text:
Is it possible for a substance to absorb visible light but still appear white?
This is a UV-vis spectrum for zinc oxide

No substance is perfectly transparent or reflective.

Most white solids seem to do that too

Your eyes are not sensitive to UV light so any absorption there will be invisible to you. The visible spectrum you give has fairly uniform absorbance across the visible. The whiteness you see is primarily due to scattering of white room lights or sunlight off the powdered zinc oxide. Light is scattered when there is a change in refractive index between two mediums, in this case air and zinc oxide. The effect is enhanced when the crystals are of a very irregular shape and so have many different scattering surfaces. Sugar and salt show the same effect when powdered  even though large crystals of each of them can be made that are clear and colourless. As the oxide has a more or less constant absorption no wavelength is very much more absorbed than any other, or alternatively, all wavelengths are equally scattered.
If you can find a solvent with similar refractive index to the oxide (and in which it does not dissolve) then the crystals will become far more transparent and effectively disappear. It is easier see this effect if you wet a piece of frosted glass with water as it will then become far more transparent. Alcohol or benzene would be better as they have refractive index closer to that of many types of glass. Zinc oxide has a value of 2 higher than any common solvent, such as diiodomethane approx 1.7.

Your impression of color is a result of physical (absorption spectrum of the substance, power spectral density of the illuminating light), biological (cone responses) and psychological (surround, adaptation, fatigue...) factors. From a purely physical point of view the posted spectra are not of a white substance (which would absorb light equally at all visible wavelengths) but of a bluish one as it absorbs less at the shorter visible wavelengths than at the longer ones. And yet you see it as white. This is because your brain averages everything in the scene and sets its white point based on that average much as the automatic white point setting in a video camera does. Thus a piece of white paper looks white when viewed under tungsten light (Coordinated Color Temperature 2400 K) as it does in shadow on a sunny day (CCT 8000 K).
There are three color receptors in the eye and their responses are approximately bell shaped curves located in the short, medium and long wavelength parts of the visible spectrum (400 - 700 nm). If those three are approximately equally stimulated by an object (or, more precisely, if an object stimulates them approximately in the same proportion as the scene) that object will appear white. Note that the spectrum of a white object does not at all need to be uniform or even approximately uniform in order to produce an impression of white. When we look at a TV, for example, what we see as a white screen is actually stimulating our eyes with three rather narrow band clumps of light energy corresponding to the primaries of the screen. It can all seem rather complex. If absorption of visible light occurs in a part of the spectrum (near 500 nm) where all three color matching functions are small and cross over one another one would not notice much change in the color of the material as each of the tristimulus values would change little. But if the absorption were at, respectively,  425 nm (peak of the short wavelength cmf) or 600 nm (peak of the long wavelength cmf) the, respectively, short or long tristimulus would change appreciably and the effect be noticeable.
The key word in your question is "appear" and the answer to it is "yes". We can go straight from your posted spectra to color coordinates in any of several systems and I'm confident that by playing around with illuminants and choosing absorbtion peaks properly we could find spectra distorted relative to yours that were imperceptibly close to the same color as the one yielded by the original spectra. These are called 'metamers'. Metamerism (i.e. producing something of one color that looks like a desired color even though the spectra are very different) is obviously the basis for color TV, color printing etc. In fact we could photograph your zinc oxide and look at it on a TV screen. It would still look white though the light coming from the screen would be missing several chunks of wavelengths.


text:
Thermal decomposition of calcium propionate
What are the decomposition products?
For example:
Calcium formate decomposes into formaldehyde and calcium carbonate.
$$\ce{Ca(HCOO)2 -> CaCO3 + HCHO}$$
Calcium acetate decomposes into acetone and calcium carbonate.
$$\ce{Ca(CH3COO)2 -> CaCO3 + (CH3)2CO}$$
How does calcium propionate decompose?

