text:
What is electron configuration used for?
What is the purpose of writing electron configuration vs just using the number of electrons off of the periodic table. I know it can be used to determine the number of electrons in each energy level, and therefore, the number of valence electrons. Is there anything else? The number of valence electrons can be found from the periodic table, so why go through that process?

By long and short hand, do you mean "1s2 2s2 2p6 3s2 3p6" versus "[Ne] 3s2 3p6"?

Yes, using the noble gases

atomic electron configurations are used in systematics of atomic spectral lines, like in atomic-emission and X-ray spectroscopy.

I feel your logic is somehow deeply flawed. Configuration is among  best forms of description of atoms, which are available. It's like asking for what is knowledge of boiling point of water used, or of color of sky.

Well, it's almost always this way.

If you know the atomic electronic configuration, you know how many electrons an element has.
If you know how many electrons it has, you'll know how many it needs to have a stable configuration (usually octet).
If you know how many it electrons it needs, you'll know how it reacts with other species.
If you know how substances react, you can predict the reactions involving this substance.
If you know reactions involving this substance, you can either discover a new reaction (knowledge of new reactions makes synthesis easier) or you could invent an unknown compound with interesting properties, which may have its own special uses.

Is there a benefit to electron configuration specifically vs finding that information from the periodic table. Bohr model, etc? Does knowing the electrons in specific energy levels give us additional info?

@Sarah Yup, it definitely does. Knowing energy levels can be a big factor in deciding if a reaction is feasible under some conditions or not. In some cases, we can even predict the color of a molecule by knowing the wavelength of light needed to excite electrons within energy bands in a molecule.


text:
Calculating total volume mixing NaCl and water
I'm (as a non chemist) trying to calculate the the volume of a NaCl solution. The problem is that I know the volume of water prior to mixing in NaCl, but I want to know the volume of the solution after.
If my little research is correct, I need to use partial molar volumes of NaCl to estimate the volume change in response to adding NaCl. I found a value of $\pu{16.61 cm3/mol}$ for NaCl (and some other literature values being similar). The initial volume of water is 12 liter, and the added mass of NaCl is 1000 g, or 17.11 moles. This results then in $\pu{284 cm3}$ increase in volume, and a total volume of 12.284 liter.
My question is to what degree the partial molar volumes are dependent on the concentration in the solution and the temperature of the solution? I would like to do these estimations of total volume for a range of temperatures from 2 degrees C to 30 degrees C, and with variable amount of NaCl concentrations. I was not able to find information on whether NaCl partial molar volumes are dependent on these factors, but it seems other aqueous mixtures are.
Thanks!

Wouldn't it be easier to look up the (surely available) tabulated values for the density of salt solutions at different concentrations and work backwards from there?

Thanks, I didn't think of that! I'll do a more careful calculation later, but a quick check on mobile gives a volume of 12.22 liter - so comparable to the partial molar volume method

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text:
Monosaccharides configurations (alpha beta, D L) identification
I am really finding it difficult to identify the correct configuration(s) of a given monosaccharide, in fact my friends and I have been trying to solve this, much to our annoyance.
In one of our classes, our teacher said, in cyclic form:

If $\ce{CH2OH}$ is below the ring its configuration is  L, and if its above the ring its D.
   If $\ce{C1}$'s $\ce{OH}$ is below the ring its alpha configuration and if its above its beta configuration .

My friends seem to agree with the above explanation, however my understanding is:

If all groups of $\ce{OH}$ and $\ce{-CH2OH}$ is reversed from D form, its L.
  If $\ce{C1}$'s $\ce{OH}$ is on the same side of the ring as $\ce{-CH2OH}$ its beta else its
  alpha.

So I want to know if these assumptions are right:

D and L in linear form:

If the last chiral carbon's OH (one before the last carbon) is on the right its D and if its on the left its L. so if you want to make for example L glucose from D, you should reverse ALL groups.

D and L in cyclic form:

It's hard mainly but can be deduced by making linear form into cyclic and cyclic into linear and comparing them or reversing all carbon branches.

Alpha and Beta in cyclic form:

If the carboxyl carbon's $\ce{OH}$ is on the same side of last ring carbons branch ($\ce{CH2OH}$ in glucose, $\ce{COO-}$ in glucoronate) it's neta, else it's alpha. So in D form, if $\ce{OH}$ is above it's beta else it is alpha. In L form, if $\ce{OH}$ is above its alpha else its beta.

Also when i see 3D linear form of Glucose it doesn't match with 2D at all: 2D view of this site isn't like what I find in books I can't find the match between 2D and 3D view neither.

I'm still confused because of following examples:
1-Alpha-L-Glucose

2-Alpha-D-Glucose

3-Beta-D-Glucuronate

4-Beta-L-Iduronic Acid

these just don't match.
also Harper biochemistry image is this:

but lehningers is this:

It’s easy to understand your frustrations, however you need to understand the basic stereochemistry and conformations of sugars.
A monosaccharide is designated D if the hydroxyl group on the highest numbered asymmetric carbon is drawn to the right in a Fischer projection, as in D-glyceraldehyde:

Note that the designation D or L merely relates the configuration of a given molecule to that of glyceraldehyde and does not specify the sign of rotation of plane-polarized light.
As I covered this explanation already on one of my posts:Are glucose and galactose cis-trans isomers of each other?

D sugars have the same absolute configuration at the asymmetric
  center farthest removed from their carbonyl group as does
  D-glyceraldehyde. The L sugars, in accordance with this convention,
  are mirror images of their D counterparts

Cyclic forms
The designation α indicates that the hydroxyl group at the anomeric center (hemiacetal or hemiketal carbon)  is, in a Fischer projection, on the same side as the hydroxyl attached at the farthest chiral center, whereas  β indicates that these hydroxyl groups are on opposite sides:

In the α anomer, the $\ce{OH}$ substituent to the anomeric carbon is on the opposite side of the sugar ring from the $\ce{CH2OH}$ group at the chiral center that designates the D or L configuration ($\ce{C5}$ in hexoses). The other anomer is known as the β form.
The Haworth projections represent the stereochemistry of sugars more realistically than do the Fischer projections, which is why they are preferred:
For a D sugar, any group that is written to the right of the carbon in a Fischer projection has a downward direction in a Haworth projection; any group that is written to the left in a Fischer projection has an upward direction in a Haworth projection.
The terminal $\ce{-CH2OH}$ group, which contains the carbon atom with the highest number in the numbering scheme, is shown in an upward direction.

Similarly  in the  α-anomer, the hydroxyl on the anomeric carbon is on the opposite side of the ring from the terminal $\ce{-CH2OH}$ group (i.e., pointing down). In the β-anomer, it is on the same side of the ring (pointing up).

References

Lehninger Principles of Biochemistry
Voet and Voet biochemistry
Biochemistry Grisham
Biochemistry Campbell and Farrel

From the given explanation above, you can easily deduce if your previous assumptions were correct:)

Why is d-glucuronate alpha when c1 oh id down while L-iduronate is alpha when c1 oh is down?

Sorry for late response. What determines whether a given sugar is d or l is the furthest assymetric (chiral carbon) and whether alpha or beta is anomeric carbon (i.e C2) so since the hydroxyl groups on d-glucuronate or l-iduronate  on C2 lie on right side (same side with anomeric carbon) they both alpha hence OH groups is dowward in haworth  projection, however  the furthest chiral carbon on d-glucuronate is to the right in fischer projection like d-glyceraldehyde but on the left side with l-iduronate (like l-glyceraldyde)

im still confused. can you compare alpha and beta in these?: Beta-D-Glucuronate,Beta-D-Glucose,Beta-L-Glucose they are different :(

"As I covered this explanation already on one of my posts" The blockquote following this line is unavailable in that answer you linked to. Are you sure you are directly quoting your answer or are there some changes?

@xavier_fakerat Did you please happen to have a look at my comment above? Thank you.

@GaurangTandon, Ok I was quoting the answer and I happen to have updated it again, but the main references are listed below the linked answer :)

Nice! It makes sense now. Can you also please comment on ketoses? Like, do they follow the same rules or are they slightly different?

In cyclic structures is this definition of alpha and beta anomer applicable only in D form or can be used in case of L forms too ?

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text:
Confusion over boiling point of gallium
I am completing a project on gallium, and I need to include its boiling point. I thought that this would be fairly simple to look up, however, it appears that different sources quote different temperatures as the boiling point, normally either ~2200 °C or ~2400 °C.
Does anyone know why two different values are presented, or which value is correct?
Below are some examples of websites quoting each figure.
2200 °C:
https://education.jlab.org/itselemental/ele031.html
http://www.rsc.org/periodic-table/element/31/gallium
https://www.livescience.com/29476-gallium.html
2400 °C:
http://www.chemistryexplained.com/elements/C-K/Gallium.html
https://www.britannica.com/science/gallium
http://www.chemicalelements.com/elements/ga.html

Wikipedia cites 2400 from https://www.researchgate.net/publication/231538496_Corrected_Values_for_Boiling_Points_and_Enthalpies_of_Vaporization_of_Elements_in_Handbooks which is article wholly about getting good values.

In order to help solve this discrepancy, it may be useful to source a graph/equation of vapour pressure versus temperature and find the temperature for a pressure of 1 bar. It is far easier to type an incorrect number than produce an incorrect graph/equation, plus whoever made the graph/equation probably put some effort into ensuring its accuracy.

This question has been recently raised in Chemistry & Engineering Letters. The CRC Handbook (2017) in section "MELTING, BOILING, TRIPLE, AND CRITICAL POINTS OF THE ELEMENTS" [1, p. 4-117] lists the following values for gallium, quoting original publication [2]:
\begin{align}
t_\mathrm{tp} (\ce{Ga}) &= \pu{29.7666 ^\circ C} \\
t_\mathrm{m} (\ce{Ga}) &= \pu{29.7646 ^\circ C} \\
t_\mathrm{b} (\ce{Ga}) &= \pu{2229 ^\circ C} \\
\end{align}

[...] values are given for the solid-liquid-gas triple-point temperature $t_\mathrm{tp}$ and pressure $p_\mathrm{tp}$; normal melting point at $\pu{101.325 kPa}$ pressure $t_\mathrm{m}$; normal boiling point $t_\mathrm{b}$; and critical temperature $t_\mathrm{c}$ and pressure $p_\mathrm{c}$. All temperatures are on the ITS-90 scale.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Geiger, F.; Busse, C. A.; Loehrke, R. I. Int J Thermophys 1987, 8 (4), 425–436 DOI: 10.1007/BF00567103.

I was hoping someone would check the CRC handbook...

@NH CRC Handbook is also handy when it comes to pinpointing up-to-date source of information, at least in the majority of cases:)

We always had a copy of the CRC handbook in our undergrad physics labs as our go-to source.

