text:
What is the difference between a radical and a neutral alone atom?
I've done a bit of googling but can't find a clear answer to this question. Is there any difference to e.g. an alone neutral chlorine atom and a chlorine radical?

A chlorine radical and a chlorine atom are effectively the same thing. This does not mean that "radical" and "atom" are interchangeable as jargons. They are used in fairly particular contexts, namely

Saying an "atom" is a reference to an instance of a particular element. They are often discussed abstractly, in terms of their constituent parts (nuclei and electrons) without coming to discuss their  surroundings or conditions within a system.
It is natural to say molecules are made of atoms, because they conserve the total number of nuclei and electrons with respect to the individual atoms, so we can rationalize them as bonded atoms. But saying molecules are made of radicals is weird.
Radicals can (very often) be polyatomic, like a methyl radical $\ce{CH3.}$
A radical has an odd number of electrons at its outer shell, it is chemically unstable, although electrically neutral. Atoms can be neutral and chemically stable, as in the noble gases, which fully comply to the octet rule.

Thus an atom being a radical and vice-verse is more of a "chemical coincidence" than anything else, kind of like a hydrogen cation is a proton. But they are definitely not equivalent.

I believe you have a typo. They have an odd not even number of electrons don't they?

Yes, I corrected it


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Small but modest displacements at negligible forces from DFT
When looking at a Gaussian log file today, I noticed the following information in the convergence criterion section after an analytical frequency calculation was performed:

         Item               Value     Threshold  Converged?
 Maximum Force            0.000001     0.000450     YES
 RMS     Force            0.000000     0.000300     YES
 Maximum Displacement     0.000190     0.001800     YES
 RMS     Displacement     0.000030     0.001200     YES

This struck me as odd. How could the forces be so low but the maximum displacement still be non-negligible? Does anyone have any hypothesis about why this phenomenon would occur? This makes me wonder if the convergence condition for the maximum displacement is unreasonably low if essentially no acting forces can result in a maximum displacement that's merely 1 order of magnitude lower than the default threshold. I believe the force units are Hartrees/Bohr and the distance units are angstroms.

It is pretty common to see this: that is why we have four different convergence criteria in Gaussian

I get that, but I'm wondering why it occurs.

I'm not sure what units those numbers are in, but geometry convergence and its criteria are a bit of a dark art.

I assume that's a displacement in atomic units, and that's just not a very large displacement at all.

@Argon What why? Why do you expect similar force constant associated to the stretch of a strong bond and eg a rotation? There is no mystery in the fact that they can have very different forces/ displacements.

For clarity I will assume Gaussian performs a generic Newton-Raphson minimization (NR), which should suffice to explain the phenomenon.
In NR, the linear problem
$$
\nabla\nabla^\ast E \Delta + \nabla E = 0
$$
is solved, where $\Delta$ is the displacement. In order to arrive at "large" displacements despite "small" (but non-zero) forces ($-\nabla E$), it suffices for the Hessian ($\nabla\nabla^\ast E$) to have eigenvalues close to 0 itself, because a small number is divided by an even smaller number. This happens when the potential energy surface is very flat.

Thanks for the reply! I realized shortly after posting this that it was an optimization detail and came to the same conclusion as you did. Indeed, the PES is expected to be very flat in my case.

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Can oxidation of hydrogen happen in room temperature?
Here is my explanation about the reaction between hydrogen or methane and oxygen.

Hydrogen molecules and oxygen molecules can collide sucessfully and react if they have enough energy. In room temperature, molecules are moving slowly, but they do have a chance to collide, and they do have a chance to have high energy level to collide successfully. Therefore, even in room temperature, hydrogen can slowly react with oxygen to form carbondioxide and water.

And as temperature increases, the chance to collide and to have enough energy gets bigger, so rate of reaction increases.
PS: the kenetic energy of particles satisfies something like normal distribution, so molecules with high energy must exist even in low temperature.
My chemistry teacher don't agree with me. why?
Which brought me to a second question: what is the definition of ignition point(the minimal temperature required to burn) if reaction can happen just at any temperature?

@Mithoron My question is NOT about organic molecules. The link you gives doesn't explain inorganic burning!

But methane is an organic molecule.

@IvanNeretin But obviously organic or inorganic is not the main concern.

Whatever the answer remove carbon dioxide from your sentence.

What you say is theoretically correct. But if you did the calculations, you would find that the reaction rate is so slow it it would be impossible to even measure it.

Your reasoning is correct. When we say a reaction occurs only with sufficient energy (activation energy), it is truly a tail of a single (pair of) molecule. But the ensemble of molecules still provides molecules of high energy, able to react, and molecules of low energy which will collide and not react. That effect is expressed macroscopically by the reaction kinetics, with a higher rate of collisions resulting in a higher reaction rate. The kinetic effect of temperature is thus continuous and there is no single temperature at which a system has the activation energy to react (as some people believe). It is indeed true that a reaction may become spontaneous in a given direction only above a certain temperature, but that is a matter of thermodynamics and not kinetics.
Despite all of this, it sort of happens that indeed some reactions do seem to begin to take place at a certain temperature while not changing its thermodynamic spontaneity. That is because the kinetic effect of temperature is not linear, it is actually much more intense that a linear model might predict, sometimes doubling the reaction rate at every 10⁰C increase - this is captured by the Arrhenius equation. Sometimes you do have some molecules reacting, but it is so slow at a the temperature you're looking it's not even noticeable. Bring the temperature up and you will see a rapid increase in reaction rate in a short temperature interval. So it looks like it was an on/off transformation.
More often than not in chemistry, the theory predict really small quantities, but the theory is also naive to assume the molecular distribution is continuous and not made of discrete entities as we know. So it sometimes happens that the idea that there no molecules reacting is in this situations actually closer to reality than the theory suggests.
A remark: molecular energy distribution is not normal generally, but follows something like a Maxwell-Boltzmann distribution.

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How to make ammonia from urea?
The chemical formula of urea is $\ce{CO(NH2)2}$ or is sometimes written as $\ce{CH4N2O}$. How would I decompose or precipitate the $\ce{CO}$ from urea to get ammonia? I know that in the olden days people used to make gunpowder (potassium nitrate) and ammonium nitrate from human urine, which was made by reacting ammonia and nitric acid, so it should be possible. (Just as a side note: this is for a high school chemistry project, I am not attempting to make explosives out of this).

Short answer: use crushed watermelon seeds (urease)
Combine urea in solution with urease, get some ammonia with carbamate intermediary, which rapidly decomposes to ammonia and carbon dioxide.
$$\ce{(NH2)2CO + H2O ->[urease] NH3 + H2NCOOH -> 2NH3(gas) + CO2(gas)}$$
Of course the ammonia gas dissolves in the water and increases the pH.
Sources:

https://www.hemmaodlat.se/research/citrullus%20lanatus%20seeds%20as%20a%20urine%20catalyst%20for%20anthroponic%20use.pdf
https://en.wikipedia.org/wiki/Urease
https://en.wikipedia.org/wiki/Ammonia_volatilization_from_urea

The following article (available freely under a creative commons license) outlines two different procedure to decompose urea to ammonia, one via thermal hydrolysis (THU), the other using a Ni catalyst under an applied electrochemical potential:

ECS Electrochem. Lett. 2015 volume 4, issue 10, E5-E7
http://eel.ecsdl.org/content/4/10/E5.full
doi: 10.1149/2.0041510eel

The following passage describes the THU process which is carried out at high temperature in water (references and more details can be found in the article):

Warner postulated that urea hydrolysis -THU summarized by reaction 1-
  consists of two steps. The first step is the decomposition of urea to
  ammonia and isocyanic ion (Eq. 6) which is irreversible at pH less
  than 5 and greater than 12. The second step is the hydrolysis of
  isocyanic ion to produce ammonia and CO2 (Eq. 7).
$\ce{ NH2CONH2 -> NH3 + H+ + CNO- ~~} $   [6]
$\ce{ CNO- + H+ + H2O -> NH3 + CO2 ~~~}  $   [7]

First hydrolyze the urea as follows to form ammonium carbonate.
$$\ce{NH2CONH2 + H2O -> (NH4)2CO3}$$
Next decompose this ammonium carbonate to yield our required product, $\ce{NH3}$.
$$\ce{(NH4)2CO3 -> 2NH3 + H2O + CO2}$$

Please use MathJax to format the formulas. You'll see links to the right of your text box when you ask a new question.

This answer could use some more background information, like reaction conditions with explanations, instead of only reaction equations.

Does said ammonium carbonate decompose readily? Rather, does it decompose on its own, and if so, how quickly? Is a catalyst needed to speed up this process?

How would I react the water and urea together? Urea and water already exist together in urine, and they don't react to form the above compound.

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Stability of resonance structure
My book states- " the one that has ...less separation of opposite charges ,...., more dispersal of charge is more stable than others"
Aren't 'less separation ' and 'more dispersal' opposite to each other?
Please explain with an example.

Why do you think that  'less separation ' and 'more dispersal' opposite to each other ?

Re-read the statement carefully. It says that the separation of opposite charges ("charge separation") destabilizes the resonating structures. The second phrase ("charge delocalisation") refers to the delocalisation of the same charge around a resonating structure.
For example, here's an example of how "charge delocalisation" increases stability:

(source)
Notice the same positive charge is delocalised throughout the conjugated system. It reduces - the unit positive charge on a single carbon atom - to a partial positive charge over several carbon atoms in the resonance hybrid, thus increasing the stability of the molecule.
Here's an example of "charge separation":

which destabilises the molecule. Remember that uncharged structures are more stable than charged structures.
I hope it helps!


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Trend in reducing property of dioxides
I am not sure of the explanation for the following statement:

The reducing property of dioxides of group 16 elements decreases from $\ce{SO2}$ to $\ce{TeO2}$.

Is the statement valid? If yes, how do I explain it theoretically?
If not, what would be the actual trend in that case?

The statement is actually valid.If you go from S to Te, the compounds of Se and Te are stable at lower oxidation states due to inert pair effect.
So, the tendency of these group 16 elements to be in higher oxidation states decreases down the group.That means, tendency of the elements getting oxidised decreases down the group and thus their reducing property(reducing others, getting oxidised itself) decreases frm S to Te.

