. That's all.

Maurice, I am afraid your last part of explanation is quite incorrect. There is no blue precipitate. Instead an orange precipitate forms.

The chemistry here can be explained from the associated products of the electrolysis of brine. To quote an education source:

Three important chemicals, NaOH, Cl2, H2, can be obtained by electrolyzing an aqueous NaCl solution (brine). This forms the basis of the chlor-alkali industry. The diaphragm cell (also called a Hooker cell)...

As such, depending on the cell design, expect also hypochlorous acid:
$\ce{Cl2 (g) + H2O (l) <-> HCl (aq) + HOCl(aq) }$
and hypochlorite with OH- as a side product of the electrolysis of the brine:
$\ce{ OH- + HOCl -> H2O + ClO-}$
However, some of the $\ce{HOCl}$ can also engage in a Fenton-type reaction (example here) with $\ce{Cu}$ (and, to a lesser extent, $\ce{HOCl}$ acting on $\ce{Cu2O}$ resulting in some possible $\ce{CuO}$ or even a basic chloride formation) generally depicted as follows:
$\ce{Cu/Cu+ + HOCl -> Cu+/Cu++ + .OHCl-}$
where the latter radical intermediate can further break down with products dependent on the pH level (here, likely over pH 7 from $\ce{OH-}$ created in the parallel electrolysis reaction) and the chloride concentration. Source: See reactions listed in Supplement Table S1 available here by clicking on Supplement File to obtain a download.
Support: With metal spoons serving as electrodes for the electrolysis of a saltwater solution, there is clear discoloration of one of the spoon electrodes (visible at 5 minutes and 20 seconds into this science experiment video).
Bottomline in time, expect similar discoloration of a copper electrode as well from the associated electrolysis products.

All points in my answer are now fully sourced.

Great answer +1, but could you please elaborate on the reactions on the copper electrodes? I tried to build a faraday motor, see several youtube movies, but observed that after a few minutes some dark corrosion occurs on the copper electrodes that obstructs the current flow. Or perhaps the amount of ions is reduced which reduces current? Should I try different metals for the electrodes?


text:
Dihaloalkane reaction with sodamide/ammonia

In the following sequence of reactions:
$$\ce{CH3-CH2-CH2-I ->[\text{Alc.}~KOH] \textbf{A} ->[Br2] \textbf{B} ->[NaNH2/NH3] \textbf{C}}$$
the product C is:

alkene
alkanol
alkyne
alkylamine

I haven't really came across a reagent $\ce{NaNH2}/\ce{NH3}$ in my whole session. Well, I have got B which is $\ce{CH3-CH(Br)-CH2-Br}.$
Then, how do I proceed next? How will my substrate  react with that reagent and what product will I achieve?

Google is your friend! Search sodamide and all will be revealed.

...or go to a chapter on Alkynes.

@SiddharthBisht Sorry, two identical naming schemes based on capital letters for the compounds and the list of multiple choices drives me crazy. I changed the latter to the numbered list for the sake of sanity.

You were correct up to this point, the strong base alcoholic KOH sets off elimination instead of substitution so you will not get an alcohol.
Accordingly your products A (presumably propene) and B are correct.
The amide/ammonia is an even stronger base than alcoholic KOH and will also produce elimination.  With a vicinal dihalide like your B, there is an added twist.  Read here to find out what happens.


text:
While treating 4-aminophenol with acetic anhydride why does the reaction stop at Paracetamol and does not undergo esterification with acetic acid?
This needs a little background so I will be showing you a question from one of my Organic Chemistry problem books:-

Now, the book shows the answer as Option (c)  which is Paracetamol.
I tried to work out the mechanism and I was pondering over the fact that the
products left in the reaction are Paracetamol and Ethanoic acid . Now Paracetamol contains an -OH(Hydroxy group) and O(Oxygen) atom can easily aquire positive charge by mesomeric stabilisation and attack Ethanoic acid to further form ester (as in Option (a)) . Can someone please help me out ?

Acetates of phenol are pretty as good acylating agents and don't survive aqueous base work up

There is no stoichiometry in your reaction. If Ac2O is in excess, then A is the answer and Waylander's comment applies to form C. If the stoichiometry is 1:1, then C>>B. Heating this mixture, possibly neat, would convert B into C. Can you figure out why?