There is a paper (1) where TG-GLC-MS with subsequent XRD analysis of solid products has been utilized for the analysis of evolved gases during the decomposition of calcium propionate at various heating rates. Detected products were $\ce{H2O}$, $\ce{CO2}$, $\ce{C2H5CHO}$, $\ce{C2H5COCH3}$, $\ce{C2H5COC2H5}$, $\ce{CaCO3}$.
The Discussion section of the paper suggests the following mechanism of formation:

Following the free radical mechanism of Hites and Biemann (2), proposed for
  the decomposition of the calcium salts of monocarboxylic acids we suggest that the initial stage in the decomposition of calcium propanoate can be represented as:
$$\ce{Ca(C2H5COO)2 -> CaCO3 + C2H^.5 + C2H5C^.=O,}$$
Combination of the two free radicals results in the formation of the major product, 3-pentanone:
$$\ce{C2H5C=O + C2H^.5 -> C2H5 - CO - C2H5}$$
The other trace products observed, 2-butanone and propanal, can be accounted for by disproportionation reactions involving the ethyl radical:
$$\ce{C2H^.5 + C2H^.5 -> C3H^.7 + CH^.3}$$
and
$$\ce{C2H^.5 -> C2H5 + H^.}$$
The methyl and hydrogen radicals so produced may react with the large excess of the original propanoyl radicals to give 2-butanone and propanal respectively.
$$\ce{C2H^.5=O + CH^.3 -> C2H5COCH3}$$
and
$$\ce{C2H5C=O + H -> C2H5CHO}$$
The $\ce{C3H7}$ may combine with any of the other radicals or may disproportionate. In any event there was no evidence for compounds larger than 3-propanone. However on close examination of the GC curves there were indications of a very small peak before that due to $\ce{CO2}$ which could be attributed to low molecular weight alkanes and alkenes.

(1) Barnes, P. A.; Stephenson, G.; Warrington, S. B. Journal of Thermal Analysis 1982, 25 (2), 299–311. DOI: 10.1007/BF01912955
(2) Hites, R. A.; Biemann, K. J. Am. Chem. Soc. 1972, 94 (16), 5772–5777. DOI: 10.1021/ja00771a039

The low molecular weight alkanes and alkenes correspond to disproportionation products, right?

@EashaanGodbole Could be, authors hasn't investigated that phenomena due to its low impact on overall picture.

Abbrev.

MEK - Methyl ethyl ketone(butanone)
CP - Calcium propanoate or propionate
CC - Calcium carbonate

It can be answered logically.  The formula for calcium propanoate is $\ce{Ca(C2H5COO)2}$ or specifically $\ce{Ca(CH3CH2COO)2}$. See, for every time, there is a common decomposition product i.e calcium carbonate ($\ce{CaCO3}$). So, remove $\ce{CaCO3}$ from $\ce{Ca(CH3CH2COO)2}$ and you get is $\ce{C2H5COCH3}$ known as butanone(MEK). In each case, there is an increase in methyl group. So, according to it, the decomposition products would be calcium carbonate and MEK.
For further reading about the reaction mechanism and condition, give a quick read at the following papers:

http://www.sciencedirect.com/science/article/pii/S0040603100003713
https://www.researchgate.net/publication/229268554_A_study_of_the_decomposition_of_calcium_propionate_using_simultaneous_TG-DTA

https://link.springer.com/article/10.1007/s11434-010-4065-8

The thermal decomposition of CP began when the carboxylic radical was
evaporated from 565 K to 759 K for which corresponding mass loss
percentage was 47.79%. The residual was subsequently decomposed to
release carbon dioxide$\ce{^{*}}$ between 843 K and 1012 K. The latter phase of
the process occurred more readily than with CC because of the loose
structure of CP resulting from evaporation of the carboxylic radical
in the low temperature zone, which could be seen directly by scanning
electron microscope. The maximum mass loss rates of this phase
occurred at temperatures of 972 K and 1012 K for CP and CC,
respectively. The Ozawa-Flynn-Wall method was used to calculate the
activation energy during the thermal decomposition process at heating
rates of 5, 7.5, 10 and 15 K/min. The result further confirmed the
multistage characteristic of CP thermal decomposition, which could be
seen in differential thermogravimetry curves. The reaction orders of
CP in the conversion range 20%–80%, calculated using the Avrami theory
were from 0.061 to 0.608, smaller than those of CC, which were 1.647
to 2.084.