So this is the trend for all types of compounds formed by these elements?
I mean, whether it be hydrides, oxides, halides etc. The reducing property always decreases from S to Te ?

You don't have chalcogen(IV) hydrides.  Otherwise yes, the +4 oxidation state becomes increasingly stable vis a vis +6 as you go from sulfur to heavier congeners.  And sulfur itself tends to favor +4 at higher temperatures; sulfate salts tend to decompose forming sulfur dioxide upon heating.


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Will NaOH react with ethanol?
I am having a doubt in here that NaOH reacts with methanol but not with ethanol.
Theoretically my instructor on chemistry says that the conjugate base of ethanol would be stronger than  NaOH.
But I can see in many websites that the reaction is given(reaction of CH4 and NaOH).
Also a side question to this is CH4,C2H6 will not react with NaOH. Is there any reason other than full saturation CH4 and C2H6.

You are mixing up alcohols and alkanes (CH4, C2H6). Please get your nomenclature right. Also this is a homework question. Please show why you think one reacts and the other not.

Compare the basic strengths of $\ce{EtO^-}$, $\ce{MeO^-}$ and $\ce{HO^-}$.

Just because one can write down the reaction equation (as you have read in those websites) doesn't mean it takes place to any considerable extent. Yes, ethanol and NaOH do theoretically react as in $\ce{EtOH + NaOH \to EtO^-Na+ + H2O}$, but can you observe this reaction in practice? The answer lies in the thermodynamics of the reaction: the equilibrium is severely shifted to the left, so the reaction as written is almost irrelevant. Conversely, methanol ionization does present a measurable equilibrium constant at appropriate conditions.


text:
AP Chemistry: Sig Figs Question
This is the official answer of a problem on the AP Chemistry 2009B Exam:

The answer has two sig figs, but I think it should have 4 sig figs. I was taught that your end answer should have as many sig figs as that of the number which has the smallest amount that is used in your calculation.
When I solved this problem, I used the second approach in the image. The only number I took from the question is 10.93, which has 4 sig figs, so my answer should end up with 4 sig figs.
Is it because 14 has two sig figs? I would think that shouldn't count.

When I read this question, I forgot that the sig figs for pH is different. For pH, the number of sig figs is equal to the number of digits after the decimal.
Since the pH given is 10.93, it has two sig figs (since two digits after decimal) which is the lowest, so the answer should have two sig figs which is does.

Remark: Linking a document ("Section E") for further reading on this.


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Using Nanodrop for UV/Vis analysis of compounds at limited amounts
I am no chemist, but a biologist who knows some biochemistry. I have some natural extracts at limited quantity, which I have been evaluating using mainly GC-MS. These extracts are rich in liquid alkaloids.
I was wondering whether a trivial "Nanodrop" could help me in some analyses  -- e.g. estimating concentration of some main compound --, via UV/Vis absorbance scans. A typical nanodrop can scan for absorbance within the range 200-800 nm.
Usually molecular biologists are the only ones routinely making use of Nanodrop, strictly for DNA/RNA purity and quantitative estimations.
Has anyone here ever tried making use of a Nanodrop micro-spectrophotometre for general analytical purposes with other compounds?
[UPDATE]
I cannot add an answer as others for some reason closed it as "too broad". It isn't, see below.

After completing my tests, I have decided to come back to answer my own question.
A Nanodrop machine has served my purpose quite efficiently, as far as I can judge. I used have used it to evaluate the degree of purity of a natural alkaloid extract from insects, using synthetic alkaloid analogues as controls. See our published paper discussing the method here, along with raw data.
We have noticed no alterations nor cross-contaminations to other DNA/RNA samples running in parallel, highlighting on the fact that the equipment was properly cleaned between each use (with 100% ethanol & distilled water).
Therefore, yes: I recommend using the Nanodrop as a cheap & quick method to scan alternative biological/chemical samples, provided the chemicals and solvents involved are compatible with the pedestal according with the manual.
Hope this discussion encourage others to dare into multidisciplinary research!

Why the downvote?

I didn't down vote this, but I suspect this isn't quite the right forum for this question. A literature search would likely yield more fruitful results than waiting for a random analytical chemist who happens to have some expertise here.

Perhaps not the "right forum" because I said I am no chemist? Sorry to intrude. I don't think many chemists are in touch with biologists and their Nanodrops, based on the literature and my peers...

You didn't interpret what I said the way I wanted you to. Doesn't matter if you're a chemist or not. This isn't the right forum because you're sitting around waiting for the right chemist to come along. Not just any chemist (as you pointed out) but someone knowledgeable on what sounds like something quite esoteric. I have a Ph.D. from MIT, but I know nothing about this topic. I would do a literature search if I could, but at that point, it's easier just for you to do it yourself.

Could you tell us what a nanodrop is?

@pentavalentcarbon it is a micro-spetrophotometre for nanoliter-scale volumes. You apply a 0.5-2.0 ul droplet onto a pedestal and read out absorbance, fluorescence, transmittance, etc, data. It was released some 20 years ago and quickly got very popular among molecular biologists for (quickly) estimating DNA & RNA purity and concentrations. Possibly as a side-effect nowadays my colleagues seem to believe that it is only fit for that purpose. I'd surprised if no-one else uses it for general analytical purposes.

We got the Nanodrop One-C in our lab. We use it for example for quick determination of tetrazine concentration (strong absorbance at 535nm) or other dyes. It works well with water or DMSO but it's not very accurate in some cases. But if you don't care about some % up or down it does the trick.

I think this is a great question.  Thanks for posting it!

@DSVA You seem to be experienced with the technique. Can you please add your comments as an answer and elaborate? Thanks!

Now a literature search is possible into this experimental idea as suggested. Please feel free to peruse the cited literature in the question update. Thanks to those who helped!

In short, i would say that you cant use Nanodrop for such work. Here is why:
From a quick google search it appears that a nanodrop is a UV-Vis spectrometer that can measure very small amounts of liquid. Looks like that  bio-sciences people (I am not a biologist or biochemist) use it to quantify nucleic acids by knowing the approximate molar absorptivity for DNA, RNA etc. at a specific wavelenth, measuring the sample's absorption at that wavelength, and applying the Beer-Lambert law assuming that the only compounds that absorb in the wavelenght that measurement takes place in their sample are the nucleic acids (which i guess is a fair assumption for such samples).
However, I am not sure from your question if you want to measure nucleic acid content in a sample also containing alkaloids as part of the sample matrix or you want to quantify the various alkaloids in your sample.
I think you must mean the latter since you mention GC-MS in which case the nanodrop seems to be a poor choice: UV-Vis spectroscopy in general is a technique with very low resolution in terms of compound separation due to the broadness of the peaks in the absorption spectra. If you had only a few alkaloids in your mixture (which i dont think you do) maybe you could use nanodrop or a normal UV-Vis spectrometer followed by a lot of calibration and absorption band decomvolution in order to identify and quantify each component. If on the other hand you have a complex mixture (which is usually the case with natural extracts) any kind of UV-Vis spectroscopy will fall short. Even NMR spectroscopy which has a much higher resolution would struggle.
This is why GC-MS seems to be the best way for such work: the GC separates the mixture in simpler mixtures (or in the pure compounds if your sample is not too complicated) and then the MS identifies and quantifies the compounds present in each fraction. So i think that you will either need to find more sample or further optimize your GC-MS method to increase sensitivity.
I hope this helps.

The part about the Nanodrop is not quite correct. We got the nanodrop One-C (drop and cuvette) which can measure every wavelength between 200 and 800nm, also full spectra and even time resolved.

Ok, the model i looked up was setup with certain wavelenths only. I should have thought that there would be more advanced versions of it! I ll edit accordingly. Thank you.

Now we're talking. Thanks for your insights, though still I haven't found someone who's experimented with that. I still see the application of a Nanodrop towards my alkaloids as in the same situation of nucleotide analyses of molecular biologists. Both are natural extracts. I'd emphasise that when molecular biologists assume the only compounds absorbing in a wavelength in their sample the nucleic acids, they are usually not right. For most DNA extracts there are intrinsic and added contaminants. Likewise their molecules have many analogues like mine; I'd like to estimate [] and purity.

@DSVA Not only models, but there are clear "opensource" instructions on how to make fully functional "Nanodrops". For instance: http://hackteria.org/wiki/DIY_NanoDrop  Ultimately I'd like to have one of those, 100% mine. First I want to make sure the effort is worth it.

You are trying to estimate how much alkaloids you have in total in your extract or which ones and in what amount of each?

AMM is absolutely right here, @Scientist: UV-Vis is a terrible technique when you have a complex mixture. There's no way you'd be able to speciate a natural extract, and the complex mixture means you have no ability to even guess at a reasonable molar absorption coefficient for the mixture. AMM's suggestion of a chromatography method hyphenated with mass spec is likely your best bet.  However, your sample size may be too small for a realistic analysis.  You've posed a terrifically challenging analytical problem here.

@hBy2Py Yes I see the point, and thanks for bringing up further insights for me to consider. I however have a natural extract which is made of >95% w/w mixture of analogues of the same compound. I expect (haven't tested all yet) that all of them have the same or very approximately same peak absorbance wavelength. I believe the method is very practical, and will test it further.

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Percentage of trans-stilbene traces in cis-stilbene H NMR
This is H NMR spectrum of cis-stilbene. There are some traces of trans stilbene. I have to determine how many trans-stilbene is present, the exact percentage and I have problem with that.
I guess that it has to do with integration of peaks but I'm still not sure is trans-cis isomer ratio 1:10 or am I wrong?

Step 1: identify which peaks correspond to cis and which to trans.

Small peaks at 7,35 ppm and 7,51 ppm correspond to trans isomer and peaks at 7,27-7,15 ppm and 6,59 ppm correspond to cis isomer.

For accurate integration you also need to set the recycle delay to a longer value than usual (e.g. ~20 s), although I appreciate this may not be possible in your case.

You have to focus on the $\ce{^1H}$ of the double bond.
They are singlets because both molecules are completely symmetrical.
As per the tables, the chemical shift of the cis isomer is $5.25+1.38-0.07=6.56$ ppm.
On the other hand, the chemical shift of the trans isomer is
$5.25+1.38+0.36=6.99$ ppm.
These theoretical values correspond to the peaks of 6.60 and 7.11 ppm.
Since their integrals are of 1 and 1.1571 respectively, the proportion of trans is of:
$\frac{0.1571}{0.1571+1}100=14\%$

Instead of using tables, a better way is to simply look up the data: cis-stilbene comes in at 6.572 and trans-stilbene at 7.15.