There are other underlying issues with your description. Firstly, what do you mean by oxygen "acquiring a positive charge", and secondly, why should that be of any help with esterification? If your oxygen is acting as a nucleophile, you probably don't want it to be positively charged.


text:
Are tetra- and pentanitrotoluene better explosives than TNT?
When reading about TNT (trinitrotoluene), I noticed that the aromatic ring has still 2 carbon atoms left for two more possible nitro groups ($\ce{−NO2}$), which would get you tetra- or pentanitrotoluene.

The amount of available oxygen in TNT is not enough for a complete oxidation to water and carbon dioxide, this is obvious when looking at a possible reaction:
$$\ce{2 C7H5N3O6 -> 3 N2 + \mathbf{5\,H_2} + 12 CO + \textbf{2 C}}$$
Thus tetra- and pentanitrotoluene should be better explosives than TNT, because they both provide more oxygen than TNT. Why they were never popular? Are they harder to fabricate than TNT, can you nitrify them from TNT? Are they maybe unstable?

The synthesis is probably neither feasible not effective enough. You'll get better bang for fewer bucks.

Pentanitrotoluen has been prepared https://patents.google.com/patent/US4451681A/en  You have to balance their explosive power against the difficulty of production and handling

Keep in mind that octanitrocubane would  have ideal stoichiometry but is much harder and more expensive to synthesize .. and ironically seems to have lower crystal density. Having talked to a few explosives chemists, there's a lot unknown (even beyond the "we're not talking" bits).

On more serious note, yes they're obviously unstable and I wonder why these guys^ bothered with patent.

@Mithoron, why are they unstable?

Read up on steric inhibition of resonance and you'll get some insight.


text:
Can x = 1 for a compound of the form x-methyl-3-hexene?
I came across a question in a textbook Chemistry Live by Declan Kennedy:

Draw the structural formula for the compound 1,3,5-trimethylhex-3-ene.

My issue is with the 1-methyl part. Isn't this compound just the same as 2,4-dimethylhept-3-ene?
I don't think it is a typo as it appears again in an updated version of the same book.

Yes, it is. Maybe there is a typo as something like 2,3,5-trimethylhex-3-ene would make sense.


text:
Which one of the following is NOT a haloalkane?

Which one of the following is NOT a haloalkane?
A) Freon
B) Teflon
C) Iodoform
D) Vinyl chloride

Every compound here seems to me as a haloalkane: Freon, Teflon, iodoform, and vinyl chloride.
My text says the answer is vinyl chloride, but how?

Hint: they certainly all contain halogens, but what is an alkane?

What is the structure of vinyl group?

One of these things is not like the others
One of these things just doesn't belong
Can you tell which thing is not like the others
By the time I finish my song? Sesame Street

I have went through each option:
A) Freon is $\ce{CCl2F2},$ which is derived from methane
B) Teflon is a polymer of $\ce{-F2C-CF2 -}$ (or) $\ce{(C2F4)_n},$ derivative of 1,2-difloroethene. Here, carbon is saturated, which means no double or triple bond is present.
C) Iodoform is $\ce{CHI3},$ which again is a derivative of methane.
D) Vinyl chloride is unsaturated thus it does not classify as an haloalkane.


text:
How to separate copper(II) chloride from 4-iodophthalic anhydride?
I am planning a reaction in which I iodate phthalic anhydride. My reaction is I2, phthalic anhydride, and copper(II) chloride as a catalyst. The solvent is 99.9% isopropanol. How do I separate the product (4-Iodophthalic anhydride) from the copper(II) chloride and remaining iodine?
Edit: I planned to boil away the solvent, and sublimate away the iodine from the remaining solid. How do I separate my product from the resulting mix? At the moment, I only have isopropanol.

I would recommend just running a normal flash column since you will likely have unreacted phthalic anhydride after running your reaction. If the reaction did have a very high yield (near 100%), a column may not be necessary. You could just add a bit of sodium carbonate to convert the iodine to sodium iodide, remove the isopropanol using a rotovap, and then use a separatory with methanol and ether (or hexane) to separate the salts (sodium iodide and copper chloride) from the product.

I'm an amateur chemist, and I don't really have a flash column, so I can't really do that, but thanks for suggesting that.

@Eli Jones: How does sodium carbonate reduce iodine to iodide? Sodium sulfite or sodium thiosulfate should do the trick.

Is there any concern that the  isopropanol  may open the anhydride?