You've got twice as much "recently" in the first sentence as you need :)

The comments section in the C&E letters addresses why this discrepancy existed.


text:
What's the standard deviation in bond angle for water molecules at a particular temperature?
Suppose I have a beaker of water at 20 °C. I want to know the standard deviation of the bond angles between the hydrogen atoms in the water molecules. How do I figure this out?
If I knew that the bond angle was the only possible degree of freedom of the molecules, then I'd be able to use the Boltzmann distribution to calculate this, if I knew the energy of the molecule as a function of bond angle. But I don't know the energy of the molecule as a function of bond angle, and also there are a lot of other degrees of freedom (e.g. the molecules can have overall kinetic energy, or they can be spinning rapidly, or the other vibrational modes of the water molecule can contain some of the energy.
Is there an easy way for me to look up the energy of the water molecule as a function of the positions of the hydrogen atoms? (I guess I could just calculate this myself using my favorite quantum chemistry software.)

related https://chemistry.stackexchange.com/questions/18777/are-bond-angles-constant

Thanks! That looks like it's answering this from an experimental standpoint, whereas I'm more interested in understanding the theory and how to compute it.

I assume you talk about condensed phase. Correct?

I tried to answer by proposing a simple model. The model didn't work as well as I hoped, but the general procedure might serve as an answer.
A harmonic model
As a first approximation, you might consider a simple model for the energy of a single water molecule as a function of its angle. One possibility might be a harmonic function
$$E(\theta) = \frac{k \theta^2}{2}$$
where $k$ is an unknown parameter and $\theta$ is the angle. This is very simple and leaves out many complexities, of course.

If you had $E(\theta)$, the energy of a single water molecule at a particular angle, and if you assume an independent particle model (which may break down for a liquid), you could apply Boltzmann weighting
$$p(\theta) = \frac{1}{U}\exp\left(-\frac{E(\theta)}{R T}\right)$$
where $p(\theta)$ is the probability of finding a water molecule at a particular angle and $U = \int_0^\pi d\theta \exp\left(-\frac{E(\theta)}{R T}\right)$ is a normalization factor.

If you had a probability distribution $p(\theta)$ describing the spread of angles at a certain temperature, getting the mean and standard deviation is easy:
$$\langle\theta\rangle = \int_0^\pi d\theta\ \theta\ p(\theta)\\
\langle\theta^2\rangle = \int_0^\pi d\theta\ \theta^2\ p(\theta)\\
\sigma_\theta^2 = \int_0^\pi d\theta\ \left(\theta - \langle\theta\rangle\right)^2\ p(\theta) = \langle\theta^2\rangle - \langle\theta\rangle^2$$
So this would be how to get $\sigma_\theta^2$.
Below is my numerical attempt.
Beware: this model is too simple.

Numerical results
I did some calculations (mostly with SymPy) with the above and I came up with the following:
$$p(\theta) = \frac{1}{U} \exp\left(-\frac{k \theta^2}{2 R T}\right)\\
U = \sqrt{\frac{\pi R T}{2 k}} erf \left(\frac{\pi}{2} \sqrt{\frac{2 k}{R T}} \right)$$
$$\langle\theta\rangle = \frac{k}{R T U}\left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right) \exp\left( -\frac{\pi^2 k}{2 R T} \right)\\
\sigma_\theta^2 = \frac{R T}{k} - \left[ \frac{\pi}{U} \exp\left( -\frac{\pi^2 k}{2 R T} \right) +
\left(\frac{RT}{kU}\right)^2 \left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right)^2 \exp\left( -\frac{\pi^2 k}{R T} \right)
\right]$$
By substituting constants and temperature ($\pi = 3.14159265$, $R = 8.314459848$ J K$^{-1}$mol$^{-1}$ and $T = 293.15$ K) and setting the average above to as close as possible to 104.45° and solving numerically the complete least squares problem, I found an approximate solution for $k$: $384.05105591$ J/rad².
Sadly, this value is associated with an average angle of around 96°, which means the model is very poor.
Using this best $k$ gives us a $\sqrt{\sigma_\theta^2}$ of
$$\sqrt{0.508770928037803} = 0.713281801280393 \text{ rad}$$
which is around 41°. Again, a too simplistic model: the best prediction gives angles as 96°$\pm$41°!

A bad model
By expanding the average in series, I found the following
$$\langle\theta\rangle \approx 1.570796325 + 0.000530047067361739 k + O(k^2)$$
The second order term above was $- 7.86976333117302\times 10^{-7} k^2$, so I dropped it.
Observe the first term is pretty much 90° in radians. All other terms decay too quickly.
By doing the same expansion for the standard deviation the following comes out:
$$\sigma_\theta^2 \approx - \frac{4.54747350886464 \times 10^{-13}}{k} + 0.411233515772252 + 0.000388544793228128 k + O(k^2)$$
The next term in the series above was $- 4.34803406790969 \times 10^{-7} k^2$ so I dropped it. Again, a (wrong) zeroth order term dominates.


text:
8-element ring?
I have come across 6-element rings, 5-element rings, but very rarely 3,4, or 7,8, etc. -element rings.
I am not sure of the reason for the rarity of other rings than 5 or 6. Is it just what I have encountered so far do not include non-5/6 rings a whole lot, or is it because of some reasons that there just are not that many non-5/6 rings around?
Edit: The rings I am referring to include both same-element (e.g. benzene, octasulfur, etc.) and different-element (e.g. glucose, fructose, etc.) ones.

related https://chemistry.stackexchange.com/questions/16859/heats-of-combustion-and-stability-of-rings

It sounds like you're asking about rings made up of different numbers of atoms, not different numbers of elements. Is that correct?

Looking at the title my first association was octasulfur $\ce{S8}$.

Rings are known in all sorts of sizes.  It's just that five- and six-atom rings are most stable under Earthly conditions.  Things are different "out there", as in three-membered ring species appearing in interstellar space (http://iopscience.iop.org/article/10.1086/501121/meta) and in the atmosphere of Saturn's moon Titan (http://www.sciencedirect.com/science/article/pii/S0032063313001864).  Might such species be precursors to some of the organic compounds we know and love here?

Not to mention Saturn have rings on its own.


text:
An elementary question worth asking
We all know that an atom starts moving faster when it's heated.But actually atom is not some new thing itself in the sense that it is just a bunch of electrons and protons. So, I think that same happens to the electrons or protons within. What does?

Electrons and protons don't move. Besides, there are discrete quantum states, and the first excited state of most atoms is already pretty high. Until you reach that temperature, nothing changes inside the atom.

I am just a ninth grader so it would be better if you can make it simpler.


text:
Is there a paper glue that is NOT dissolved by acetone?
Short version of the question: I need a spray-on glue that will stick paper to wood that isn't dissolved by acetone.
Long version: I am trying to glue paper on wood. It's a big sheet of paper and I need it stuck on throughout because then I cut the wood-with-paper-on-it into pieces with a laser cutter. Therefore, I'm using a spray glue that can cover the whole sheet of paper quickly and evenly. I have tried a couple different glues. After the pieces are cut, I want to spray an acrylic coating over the resulting pieces to seal them.
The process works except for one problem: if the acetone in the acrylic spray seeps through, it dissolves every glue I have tried. I try to go light with the acrylic, but more than 50% of my pieces end up with the paper coming loose from the wood. I can glue it back on later, but that means that the positioning of the paper is way less precise and the edge seal isn't there, making it susceptible to water damage.
Are there any glues anyone knows of that meet this challenge?
(I suppose, alternatively, is there an inexpensive sealant that doesn't use acetone? But I know the acrylic will hold up to the abuse these playing pieces are going to take, so I am hesitant to swap out that part of my process.)

Silicate glue (AKA "liquid glass") might do.

Mucilage and casein type glues (e.g. Elmer's) are two types of water based glue that are more likely to be resistant to the acetone in the acrylic varnish spray. But they do not come in spray-on form. However, if you diluted them with water and applied them with a compressed air sprayer you should get an even coating. Alternatively you could keep the spray adhesive as is and apply a water-based acrylic coat at the end--again by using a sprayer if necessary.

@iad22agp I'm going to attempt thicker paper for the printing as my first "work around," but I'll fall back to this if I don't get any better posts. Thanks for the hint.

Something you can try to use is tape to put under the glue. This may stop the acetone from seeping through. Not quite sure but maybe if you really need it done. Also, Elmer's glue does offer a spray on version regardless of what some say on your comments.

A spray-on Elmer's? I've never seen such a thing. I just now tried a search experiment... it's funny -- if I search for "spray on glue", the Elmer's (or any non-acetone glue) doesn't show up even in 2 pages of search results... but (thanks to your answer) if I search for "spray on Elmer's" explicitly, there it is! This may be exactly what I'm looking for! Thank you.

Your welcome! Good luck


text:
Computational determination of unique adosrption sites
I'm not sure it's a chemistry question or physics question or math question. Since I'm active on this site, maybe someone here can point me in the right direction. I'm trying to enumerate all the possible unique adsorption site on a crystal structure. For common FCC(111) and HCP(0001) surface, we have four possible adsorption sites, namely top, bridge, fcc, and hcp. Now, what if I have a binary alloy structure like CuRe (AB/ L10) or Cu3Re (A3B/ L12) or more complex structure like perovskite or zeolites. My idea for finding adsorption site on a simple binary alloy is:
1) Identify all the top site (that's easy, as I have my geometry in fractional coordinate).
2) Find all the bridge site using the formula $\frac{x_1+x_2}{2}$ in x direction and similarly all the bridge site in y direction.
3) Then use traigle like algorithm between three top site to find three fold hollow site and so on.
4) Now calculate Coulomb matrix$^*$ for all the site within the given neighboring radius and layers. So, if I want to include two nearest neighbor and 2 layers, it will calculate Coulomb matrix for all the sites within that range. Now, sort the Coulomb matrix in descending order. If I have repeated Coulomb matrix for any two sites than they are identical sites.
From my analysis, it should be enough to find all the unique adsorption site without using complex topological formulation; however, I would rather be right than go for this simple formula. My idea about topology related mathematics is on the surface level. But I'm ready to invest time on that now.
My question is two fold:
1) Is this method good enough?
2) If not, what do I need to study to find a better method (preferably with some good reference sources)?
$^*$Coulomb matrix:
$$CM=\frac{Z_1 \times Z_2}{r_{12}}$$
Here, $Z_1=$ 1 for bridge/fcc/hcp site and molecular weight for top site. And $r_{12}$ is the distance between those two sites.


text:
What is the difference between autocatalysis and self-replication in terms of chemical terminology?
I want to know does any difference between the meaning of autocatalysis and self-replication exist in modern chemical terminology?
E.g. I want to know can both terms be used interchangeably for entities requiring the presence of themselves in a reaction environment to catalyze the formation of themselves or is there any experimentally conceivable differences between molecules capable of autocatalysis and the self-replication of entities (therefore the category includes both molecules and larger complexes of molecules) in terms of the science of chemistry?
I want to know is the definition of autocatysis versus self-replication a problem of clarification of definitions-e.g. a problem of philosophy or is there any real-world experimentally identifiable difference between the 2 process in terms of chemistry?