@orthocresol Very correct, but the question looks like the OP is learning to interpret NMR, so I thought it was better to do it with the tables

Thank you very much! Yes, I'm still learning to interpret NMR. :)


text:
Is azirene an antiaromatic compound
I believe that a compound which is planar and having parallel p orbital with $4n\,\pi$ electrons is antiaromatic, but azirene has a lone pair in a $\text{sp}^3$ hybrid orbital.
Shouldn't it be non-aromatic rather than antiaromatic?

See this.

https://chemistry.stackexchange.com/questions/51467/is-the-cyclopropenyl-anion-antiaromatic

Wikipedia, in a short article, indicates that there are actually two types of azirenes:

1H-azirenes, where there is one ligand on each ring atom as shown in the question.  Since all the ring atoms have ligands the lone pair contributed by nitrogen enters the conjugated pi system, which then has four conjugated electrons and is antiaromatic.  Such a molecule is thereby unstable and prone to rearrange into the alternative structure that follows.
2H-azirenes have two ligands on one carbon, saturating that atom so we no longer have a conjugated ring.  The other ring carbon has one ligand and is still unsaturated.
The nitrogen atom is left without a ligand, and its lone pair now replaces the missing ligand in an outward pointing in-plane orbital.  The only pi bonding is an isolated bond between the nitrogen and the remaining unsaturated carbon.  These compounds are non- (not anti-)aromatic and may be isolated.


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Prevent/Slow down oxidation of organic material
I'm having the hobby to draw paintings with the tea I drink trough the day. Usually I pour a tablespoon of the hot tea away, wait until it dries and gets more concentrated, so the colour is more intense.
Most of the colours are dark greens or variants of brown. Mostly due to the oxidation.
To get blue and red I now drink tea of herbs and flowers with such colours, and have to pour the sip away early to preserve the brighter tones.
Unfortunately after some time more than 80% of the dried tea becomes brownish.
I added some cold water to slow down the oxidation, but not much success, all I got was a small blue ring around a brown area.
Next to my office is a small laboratory where I asked what to use, but no idea from that side.
As I'm not a chemist I just did a simple calculation: The opposite of oxidation is reduction, so I need some Reducing agent. Used Ascorbic acid with not huge success.
Against rust and other effects of oxidation I can use industrial alcohol, so tomorrow I'll get a bottle of that. But as said, I'm just guessing here.
What can I actually use to hold or slow down the oxidation-process until the tea is dry?

If it is just these herbal or floral teas which are oxidising you may wish to try adding some sort of colorless antioxidant compound to the teas. What I would attempt is blending your herbal and standard teas together. There are boat loads of antioxidants in tea. If your herbal teas are lacking in antioxidants, a splash of green tea may help.

Sulphites are used in wines to reduce/prevent oxidation so could be worth trying

@Waylander - thanks for the idea. Can you recommend a certain chemical?

If the original tea turns to the wrong colour - why would a tea that oxides itself support another tea which has an even stronger oxidation?

Sodium sulphite (sulfite) looks a good candidate, used in food processing so safe to ingest in small quantities

@Waylander - I give it a try


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Weak adhesives, has some critical chemical been banned?
I notice that in the last 10 years or so, the adhesives I encounter in everyday life have gotten noticeably weaker, is this because some critical chemical has been banned or regulated out of consumer products?
For example, the typical duct tape, packing tape, masking tape, and scotch tape that I can buy in stores now are all noticeably weaker and less adhesive than the same products 10-15 years ago. For example, duct tape used to a have thick, white adhesive that was extremely sticky, but duct tape that you would find in a hardware store now has a clear adhesive that is much less sticky.
I know that there other consumer products that have become inferior due to chemical banning; for example most plastic wrap is now made of HDPE plastic instead of the original polyvinylidene chloride, so it is greatly inferior to the original Saran wrap. Has the same thing happened to tape adhesives? If so, what is the operative chemical that has been banned?

How do you know that in the past tape was really stronger? How do you know that this is not just an impression you may have?


text:
Possible to Calculate Electronic Charge Distributions/Densities for Organic Molecules?
I was wondering if there are any programs that can calculate the electronic charge distributions around an organic molecule, possibly in the form of a volumetric charge volume density? I am reading up about using DFT programs that (correct me if I'm wrong) I believe may be able to tell us something about finding the charge density.
I'm working on a project that involves obtaining the electric field values surrounding an organic molecule. My current approach involves finding a function for the molecule's electronic volume charge density and doing a volume integral to get the electric field.
Is this a potentially valid method? If not, what else should I look into?

Even I'm searching for it.

It's possible to calculate the electrostatic potential at points around a molecule. Is that what you're after?

The problems is that programs let alone do very little.  Yes they do exist and for sure there are specialists or  chemists knowing enouggh to help, here. .

Every DFT program computes the electronic density, i.e., the negative charge density due to electrons. The positive charge density is a sum of Dirac distributions over $\mathbb{R}^3$ given by the nuclear positions and their charges. Are you looking for the electrostatic potential? If you have Gaussian and molden, you can visualize the electronic density and ESP given the right input.

It is not clear what you want to analyze, so you may want to check out electrostatic potential maps: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Electrostatic_Potential_maps

If I understand correctly, you want the total electric field at some point $\mathbf{r}$ due to a molecule. This will have at least three components. The first due to the electric field from the nuclei. This can be calculated classically from the field due to point charges. The second due to the dipole moment of a molecule. This can also be calculated completely classically (though you will have to use some electronic structure calculation to get the dipole and direction). Finally, you will need to find the total charge density of the electrons.
The charge density can be found using either DFT or any post-Hartree Fock method such as MP2 or CCSD(T), etc. When using something like MP2, you will get the charge density from the molecular orbitals. You can do this manually, but one code I know of that will give you the relevant information from the HF orbitals is NWChem. You will want to use the dplot command. This allows you to print out electron densities for each orbital using a grid. You will want to use a fine enough grid to capture the actual spatial changes in density. Notice that you also provide limits i x, y, and z. This is because, in principle, the charge density extends to infinity in all directions. Thus, you will want to choose a cutoff where the density is sufficiently small as to be negligible. Then, you can again calculate the field due to the electron density from classical equations which can be found online or in an E&M textbook. You can either calculate the field discretely from the points on the grid, or fit it to some functional form and then do a true integral over the density.
Another way to do this is to use a code which will give you approximate charges per atom from some type of population analysis such as Mulliken charges. You can do this with Gaussian using population keyword. This will aggregate the nuclear charge and the electronic charge into one field. Fair warning, however, Mulliken charges (and all population analyses) are approximate and not even necessarily physical.
There may be other ways to do this from DFT, but I surprisingly cannot find any software packages that do this directly.

We've decided to follow your suggestion and try out the DPLOT functionality of NWChem, and I had a few followup questions about NWChem specifically:

Is there a way to be able to visualize the electron density from the output Gaussian file? We have tried avogadro, but it seems tojust show a ball-and-stick model of the molecule.

Is there an easier way to input the structure/geometry of the molecule for NWChem? Any way we can translate a typical 3d model of a molecule into the input for NWchem?

Thank you so much for the pointers, it has been of great help to us!

@phill I have only ever used the molden option to visualize orbitals using Molden. Molden will then calculate electron densities from these orbitals. I do not know what you mean by your second question. There are various programs which output xyz coordinates from a ball and stick model and then these can be used in a geometry optimization. I would suggest asking another question about this on SE as I usually don't use these programs but I know there are people here who could answer that question.


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What metal is used to back this mirror?
My father is attempting to remove the mirror from a piece of glass. We stripped the black paint from the back and exposed the shiny surface with acetone. In the past he has been able to remove aluminum mirror with lye, but that didn't react. We also attempted bleach in case it was actually silver but that didn't react either. Is there another metal that could be used to back a mirror? Are we doing anything wrong? Also adding hydrogen peroxide to the lye made the mirror bubble but no black silver oxide or white aluminum oxide.

I'm not sure if this will help, but https://en.wikipedia.org/wiki/Mirror#Types_of_glass_mirrors has several other types of glass mirrors in addition to aluminum and silver backed mirrors.

Try to use HNO3 on it... Or maybe there is still a thin plastic coating over the metal that was resistant to the acetone...

The back side of a mirror is typically made of glass, which is coated with a thin layer of metal (such as aluminum or silver) to create the reflective surface. This reflective coating is applied to the back of the glass, and it is this side that is usually polished to achieve a smooth, flat finish.
Removing the reflective coating from the back side of a mirror can be done through a chemical reaction that dissolves the metal layer. One way to achieve this is by using a mixture of nitric acid and hydrochloric acid, also known as aqua regia. This mixture can dissolve many metals, including the ones commonly used in mirror coatings. You can learn more about mirrors here.


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Why are ammonium chloride and ammonium hydroxide used in calcium salt detection?
On adding $\ce{ (NH4)2CO3 }$ solution to $\ce{ Ca^{2+} }$ aqueous salt solution, $\ce{ CaCO3 }$ is formed as a white precipitate, which confirms the presence of the cation.
While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ Ca^{2+} }$ it read:

Experiment: To the original solution added a small amount of $\ce{ NH4Cl }$, and an excess of $\ce{ NH4OH }$, followed by solid $\ce{ (NH4)2CO3 }$
Observation: white ppt

Why do we need to add $\ce{ NH4Cl }$ and $\ce{ NH4OH }$?

You want to fix the carbon in solution so it doesn't volatize as $\ce{CO2}$, since the equilibrium $$\ce{H2CO3 <=> CO2 + H2O}$$ favours the product side. By adding a base ($\ce{NH4OH}$), you shift the water ionization equilibrium, which in turn shifts the carbonic acid dissociation equilibrium. $\ce{NH4Cl}$ serves to regulate the $pH$ as a buffer. With your carbonate in solution, it is then able to precipitate as a solid by adding calcium.
Using a strong base would hinder the solid precipation by favouring the solubilization of ions, justifying the use of ammonium hydroxide.