@user55119 I agree that sodium thiosulfate would probably be better for neutralizing a bunch of iodine, but since only a small excess of iodine will be present, I read that sodium carbonate will convert iodine to sodium iodide and sodium iodate. I'm not sure how effective this is, so I can't guarantee this method, but it would probably be better than dealing with the messy sulfur byproducts using the other method.

I read that isopropanol does not open the anhydride.

I just realized I can add sodium hydroxide, and filter off the resulting insoluble copper(II) hydroxide and sodium chloride, then boil down the result. If you can think of a better answer, feel free to suggest it to me.

The sodium hydroxide would probably destroy your phthalic anhydride (hydrolyze the anhydride moiety), and your product would still be contaminated with iodine and leftover starting material.

Okay, then what should I do? I don't have any ether. Also, would sodium acetate work instead of sodium hydroxide?

I just mentioned ether because it was a common nonpolar organic solvent. You could try something like gasoline, but I cannot guarantee how effective this will be. Also, be safe while removing your solvents from the product! Since you probably do not have fancy equipment, you should only use small amounts of solvent while outside and away from any open flames.

I would recommend trying to use either sodium carbonate (you might have to heat the solution a bit for neutralization) or sodium thiosulfate. Sodium carbonate is sold commonly as washing soda, and sodium thiosulfate is also cheap and readily available. Sodium acetate might result in some transesterification of the anhydride. Also, I haven't read of sodium acetate being used to neutralize iodine.


text:
How do structural differences between neostigmine and TL-599 contribute to differences in toxicity?
Stevens and Beutel studied the activity of several carbamate anticholinesterases. Among other things, they found that the (4-trimethylammonio)phenyl dimethylcarbamate iodide (The para-analog of neostigmine) is significantly less active than neostigmine (Subcutaneous LD50 in mice of the para-analog was 120 mg/kg compared to 0.45 mg/kg for neostigmine). However, ((4-trimethylammonio)-3-isopropyl)phenyl dimethylcarbamate iodide, also known as TL-599, was more toxic than neostigmine, having an LD50 of 0.075 mg/kg. How exactly do the differences in the structures of TL-599, neostigmine, and the para-analog of neostigmine contribute to their differences in toxicity?
Reference:
Stevens, J. R., & Beutel, R. H. (1941). Physostigmine Substitutes. Journal of the American Chemical Society, 63(1), 308–311. doi:10.1021/ja01846a078


text:
Is the correct zeroth-order wave function $\phi^{(0)}$ an eigenfunction of the perturbation Hamiltonian $H'$?
Consider the perturbation treatment of He $1s2s$ excited state configuration. The quantum mechanical Hamiltonian for the He-atom is:
\begin{aligned}
\hat{H} = \sum_{i=1}^2 h_i + \frac 1 {r_{1 2}} \\
h_i = -\tfrac 12 \nabla_i^2 - \frac 1 r_i
\end{aligned}
In the following perturbation treatment, the unperturbed Hamiltonian $\hat{H}^0$ is $\sum_{i=1}^2 h_i$, while the perturbation Hamiltonian $\hat{H}'$ includes the inter-electron repulsion $r_{1 2}^{-1}$.
The four unperturbed degenerate wave functions $\psi^{(0)}$ are shown below as Slater determinants:
$$
\begin{aligned}
&\psi_1^{(0)} = \left|\begin{array}{} 1s(1) \overline{2s}(1) \\ 1s(2) \overline{2s}(2) \end{array}\right| \equiv \left| 1 \overline{2} \right\rangle & \psi_2^{(0)} = \left| \overline{1} 2 \right\rangle \\
&\psi_3^{(0)} = \left| 1 2 \right\rangle &\psi_4^{(0)} = \left| \overline{1} \overline{2} \right\rangle
\end{aligned}
$$
The correct zeroth-order wave functions $\phi^{(0)}$ for the perturbation Hamiltonian $\hat{H}'$ are linear combinations of unperturbed eigenfunctions such that $\phi^{(0)} = \sum_i c_i \psi_i^{(0)}$. Solving for $\phi^{(0)}$ using the linear variational method is equivalent to diagonalizing the perturbation Hamiltonian $\hat{H}'$ in a basis of $\psi^{(0)}$. The eigenvectors and eigenvalues of the $\mathbf{H}'$ matrix, whose elements are $\left\langle \psi_m^{(0)} \middle| \hat{H}' \middle| \psi_i^{(0)} \right\rangle$, are the correct zeroth-order wave functions and the corresponding first-order corrections to the energy, respectively.
Because $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$ (the Condon-Slater rules), $\psi_3^{(0)} = \left| 1 2 \right\rangle$ is already the eigenvector of the perturbation Hamiltonian $\hat{H}'$ with the eigenvalue $E_3^{(1)} = \left\langle  1 2 \mid \mid 1 2 \right\rangle = \left\langle  1 2 \mid 1 2 \right\rangle - \left\langle  1 2 \mid 2 1 \right\rangle$. Could you suggest whether it is correct to go on to state the eigen-equation of $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ ?