The two terms represent very distinct concepts and are everything but interchangable.
Autocatalysis means that the product of a reaction catalyses the same reaction. But there is no focus on the catalysed reaction being exactly that or only that product catalysing it. As an example, take acidic ester hydrolysis e.g. with ethyl acetate:
$$\ce{CH3COOC2H5 + H2O ->[H+] CH3COOH + C2H5OH}\tag{1}$$
This is a reaction catalysed by acids that produces an acid in its course. This acid increases the solution’s overall acidity and thus catalyses the reaction that created it. But this feature is in on way specific to the acid we have generated. For one, we might as well add sulphuric acid to the reaction mixture to generate the same catalytic effect. For two, the acid produced in this reaction would also catalyse the hydrolysis of methyl formate. So neither is a specific structure required for catalysis nor is the catalysis restricted to itself.
Self-replication again means that a product is able to catalyse its own formation but only its own formation. The catalysis must be specific to a certain (set of) reactant to produce a certain (set of) product. Similar but different compounds bearing the same functional groups should not be able to effectively catalyse the reaction in question and the reaction product should not be able to catalyse any other reaction equally efficiently.
Typically, self-replication is achieved by having the product molecule act as a template which brings two reactive moieties closer together and allows for a more efficient reaction to itself. For example, you might consider the trimeric complex shown in the figure below:

Figure 1: Self-replicating system as described by
Terford and von Kiedrowski[1]
Here, the amine and the aldehyde shown on the right-hand half of the figure can condense to form an imine. This process can, of course, be catalysed by acids or bases. However, the condensation product shown on the left-hand half of the figure is able to form hydrogen bonds to the two compounds in such a selective way that the reactive moieties are brought close together. This eases the creation of the imine because the amino group is already in the perfect position of attack.
The imine product is therefore able to catalyse its own formation by templating but not the formation of closely related systems; and there is no other compound or functional group that would have an equally good catalytic effect on the reactants. This is not just autocatalysis but also self-replication.

Reference:
[1]: A. Terford, G. von Kiedrowski, Angew Chem. Int. Ed. 1992, 31, 654–656. DOI: 10.1002/anie.199206541.


text:
Why are aqua ion complexes of transition elements not solid but a solution?
Water ligands form dative covalent bonds with the central transition metal ion. That would mean a molecule is formed, just like sulfur dioxide or any other.
The transition metal ion does not form dipole ion attraction with water as does sodium chloride, but forms a covalent bond. If so, why is the resulting mixture a solution, and not a solid?

You have it all backwards. Solutions of transition metal salts are not all that different from the solution of NaCl. Besides, a hydrated transition metal ion may be a part of a solid all right.

For some reason it keeps getting repeated that transition metals dissolved in water were conceptionally different in any way to sodium chloride dissolved in water. This is not the case! Sodium also undergoes complex formation with water molecules. The difference to many (but not all) transition metals is that the complexes are structurally highly fluctional and thus not well defined.

The transition metal ion does not form dipole ion attraction with water as does sodium chloride

Actually it does. What we refer to as the complex is only the innermost hydration shell. This comples is still positively charged so the water molecules around this complex will still tend to orient themselves oxygen-inwards to build up dipole-dipole interactions (and hydrogen bonds) with the proper complex. Likewise, you should not consider dissolved sodium ions as in sodium chloride any different. They, too, have a hydrate shell which can be described as a coordination complex. The key difference is that the hydrate shell is very labile and no precise geometry can be given. There is little energy cost associated with water molecules diffusing away or attaching back. But other than that, the two are the same.
Whether or not the complex ion (e.g. $\ce{[Fe(H2O)6]^2+}$) is in solution or in a solid depends on what you did with the compound. A lot of hydrates are actually cationic aquacomplexes; for example, solid $\ce{CuSO4.5H2O}$ can be described as $\ce{[Cu2(H2O)8(\mu{-}SO4)2] . 2H2O}$ instead, showing that we have complexes in the solid state, too.


text:
Is it possible for Cytosine and Hypoxantine to make three hydrogen bonds?
I think the O from Hypoxanthine can bond with one of the Hs in NH2 from Cytosine and the H that is connected to N in Hypoxanthine with the N in Cytosine but is it possible also for the H connected to C in Hypoxanthine to bond with the O in Cytosine?
[Hypoxantine]
[Cytosine]
I have almost the same question between Xanthine and Adenine.(please note that I'm not talking about Xanthine and Adenine inside RNA, I know they can't bind at all in RNA because of their sizes)

I strongly believe that it won't be possible for these compounds to form three hydrogen bonds.
Revisiting the basic definition of a hydrogen bond:

A hydrogen bond is an electrostatic attraction between two polar
  groups that occurs when a hydrogen (H) atom covalently bound to a
  highly electronegative atom such as nitrogen (N), oxygen (O), or
  fluorine (F) experiences the electrostatic field of another highly
  electronegative atom nearby.

Such electronegative atoms as oxygen  and nitrogen with free lone pairs are potential hydrogen bond acceptors.
Hydrogen atoms attached to very electronegative atoms like O and N have strong partial positive charge and these are potential hydrogen bond donors
These two crucial requirements for hydrogen bond formation, have thus been summarised:

In order for a hydrogen bond to occur there must be both a hydrogen
  donor and an acceptor present. The donor in a hydrogen bond is the
  atom to which the hydrogen atom participating in the hydrogen bond is
  covalently bonded, and is usually a strongly electronegative atom such
  as N,O, or F. The hydrogen acceptor is the neighboring electronegative
  ion or molecule, and must posses a lone electron pair in order to form
  a hydrogen bond

Here is an impression of how these two would bond:

Note the "X" illustrating the missing requirement(s) to form a 3rd H-bond

missing potential donor group (electronegative atom) just C present.

Acknowledgements

Hydrogen Bonding
Voet & Voet Biochemistry: Genetic code

.

There is evidence for C-H...O and C-H...N hydrogen bonds, see e.g. the book by Desiraju and Steiner 1. These bonds are much weaker than conventional hydrogen bonds. Also, in general, hydrogen bond strength depends on the distance and the angle of donors and acceptors. In the given case, the hydrogen and oxygen atoms you are asking about are to far apart, and the atoms Y-H X (with Y the electronegative bonding partner of the hydrogen, H the bound hydrogen and X in acceptor) are not arranged in a line, which would be optimal.
So there are two strikes against a strong hydrogen bond here: A weak hydrogen bond donor in C-H, and suboptimal geometry.

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Calculating percentage of use

In the following reaction, $\pu{14g}$ of $\ce{KClO3}$ produces $\pu{1.4g}$ of $\ce{KCl}$
  $$\ce{2 KClO3 -> 2 KCl + O2}$$
  What is the % of use?

So first we balance the equation:
$$\ce{2 KClO3 -> 2 KCl + 3 O2}$$
Next find that $\pu{14g}$ of $\ce{KClO3}$ is $\pu{0.114 mol}$ and $\pu{1.4g}$ of $\ce{KCl}$ is $\pu{0.018 mol}$. We can find how much mols of $\ce{O2}$ we have left (according to conservation of matter) but that doesn't add up with the stoichiometric coefficients, and neither does the previous calculation. So how do I continue from here?

Note that "% of use" is what is typically called "yield" in chemistry, and "to balance the equation" is preferred over "to even the equation". Regarding the problem, you don't have to take care of oxygen at all, this brings nothing. You know the initial amount and the final amount, that's enough to calculate the yield (%).

I thought the formula addresses the total amount of matter. Alright, thanks

@TAR86 Please check out the examples in this post.

The question asks you to calculate the yield, i.e. the amount of reactant that was transformed into product. In your example, obviously, not all chlorine atoms that were present in $\ce{KClO3}$ end up in $\ce{KCl}$ since there are $\pu{114mmol}$ of the former but only $\pu{18mmol}$ of the latter although they should be equal (both their stoichiometric coefficients are the same).
To calculate the yield or ‘percent use’ as you wrote, take the amount of product, divide it by the amount of reactant and if necessary multiply by stoichiometric coefficients accordingly. So in your case
$$\text{yield}=\frac{\pu{18mmol}}{\pu{114mmol}}$$
Of course, you need to convert the result to a percentage.


text:
Xe–F bond length in xenon fluorides

The order of bond length of $\ce{Xe-F}$ in $\ce{XeF2}$, $\ce{XeF4}$, $\ce{XeF6}$ is?

My thoughts:
As hybridization of $\ce{Xe}$ in $\ce{XeF2}$ is $\ce{sp^3d}$, $\ce{XeF4}$ is $\ce{sp^3d^2}$, $\ce{XeF6}$ is $\ce{sp^3d^3}$.
%$\ce{s}$ decreases from $\ce{XeF2}$, $\ce{XeF4}$ to $\ce{XeF6}$, so bond length should increase, but my book says the opposite, i.e. $\ce{XeF2} > \ce{XeF4} > \ce{XeF6}$.

Downvoting for incorrect use of d orbital participation for main-group compounds.

@Jan I don't think a misconception in a question (and a misconception based on something which continues to be taught in general chemistry courses in the US) warrants a downvote.

Something seems fishy... Multiple sources say that the Xe-F bond is strongest in XeF2 and weakest in XeF6...

The explanation for the trend in bond strength of Xe-F in the xenon fluorides is the steric crowding about xenon, resulting in weaker bonds when there are more fluorine atoms packed around the central xenon atom.

Yes, your book is correct.
To rationalize it, throw away hydridization and electron exitation and work with concept of three-center four-electron bond. The idea is that in $\ce{XeF2}$ the real structure is a mix of $\ce{F-Xe+...F^-}$ and $\ce{F^-...Xe+-F}$, so the bonding is half-covalent and half-ionic. By packing more fluorine around the xenone atom, we force xenone into state with higher effective atomic charge, so the ionic part of the bonding becomes stronger as $\ce{Xe^{3+}}$ pulls fluorine anions much stronger.
same logic applies to many other cases, in particular sulfur oxides and sulfur fluorides.
For context, I quote experimental bond lengths from paper. Note, the paper itself quotes the values from other sources with slightly different values from different sources.
XeF2 ... 1.977
XeF4 ... 1.94
XeF6 ... 1.94 & 1.850 (two bond types)

Your resonance structure doesn't seem correct. In your structure, there is a negative formal charge on fluorine and the Xe-F portion is neutral, giving an overall negative charge to the molecule. However, the XeF2 molecule is neutral.