There is a thing to note. Applying further heat (going beyond the temperature in which CP completely decompose to MEK and CC), calcium carbonate decompose to calcium oxide($\ce{CaO}$) and MEK decomposes to emit irritating and harmful gases as noted by pubchem, $\ce{^{*}}$most probably carbon dioxide and carbon monoxide .

When heated to decomposition it emits acrid smoke and irritating fumes.
Lewis, R.J. Sax's Dangerous Properties of Industrial Materials.
9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p.
637

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text:
A question about acidity
Since everyone knows an acid would dissociate(partially or fully) in its aqueous solution and liberate $\ce{H+}$. We also calculate the $\pu{pH}$ of the solution using $[\ce{H+}]$ and this$(\ce{H+})$ is what gives the solution acidity. Now consider we have just got a strong acid i.e pure acid without having any water in it, for example, $\ce{H2SO4}$ and no water in it, would the sulphuric acid dissociate in this case i.e without the aid of water? If it does not dissociate would this substance be acidic, since there would be no $\ce{H+}$ in that case?

Depends on the definition of acid that you use - https://en.wikipedia.org/wiki/Acid#Definitions_and_concepts Undissociated H2SO4 isn't an Arrhenius acid (unless you count the tiny amount of self-ionisation), but it is still a Bronsted acid

I think if we had FAQ section or sth like that, this topic would be there.

@Mithoron Excuse me, my question is different.

https://chemistry.stackexchange.com/questions/53580/is-there-any-notion-of-ph-out-of-solution

Not really. Like lots of things with polar bonds, sulfuric acid can undergo ion transfer reactions in the liquid and make ionic species in solution.  And it is indeed protons that get transferred just like water.  But as with water, at least under ordinary conditions, these ionic species are only a minor portion of the liquid which is mostly $\ce{H2SO4}$ molecules.
Strong acids such as sulfuric acid are indeed acidic, capable of acting as acids.  But they still need a base with which they can react, to actually act as acids.  In the case of diluting sulfuric acid, water serves as such a base.

So if you put your hand in such sulphuric acid you will get shocked right?

Your hand contains things like water that can act as bases .

First of all, sulfuric acid is just hydrated sulfur trioxide ($\ce{SO3}$). Sulfur trioxide is obviously not an Arrhenius acid, but it is Lewis acidic.
You're specifically asking about something like gas phase acidity. It is possible to give away a proton but without a solvent to stabilize the ions, gas phase acidity is much lower than aqueous acidity.

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text:
How does a hydrophobic environment help in bond formation?

The glucose-induced structural changes are significant in two respects. First, the environment around the glucose becomes more nonpolar, which favors reaction between the hydrophilic hydroxyl group of glucose and the terminal phosphoryl group of ATP.

I am unable to understand the chemistry behind this. How is a hydrophobic environment helping bond formation between two polar molecules?

Source: Biochemistry 8th edition by Jeremy M. Berg John L. Tymoczko Gregory J. Gatto, Jr. Lubert Stryer

Basically forcing them to themselves

What is the physical reason for that?

Whatever polar group will be less solvated in a hydrophobic environment. May I make it too simple, but alike oil forms droplets in water, or viceversa. Molecules find each others, and droplets do finally the same. In hydrophilic environment the reacting polar groups are shielded by the surrounding medium molecules, e.g. water.

But how does that lead to covalent bonding?

The major reason of the ‘hydrophobic environment’ is to block the enzyme-complex from accessing the solvent (water), which might otherwise enter the active site and hydrolyse ATP.
The reduction of polarity also helps to speed up the subsequent nucleophilic substitution.
Note: Hexokinase is a soluble, cytosolic protein
The reasoning behind this statement comes from the X-ray structural studies of yeast hexokinase (HK) by Thomas Steitz.