I am unable to understand the equilibrium you've mentioned. I have two doubts: (1) Adding $\ce{(NH4)2CO3}$ to the solution produces $\ce{CO3^{2-}}$ ions and not $\ce{H2CO3}$, so why have you taken the latter in the LHS of your equilibrium? (2) Re: "favours the product side" In any aqueous solution, water is in excess. So, shouldn't the equilibrium favor the reactant side instead? Thanks for your answer!

When we're dealing with equilibrium chemistry, having a certain species in solution means having all species that the particular species will allow from association or dissociation with other ions. The amount of each species is then dictated by the set of equilibrium constants.  Regarding the excess of water, it is true that water always wins the concentration fight in solution as a solvent, but it wins so hard it doesn't even enter the discussion: water is ommited from the equilibrium constant expression because its concentration remains essentially constant.

Carbon dioxide still prevails with respect to carbonic acid: put a $\ce{CO2}$ atmosphere over liquid water and much of it will dissolve and convert to the acid form, but most of it will stay as $\ce{CO2}$.


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How to define equivalents here

$3g$ of activated charcial was added to $50 ml$ of acetic acid solution ($0.06N$) and filtred after an hour and it was found to be $0.042 N$ .The amount of acetic acid adsorbed (per gram of charcoal) is

The solution for this question is exactly given as I have written below
No of equivalents before adsorption $=50\times0.06 =3$
No of equivalents after adsorption $=50 \times 0.042=2.1$
No of equivalents left $=0.9/1000 \times 60=54mg$
Acetic acid per gm $= 54mg/3=18mg$
Now I have some doubts regarding the question and it's answer

Why is normality being decreased here?
Why are they taking volume to be same when it generally increases when we add solutes?
What does equivalent mean here? According to Wikipedia, it is said to be "the mass of one equivalent, that is the mass of a given substance which will 1)  combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valence; or  2) supply or react with one mole of hydrogen cations (H+) in an acid–base reaction or 3) supply or react with one mole of electrons (e−) in a redox reaction." The equivalent talked in the question doesn't meet any of this criteria so what is its role here. Can somebody please explain me this?

Number of equivalents and normality are out-of-date concepts in chemistry. For acids, they are just like number of mols and molar concentration (or molarity), with an extra factor indicating how many hydrogens can be effectively ionized in solution. Because acetic acid ($\ce{CH3COOH}$) is monoprotic, you can just exchange the terms by number of mols and molarity and you are good.

Why is normality being decreased here?

Some of the acetic acid adsorbs - meaning it bonds - in the charcoal. So the concentration in solution is decreased. Pay attention to the fact that the charcoal remains solid and does not dissolve.

Why are they taking volume to be same when it generally increases when we add solutes?

As I wrote above, the charcoal stays solid, so it has no effect on the solution volume. The loss of acetid acid in solution might have a slight effect on volume, but that is mostly negligible.

So as long as charcoal is solid and not dissolved in the solution,the volume will remain same?

You should question your own intuition. What happens to the volume of a liquid when you add a solid object into it?

If it is floating then no change and if it sinks then the solution will rise but it will still remain same in my opinion

The liquid level will rise but the volume will remain the same as you say (like if you add the same volume in a narrower container) so no problem here. The more interesting question would be: what if the solid can dissove in the solution or if it is a liquid that is miscible with your solution, then what happens with the volume? How would you calculate concentration and prepare the solution in practice?

if the solid is dissolvable then some of bonds of the liquid will be replace by the bonds with the solid so volume will change at that case


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Can carbon be the central atom of a complex? Can it form a quadruple bond with itself?
Why does $\ce{[C(NH3)2]}$ not exist as a complex? Here, $\ce{C}$ is carbon and the configuration of $\ce{C}$ in ground state is $\ce{[He] 2s^2  2p^2}$.
Can't the two $\ce{NH3}$ molecules pair the two $\ce{p}$ electrons of carbon and form coordinate bond with carbon?
Why does $\ce{C2}$ (one carbon atom bonded with other through quadruple bonds) not exist? Two $\ce{2p}$ orbitals of $\ce{C}$ should form π bonds and one $\ce{2p}$ and one $\ce{2s}$ should form σ bonds with the other $\ce{C}$ (without hybridisation).
I'd appreciate an answer for a beginner.

https://chemistry.stackexchange.com/questions/594/bonding-in-diatomic-c2-a-carbon-carbon-quadruple-bond?s=1|39.4189

Asking multiple unrelated questions at once is considered a bad practice, but let's forget that for a moment.
Your first molecule has more or less satisfactory electronic configuration, but that's not enough for the long-term survival. N would be more than willing to relinquish a proton, and C to accept it, and thus in no time you will have $\ce{H2N-CH2-NH2}$, which then may or may not transform into something else.
Your second molecule has been discussed here before on multiple occasions. Long story short, it does exist, but doesn't have a quadruple bond. (Such bonds are known, but not in $\ce{C2}$; they have a delta component, and that requires d orbitals). The problem with your description is that the sigma-oriented 2p orbital of one carbon will interact not just with the similar 2p, but also with the 2s orbital of the other. If not for this fact, it would work out pretty much as you expected.

We may regard oxyanions as complexes with oxide or oxo ligands; indeed such an approach may offer advantages in understanding the electronic structure of some of these ions.  In that sense carbonate ion is indeed a complex with carbon as a central atom.


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How could I easily quantify the amount of EGCG in a sample of green tea?
I am currently conducting an experiment to study the effects of pH on the rate of thermal decomposition of Epigallocatechin Gallate (EGCG) in green tea.
One of the key steps in my procedure is quantifying this EGCG content in the green tea. What are some of the methods which I could use to determine and quantify this EGCG content?

To analyze specifically this component, you will have to use HPLC.
In case you can use a more generic approach, you can analyze the total phenolic content with the Folin-Ciocalteu method.
The Folin-Ciocalteu method is based on a reaction with polyphenols that gives a blue color.
This color can then be measured in a UV-Vis spectrophotometer and the concentration can be established.
This method has been used to quantify polyphenols in many plants, as for example here.

Can I get a more detailed breakdown of the Folin-Ciocalteu method? There isnt a HPLC in the lab im working in

It is based on a reagent for polyphenols and a UV-Vis spectrophotometer. But remember that it will give you the concentration of all the polyphenols not only ECGC. I shall edit to provide e reference

Thanks a lot. I read up on this method previously and one of the things I could not get my head around was the reason for the addition of Sodium Carbonate to the solution.

If it has been useful you might consider accepting the answer


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Is it right to compare the molarity of a substance with the probability of its molecules reacting?
I was watching this video on Khan Academy (about the equilibrium constant), and the analogy that they made really got me thinking. I wanted to verify whether it was completely correct, or just an analogy to explain reaction rates to high school students. I'm well aware that this is an equilibrium constant, and the probability of molecules reacting relates to kinetics. However, this is how it was explained in the video, and I'm trying to make sense of it.
Would anyone please guide me as to whether this analogy, of the molarity and the probability of a molecule reacting, is really true and mathematically correct or whether its just a simplification for students? (I'd recommend that you watch the video first before trying to answer.)
Also, please note that I'm not saying that what Khan Academy is teaching is wrong, I'm just trying to clarify whether what they're teaching is a simplification, or a scientifically accurate explanation.

Probability of colliding and reacting is indeed higher as concentration is higher.  Note: do not mix thermodynamic (equilibrium of the reaction) with kinetics (how fast equilibrium is achieved) as it seems you perhaps do  so comparing title and body of the Q.

As Alchimista points out, two different principles are at work here. Concentration, or how many of the species are available, and are these species able to react when they do meet.
The transcript of the video at 3:32 states:
'some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent on them just being there. You have to worry about their kinetic energies...'
So, Molarity will not affect the probability that any two molecules that meet will react, but it does absolutely affect rate of reaction. If the reaction is possible (there is sufficient activation energy) then the more concentrated the substance, the more likelihood of reactable species meeting, and therefore a greater rate of reaction.

In this video the author is first explaining the origin of the Keq constant based on the rates of the forward and backward reaction and then moves on to explain why the rate law expression is of the form rate=k[A]$^a$[B]$^b$.
No problem with the first part but disagree with the second especially when he says that [A]$^a$[B]$^b$ expresses the probability of having a of A molecules and b of B molecules in a small area at the same time:
Rate = kProbability(Having a of A molecules and b of B molecules in a small area at the same time)
I am no mathematician but the probablity of anything has values between 0 and 1 which is obviously not the case for [A]$^a$[B]$^b$. Only this makes the video wrong.
He also says that:
"The probability of having V molecules in some volume = [Concentration]xvolume"
which is incorrect as well. In a solution of constant concentration you will always have (Conc. x volume) amount of molecules evenly spread in that volume. There is no probability involved. If there is it would be the probability of having (Conc.x volume) molecules of compound in that volume which would be 1 since that is how we define concentration.
In order to understand the form of the rate law equation you should look into collition theory (e.g. http://umich.edu/~elements/03chap/html/collision/index.htm) where you will be able to see exactly how the probability of two coliding molecules reacting or not is properly used.
Note that that the probability of two molecules colliding and reacting is higher with increasing concentration, as Alchimista commented, and this is one of the principles collition theory relies on but not in the way it is presented in the video.

I think that the video is getting at 'In order to have a reaction occur, particles need to actually collide, which means be in a region of the overall volume.' So, the greater the concentration, the greater probability of them being in the area of interest. Imagine a 1 mm cube that we can observe in great detail. And I think 'relative probability' is what is meant.

What you say is correct in qualitatively explaining why increasing the concentration of the reactants of an elementary reaction will lead to an increase in rate. I appreciate the attempt in the video and that it has some parts that are valid, but you can not derive the rate law equation in the way it is described there. It seems like false logic to me and anyway there is already collision theory that does a fine job!


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reactions of tertiary alkylhalide with aqueous potassium hydroxide?
When a tertiary alkylhalide reacts with  aqueous koh then why #E2#reaction occurs .As tertiary alkylhalide undergo #E1#reaction What makes the difference?