This happens because higher number of fluorine atoms induce a higher positive oxidation state on Xe which causes the Xe atom to shrink in size resulting in a smaller bond length.
XeF2 has Xe in +2 oxidation state while XeF6 has it in +6 oxidation state
link


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In C2 molecule, both bonds of the double bond are pi bonds. But isn't the rule that there should exist a sigma bond before a pi bond is formed?
C2 molecules have been found in vapour state. Their double bonds are made of two pi bonds because four electrons need to be accomodated in each bond. So this is against the rule that there must be a sigma bond before a pi bond is formed in the double bond. Why the anomoly?

Please reopen this question. This question wasn't about some high level research biradical stuff, but rather simple high school MOT theory application.

@GaurangTandon I am against reopening this question, especially for this reason; the C2 molecule has nothing to do with high school chemistry, it can only be understood with 'high level research biradical stuff' and there is still (and will probably forever be) a lot of debate about the bonding situation. Some say it's only a single bond, some say it's a double bond, there's arguments for the triple bond, and the believe for the quadruple bond. If you don't understand why this is the case, don't try apply 'simple high school MO theory'.

@Martin-マーチン I completely agree. The problem is that this is actually asked in Indian high school chemistry examinations(like JEE). Too bad we have to just remember these and move on without giving any thought to it.


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rovibrational spectroscopy: why do intensities of rotational lines increase with J?
how and why do the intensity of rotational lines change with J value?
in the simplest terms possible please


text:
Could UV lasers be used to split CO2 and H2O?
I was wondering whether intense UV light, tuned to the correct wavelength, could be used to split carbon dioxide and water molecules as a first step towards synthesizing liquid fuels.

Sure, though it might not be the most efficient way of doing things.


text:
Carbocation formation in electrophilic addition of hydrogen halides to alkenes
Question:

My attempt:

I made the carbocation on the carbon labelled $1$ because I think that the positive charge would be slightly more stable on $1$ compared to $2$ due to the inductive effect of $\ce{Cl}$.
But my teacher made the carbocation on $2$ giving the explanation that after rearrangement, the positive charge will be stabilised by resonance with $\ce{Cl}$.
His approach:

Now, my question is, comparing the stabilities of the carbocation intermediates without rearrangement, the positive charge will be stable on $1$ compared to $2$. After rearrangement too, I think that the tertiary carbocation will be more stable than compared to the one stabilized by resonance with $\ce{Cl}$.
So, what is wrong in my reasoning?
Or as suggested by Mithoron, will there be intramolecular attack by $\ce{Cl}$?

https://en.wikipedia.org/wiki/Neighbouring_group_participation

@Mithoron I don't understand how will be NGP here? If it was that way, the carbocation will be formed on $3$ without $\ce{Cl}$ to stabilise it. It would be very unstable. It would also lead to a different product, $\ce{Cl}$ bonded to $2$.

Protonation on "1" should be preferred because it leads to cyclic chloronium cation and  indeed a product is different then both of your suggestions.

@Mithoron I don't think that due to the following reasons. 1) The cyclic chloronium transition state will be 4 membered. It would be highly unstable. It is not even aromatic. 2) The resulting cation after rearrangement of $1$ carbocation will be a tertiary carbocation. So, the intermediate will be much stable compared to the one on $3$ which will be there if it did NGP. This carbocation is primary with nothing to stabilise it.

I think it 1,2 - Hydride shifts would be faster than the $\ce{Cl}$ attacking from a distance of 2 carbons in between. $\ce{Cl}$ attacking will require more work compared to $\ce{H}$ as $\ce{Cl}$ is much heavier than $\ce{H}$.

3-membered, exactly your second scenario, but no hydride shift.  Instead chloronium and then ring opening.

@Mithoron Is Hydride shift faster than the intramolecular attack of $\ce{Cl}$? Are the rates comparable? If yes, then I think we will get a mixture of different products. I don't know which will be major or minor.

@Mithoron I've edited the question. Is this what you were suggesting?

Is there a way to predict when these kinds of nucleophilic attacks occur?

The rate determining step is carbocation formation. After which, depending on conditions, the trapping of the cation may or may not be slow enough for [1,2]-shifts.
The [1,2]-shifts are probably under thermodynamic control if there is time for them to take place. This means that if there's time for the shifts, we're hunting for the most stable cation. Otherwise, you trap the most readily formed cation in a kinetic sense (which is not necessarily the most stable).
Assuming shifts are fast, I think your solution is better because the tertiary cation is more stable. Dichloromethane doesn't undergo substitution as easily as t-butyl chloride, so I would have to argue that the tertiary carbocation is more stable.

I think in this case, both kinetic and thermodynamic factors help the formation of carbonation at $1$.

You've already assumed the kinetic factor doesn't matter because it would mean that you would trap the cation where it's first formed without any migration.

As suggested by Mithoron, is there any chance of intramolecular attack by $\ce{Cl}$?

That's actually a really good suggestion, and probably better than either of the proposals you've presented to pick from. The chloronium ion is the accepted intermediate for halogenation of alkenes. This is an interesting way to access the intermediate.

Given the bond angles involved, I'm not convinced that the intramolecular attack is very likely.  Even though the chloronium ion is the accepted intermediate in the chlorination of alkenes, the mechanism of formation is completely different.

@MichaelLautman I would still consider it by microscopic reversibility arguments. I don't have access to literature searches anymore, but if this is possible, I would suspect that it would have been reported more than once. Want to look?

@Zhe . I just might.

@Zhe Can you provide the source of "Dichloromethane does not undergo substitution as easily as t-butyl chloride"?

@ApoorvPotnis Experience? I mixed dichloromethane and alchoholic solvents but there's been minimal reaction. On the other hand, t-butyl chloride reacts fairly readily with water/alcoholic solvent mix.

@Zhe "The chloronium ion is the accepted intermediate for halogenation of alkenes". Are you talking about addition of $\ce{Cl_2}$ or $\ce{Br_2}$?

@ApoorvPotnis Yes

@Zhe Do you think that this question of a worth of a bounty by me?

@ApoorvPotnis No, but it's your reputation, and you should use it as you see fit.


text:
Is it correct to say that ethenol is an alcohol?
IUPAC Gold Book defines alcohols as:

Compounds in which a hydroxy group, $\ce{–OH}$, is attached to a
  saturated carbon atom $\ce{R3COH}$.

However, Wikipedia says that:

Vinyl alcohol, also called ethenol (IUPAC name), is an alcohol.

Is Wikipedia wrong about this?
And if ethenol is not an alcohol, what would be the correct name for compounds in which an $\ce{–OH}$ group is attached to an unsaturated carbon atom?

Ethenol is not a thing at all, which pretty much makes the question void. Unsaturated carbons don't survive near -OH, unless we're talking about phenols.

Why is it not a thing? It's the enol tautomer of acetaldehyde... @IvanNeretin

A minor tautomer, so to say.

@IvanNeretin No argument there, but seems excessive to say "not a thing at all."

One could use a similar argument to say that tennessine is not a thing at all. Still there are a lot of scientific articles published about it.

A friend of mine just pointed out that compounds with their –OH group bonded to a $sp^2$-hybridized cardon should be called enols. (https://en.wikipedia.org/wiki/Enol) So, should the Wikipedia article about vinyl alcohol be fixed to reflect that?

related: https://chemistry.stackexchange.com/questions/729/differences-between-phenols-and-alcohols

The golden book states that ethenol is an alcohol:

Enols
Alkenols; the term refers specifically to vinylic alcohols, which have the structure $\ce{HOCR'=CR2}$.

However, the modifier "vinylic" is rather important, so I would never omit it.

So, should Wikipedia be edited? What would be a good wording? Would it be good to change it for example to: "Vinyl alcohol, also called ethenol (IUPAC name), is an enol. It's not an alcohol according to the IUPAC definition, because the $\ce{–OH}$ group is not attached to a saturated carbon atom."

I personally think that the IUPAC definition is not really on point, here's why: one needs to distinguish carbon acids, thiocarbon acids etc. and alcohols, and that's not easy to word. For wikipedia, the wording "Vinyl alcohol, also called ethenol (IUPAC name), is the prototypical vinylic alcohol" may be sensible.

The distinction "attached to saturated carbon atom" is quite on point. Also phenol is not an alcohol. Enoles are chemically even much more different from alcohols than phenols.


text:
How to make an alkaline wax mixture?
I want to treat a wax so that it has an alkaline pH of at least 10. I tried just mixing sodium hydroxide with beeswax but the sodium hydroxide reacted with the beeswax and changed its properties. I want the wax to retain its natural properties and not be chemically altered by the pH-increasing component. How can I do this?
If it matters, the ideal wax for my purposes is carnauba wax which is a wax that comes from the leaves of a tree, but beeswax could work as well. It would be nice to have a method that would work for any kind of wax.

Possibly adding an organic base that is miscible with the wax e.g. triethylamine

The following link from a very old text (pages 279-280) suggests boiling the wax in an alkaline solution. This method may be specific to that wax described as a new vegatable wax from a laurel like shrub from the forests of Para and Bahia. Please explain the application. Why do you want the wax to be alkaline? What are you going to do with the wax?


text:
CASSCF basis set
I have a question about the CASSCF method. According to what I have read, it is easiest to build the CASSCF active space by selecting specific localized orbitals from a small-basis DFT or MP2 calculation. With near-minimal basis sets, it is easy to identify important active space orbitals visually (e.g., $\sigma$, $\sigma^*$, etc.)
For large basis sets, it appears to be straightforward to visually identify the active space orbitals that originate from occupied localized orbitals. However, identifying active space orbitals that originate from unoccupied localized orbitals (e.g., $\sigma^*$) is more difficult because of the large number of virtual orbitals.
Does this mean CASSCF calculations are not run with orbitals computed from large basis sets?

No, this does not imply that CASSCF calculations are not run with large basis sets. This would mean that quantitative results were inaccessible with this method.
Instead you take your inital guess orbitals (these can be produced in various ways, DFT, MP2 or another CASSCF calculation) and project them on a larger basis set. Thus you can set up the CASSCF run relatively easy and still get decent results.


text:
Nitrobenzene and other monosubstituted benzenes 1H vs 13C NMR assignments
The reported 1H NMR spectrum for nitrobenzene has three signals corresponding to H2/H6, H3/H5 and H4, respectively. The signals (ppm) appear at 8.25 (H2/H6), 7.71 (H4), and 7.56 (H3/H5). This is, the ortho protons are the most deshielded, followed by the para, and, finally, the meta. This makes sense when rationalizing the inductive effect of the nitro group by resonance, which makes the ortho and para carbons deficient in electrons, hence deshielding them. Furthermore, the ortho protons are closer to the nitro group, hence they are the most deshielded of all, so the from more to less deshielded: ortho > para > meta.
However, the reported 13C NMR spectrum shows signals at 148.3 (ipso), 134.7 (para), 129.4 (meta), 123.5 (ortho). So from more to less deshielded, the positions are now ipso > para > meta > ortho. Why is the trend so different for the carbon-13 NMR?
Also the trends for other monosubstituted benzenes are:
Benzonitrile. 1H NMR: ortho > para > meta; 13C NMR: para > ipso > ortho > meta. Here the trend changes a bit as well, but now only between ortho and para.
Benzaldehyde. 1H NMR: ortho > para > meta; 13C NMR: ipso > para > ortho > meta.
Aniline (EDG). 1H NMR: meta > para > ortho; 13C NMR: ipso > meta > para > ortho. Here the trend is inverted, but why would the para still be more deshielded than the ortho as seen in all the other cases?