Comparison of the X-ray structures of HK and the glucose–HK complex
  indicates that glucose induces a large conformational change in HK.
  The two lobes that form its active site cleft swing together by up to
  11.5 Å so as to engulf the glucose in a manner that suggests the closing of jaws.

This movement places the ATP in close proximity to the group of $\ce{-C6H2OH}$ glucose such that water is sequestered from the active site.
Water is certainly small enough to fit into the phosphoryl acceptor group’s enzymatic binding site. Moreover, phosphoryl transfer from ATP to water is more exergonic than it is to glucose.  If the catalytic and reacting groups were in the proper position for reaction while the enzyme was in the open position, ATP hydrolysis (to yield ADP + Pi) would almost certainly be the dominant reaction.
However, the large conformational change of the enzyme induced by glucose causing the exclusion of water from the active site making the phosphorus atom more accessible for the nucleophilic attack of the C6-OH group of glucose ensures that HK catalyses the phosphoryl transfer to glucose rather than ATP hydrolysis and also reduces the polarity of the active site, which help accelerate the nucleophilic reaction process.
This substrate-induced conformational change in HK is responsible for the enzyme’s specificity.
References

Lehninger Principles of Biochemistry
Voet and Voet Biochemistry


text:
Is it possible to have negative corrorion rate and weight loss, meaning specimen [steel] gained weight after immersion in electroyte solution?
Regarding this article, I have read that the corrosion rate and weight loss of a steel specimen after immersion in different corrosive media is negative. I'm surprised that this is really possible. Is there a chemical explanation for this?

You mean like electroplating?

Immersion in saltwater and other corrosive media for corrosion protection.

In the data cited in your reference, corrosion is measured by weight loss or gain. In the case of the steel coupon immersed in sodium carbonate, the weight gain is not due to "negative corrosion", but likely due to a tightly-adhering film of oxides that was not removed in cleaning. The oxides are made from the iron of the coupon plus oxygen from water, which is greater than the mass of the iron alone. (Normally, cleaning the coupon removes loosely adhering oxides, so there is a weight loss.)
According to Toyin, "the corrosion inhibition efficiency of $\ce{Na2CO3}$... indicates that a protective film is formed on the metal surface. The passivation of the mild steel may be caused by the formation of $\ce{Fe2O3}$  and $\ce{Fe3O4}$."
That said, it does appear that high carbon concentrations may be effective in greatly reducing the corrosion of mild steel.

If that's the case, does it mean that the higher weight gain, more corrosion take place?

Perhaps... it depends on how well the coupon is cleaned after the test: if the cleaning removes flocculent iron oxides (often reddish), then the weight would drop with corrosion, but if the (black) oxide adheres well and cannot be removed, then the weight might increase, depending on how much Fe dissolved before the passivating coat adhered.


text:
Handling reactions in chemistry
Q:1  So far, after studying chemistry for two years I am unable to figure out any well-defined methodology of handling a particular reaction either in General or Organic chemistry. So I would appreciate If some one could point out some useful source(book etc) where I could look deeply into the back ground of the chemical reactions.
Q:2  If we are given for example two or three reactants is it possible to determine how they will react?

Handling reactions? Search books for practical organic chemistry

Welcome to ChemSE.  Your first question is rather broad; both because of the large body of books (and sources, in general) about Chemistry, and because it is not further specified what you already learned by self study.  Similar questions already were addressed on ChemSE, like here, or here; and more generally surveyed in a synopsis here.  However, none of them aims to be exhaustive, none of them considers what kind of self-study fits you best.
Regarding the second question by you, reactions with more than two starting material are not rare. In the Friedel-Crafts acylation, for example, you have both an arene and an acylchloride reacting with each other to yield a ketone, however, often you need a Lewis acid (like $\ce{AlCl3}$) promoting the reaction:

(source)
which however rarely is seen as a complex reaction in terms of the number of reagents involved.  There are, however, reactions summarized as multi-component reaction, where, like in the Ugi reaction, several reagents are present at the same time in the same reaction vessel to yield the product.

(source For clarity, the starting materials are colour encoded.)
Even here one was able to postulate a reaction mechanism:

(source).