No, writing $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ will be incorrect. I will give you two explanations for the same, one intuitive and other mathematical.

Intuitive Explanation:
Assuming that you can write $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$. But, $\hat{H}_0\psi_3^{(0)} = E_3^{(0)} \psi_3^{(0)}$, and in an overall sense, you can write
$$\hat{H}\psi_3^{(0)} = (E_3^{(0)}+E_3^{(1)}) \psi_3^{(0)}$$
This essentially means that electron-electron repulsion is keeping the wavefunction same as that of single electron Hamiltonian. Now, since in single electron Hamiltonian, each electron does not "see" the other electron, and they are allowed to be in the same spatial position. Now, on bringing the $1/r_{12}$ term inside the Hamiltonian, the energy goes to $\infty$ for such a case, and hence a singularity is created in the system.
So, it means that the $\psi_3^{(0)}$ is not an eigenvector of total Hamiltonian, $\hat{H}$. Since $\psi_3^{(0)}$ is defined to be eigenvector of $\hat{H}_0$, so the initial assumption that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ is incorrect.

Mathematical Explanation:
You have given $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$, which is correct. Along with that, we know  $\left\langle \psi_{m}^{(0)} \middle|  \psi_n^{(0)} \right\rangle = \delta_{mn}$. However, you could write the given statement $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ provided $\psi_n^{(0)}$ would be a complete basis set (for explanation, see below). However, $\psi_n^{(0)}$ is not a complete basis set, since for a He atoms, there are infinite number of ways to put two electrons in an orbital, since there are infinite number of orbitals. Since you are treating the orbitals as hydrogenic, it is not guaranteed that the orbitals will be orthogonal to each other once the $1/r_{12}$ term is pulled in.
Even for a hypothetical case if we consider only 1s and 2s orbitals are present and are orthogonal to other orbitals even after the presence of the $1/r_{12}$ term, still they would not be a complete basis set. The reason is simple: You have considered only $1s^12s^1$ configurations, and not $1s^2$ and $2s^2$.

For complete basis, proof:
Assuming $\psi_n$ is an othronormal and complete basis set of dimension $M$, one can
write: $$\hat{H}'\psi_3^{(0)}=\sum_{i=1}^M c_i \psi_i^{(0)}$$ Now,
$$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)} \Rightarrow C_m=0
> \forall m\neq 3$$ $$\Rightarrow \hat{H}'\psi_3^{(0)}= c_3
> \psi_3^{(0)}$$
However, this would have been true if, along with other conditions,
$\psi_n^{(0)}$ would have been a complete basis set.

So, from the given information, it is impossible to conclude that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ mathematically, and from a chemistry standpoint, $\hat{H}'\psi_3^{(0)} \neq E_3^{(1)} \psi_3^{(0)}$.

Hmm... I don't agree with your "intuitive explanation". The energy doesn't just go to infinity simply because there is a nonzero probability density corresponding to $\vec{r} = \vec{r}_1 - \vec{r}_2 = \vec{0}$. According to that logic, a 1s orbital should not exist either, because it has nonzero probability density at the nucleus, which should cause infinite attraction. Basically, the integral in question is not $\int_0^\infty (1/r) ,\mathrm{d}r$ (which would explode); it's $\int_0^\infty \psi_i^* (1/r) \psi_j,\mathrm{d}r$, and those $\psi$'s depend on $r$ as well.

@Mitradip Hi, I mostly agree with your mathematical explanation. I was wondering if it would be appropriate to argue that the number of atoms is a very large number, and the number of orbitals is very large and approaches infinity. Here's a link: https://en.m.wikipedia.org/wiki/Electronic_band_structure Thank you for the answer.

@orthocresol Thank you for the comment. Since $\psi_3^{(0)}$ is derived from one-electron hamiltonians, there is no such term in the wave function that depends on $|r_1-r_2|$. Now, based on your argument, the wavefunction should vanish at $|r_1-r_2|=0$, irrespective of the value of $r_1$ or $r_2$. This essentially means the one-electron wavefunction should go to zero always.