Looking at the electron density is very important for 1H chemical shifts (dominated by diamagnetic shielding), but less so for 13C shifts (dominated by paramagnetic shielding). In general it's fairly common for aromatics to have "counterintuitive" patterns in 13C shifts, like the ones you point out, simply because 13C shifts don't play by the same rules that 1H does. I'm not sure if I can find a good explanation for aromatic rings, but if I do then I'll come back to this.

I kinda thought it was something along those lines, but couldn't find anything concrete on it, or something I could understand. I'm a PhD student but NMR and PChem aren't my strong suit. Also the question came from a general OChem 1 homework (I'm a tutor), so I was looking from an "low level school answer", but it might as well be the professor didn't even bother to look at the answer himself.

I don't really think there's a good explanation at that level. Tried to find an explanation, but can't really find anything except that "at the ortho position, additional steric effects may operate" - the answer is probably out there but I'm not sure where to find it right now.

I can only find the pyridine/PhLi examples, but I'm having a hard time bringing that to the examples I'm showing. For now at least I know "why", but I still hope someone can explain why. https://www.chem.wisc.edu/areas/reich/nmr/06-cmr-02-shifts.htm

If you want, I can write up a step-by-step derivation of where the diamagnetic and paramagnetic terms come from. I can also provide a qualitative physical description of the origin of these terms, but as orthocresol notes, there is not really a good way of arriving at these conclusions for a general aromatic molecule. The energy differences at play in these systems are so small that qualitative arguments will often be insufficient to get the order right. Let me know what if any of these will help you to why.

If you can provide a qualitative description it would be much appreciated. I am not all that familiar with the paramagnetic term (or maybe I am but I don't know that's the name). I feel the derivation would go over my head, but if you're bored and want to go for it, who am I to stop it? Maybe someone is interested in that.

As I wrote in the comments, there is not really a good way of arriving at these conclusions for a general aromatic molecule. The energy differences at play in these systems are so small that qualitative arguments will often be insufficient to get the order right. The diamagnetic term, which can be qualitatively rationalized by arguments based on electron density, for non-H atoms, is dominated by the paramagnetic term, which is, in some sense, an orbital response term due to the application of the magnetic field of NMR. Qualitatively reasoning about the paramagnetic term is possible in some special systems like PhLi or pyridine, but the accuracy of these qualitative arguments is insufficient for general substituted benzenes. The paramagnetic term can be calculated by electronic structure packages, but I assume that is not really the question.
I wrote some lecture notes a few years ago for a lecture I gave on the subject of diamagnetic and paramagnetic terms in NMR. I tried to pare it down a bit but they are still rather long. I will start with the qualitative quasiclassical conclusions and then have the derivation.

Qualitative Explanation of Diamagnetic and Paramagnetic Terms
The diamagnetic and paramagnetic terms in NMR shifts arise from considering how the electron motion is affected by an applied field. The terms diamagnetic and paramagnetic only indicate the sign of the magnetic response, opposed to the field and with the field, respectively (it does not have anything to do with unpaired electrons, which is the place most chemists have heard these words before). The diamagnetic term can be thought of being like Lenz's Law in classical electromagnetism; applying a field induces a current which opposes the field. This is sort of (with excessive handwaving) why electron density correlates with how shielded the nuclei are: with less electron density around, the amount of current is less, the induced field opposing the applied field is less, and the effective field is larger.
For the paramagnetic term, think of an electron moving (remember this a quasiclassical picture, but it's not wrong) in a $p$ orbital. It has angular momentum $l=1$, including an $m_l$ value of $+1$ which would generate a huge field and an $m_l$ value of $-1$ which generates the same field and in the opposite direction (also $0$ but that's not important here). Hence, in the absence of a field there is no field due to this orbital angular momentum being ``quenched'', that is $m_l=\pm 1$ are represented equally in the ground state wavefunction and exactly oppose each other. If a field is applied though, the $m_l$ aligned with the field (let's say $+1$) is stabilized while the $m_l=-1$ is destabilized and so there is more of one than the other in the ground state wavefunction at that angular momentum is unquenched. This results in the addition of a field in the same direction as the applied field, making it larger. This is the origin of the much larger range of shifts for $\ce{C}$ as opposed to $\ce{H}$. Calculating the magnitude of the paramagnetic term is not super straightforward and so qualitative arguments to try to rationalize the order of simple aromatic molecules is hard to do (examples like PhLi or pyridine are possible because there is an obvious paramagnetic perturbation, the lone pair). The amount of paramagnetic shifting tells us something about the low-lying vacant orbitals of the system and so can be a useful tool for predicting reactivity in some cases, but predicting those shifts without calculating them explicitly is not readily possible.2
In hydrogen, which is essentially only an $s$ orbital (which has an angular momentum of 0), there is no paramagnetic term (because there is no orbital angular momentum to unquench) and all of the shifts are due to the diamagnetic term. For heavier elements, like C, where $p$ orbitals get involved, there is a paramagnetic term and it is dominant.