May be ,but I can't understand from it that what happens on reaction of alkylhalide with aqueous koh? because its reaction with alcoholic koh is only what I get

There is a significant difference between two mediums.
For Substitution reactions we need a $Good Nucleophile$, not a $Good Base$.Because we want the attacking species to preferably attack the electrophilic atom(In most of the cases it's Carbon).If it's a good base it will take the proton attached to the Carbon which will not help the Substitution reaction. The medium is aqueous KOH(amount of water is large compared to OH-), in which $H_2O$ is the dominant Nucleophile,as it is not so strong base,so it attacks the Carbon and in the deprotonation step, OH- acts as a base to take H+(as, water is a weak acid, so, conjugate OH- is a strong base).

Now, if you consider alcoholic KOH medium(generally $EtOH$ medium),There is predominant OH- and little amount of $EtO-$ which are both Strong bases. For,the eliminantion reaction we need specifically a proton acceptor. Thus, highly basic medium supports elimination.
Now, if we consider Substitution and Elimination in case of Tertiary alkylhallide, there is always competition between $S_N1$ and $E_1$ .So, the reaction medium and maintaining $Temperature$ is very important. If your solvent medium acts as a good Nucleophile(bad base) then $S_N1$ will occur.But if your medium is highly basic,$E_1$ occurs.
Sometimes, there are solvent mediums such as $MeO-$ or $EtO-$( in some cases) which can act as both Nucleophile and Base.Then,
lowering the temperature follows $S_N1$and at higher temperature $E_1$ occurs for tertiary alkyl hallides.

Ok ,I understood what are you saying but there is one doubt in my mind ;Can tertiary alkylhalide undergo $E2$ reactions ,by reacting with aqueous koh? As my book says so?

Aqueos KOH is a solvent which is good at nucleophilic substitution, you need a basic medium. So, I think it's difficult to carry Elimination. More over, In polar solvent Tertiary carbocations are highly stable.So it will definitely choose $E_1$ reaction pathway. Aqueous KOH is also not suitable for $E_2$. You need a highly non-polar basic solvent to do that ,but it is still difficult.


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Is pyridine a reducing agent? If so, is it a strong or weak reducing agent?
My question relates to Darzen's procedure to prepare monohalides by reacting an alcohol with thionyl chloride in the presence of pyridine.
Pyridine is the reason why this reaction follows an SN2 mechanism. That means pyridine is a reducing agent since it abstracts the hydrogen from electropositive oxygen (while the reaction is in process) before the electronegative chlorine (from $\ce{SOCl2}$) can. So my question is,

Is pyridine a reducing agent? If so, is it a strong reducing agent or a weak one?

Pyridine does not act as a reducing agent here. Neither does anyone else. This is not a redox reaction.

Pyridine has two roles in this reaction but neither of them are as reducing agent.
Firstly it mops up $\ce{H^+}$ (it does not deprotonate OH, it is not a strong enough base for that).
Secondly it acts as a nucleophilic catalyst by reacting with thionyl chloride to give $\ce{[pyridinium - SO2 - Cl]^+}$  which is more reactive than thionyl chloride towards ROH.
http://www.chemtube3d.com/Pyridine%20-%20Nucleophilic%20Catalyst.html

That made a little bit more sense, thank you!


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Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)?
Why 'Enthalpy change' (∆H) is equal to 'Heat transfer at constant pressure' (Qp)?
∆H = ∆U + ∆pV, here only expansion work done by the sustem is added. If non expansion work is done on the system then it should be,
∆H ≠ Qp.
What's wrong? Explain.

Comments are not for extended discussion; this conversation has been moved to chat.

Enthalpy is a intensive property of the material that has nothing to do with the specific process that the material is subjected to.
It just so happens in a constant pressure condition, involving only P-V work, that the change in enthalpy is equal to the heat added.
If non expansion work is done on the system, then ∆H≠Qp. It should be mentioned while writing '∆H = ∆U + p∆V =Qp', that 'it is applicable when only p-V work is done'.
Thanks to @ChesterMiller

Why are they equal? $$dH=TdS + VdP$$ but $dQ_{rev}=TdS$. Thanks in advance.

In,Thermodynamics,Work is defined as
$W$= $-\int_{V_1}^{V_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V_1=V_2$.

Coming to the enthalpy,you are right.
$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,
$\Delta$H=$\Delta$U + $P$$\Delta$V + $V$$\Delta$P.
At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\Delta$V.But,according,to first law of Thermodynamics,$Q$= $\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So,$W$= $-p\int_{V_1}^{V_2}dV$= -$p$($V_2-V_1$) = -$p$$\Delta$V.So, first law in constant presuure becomes,$Q_p$= $\Delta$U - (-$p$$\Delta$V)= $\Delta$U+$p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.

Your formatting is terrible. Use correct punctuation and check out https://chemistry.meta.stackexchange.com/questions/443/should-we-correct-math-formatting


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What is the greenish substance formed after the electrolysis of two graphite electrodes in a sodium sulphate solution?
Today at school we tried the electrolysis with two graphite electrodes in a sodium sulphate solution. We put the electrodes in two test tubes and filled them with some more sodium sulphate so that the solution could react with the electrodes and after that, we quickly turned them over and put them in a beaker filled with sodium sulphate. After that, we added some bromothymol blue in it to see where the hydrogen and oxygen would have gone. Afterwards, we turned the power on and saw that it was producing gas and everything was normal at first (1st image).
Some time after that we didn't see the volumes between the two test tubes changing so we waited some more. Later, we saw that the solution started assuming a green color and something started forming on the surface (2nd image).
It became more and more green so we decided to turn the power off. Then we noticed that on the bottom of the beaker was something yellowish and brownish (3rd image). We checked if the wires were consumed by the solution or if they had reacted with it but nothing was wrong with them. I have no idea what it is and my teacher told me and my lab group to find an explanation to that. What is the problem here?
List of the materials used:

1 beaker
2 test tubes
two electrodes connected to wires (graphite)
$\ce{Na2SO4}$ (sodium sulphate)
bromothymol blue

I think that what possibly happened here is that the copper wire that connects the cable with the graphite electrodes is exposed in the solition (if you look carefully in Fig.1 there seem to be a lot of bubbles forming near the red cable insulation at the bottom of the tube which escape out of the tube). So copper is oxidising to give copper cations which give the coloured material in the solution.
This is just a theory though. To confirm, repeat the experiment first with copper electrodes then with your graphite electrodes but ensure that only the graphite is in the solution (so get rid of the tubes). If using the copper elecrodes gives the same result as above (blue green stuff) while using the graphite electrodes just gives O2 and H2 it means that the above theory is likely. Finaly you could isolate the greenish solids by filtration, dry them and do a flame test. If the flame becomes green it suggests presence of copper further confirming the above.
The only thing that doesnt fit with the above is that Cu2+ ions are blue and soluble in solution not green, so maybe the green stuff also (or only) result from some reaction with the indicator. Hence, it would be good to repeat the experiment with and without the indicator to see what happens (that makes 4 experiments in total).

There are hydrated forms of cupric sulfate that are green and insoluble and there is probably some black copper oxide in there as well.  There may be some iron as well in the connectors.


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Sigma and pi coordination in complexes (symmetry)
According to the Miessler Tarr and many molecular orbital diagrams for octahedral complexes on the internet will σ-donors/acceptors only influence the $e_g$-orbital, while π-donors/acceptors influence the $t_2g$-orbitals. This can also be shown using the corresponding character table, but it's been a while since I used these, so I will just believe its correct. Still, I was wondering, is there no case of it ever mixing, for example, that a π-donor/acceptor does not coordinate with both lobes but in a σ-fashion? I thought about this case, and there is also an example mentioned in the book, where they show how a σ-donor and π-acceptor-ligand (I suppose they mean one with both abilities) would look like. Here they draw the lines from the π-orbitals or rather π*-orbitals in this case to the $t_2g$-orbital and the σ-orbitals are connected to the $e_g$-orbital.
One example where I know this behavior, just vice versa would be chloride as a ligand, and I have also seen schemes, where it was drawn to coordinate in a π-fashion to the metal's d-orbitals. But in this case, you have both properties. What about a hydride for example, as pure σ-ligand? Would it only influence the $e_g$-orbital or is there a way it could also somehow weakly influence the $t_2g$-orbital?

Of course - many ligands have both sigma and pi effects. Nearly all ligands do, in fact.

Yes, but would a pure σ-donor/acceptor be able to also perhaps really weakly influence the $t_2g$ orbitals if it had no π-abilities at all?


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Lowering the boiling point of wax
Wax comes in all kinds (paraffin, bee wax, palm oil wax, gel wax, ...) but since it's usually a mixture of different carbonated chains, its physical properties vary a lot from one manufacturer to another.
A good candle is usually one made of a wax with a high melting point, such that it takes longer to burn.
I am trying to make a bad candle, more specifically I am trying to make a candles that produces as much wax vapor as possible. To make it happen, I am looking for a wax that has the lowest possible boiling point.
How can I lower the boiling point of a given wax (paraffin wax for instance since it's the cheapest and easiest to buy) by adding other substances inside ? One can easily find on the internet how to increase the BP (by adding stearin for instance), but not the reverse.
Ideally the additional substance should not be added in too great proportion, since that would lower the wax composition and thus the amount of wax vapor emitted.

How about mixing your paraffin wax with some paraffin oil which in practice is liquid wax (just shorter parafinic chains) in the appropriate ratio to get whatever melting point you like?
The boiling point and the amount of wax vapours is a different story since as Karl said wax is a mixture so even if it could be boiled (most likely under vacuum, otherwise it will autoignite) it will fractionally distill so it wont have one boiling point really. Also the amount of wax vapour will be small in all cases since the wax will burn to CO2, CO and H2O (however you could have some lower boiling point hydrocarbons flying around if you mixed your wax with them but that wouldnt be wax vapours then).
If you really wanted to lower the boiling point of your liquid you would need to mix it with something that will form a low boiling azeotrope mixture. No idea what that would be for wax.
And lastly, taking James' thought below a step further, if you would like a really really "bad candle" just use one of those old oil lamps. The way they work is exactly the same as a candle but the waxy bit is a liquid so it needs to be in a container. Then you have a wick just like candles do. Only difference is you dont need to melt the wax first in order for it to be sucked in the wick since it is already a liquid.