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Exact meaning of "degree of crosslinking" in polymer chemistry
In the ion-exchange chromatography literature, a term "degree of crosslinking" appears frequently. Ion exchange resins are typically made from divinylbenzene and styrene. One of the stated characteristics is the "degree of crosslink" which is often stated as 5% to 55% crosslink with DVB.
The term degree of crosslink does not appear in the IUPAC's glossary; yet it is a very well established terminology. What is the physical significance of, say, 55% crosslink when the manufacturers quote this number? Is it simply the mole percent of DVB present in the polymer?
I am looking for its exact definition from a reliable source but it seems to be elusive.

Normally, in practice the amount fraction of DVB in the feed is used as a nominal number for the degree of crosslinking. IUPAC provides a definition in a somewhat unexpected place [1, p. 2352]:

Crosslinking: Property of a solid support prepared from polymeric materials with interconnected strands. Often results from the inclusion of multifunctional monomers in the polymerization reaction, e.g. divinylbenzene in polystyrene production. In such cases, the degree of crosslinking is often quoted as the proportion of the multifunctional monomer in the reaction mixture. The extent of crosslinking is important for physical properties of the solid support, such as the propensity to swell in different solvents [18].

Following reference [18], one can find further clarification [2, p. 2278]:

Care is therefore required in defining or interpreting the DVB content of crosslinked polymers since this might be quoted as a percentage of technical DVB used to make the
polymer, or the figure can be adjusted to reflect only the content of actual DVB isomers present. Since for convenience the nominal crosslink ratio or degree of crosslinking of a polymer network is often quoted as the mol% of crosslinker used to prepare the network, defining the actual percentage of DVB isomers employed is more informative since this equates with the nominal crosslink ratio (Fig. 4). Bearing in mind that there is no unambiguous method, and certainly no simple and rapid method, for determining the real crosslink ratio in a polymer network, the nominal figure based on the actual DVB feed is a
very useful parameter.

Fig. 4 Polymerisation of styrene and divinylbenzene to form an infinite polymer network.

Further, the paper lists several reasons as to why the real level of crosslinking is always accompanied with a level of uncertainty:

The effective crosslink ratio significantly decreases with greater quantity of DVB being fed due to growing number of unreacted vinyl groups.
Entanglement crosslinking occurs at higher rates of polymerization in slowly-agitated systems and results in additional mobile crosslinking.

Fig. 5 Permanent entanglement crosslink

Additional crosslinks can also be created as the result of intramolecular side-reactions.

These factors are tricky to estimate quantitatively; a method allowing to do so would be a single pulse excitation (SPE) 13C solid state NMR.
References

Maclean, D.; Baldwin, J. J.; Ivanov, V. T.; Kato, Y.; Shaw, A.; Schneider, P.; Gordon, E. M. Glossary of Terms Used in Combinatorial Chemistry. Pure and Applied Chemistry 1999, 71 (12), 2349–2365. DOI: 10.1351/pac199971122349.
Sherrington, D. C. Preparation, Structure and Morphology of Polymer Supports. Chem. Commun. 1998, No. 21, 2275–2286. DOI: 10.1039/a803757d.

Sherrington is missing out: NMR is a very straightforward way to determine the actual degree of crosslinking. Sufficiently swollen networks give useful spectra as-is, otherwise there is CPMAS. Of course the unavoidable (and extremely hard to determine) network heterogenity makes the number nearly as ambigous as the molar ratio of crosslinker in the educt. ;)

Does the paper give a reference for those "entanglement crosslinks"? I'm sure they exist, but can they be shown to exist, i.e. do they have any effect on the network?

@Karl Why, entanglement crosslinks is a well-known phenomenon, I guess, since 1950s. Also studied e.g. with NMR (https://doi.org/10.1021/ma960650g). One of the latest studies on contribution of entanglement crosslinking to mechanical properties: https://doi.org/10.1021/acs.macromol.0c00682.

Thanks! I knew they are relevant for post-crosslinked material (vulcanisation etc.), but found it hard to imagine how they occur in a solution polymerisation. On second thought, emulsion polymerisation of PS with not too high DVB content should be a nice model system to study that.

@Karl Entanglement cross links occur frequently in DNA, and there is a dedicated enzyme (topoisomerase II) to deal with them.