Derivation of the Diamagnetic and Paramagentic Terms
Classical electromagnetism tells us that the magnetic field $\textbf{H}$ generated by a charge $q$ moving with velocity $\mathbf{v}$ at a point $\mathbf{r'}$ away is
$$\textbf{H}=\frac{q}{c}\frac{\mathbf{v}\times\mathbf{r'}}{r'^3}$$
Since we care about the field at the nucleus, we set the coordinate system origin at the nucleus and consider the field at the origin of a charge a distance $\mathbf{r}$ away. Then $\mathbf{r'}=-\mathbf{r}$ and we have
$$\textbf{H}=\frac{q}{c}\frac{\mathbf{r}\times\mathbf{v}}{r^3}=\frac{q}{mc}\frac{\mathbf{r}\times\mathbf{mv}}{r^3}=\frac{q}{mc}\frac{\mathbf{L}}{r^3}$$
We'll need this later. Let's consider though: the above implies that $\textbf{H}$ is on the order of $\approx \beta/r^3$, where $\beta$ is the Bohr magneton and $(1/r^3)$ for a 2p orbital is about 0.25 Å and so $H$ is about 600,000 gauss, which is massive! This should bring up the question: if non-zero $\mathbf{L}$ leads to huge magnetic fields why don't these fields completely destroy our NMR signal? Are there magnetic fields present in diamagnetic molecules when there are no magnetic fields present? I've sort of already hinted at the answer. No, there aren't such huge fields. Loosely, this is because the field generated by the electrons circulating in the $m_L = +1$ level exactly cancels the field of the electrons in the $m_L=-1$ because both are equally represented in the ground state (as they must be without a magnetic field present). This quenching of angular momentum occurs in all cases where the orbitals are real valued, such as if they are solutions of a Hamiltonian in the absence of a magnetic field (so nothing with unpaired spins, which carry a magnetic field). If we turn on an $\textbf{H}_0$ field though, one of the directions of circulation will be favored over the other (the $m_L=-1$) and the wavefunction will adjust so that the ground state has a slight net circulation in the favored direction. In this way, an applied magnetic field unquenches a small amount of the orbital angular momentum. This gives rise to the paramagnetic contribution.
Let's develop this mathematically. First, we should show that our semi-classical picture is a valid way to consider this quantum mechanical system. Fundamentally, a magnetic field $\textbf{H}$ is given by the curl of some vector potential $\textbf{A}$. We ultimately have two fields to consider: $\textbf{H}_0$, the applied field, and $\textbf{H}_n$, the field generated by the nucleus of the atom, each with their associated vector potentials.
\begin{align}
\textbf{H}_0&=\nabla\times \textbf{A}_0\\
\textbf{H}_n&=\nabla\times \textbf{A}_n
\end{align}
We're going to neglect the magnetic fields of the nucleus for this discussion. Obviously, they are important, since the perturbation of the nuclear spin energy levels are mediated by the coupling of the nuclear magnetic fields to the electronic magnetic fields, but we first need to determine what those electronic fields are.
As you may recall from classical electromagnetism, there is more than one vector potential that produces a given field. Specifically, if $\textbf{H}=\nabla\times\textbf{A}$, another vector potential $\textbf{A}' = \textbf{A}+\nabla\phi$, where $\phi$ is a scalar function, gives the same field, since the curl of a gradient of a function is always 0. The transfomation $\textbf{A}\rightarrow\textbf{A'}$ is called a gauge transformation. We need all physical results that we calculate to be gauge-invariant.
So now, how does an applied magnetic field show up in our Hamiltonians. It appears by changing $(\hbar/i)\nabla\rightarrow(\hbar/i)\nabla-(q/c)\textbf{A}$. So the Hamiltonian becomes
$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}\right)^2+V$$
If we were to calculate expectation values of $\hbar/i\nabla$ with different gauges, we would find that they aren't equal and so $m\mathbf{v}$ is no longer $\hbar/i\nabla$ but is $\hbar/i\nabla-\frac{q}{c}\textbf{A}$. Similarly, $\mathbf{L}$ is now
$$r\times\left(\frac{\hbar}{i}\nabla-\frac{q}{c}\textbf{A}\right)$$
We define something that will be useful to us throughout this discussion, the current density $\mathbf{j}(\mathbf{r})$.
\begin{align}\label{current}\tag 1
\mathbf{j}(\mathbf{r})=\frac{q}{2m}\frac{\hbar}{i}[\psi^*\nabla\psi-\psi\nabla\psi^*]-\frac{q^2}{mc}\textbf{A}\psi^*\psi
\end{align}
Note that it is a function of position and real-valued. It is also $q$ times the quantum mechanical probability current and gauge-invariant. If we assume $\psi$ is a solution of the time-dependent Schrodinger equation, then
$$\text{div} \mathbf{j}+\frac{\partial \rho}{\partial t} = 0$$
where $\rho = q\psi^*\psi$ (that is, the charge distribution). For pure states, $\psi^*\psi$ is time-independent and so $\text{div} \mathbf{j}=0$, and hence behaves as a classical current density. One can do some more math and show that it is possible to express the chemical shift, a quantum mechanical phenomenon, entirely classically as the coupling of the the nuclear magnetic moment to the effective magnetic field at the nucleus due to the circulating density $\mathbf{j}_0$.
\begin{align}\label{chemshiftE}\tag 2
E_\text{pert}=-\mathbf{\mu}\cdot\left[\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}\right]
\end{align}
So if we can calculate $\mathbf{j}_0$ we have done the hard part of finding the chemical shift. Let's turn to that now.
Now, the Hamiltonian for our electron is
$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}_0\right)^2+V$$
We multiply out the binomial and rearrange to get
$$\mathcal{H} = \frac{p^2}{2m}+V-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})+\frac{q^2}{2mc^2}A_0^2$$
Let's assume we have the eigenfunctions of $\mathcal{H}_0=\frac{p^2}{2m}+V$ in the absence of the field (i.e. $\mathcal{H}_0\psi_n=E_n\psi_n$). Our goal is to calculate $\mathbf{j}_0$, which has a term linear in $\textbf{A}_0$, which looks like
$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$
That's one choice of gauge, but in any gauge $\textbf{A}_0$ is proportional to $\textbf{H}_0$. So if we want $\mathbf{j}_0$ right to first order, we need $\psi$ at least accurate to first order in $\textbf{H}_0$ (the $\psi^*\psi$ in the last term of $\mathbf{j}_0$ can be the unperturbed wavefunction since that term already has a $\textbf{A}_0$). So let's do some first-order perturbation theory! In case you forgot your first-order PT, we first need to define the perturbing Hamiltonian. We'll use the linear part of our Hamiltonian above.
$$\mathcal{H}_1=-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})$$
Our first-order corrected ground-state wavefunction (this is just standard first order PT) is
$$\psi'_0=\psi_0+\sum_n{\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}\psi_n}$$
where $E_0$ is the ground state energy, $E_n$ is the energy of the $n$th excited state $\psi_n$. Let's define
$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}$$
So that
$$\psi'_0=\psi_0+\sum_n{\epsilon_{n}\psi_n}$$
Substituting this $\psi'_0$ into the equation for $\mathbf{j}$ (equation \ref{current} above), we obtain
$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)+\frac{\hbar q}{2mi}\sum_n\left[{(\psi_0^*\nabla\psi_n-\psi_n\nabla\psi_0^*)\epsilon_n+(\psi_n^*\nabla\psi_0-\psi_0\nabla\psi_n^*)\epsilon_n^*}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$
The first term
$$\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)$$
is the current flowing in the absence of a magnetic field. If our angular momentum is quenched as we described above (as in closed-shell molecules), then this is 0. If we take the $\psi_n$ as real, then
$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}\sum_n\left[{(\epsilon_n-\epsilon_n^*)(\psi_0\nabla\psi_n-\psi_n\nabla\psi_0)}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$
We're gonna see that the first term is the paramagnetic current, related to $\epsilon_n$, which recall is the matrix element coupling the ground state to excited states by the angular momentum operator, and the second term is the diamagnetic current, always present and independent of $\textbf{L}$. Let's go back to that choice of the vector potential $\textbf{A}_0$ we made before
$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$
Subbing that into the perturbing Hamiltonian gives
\begin{align}
\mathcal{H}_1&=-\frac{q}{2mc}(\mathbf{p}\cdot\frac{1}{2}\textbf{H}_0\times\mathbf{r}+\frac{1}{2}\textbf{H}_0\times\mathbf{r}\cdot\mathbf{p})\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot(\mathbf{r}\times\mathbf{p}))\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot\mathbf{L})
\end{align}
Where that $\mathbf{L}$ is in the absence of a field. Then we have that
$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}=\frac{qH_0}{2mc}\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle$$
where I assumed that $\textbf{H}_0$ is some magnitude $H_0$ along some unit vector direction $\mathbf{k}$. Now, going back to our chemical shift energy (equation \ref{chemshiftE}), we want the field that this current $\mathbf{j}_0$ generates. This is
$$\textbf{H}=\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}$$
Substituting $\mathbf{j}_0$ and $\mathcal{H}_1$ in gives
$$\textbf{H}=H_0\frac{q^2\hbar}{2m^2c^2}\sum\frac{\langle 0|\sum_j\frac{\mathbf{L}_j}{r_j^3}|n\rangle\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle+\langle 0|\mathbf{k}\cdot\mathbf{L}|n\rangle\langle n|\sum_j\frac{\mathbf{L}_j}{r_j^3}|0\rangle}{E_n-E_0}-\frac{q^2}{2mc^2}H_0\langle 0|\sum_j\mathbf{r}_j\times(\mathbf{k}\times\mathbf{r}_j)|0\rangle$$
where $\mathbf{k}$ is the direction of the applied field and the $j$ subscripts denote electron number in the wavefunction.
For an $s$ orbital, all matrix elements with $L$ are 0 and so the first term is 0. This leaves only the second term which is opposed to the applied field. For anything with higher angular momentum than $s$ orbitals, the first term will be nonzero and aligned with the field. This term arises due to unquenching of the orbital angular momentum in the presence of a magnetic field and hence the direction of the paramagnetic field was determined by the projection of angular momentum which aligns with the field, rather than arising out of Lenz's Law. Said yet a different way, the orbital was already a magnet, it's just that we couldn't see it before. The applied field does not induce the quantum mechanical current flow or paramagnetic field (in the sense of classical EM), merely reveals it by mixing more of the aligned magnet into the ground state wavefunction than the anti-aligned one.
Putting some reasonable numbers into these equations gives that the paramagnetic term is larger at distances less than 1 Å from the nucleus, while the diamagnetic term dominates outside this range. As we mentioned above, the distance important for chemical shifts (average $r^3$ for a $2p$ orbital) is about 0.25 Å and so the paramagnetic term will be larger for NMR related problems.  On the other hand, these paramagnetic terms play practically no role in macroscopic magnetic moments of molecules.
References:

Slichter, Charles P. Principles of Magnetic Resonance (Harper and Row Publishers, 1963).
Gordon, C. P., Yamamoto, K. , Liao, W., Allouche, F., Andersen, R. A., Copéret, C., Raynaud, C., Eisenstein, O. Metathesis Activity Encoded in the Metallacyclobutane Carbon-13 NMR Chemical Shift Tensors. ACS Cent. Sci., 2017, 3 (7), 759–768. DOI: 10.1021/acscentsci.7b00174

I absolutely like what you've written, but if you could add a small paragraph which directly addresses the question (similar to what you wrote in your comment), that would be great and I would be more than happy to assign the bounty when the bounty period ends. I was hoping that there was a direct rationalisation out there, but I increasingly realise that not every question has a well-defined answer...

Added. The diamagnetic and paramagnetic terms can be calculated by an electronic structure package and will probably give the qualitatively correct ordering. The problem with chemistry in general is that qualitative arguments are often only vaguely correct, but the level of accuracy needed to exactly nail down things requires explicit computation. So it has a well-defined answer, just not an easy one. For example, you could make a qualitative argument about the paramagnetic term that $sp^2$ carbons will be more down field than $sp^3$, but distinguishing among $sp^2$ is less straightforward.

You can also, armed with a calculation, point to specific orbital interactions that account for the origins of the paramagnetic term. Of course, this description is not unique and will depend on your orbital basis, but if you really wanted to interpret the origin of the paramagnetic term, not all hope is lost.


text:
Phosphorous acid vs phosphonic acid?
I want to know the difference b/w phosphonic acid and phosphorous acid.
The formula for both is H3PO3.
Websites give me conflicting answers.
Which one of them is dibasic and which one is tribasic?

Strongly related: https://chemistry.stackexchange.com/questions/40506/what-does-the-prefix-meta-mean-in-inorganic-chemistry/40517

https://chemistry.stackexchange.com/questions/72954/how-did-hypophosphoric-acid-get-its-name/72960#72960

Somewhat related but unsourced: https://chemistry.stackexchange.com/q/50427

The difference is that phosphonic acid ($\ce{HPO(OH)2}$) is an existing, shelf-stable compound while phosphorous acid ($\ce{P(OH)3}$) is extremely short-lived if at all and immediately tautomerises to phosphonic acid.
All the -ous acids are theoretically produced in the same manner: take the -ic acid and remove one unprotonated oxygen, leaving an additional electron pair with the central atom which reduces its oxidation state by two. So while sulphuric acid is $\ce{SO2(OH)2}$, sulphurous acid is $\ce{SO(OH)2}$.
Likewise, you can always theoretically generate the -onic acids by tautomerisation from the ‘parent’ -ous acid; one proton that previously was attached to an oxygen bonds to the central atom‘s newly-formed (when coming from the -ic acid) lone pair. Thus, sulphonic acid is $\ce{HSO2(OH)}$.
In case of sulphur, while both acids have been observed and crystal structures of both have been determined to the best of my knowledge,[citation needed] the two can still tautomerise. In the case of phosphorus, only the phosphonic acid is shelf-stable and all isolated salts have been determined to be phosphonate salts.[citation needed]
In both cases, however, organic esters have been generated in both ways. Thus, you can have e.g. triethyl phosphite $\ce{P(OEt)3}$ or diethyl ethylphosphonate $\ce{EtPO(OEt)2}$.

As already mentioned by Jan, phosphonic acid actually the stable, existing compound while phosphorous acid is unstable, short-lived and quickly tautomerizes to phosphonic acid. Wikipedia supports this fact:

$\ce{H3PO3}$ is more clearly described with the structural formula $\ce{HPO(OH)2}$.
  In the solid state, $\ce{HPO(OH)2}$ is tetrahedral with one shorter $\ce{P=O}$
  bond of 148 pm and two longer $\ce{P–O(H)}$ bonds of 154 pm. This species
  exists in equilibrium with an extremely minor tautomer $\ce{P(OH)3}$. IUPAC
  recommends that the latter be called phosphorous acid, whereas the
  dihydroxy form is called phosphonic acid. Only the reduced
  phosphorus compounds are spelled with an "ous" ending.

$\ce{H3PO3}$ is described by the name of phosphorous acid but the name is actually quite misleading. As mentioned in Wikipedia, $\ce{H3PO3}$ actually has the structural formula $\ce{HPO(OH)2}$(dihydroxy form) which is actually called phosphonic acid and for which the acid is diprotic and not triprotic as might suggested by the formula. The tautomer of this acid is $\ce{P(OH)3}$ which is called phosphorous acid. This is unstable and quickly converts to its dihydroxy form.
Sometimes things may look very obvious but if you look for the story behind it, you may find that 'it was not that I thought of'.


text:
Solid with Boiling Point less than 100 °C
For an experiment I require a substance which is a solid at room temperature and boils at a temperature lower than 100 °C. Also, this solid should preferably be soluble in water. Are there any such compounds?

If you're going to edit your question in response to an answer, you should make it clear how you've changed it.  If you don't, then you can make an answer look bad, as you have to mine.  That is at best careless, at worst unethical.