Whenever you just mix two substances (dissolve in each other), you lower the melting point and raise the boiling point. That's a fundamental principle. If it was the other way round, the substances would not mix, or at least not mix well.
(Azeotropes of compounds with very different polarity (water/ethanol) are the exeption confirming that rule, but the acchievable raise in Bp. won't be large. And I doubt you find one you can make a candle from.)
That sounds bad, but: Paraffin wax already is a mixture, so the desired effect might be reached by fractionating it, instead of adding sth. else.
You might try to slowly crystallise a molten sample of paraffin wax, and pour off the residual liquid when the sample is about half solid. Now the residual solid should have a higher melting point than the original wax.
(Note: This is purely my educated guessing, no idea if the effect is large enough or if some secondary kinetic effect might completely preclude it.)

Two points: melting point and boiling point. Hmmm. If you lower the boiling point, you will likely also lower the melting point, so your candle will be really bad - it will drip away.
Well, let's take a low melting "wax", like petrolatum (Vaseline). Mix in a stiffening agent like fumed silica or cellulose fiber (it will burn off when the flame hits it). The wick should be thick to hold the melted petrolatum.
This might resemble a torch, with a huge flame. To moderate it somewhat, you could blend paraffin with the petrolatum.


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How to determine if a reaction proceeds via SN1, SN2, E1 or E2 mechanism?

  Give the main product and reaction type: SN1, SN2, E1 ,  E2.

As the 1-bromohexane is primary and the nucleophile a strong unhindered base, the reaction should be a SN2 reaction. However, the solvent $\ce{EtOH}$ is a polar protic solvent which favours SN1/E1 reactions. In addition, the temperature of the reaction is quite high and high temperatures favour the formation of elimination reactions, thus the reaction should be E1.
But an E1 doesn't require a strong base, or else an E2 reaction would take place at the beginning. So does that make this reaction an E2 reaction? Am I thinking correctly on how to consider all the different parts of the reaction?

$\ce{EtO-}$ is a strong base, as well as a strong nucleophile, and the temperature is not too high. Therefore a mixture of products is most likely to be formed, where substitution will lead to the major and elimination will lead to the minor product.
Also, the reactant is a primary halide. So, in substitution, SN2 will be favoured, whereas in elimination E2 will be favoured.
So, final product will be,

Ethoxy hexane (via SN2) (major)
Hex-1-ene (via E2) (minor). (Maybe, a very little amount of hex-2-ene may also form by very very small fraction of E1).

Why is Sn2 major? I'm having difficult time understanding how a polar protic solvent can result in a Sn2 reaction. Isn't elimination reactions favoured in polar protic solvents?

@J.Se take a look at this, will try to post a bit more detailed explanation.

As a general rule, carbocations do not form in the presence of strong bases.  This means that $\ce{S_N1}$ or $\ce{E1}$ mechanisms are ruled out immediately.
It is important to look at all of the factors that are at play in the reaction: the alkyl halide, the nucleophile/base, the solvent and the temperature.  It is the cumulative effect of these that determines the major product, not any one of them alone.


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Why did permanent marker come off of glass after rubbing it with my fingers?
No, my fingers don't look stained but I accidentally made a stroke of permanent marker on glass and I was looking for nail polish remover to use, couldn't find it. So I started rubbing the stroke of permanent marker with my fingers and it started coming off. I was able to get it off with 3 fingers, with the thumb getting most of it off.
I wonder, is it because permanent marker dissolves in skin oil that I was able to get it off? Or was it the small amount of sweat? Or both? Or was it just that I was rubbing my fingers on the permanent marker and the shear force got it off?
I mean yes, this was just a few minutes after the accidental stroke that I got it off with my fingers but that would have been plenty of time for it to dry, I think.

What kind of permanent marker?  The type of ink deposited makes a great deal of difference

It is a Sharpie permanent marker that I accidentally got a stroke of on the glass and was able to get off by rubbing with my fingers.

Glass is a very smooth surface and therefore things adhere to it less effectively such as the ink in your permanent marker. Maybe your marker is not of a very good quality or maybe you didnt leave it long enough for the ink to stabilize on the surface.
Why dont you conduct an experiment to try your various hypotheses by varying one factor while keeping the rest constant? For exaple:
1) Try different brands of markers on the same glass surface leaving them to dry for the same time and then rubbing them in the same way.
2) Use the same marker but leave it to dry for different amount of time and then rub it in the same way each time.
3) Use the same marker on different surfaces and rub in the same way.
4) Try with different amounts of dilute saline solution to simulate sweat or cooking oil to simulate skin oil etc.
etc.
Or even better: Try at glass surfaces that have been etched with different sandpaper grits.
Your experiment is reproducible though, i tried as well and small amounts come off easily.


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How are these molecules chiral and achiral?

I always thought that a molecule couldn't be chiral if it had a plane of symmetry. Doesn't the chiral molecule here have symmetry diagonally across the OHs?
The achiral one doesn't look symmetrical to me, so I also don't know how it is achiral.

Both of your structures (Fischer projections) are chiral. Read this to help you understand such structures. https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

The one on the left has point symmetry (in the Fischer projection, not in reality), but point symmetry anyway is not relevant for chirality.

Regarding the first molecule: I understand that the simplistic Fischer projection - with all its bond depicted by solid lines - can be very confusing at first glance in determining optical activity. You may believe it to be optically inactive, hoping for a center of symmetry to exist, but it is not so.
You must keep in mind how the molecule actually exists in 3d! Recall the rules of Fischer projection, the horizontal bonds project outward, while the vertical bonds extend inward. Hence, the molecule would actually look like this:

Now, I hope you can tell, there is neither any center of symmetry nor any plane of symmetry here. Hence, the molecule is chiral.
PS: Your second molecule is chiral by similar logic, as also mentioned in the comments, and as you yourself noted.


text:
Are all degenerate d-orbitals identical?
This discussion spun off from the comments to another question. The basic idea of that question was that electrons don't have a preferred order of filling in $p$ orbitals, i.e. the first electron will fill in the $p_x$ orbital just as well as it will fill in the $p_y$ or $p_z$ oribtal. This is because in the absence of electromagnetic fields, all p-orbitals are degenerate, and the coordinate axes are things we arbitrarily assign, so, the degenerate $p$ orbitals are all identical.
I wonder if the same can be said about $d$-orbitals. Four of them ($d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{x^2-y^2}$) feel identical to me, as they all have four perpendicular and coplanar lobes (but I could be wrong). The fifth one - $d_{z^2}$ - doesn't look identical to me. Its positioning of the lobes is wildly different from the other four.
This makes me wonder,

Are all degenerate $d$-orbitals identical?

By identical, I mean that the electron filling order has no preference. (but this might not be the best definition of identical orbitals please mention if you have a better one)

Remarks:

I didn't mention $f$-orbitals because they have even ridiculous orbital structures (4 types have six lobes, 2 have 8 lobes, and 1 seems to be a advanced version of $d_{z^2}$ with two rings. Crazy!), but in case their answer is remotely similar, please consider mentioning them in your answer for completeness.
I have talked about atomic orbitals and not molecular orbitals (just in case that made the answer different).

Yes, they are identical.
One thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.
$p_{x}$ has the same symmetry as the function $f(x,y,z)=x$.
Likewise, $d_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.
What about $d_{z^{2}}$? You should note that $d_{z^{2}}$ is really $d_{2z^{2}-x^{2}-y^{2}}$.
Imagine taking $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.
What you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.
Why don't we just use $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other "normal"-looking orbitals. $d_{x^{2}-y^{2}}$ summed with $d_{z^{2}-x^{2}}$ is $d_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See here.
You don't seem to have a problem with $d_{xy}$, $d_{zx}$, $d_{yz}$, and $d_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.
EDIT:
Based on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.
This analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.

Thanks for your answer! That's quite a lot of things you touched upon. I tried to interpret them, please let me know if I am wrong. 1. $f(x,y,z)=x$ represents $p_x$ orbital and it is symmetric about $y$ and $z$ axes. 2. Renormalization on Wikipeda refers to QFT, QED and perturbations makes no sense :( Do you have a simpler alternate word/meaning for what is its importance in this context? 3. "It's really two of the other looking orbitals put together." Are they joined head-to-tail? Why don't they have their tail at the origin (like all others)?

This doesn't appear to actually answer the question "does electron filling order exhibit a preference for particular d-orbitals?", or at least not explicitly.

$f(x,y,z)=x$ only has the same symmetry. That is not the $p$-orbital. Normalization simply means that the sum of the square of the wave function over all space should be 1. If I add two orbitals together, I need to normalize again because the sum of the squares of the sum over all space is probably not still 1. I don't understand what you mean by "tail" or "head-to-tail"... @GaurangTandon

@OrangeDog That's true, but the OP seems to think $d_{z^{2}}$ is different but doesn't have an issue with the other 4 being degenerate. I was hoping that by showing that $d_{z^{2}}$ is exactly like the other 4, that disconnection would go away.

@Zhe then your answer would be greatly improved by starting it with Yes.

@OrangeDog I added something at the end. Thanks for your feedback.

@Zhe Regarding "head -to-tail" The $d_{z^2}$ has a torus. You said that "it's really two of the other looking orbitals put together." But, orbitals in general have an appearance like this https://i.stack.imgur.com/c65Yg.png The torus neither shows the tail nor the head. So, how do you justify this?

@GaurangTandon Remember that the orbital is a not a solid blob with a boundary. The surface you look at is the boundary surface drawn at some probability threshold. If you sum the wave functions from two of these orbitals at 90 degrees to each other, you get a torus.

Two things: (1) could you please add this to your answer above. It will make sense to many more people then. (2) could you please add some elaboration in your answer for $f$-orbitals being degenerate?  (as I requested in my question above) Thanks a lot ^_^

@GaurangTandon I don't have much to say about $f$ orbitals. I've never done this analysis, but I assume it's pretty much the same as above except that you should find the full names for the functional labels on all the orbitals.