My bad check edits

Ammonium bicarbonate seems like a winner.  It:

Is a solid at RTP.
Is highly soluble in water.  At RTP, the solubility is about 3 moles of $\ce{NH4HCO3}$ per liter.
Decomposes at high temperatures to a gas.  Well, actually, two gases: $\ce{CO2}$ and $\ce{NH3}$.

This decomposition is not truly "boiling", but it does involve a transformation into gases.

You haven't specified a melting point, and by 'turns into a gas' you haven't specified boiling.
Iodine sublimes readily and is also sparingly soluble in water.
If you really mean 'solid at RTP, boils below 100 and readily soluble in water' then clearly Iodine doesn't meet that more stringent requirement.

You could add iodide to the water.


text:
Metallic Bonding
Since in metallic bonding, metals form ions and are surrounded by delocalized electrons, does that mean metal atoms always react in this ionised state? Can metal atoms lose valence electrons (like $\ce{Na}$ to $\ce{Na+}$) in chemical reactions, when they have already lost all their valence electrons in metallic bonding?

When the metal reacts to form an ionic compound, the electrons transferred from the metal actually come from the metallic bonding rather than from the ions proper.  The metal ions are in effect spectators, while the metallic bonding between them is broken in favor of forming stronger ionic bonds.  Breaking bonds to make stronger bonds is no different from reactions between nonmetallic substances.

While this may be attractive rationalisation, bonds cannot be treated as separate entities. In reality, there are no ions or free electrons in metals.

Is the O.P. ready for such subtlety?

@Mithoron If there are no free electrons in metals,then how do it forms a delocalized electron cloud???

Yeah, it seems he isn't, but posts should be useful for as broad group of users as possible.

This is more or less accurate: https://chemistry.stackexchange.com/a/8082/9961


text:
Deciding which α-Η gets abstracted in aldol condensation
Consider the base catalysed aldol condensation of methyl ethyl ketone. The first step involves the abstraction of a proton by the base. However I am having trouble deciding whether the proton will be abstracted from the $\ce{-CH2-{}}$ group or the $\ce{-CH3}$ group.
I think it should be from the latter as the $\ce{CH^-}$ ion so formed in the former experiences undesired $+I$ effect by the two groups neighbouring it; but I'm not sure.

There have been tons of articles, review articles and even books written on the subject. The tl;dr version is that it depends on the conditions you use.
In general, the less-hindered enolisation (in your case: enolising towards the methyl) will be more rapid. This leads to the possibility of kinetic control to selectively generate the terminal enolate by very strong bases and low temperatures such as LDA or even BuLi if you’re daring (both at $\pu{-78^\circ C}$).
On the other hand, the more substituted enolate is more stable thermodynamically as you noted. Thus, if you use a weak base that binds and deprotonates reversibly and if you use relatively high temperatures or longer reaction times, the more stable enolate will be formed. An example for these conditions would be an alcoholate base at room temperature.
Unfortunately, I can give you no simple rules. You will have to check the literature or your lecture notes to see which combinations of reagents give which selectivity.

Why make LDA from nBuLi when you can just use nBuLi, amirite... Anyway, I don't think OP said that the more substituted enolate is more stable; they were implying that the more substituted enolate was less stable, because the carbanion resonance form would be a secondary carbanion, vs a primary carbanion when abstracting a proton from the methyl group.

@orthocresol Now that I reread OP’s statement, yes they did say that. However, since the charge is on the oxygen in the canonical structure, I’ll just stick with it ;)


text:
Electron donating ability of substituted amines on aromatic ring
How would you arrange the conjugate bases of the following amines in order of electron donating ability as EDGs as substituents on an aromatic ring?
$\ce{NH3}$, $\ce{(C2H5)(C2H4CN)NH}$, $\ce{(C2H4OH)2NH}$, $\ce{(C2H5)(C2H4OH)NH}$
I would think that the order would be:
$\ce{(C2H4OH)2NH}$ > $\ce{(C2H5)(C2H4OH)NH}$ > $\ce{(C2H5)(C2H4CN)NH}$ > $\ce{NH3}$
based on the reasoning that alkyls cause an inductive effect which increases electron density.
I think that pKa would be able to measure the electron donating ability of the conjugate base.
However, I found that $\ce{NH3}$ has a higher pKa (9.25) than $\ce{(C2H4OH)2NH}$ (8.96), which would mean that the conjugate base of $\ce{NH3}$ is a stronger base, which I don't understand.
Is my reasoning for the order incorrect? Or is the usage of pKa to measure electron donating ability of the conjugate base incorrect?

Hmm. Neither the amines nor their conjugate bases can be substituents on a ring; that role is reserved for the corresponding amino groups. I’m also not sure whether the $\mathrm pK_\mathrm a$ values have anything to do with it.

Ah my bad, I am trying to refer to the corresponding amino group.  If pKa values don't have anything to do with electron donating ability of the amino groups as EDGs, based on what is the order of the strength of EDGs determined?


text:
Why do acid base reactions cause the reacting or forming molecules to accelerate? (Basically, why do acid base reactions produce heat?)
My question is hopefully pretty clear, but to rephrase my question:
Why do acid base reactions produce heat? We know that heat is merely the transference of momentum (or velocity) between molecules. So, if that is the case, and acid base reactions do produce heat, then the reaction of (for example) H₂O and NaOH MUST be resulting in an increase of the molecules' velocity. The question is, why would there even be an increase of molecular velocity during acid base reactions?
Note: Using terms like 'potential energy', 'enthalpy', 'electrical potential', 'lower energy state', and 'energy' are forbidden, because the question looks for an answer that is purely based on 2 observable rules.
1- An object will accelerate if it is subject to a force
2- Electric charges can apply forces on each other.

related https://chemistry.stackexchange.com/questions/40885/heat-dissipation-of-chemical-reaction

You aren't, in fact, in any position to forbid us anything.

also https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding?s=1|58.4810

@Mithoron 2 of the very links that you suggested contain nothing of the answer that I am looking for. The third one, which is the longer of the 3, probably too does not have it, meaning, I have not read it. If you are not  able understand the question I asked and stated clearly, and the kind of answer that I asked for, then you really need to doubt that intelligence of yours. I forbad people from giving me an answer containing the terms I listed. This basically meant that I just DO NOT WANT such answers. If you think I am wrong to ask this, then argue with me as to why.

Watch your language - asking somebody to “doubt their intelligence” is extremely inappropriate and will not be tolerated here. Consider this a warning.

@EmirhanG. The warning was about being insulting.  You're continually justifying yourself when all you needed to do was not be insulting in the first place.  We have a very strict Be Nice policy throughout Stack Exchange that strongly discourages such behavior.  Ortho was not acting on "Ego," but instead emphasizing that moderators are well within their rights to suspend an account if a member becomes problematic.  You're teeter-tottering on the edge right now, so please choose your words and your manner carefully from now on.

Guys, ok, you are right, I was being rude. But how else was I supposed to complain about the fact that my question was being bombarded with undeserved dislikes and that it was being tagged as a duplicate while the person who was doing that clearly had not even understood my question and the answer that I had asked for? He just jumped into conclusion like that. That is ruder.

Strong acids and bases like $\ce{HCl}$ and $\ce{NaOH}$ dissociate fully in water. Let us consider dissociation of $\ce{HCl}$. $\ce{HCl}$ in water dissociates as this.
$$\ce{HCl <=> H+ + Cl-}.$$
Now, initially, the $\ce{H}$ and $\ce{Cl}$ are covalently bonded to each other. If we want to break this bond, we have to separate the positive and negative charges and this needs energy. Water molecules are in constant motion. They have kinetic energy. When you add $\ce{HCl}$ to water, some of the water molecules collide with the $\ce{HCl}$ molecules and break them. Now, the potential energy of the $\ce{H+}$ and $\ce{Cl-}$ system increases(becomes less negative). Thus the $\ce{HCl}$ molecules absorb some of the kinetic energy from the water molecules and some of this energy is stored as potential energy.
But as $\ce{H2O}$ is a polar molecule, the water molecules get attracted to the $\ce{H+}$ and $\ce{Cl-}$ ions. The water molecules hydrate the ions and this decreases potential energy of the system(water molecules + ion). This decrease in potential energy is marked by an increase in kinetic energy of the hydrated ion and the surrounding water molecules.
So, as dissociation takes kinetic energy of the system and converts it into potential energy, this process should decrease the temperature of the system. Similarly, hydration should increase the temperature of the system. To decide which effect is more dominant, we have to take into consideration $\Delta H_\text{Lattice Energy}$ and $\Delta H_\text{Hydration Energy}$ to decide which of the overall process of dissolution of $\ce{HCl}$ is exothermic or endothermic.
Similarly, using Hess' Law you can calculate if the acid-base reaction will be exothermic or endothermic.

The first paragraph of your answer was pretty nice, but the second paragraph used some unnecessary and (personally) disliked terms, which I had forbidden to be used in the answer.
Also, your answer suffices only to explain the production of heat when 'separate solutions of an acid and a base get mixed, but not to explain the production of heat when just an acid or a base is dissolved in water. A few weeks ago in the lab, I had dissolved some KOH in water and the solution had gotten pretty hot.

I used the energy terms to supplement the force-momentum explanation. When you'll be doing calculations in chemistry, the the energy or enthalpy changes are easier to measure and calculate. So I added a paragraph on it.

I've edited the answer. Please see it again.

What happened to covalent bonding? If this is just about ions attracting each other and accelerating, then salt water should spontaneously heat up, since all the ions inside there are constantly being accelerated. And also the NaCl should spontaneously precipitate, since the Na+ and Cl- ions want to be next to each other.

@orthocresol Yes. I made a mistake. There is no ionic bonding between $Na$ and $\ce{Cl}$ in solution or it'll precipitate.

Apoorv Potnis, could you elaborate on the last paragraph? I couldn't understand you.

How do you extend this type of explanation to the heat of combustion?

@orthocresol I've edited the answer. Is it ok now?


text:
Why do Alpha particles not collide with electrons during alpha decay?
Alpha particles are positively charged, so my question is that during alpha decay, when these are released from the nucleus, why doesn't it ever occur that they collide and grab an electron or two since they are essentially a Helium nucleus?

Why do you think it never happens?

I always thought this was because the alpha particle was ejected with too much energy to grab any electrons from its parent element's electron cloud. The alpha particle needs to be slowed down before it can grab electrons. I don't know if this is so, just what I assumed (presumed?). Or to put it another way, by the time any electrons could "feel" the alpha particle's influence, it was long gone.

Actually they do collide.  But, the emitted alpha particle carries much more energy than the binding energy of the electron(s) in a helium atom or $\ce{He^+}$ ion.  Therefore such a collision scatters the electrons (which carry away some of the original energy of the alpha particle) instead of forming an atom.


text:
Why is double my half-equivalence point not equal to my equivalence point?