By identical i guess you mean to be able to be transformed to one another by a symmetry operation. So:
I think it is the same reason why  the orbitals 2s, 2px, 2py, and 2pz are all degenerate in the H atom. You can clearly see the symmetry between the 3 p orbitals but how can the spherical 2s be symmetric with the rest? The answer is that the Coulomb potential has a  hidden symmetry that shows up in spaces of dimension higher than 3. Hence, in the same way that a 2 dimentional being would not be able to see the symmetry between the 3 p orbitals (the px and py will look like two disks that touch when projected to a plane while the pz will look like a simple disk)  we cant see the symmetry between the s and p orbitals in the H atom (or dz2 and the rest ds) since we are only 3 dimentional beings. Ref: Molecular quantum mechanics by Atkins, p.95. For this reason you need to look at the math behind this. Unfortunaltely, being no physical chemist i wouldnt be able to show the math behind this (or it would take me a while figuring it out).

Nice! But, I got confused at this line: "The answer is that the Coulomb potential has a hidden symmetry that shows up in spaces of dimension higher than 3." Do you mean to say that all these orbitals are not 3d but actually 4d orbitals???

No, they are not 4d. When i say dimention i mean it in a mathematical way and not changing the principal quantum number which is what you have done. For example when you plot an xy chart you are ploting in two dimentions. When you plot an xyz 3D graph (it is a surface now) you are ploting 3 dimentions. However you can plot in more than 3D, like xyzv, but now you wont be able to visualize the result bacuse this is how far our perception goes (in the same way you cant see in IR or UV). Have a look at the ref above as well for more details. It is quite a read though!

Sorry I didn't make my comment clearer I actually meant "...are not 3 dimensional but actually 4 dimensional orbitals???"

Sorry, i see what you mean and it is a good point. I think that the equations that describe the orbitals must be multidimentional as you say. In the answer by Zhe you can see that the function that describes the dz2 orbital is of the type f(x,y,z) which is already 4 dimentions. However i dont know the math behind it being a synthetic chemist. Maybe a physical chemist would like to jump in at this stage?


text:
Why dichlorobenzene doesn't show geometrical(cis/trans) isomerism?
As it has restricted rotation due to the presence of a double bond, it should show geometrical isomerism. Then, why doesn't it show cis/trans isomerism?

The trans form will not be formed because it leads to too much strain in the small ring ring. 2. It will lose aromaticity.

This doesn't make any sense. Are you sure you have the right molecule? Dichlorobenzene is basically a benzene ring, there is no real isomerism to speak of. Do you mean something like azobenzene?

Geometrical isomerism is shown by linear as well as cyclic chain. So why can't we talk about benzene ring? And yes, the question is correct

Benzene, as an aromat, does NOT contain double bonds like 2-hexene, or any other olefin. Read two pages further in your textbook.

I have read my textbook but benzene ring wasn't mentioned there that's why i asked this question

The questions are a bit similar except the fact that we are discussing about a different compound

1-bromo-2-chlorobenzene is in principle almost the same compound, at least in the sense of stereochemistry, it behaves exactly the same. That why I closed it as duplicate.

There are dichlorobenzene isomers but not cis-trans. You can have ortho (where the Cl atoms are close together), para (where the Cl atoms are opposite) and meta where Cl atoms are one C atom apart. Lets examine the ortho-dichlorobenzene which part of it "looks" the most like cis-1,2 dichloroethene (which also has a trans isomer). Is there any way you could make the Cl atoms in ortho-dichlorobenzene to be as far apart as in trans-1,2 dichloroethene without breaking any bonds? The answer should be no and this is why cis-trans isomerism in common substituted aromatic rings makes no sense.


text:
How to calculate the electrochemical potential of a cell using known half-cell potentials?

What is $E^\circ$ for the half-reaction representing the oxidation of $\ce{HOCl}$ to $\ce{ClO3-}$ given the equations below?
  \begin{align}
\ce{ClO3- +6 e− + 6 H+ &= Cl− +3 H2O},& E^\circ &= \pu{1.446 V}\\
\ce{HOCl + 2 e− + H+ &= Cl− + H2O},& E^\circ &= \pu{1.481 V}\\
\end{align}

It seems too simple, but would the answer be to take
$$E^\circ = \pu{1.481 V} - (\pu{-1.446 V}) = \pu{2.927 V}?$$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site.  I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

For adding up equations, you can't directly add $E^0$,unless you know that the no. of electronic transfres are same in both the reactions given.So, it is safe to calculate by $\Delta$$G^0$ calculation,
For $HOCl$ reaction , you have,
$HOCl + 2e^- + H^+ = Cl^- + H_2O$ --  $\Delta$$G^0$= -$nFE^0$ = $-2*F*1.481$
By inverting the First reaction, sign of  $E^0$ will be changed and thus,
$Cl^- + 3H_2O = ClO_3^- + 6e^- + 6H^+$ ---  $\Delta$$G^0$=-$nFE^0$ = $-6*F*(-1.446)$
Adding the two reaction ,
$HOCl + 2H_2O = ClO_3^- + 4e^- + 5H^+$ --  $\Delta$$G^0$= - $(4*F*E^o)$ $\Delta$$G^0$ will be added and for the whole reaction,
$E^0(total) $ =  $\Delta$$G^0$(total)/$nF$= $-F*(2*1.481-6*1.446)/(-4*F)$)=$-1.4285$ $V$.So,  $E^0$ for the oxidation from $HOCl$ to $ClO_3^-$ is $-1.4285$ $ V$.Thus, the answer is different from simply adding the $E^0$ for reactions which is not correct for all the cases(is correct when electron transfer is same for both reactions).


text:
Electron Energy Levels?
New to chemistry; In my book it talks about electrons in atoms moving from one energy level or shell to another and denotes this by n. How does this exactly happen, do electrons move to different energy levels when they bond with different atoms? It also talks about something along the lines of photons. Also please explain ground level.  It says Hydrogen is in ground level when n=1 but what about other atoms where n>1? For instance, carbons has electrons where n>1 does this mean all the electrons other than the first shell are NOT in ground level? Please explain in layman terms as Im not that fluent in this discipline.

You ask too many questions at once. The supposed answer is so long that nobody wants to write it, so have a comment instead. Electrons do not move to different levels when they bond with different atoms; nor do they remain on the same levels. Yes, it is complicated. Forget about bonds for now. Look at a single atom. Yes, some electrons of carbon are not on the level with n=1; they would like to go there, but the place is already occupied by other electrons.

Please explain in layman terms as Im not that fluent in this discipline.

Introduction: According to Bohr's model, an atom is surrounded by shells. They are depicted by $K,L,M,...$  and are denoted by $n=1,2,3,...$ respectively.
When an electron occupies ground state level, it means that the electron occupies the lowest energy state possible. Since the lowest energy occurs closest to the nucleus, each atom tries to occupy the lowest $n$ possible (while also satisfying the rule that in each shell the maximum number of electrons can only be $=2n^2$)

For instance, carbons has electrons where n>1 does this mean all the electrons other than the first shell are NOT in ground level?

No, note that different electrons have different ground levels. An electron may have ground level $n=2$. If it gained extra energy and moved to $n=3$, we would consider it to be in excited level now. But, an electron that was already in $n=3$ is not considered to be in excited level.

It also talks about something along the lines of photons.

Note that a photon with wavelength $\lambda$ has energy $E=\frac{hc}{\lambda}$. A photon beam (having several such photons) when incident on an atom can excite its electron to higher energy levels. You'll study this in detail in the chapter on "Atomic structure" (topic "Atomic spectra") in higher classes. Do not worry about this for now.

Do electrons move to different energy levels when they bond with different atoms?

This is a very complicated topic. It will unleash hybridiation, MOT, and too many other things. It's like opening a dam of Niagara falls. Please don't confuse yourself with bonds and excitation for now.

Note: this was a pretty dumbed-down explanation of what all you asked. None of it will actually hold true in higher sciences, but, it will work for the class you're in currently.


text:
Why does [Co(NO2)6]4- ion have 3 unpaired electron as opposed to 1?
Why does $\ce{[Co(NO2)6]^4-}$ ion have 3 unpaired electron as opposed to 1?
The book says that it has 3 unpaired electrons. I thought $\ce{NO2-}$ was a strong field ligand, making $\ce{Co^2+}$ a $\ce{d^7}$ low spin ion. Six electrons go in the $\mathrm{t_{2g}}$ orbital and 1 in the $\mathrm{e_g}$ orbital.
Please explain the bonding in this compound (preferably with a suitable MO diagram to go with it).

You can't predict which ligand is strong or weak, there are many factors which can change this.

An easier method - instead of getting tangled in the mess of strong vs weak field ligand - would be to make deductions based on the compound's calculated paramagnetic moments. Unfortunately, a simple Google search yields too little information. Perhaps, the chemists with more efficient database searches can help out here?

According to Elliot, Hathaway & Slade (1966) cited by this paper, Co(II) with nitrite ligands is low spin.

So maybe it is just a typo. For d7 low spin, you would expect one unpaired electron.


text:
Why does hydration of ions depend on their charge to area ratio?
The following is written in my book Concise Inorganic Chemistry by J. D. Lee.

Hydration energy, hydrated radius and hydration number of a particular ion depends upon the charge per unit area. Hence their order is the same for the following cations in the given examples:
$\ce{Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)}$

And other similar examples are also given. My question is, why does the charge to surface area ratio matter? The force felt by the water molecules due to any of these cations should be the same due to the shell theorem. Also, a cation of bigger size can accommodate more water molecules in its vicinity due to its size. Then still, why does an increase in charge to surface area ratio correspond to an increase in hydration tendency?

"My question is, why does charge to surface area ratio matter?" (+1) I've always wondered that myself, but never quite figured it out :(

Related: https://chemistry.stackexchange.com/questions/2883/what-is-charge-density

Could be that hydrogen bonds need polarity to work and charge density simply means that the polarity is high.

@Avatar I am unable to understand what do you mean.

Many people have done experimental, theoretical, statistical, molecular dynamics, etc. studies of solvation free energies of ions. I encourage you to just search solvation free energies of ions on google scholar and you'll find a bunch of very interesting and quite readable papers.
One of these papers ref. [1] (here for pdf) provides a lot of data and a very simple model which can basically be used to answer your question. In this paper, they derive a simple model which has only one fitting parameter. From this, they can determine the effective cavity size of an ion in solution. This cavity size is given by $r+\Delta r$, where $r$ is the radius of the ion and $\Delta r$ is the thickness of the shell.
Essentially, the value of $\Delta r$ is determined by the fitting parameter, and also determines the number of water molecules within the cavity of the ion. A portion of the data table is shown below with actual experimental values on the far right and parameters determined from the model in the other columns.

Just focus on the alkali metals which you list in your question. Notice as we go further down, the value of $\Delta r$ approaches zero and the actual radius of the ion is also the size of the cavity. This, however, is very much not true for the smaller ions. The Hydrogen ion being an extreme example where a solvation number of 12.0 is obviously unphysical. This, however, does provide a clue for a reasonable interpretation of $\Delta r$. $\Delta r$ is a measure of how far away a molecule must be for it to not "notice" the effect of an ion in solution. Thus, somewhat counter-intuitively, smaller ions interact effectively with a larger number of water molecules. The reason for this is that as the ions get larger, solvating them requires a greater reorganization of the hydrogen-bonding network in liquid water. This means that one must lose hydrogen bonds in order to solvate the ion.
Thus, the reason the charge to area ratio dictates the solvation energy and the solvation number is because a smaller ion will bring the same charge into solution, but due to its smaller size, will only require a small reorganization of the hydrogen bonding between waters, which minimizes losses in already stable interactions. This lack of reorganization is what leads to large values of $\Delta r$ and hence large values of $n$ in the model described above.
Let me be clear, $n$ is not simply the number of water molecules pointing towards an ion. This would not be measurable experimentally, or well-defined theoretically. Rather, $n$ will always be some kind of average value which tells you the effective number of water monomers which energetically benefit from the presence of the ion in solution. So, now the picture is clear. The smaller cations do not cause too large of a first hydration shell, so that the second hydration shell is closer to the ion than for a larger ion. Hence, the free energy and the solvation number are larger for smaller ions of the same charge.
As a bonus, ref. [2] is an interesting paper which approaches this problem from molecular dynamics. They give plots showing the relative effects which van der Waal's forces and electrostatics have. Note that the smaller ions are the only ones which are stabilized effectively by van der Waal's interactions as these scale closer to $r^{-5}$ rather than $r^{-1}$

References:
[1] Marcus, Y. (1991). Thermodynamics of solvation of ions. Part 5.—Gibbs free energy of hydration at 298.15 K. Journal of the Chemical Society, Faraday Transactions, 87(18), 2995-2999.
[2] : Grossfield, A., Ren, P., & Ponder, J. W. (2003). Ion solvation thermodynamics from simulation with a polarizable force field. Journal of the American Chemical Society, 125(50), 15671-15682.

This is a great reference, but it should be pointed out that the data in the pasted table actually contradict the statement in the question that the hydration energy is proportional to the ratio of charge to area. If that were true, the hydration energy of the alkali metals would show an approximately linear correlation to $\frac{1}{r^2}$, whereas the data show a closer match to $\frac{1}{r}$.


text:
Why does nitrogen have a maximum covalency of 4?
As nitrogen has $1$ lone pair and $3$ electrons, either it should have maximum covalency of $5$ or $3$. But why does it have a maximum  covalency of $4$ instead?
Why did it leave $1$ electron? Why did it have to break a lone pair?

Recall that covalency is the number of shared electron pairs formed by an atom of that element. Nitrogen's maximum covalency is indeed $4$. And no, it does not break up its lone pair.
I'll give you a simple example. Have a look at the Lewis structure of the ammonium ion:

(source)
Notice that nitrogen's octet is complete as soon it bonds with three $\ce{H}$ atoms (aka forms ammonia). The fourth covalent bond is actually a coordinate covalent bond, formed when that nitrogen atom's lone pair gets donated to a proton.
This is also the maximum covalency for the nitrogen atom, since it has no more unpaired electrons that could be paired up with other atoms, to form more covalent bonds.
Homework exercise: Can you now deduce the maximum covalency of nitrogen's elder brother, oxygen?

Caution: extra stuff ahead this will help those with knowledge beyond high school level. If you're at or below high school level, go back home and play with your cat.
After-thought: Valency is a useless term that doesn't help you do any chemistry. Different sources define it differently (two definitions on Wikipedia). It just helps fill high school textbooks with more pages, but it becomes irrelevant with the introduction of better terms like coordination number, which actually tells you something about the structure of the molecule. While "valency" maybe  useful in basic beginners course at getting a grip, there are limitations to this point of view.
Here's an example of such a contradiction. You may expect elements of period $3$ and above to display higher covalencies. One such example is phosphorus. Though it belongs to the same group as nitrogen, it can form compounds like $\ce{PCl5}$, (apparently) increasing its maximum covalency to $5$ instead. The reasons for these are usually attributed to hypervalency/octet-expansion, but they are wrong and obsolete concepts, having been superseded by newer concepts. In fact, $\ce{P}$ still has a covalency of four in $\ce{PCl5}$, since there are only four shared pair of electrons (the non-bonding electrons don't count). This reinforces the idea that coordination numbers are a better and more useful term than valency. ($\ce{P}$ now has a coordination number of $5$ in $\ce{PCl5}$)

It must be noted that as generally only valence electrons participate in bonding, the 1s electrons cannot be donated to form 5 bonds from 5 available orbitals.

Gaurang, your "homework exercise" is rather difficult.  Certainly oxygen can share 3 pairs as H3O+, but can it share 4 pairs? https://www.sciencedirect.com/science/article/abs/pii/0009261487804906 ; https://pubs.acs.org/doi/pdf/10.1021/ja00265a031

@DavePhD Yes, you're right, thanks for bringing it here. I had later also  come across this Zinc acetate. So, tetravalent oxygen does exist. Though, I was only thinking from an elementary point of view.

Hi Gaurang, does Oxygen shows a maximum covalency of four (2 cb, 2 coord-cb) or three (as 2 coord-cb isn't stable due to high EN of O) ?

Gaurang, why exactly doesn't it break up its lone pair? (Disregarding the octet rule as it's kinda useless)

(continued from my earlier comment) I've worked out a reason; steric hindrance. Apparently nitrogen can't even fit five fluorines around itself. Is this correct?

nvm, got it. Nitrogen can't hybridise to form 5 bonds, since it literally has only one s and three p's in its ground state energy level. Since exciting electrons to 3d is a tall order, it's stuck with having to form three covalent bonds and a coordinate one, if pressed.

If maximum number of bonds is 4 for then how could it attain +5 oxidation state like in $HNO_3$? Sorry if this question is silly I am new to this.

Oxygen can form four covalent bonds. It takes a superacidic environment to render these bobds with hydrogen, but basic zinc acetate, described here, qualifies with an oxygen bonded to four zinc ones.

The nitrogen has an electronic configuration $\ce{1s^2 2s^2 2p^3}$, therefore it could be expected to make 3 covalent bonds accommodating 3 electrons in orbitals $\ce{p}$. However, there is less repulsion of the electron domains when the $\ce{2s}$ orbital hybridizes with the $\ce{p}$ orbitals to form 4 $\ce{sp^3}$ orbitals. In one of the orbitals there will be a pair of paired electrons that can make a coordinate bond. The other 3 orbitals will make 3 covalent bonds accommodating 3 electrons.
Note:
In the ammonia molecule the electron pair alone is in an orbital $\ce{sp^3}$, different from the isolated atom.
There is no distinction between the bonds (covalent and coordinate) in the ammonium ion.

[...] there is less repulsion of the electron domains when the 2s orbital hybridizes [...] Hybridisation is a mathematical concept, what you are probably trying to say is that s and p orbitals mix, which is similar, but different. (You can correctly describe the molecule without hybrid orbitals, you cannot describe it correctly without sp mixing.) Also, nitrogen does not form four sp³ orbitals, that is an approximation, as the lone pair will still have a slightly larger s contribution. Nevertheless, this answer is correct in its essence.

Nitrogen has maximum covalency 4, because we know that maximum electrons a shell can accommodate is $2n^2$, and hence it can have only 8 electrons. So covalency cannot exceed more than 4.
`

This doesn't seem to make sense. By your reasoning, all elements with the same n should have the same covalency.

@Dmitry You omitted maximum, and that is the key word, therefore the statement is correct (in the confines of the definition of [co-]valency).


text:
Effect of rain on water and soil samples contaminated by petroleum
I am about to perform research on the impact of petroleum fire disaster on the soil and water sources within the region of disaster, using heavy metals and total petroleum hydrocarbons as impact markers.
My supervisor suggested I collect samples during raining season and dry season and compare.
My question is, would there really be any significant difference between the  two samples in the rainy and dry season?

https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

What is "d region"?

@pentavalentcarbon probably "the region"

It has much sense since the rain will modify the concentrations of both the heavy metals and the petroleum.
The initial quantities will be redistributed.
One part will penetrate deeper in the ground, another will be moved due to water runoff and what is left will be measured in your samples.
If possible, you should take the first samples as soon as possible and the the next ones towards the end of next season.
It is also important to get the value of the total precipitation during the rainy season.
This will serve to estimate how long it will take to eliminate the effects of the disaster.


text:
Rearrangement of an oxyallyl cation
I am supposed to predict the product of rearrangement of this cation, having been told that it's an oxyallyl cation:

I did have a look at the Wikipedia entry (since it mentions "oxyallyl cation") and not even once does it mention such a compound as I have shown above. This question is from my high school text book chapter on "Aldehydes and Ketones", and I'm not sure I'm even supposed to know what to do here in the first place.
Anyways, I did guess that the negatively charged oxygen atom will donate its lone pair to the positively charged carbon atom1, forming a three member hetero-ring. But, my book gives the correct answer as:

I'm not sure where they dumped the second oxygen atom (and from where they brought an extra carbon atom) to form this all-carbon ring. Also, is that $\ce{-OMe}$ at the ortho position only for ornamental purposes, or does it affect the mechanism in some way?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; page 218 in solution manual, Q91 chapter on Aldehydes and Ketones

1: Yeah I know that all these atoms are in conjugation and the charges are only partially negative/positive.

Count carbon and oxygen atoms in the book's answer versus the original molecule, and infer that the book got it wrong.  One of the carbon atoms in the book's answer should be oxygen.  Question for you:  Which one?

Thanks for confirming that the answer is wrong. The correct answer should replace the $\ce{-CH2}$ at the vertex of the triangle with an $\ce{-O-}$. Is that what you intended?