Tris pKA = 8, therefore at pH = 8, the volume is 4.5mL. However the equivalence point simply can't be at 9mL, looking at this graph. It should be between approximately 9.5 and 10.5, no?

You were titrating diluted tris?

Yes, 100mL of 0.1M tris.

It's rather weak base and conc. was low so this equiv. point simple is only slightly visible.

But how do I determine the actual value?

https://chemistry.stackexchange.com/questions/28054/calculating-ph-for-titration-of-weak-base-with-strong-acid


text:
Finding CML files for small molecules
I'm trying to generate a figure for a physics textbook to show examples of the vector addition of dipole moments in small molecules. For this purpose, I would like to render pictures of a bunch of small molecules, such as CS2, formaldehyde, benzene, SF6, and chloroform. I've got avogadro running on my linux box, but I'm finding it very time-consuming and confusing to construct the molecules using the GUI. Is there anywhere that I could find pre-made CML files for small molecules like these?

Do you have access to chemoffice? Chem3D can produce 3D structures from formulas and if you run the optimization in there it should be good enough for this application. Other than that: how many molecules are we speaking about? With the right software making each of those takes less than a minute each including low level geometry optimization. If it's not too many I'm happy to help, we can speak in chat.
Another option: you can open the 3D model in jmol from their wikipedia page and extract the *.mol file which to my knowledge can be opened by Avogadro.

@DSVA: Thanks for the ideas, and for the offer of help. This pointed me to the info that I'll give in an answer to my own question.

DSVA's comment pointed me to a database that contains the information I need. The wikipedia articles contain links to an app called jsmol, but that app appears to be getting its data from a database called ChEBI. You can search the database using search terms such as "formaldehyde" or "sf6." The entry for each molecule contains a link to a .mol file, which avogadro can read.

Another location, if you're ok with the mol2 format (which Avogadro can open), PQR has links to download structures on every page. For example, on the cadaverine page:

This will trigger a browser download dialog box for a mol2 file.

Consider a blend of Wikimedia Chemical Structure Explorer and OpenBabel.

Wikipedia includes many entries about chemicals, and often, the property box right on top of the page (left, or right, depends on your script) include the condensed structure representation as a SMILES string:

(credit, the Arabic entry about toluene)
With above mentioned freely available, open source structure explorer which you may run without installation from a web browser (further description, e.g., in J Cheminform 7, 10 (2015)) you may query all of these entries by a (sub)structure search, a keyword, or a combination of both to access the English article and find the Wikipedia page on the right hand pane:

(screen photo of the explorer)
As by today, the public index contains 18484 entries.

With the help of OpenBabel, you may convert the SMILES string into the CML format (the the documentation presents many additional input/output formats, too).   After installation of OpenBabel, the program may used from the CLI / cmd.exe, or from a GUI.  Then, you have at least two options:
a) You save the SMILES string from wikipedia in as file, for example toluene.smi.  If you opt for a conversion from the CLI, the command
obabel -ismi toluene.smi -ocml -O toluene.cml
writes the new file toluene.cml.
b) You paste the SMILES string into the command on the CLI when writing
obabel -:CC1=CC=CC=C1 -ocml -O toluene.cml
to generate the .cml file right in the current working directory. (No, do not set a space between the colon and the SMILES string.)
The documentation lists many more examples and showcases the use of the GUI.  OpenBabel may work on multiple input files at once (e.g., obabel *.xyz -opdb -m to convert all .xyz of a folder into .pdb) and interacts well with other programming languages, too (introduction).

If you have Microsoft Word installed on a Windows machine you can use our open source Add In called Chem4Word to search the ChEBI database and import structures.
Once imported they can be optionally edited, then exported as CML.
You can find Chem4Word here https://www.chem4word.co.uk/download


text:
Formulation solubility?
Creatine is a popular sports/fitness supplement, generally sold as a powder to be mixed with water (usually like 5-10g creatine per serving). However, it is not very "soluble" in water -- barely 13.3g/L at 18°C. To enhance solubility, some manufacturers sell it as a creatine-HCl formulation -- bringing it up to 88.8g/L!
This formulation, of course, is also more expensive. It makes sense why lowering the pH (making more of the creatine molecules positively charged) would make creatine more soluble, but I'm not sure if there is anything "special" about the HCl component here.
Namely, at home I have regular creatine monohydrate powder (not HCl) and a bag of anhydrous citric acid. I haven't measured the exact increase in creatine solubility after the addition of citric acid, but solubility does increase a little bit and the solution is quite sour as expected.
What I am wondering, is there anything "special" about the creatine-HCl formulation that would greatly bump up creatine solubility that could not just be done with regular creatine (non-HCl) powder and some citric acid powder to lower the pH? I know citric acid is not as "strong" an acid as HCl, but regardless whether it's HCl or citric acid that's used, the final solution pH still has to be safe for oral consumption. And adding citric acid already makes things really sour!

Strong acid + weak base doesn't necessarily equate to strongly acidic mixture. Well, it does, if you add too much HCl. But I'd assume that it is measured quite carefully.

@orthocresol Ahh you're right, good point!  I've edited my question now

Nah, citric acid should do just fine.

Related https://chemistry.stackexchange.com/questions/20309/why-formulate-drugs-as-hcl-salts-when-hcl-is-in-stomach-acid?rq=1 https://chemistry.stackexchange.com/questions/85032/difference-between-sildenafil-and-sildenafil-citrate

@Mithoron Ahh, so there really is nothing special about HCl here that couldn't also just be done with citric acid solution instead. Saves me quite a bit of $$$ too, since, at least for me, simple anhydrous citric acid is quite more available...


text:
The difference between bonding and unpaired single electron radicals
I was wondering as to why would two unpaired single electron radicals form a bond. So, I tried to compute the added ground state energies of two particles in two distinct, yet identical boxes, The energy of ground states would be $E_1=\frac{\hbar^2\pi^2}{2mL^2}$ each, so the total energy of the system would be $E=\frac{\hbar^2\pi^2}{mL^2}$. But when I compute the ground state energies of putting both the particles in the same box. (I am keeping the box length the same, just trying to look at it like an approximate model). Then I have a ground state energy of $E=\frac{5\hbar^2\pi^2}{2mL^2}$. So, the ground state energy of having two electrons in a box is higher than the combined energy of the system having one electron each in a separate but identical box. How does one explain bonding to himself. Is there any literature on it, or am I going wrong somewhere? I would appreciate if I was directed towards some literature on it.

An electron in a molecule (say, H2) doesn't experience the same potential as an electron in an atom (hydrogen atom in this case), because there are now two nuclei to deal with, as well as a second electron.

@orthocresol: Is there some QM model which explains this in a better way??

I don't know how much simple models can accurately reproduce molecules - maybe they can to some extent, but I genuinely don't know. The "real" explanation is really quite involved and I don't entirely understand it myself, but you might want to take a look at this page which seems to be a good overview of the topic and has some key references. http://www.users.csbsju.edu/~frioux/h2-virial/virialh2.htm

Regardless of what model you use, though, you should probably put both electrons in the lowest energy state (one can be spin up and the other spin down. The standard QM particle in a box doesn't include spin). In your particle in box treatment, that results in the same energy for bound vs. unbound, which is not a huge surprise - if you assume the system doesn't change at all upon bond formation, that is equivalent to saying that both atoms have zero interaction with each other. Clearly, it is this change in the system that leads to bonding.

@orthocresol Thanks a lot for the article, I'll take a look at it and then I can post questions about the parts I don't completely understand :)


text:
Why is that in zwitterions amino acids or compunds with amine and acid group are mentioned but not compounds with alchol and acid group?
Does it have something to do with the pkb of the alcohol group at ph 7 which makes it not protonate?

That is precisely the reason why. Zwitterions are defined as molecules that contain both a positive and a negative charge within the molecule. Alcohol groups won't deprotonate at neutral pH = 7. The pKa of a typical alcohol is about 16, so you need a pH of 16 to deprotonate an alcohol (or a very strong base). Keep in mind a pH of 16 is difficult to achieve in water since for water, pKw = 14.
There are some molecules with an alcohol sidechain that can deprotonate at a reasonable pH. These are typically phenols. Take tyrosine as an example

The pKa for the carboxyl group is 2.2, the amino group is 9.21, and the hydroxyl group's pKa is 10.46. At neutral pH, the carboxyl group will be deprotonated (negatively charged), the amino group will be protonated (positively charged), and the hydroxyl group will be in its normal protonated form, making the molecule a zwitterion. A pH of 11 will be enough to deprotonate the hydroxyl group, as well as the amino and carboxyl groups to give the molecule an overall negative two charge.

As you surmised, zwitterions contain groups that are proton ($\ce{H+}$) donors and groups that are proton acceptors. The $\ce{C-O-H}$ group in alcohols, though, is covalently bonded, and is close to neutral. Zwitterions are usually defined by groups with stronger basic or acidic properties.
That said, alcohols can be amphoteric, i.e. act as either base or alkali. For example, 2-methyl-2-butanol reacts with Lucas' reagent as if alkaline, losing its $\ce{OH-}$ group, but reacts as if it were an acid, surrendering $\ce{H+}$ with potassium to form potassium 2-methyl-2-butoxide.


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How can chlorine form ClO2?
Doesn't it have only have one bonding electron?
Then how can it bond with 2 oxygens?
The Lewis structure doesn't make much sense to me.

Not all molecules have reasonable Lewis structures (nor must they), consider $\ce{NO}$. In this case, some of the lone pairs participate in the bonding, if memory serves.

In this case most notably, you are dealing with a free radical.

Related: https://chemistry.stackexchange.com/questions/37381/speculation-about-chlorine-dioxide-isomer


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Pd-catalysed hydrogenation of imines
Hydrogenation of imines to amines by $\ce{H2}$ on $\ce{Pd/C}$ has been used e.g. in the synthesis of Pumiliotoxin. By contrast, carbonyl groups are rarely reduced in this way, which I though was due to the fact that if a carbonyl is adsorbed, the interaction is through O’s lone pairs, rather than C=O $\pi$. However, the lone pair on the imine N should be of an even higher energy than in carbonyl and so should adsorb more easily. Why then does the hydrogenation of imines happen successfully?

I feel like the polarity of the bond might play a role, but not sure

The HOMO should be the $\ce{C=X}$ bond either way …

@Jan So why does the adsorption of carbonyls occur at O? And does it actually?

@substitutedbutadiene I wouldn’t be sure, I would have to look through the literature. Heterogenous catalysis is really not my strong point … ^^'

Depending what you look for and where you get the information,  I was under the impression the typical C=O is around 100kJ/mol stronger than the typical C=N bond.  Hence, the latter is easier to break.  Stronger adsorption is backed up by the fact that the resultant amines,  in certain circumstances,  can poison catalysts, whereas alcohols do not tend to.