The postulated mechanisms of multicomponent reactions may be debatable.  About the more common multicomponent reactions à la Biginelli, Hantzsch, Mannich, Passerini and Ugi, there is a recent review by Alvim, Júnior and Neto in RSC Adv., 2014, 4, 54282-54299 (doi 10.1039/C4RA10651B).


text:
Is it true that un-dissociated HCl exists in dilute hydrochloric acid?
My book says that the equilibrium constant for $\ce{HCl + H_2O}$ is so high that the reaction goes to completion. But that does not make sense since the existence of an equilibrium constant for this reaction means that there is some reactant in the equilibrium mixture. I think there is only a very small amount of reactants in the mixture. What is the truth?

Hint: while we are more used to classify acids in strong (like the example mentioned by you, $\ce{HCl}$), or weak ones  (like $\ce{HOAc}$) by means of their $pK_{\text{a}}$ value, this very constant describes the dissociation equilibrium constant of $\ce{HA <=> H+ + A-}$, even if in case of $\ce{HCl}$ the dissociated form is dominating.  However it allows you to estimate how much $\ce{HA}$ is still present, too.

Once the concentration is low enough, you're no longer asking the right question. You should be approaching this problem probabilistically and asking the probability that such $x$ molecules of undissociated acid exist at within a given time window.

Your book is right, for all practical purposes. Also keep in mind that the equillibrium is highly dynamic. A finite concentration of HCl molecules will be present, but their average lifetime is in the microseconds (or less).

see https://chemistry.stackexchange.com/questions/23572/why-doesnt-hcl-form-when-you-dissolve-nacl-in-water


text:
Convert CIF file to SDF file using free Linux-compatible software other than Avogadro or OpenBabel?
I am trying to convert http://crystallography.net/cod/2236556.cif to a file that I can open in Avogadro without some bonds disappearing. See this crystal structure is of tianeptine hydrochloride, yet many of the bonds are not shown when I open the CIF in Avogadro.
This is what the CIF is meant to be rendered like (shown in PyMOL):

while this is what I see in Avogadro (also displayed like this in Avogadro 2):

note how so many bonds are not shown properly. I suspect if I could convert the molecule to SDF or CML or some other 3D chemical format I'd be able to fix this issue and Avogadro will be able to render the file properly.
I have tried doing the file conversion with OpenBabel but the files generated look, when viewed in PyMOL or Jmol (which render the CIF fine), just like the CIF does in Avogadro with all the disconnected atoms.

Similar issue has been discussed here, where use of Mercury from CCDC suit is suggested. I can confirm that SDF/MOL-file generated with Mercury behaves as expected in Avogadro with proper connectivity. The problem is, Mercury is not open-source, though it can be run under Linux, so I'm not sure if it solves anything for you.

I actually had it already installed just didn't realize it could do that. Is it possible to automatically determine bond order in it too?

I don't know for sure, it seems like MOL-file generated with Mercury doesn't have bond order info. I suspect it's more software-dependent (how it treats certain interatomic distances).

Write it into an answer and I'll accept it. I've found another way to determine bond order automatically so I'm happy as far as that goes.

I can confirm this was actually an inadvertent bug in Open Babel 2.4.x. I'm currently fixing the behavior, which will go into the newest releases of Open Babel and Avogadro.

It seems like OpenBabel and Avogadro sometimes don't ideally parse certain CIF-files' connectivity data. As OP mentioned, PyMOL and other crystallographically-oriented programs (I personally like Olex2, it also run under Linux and is free) do it on-the-fly:

As @GeoffHutchison suggested in a similar thread on binding atoms within a unit cell, Mercury does a good job at providing MOL/SDF files from CIF data. As of now, there is an installation script for Linux for Mercury 3.9. Avogadro handles such MOL file from Mercury way better:

2236556
  --CCDC--071617    3D

 56 57  0  0  0  0  0  0  0  0999 V2000
    1.3664    7.7013    4.0020 Cl     0
    6.6663    8.7266    4.4355 S      0
    6.1525    9.4153    5.5710 O      0
    7.7255    7.7726    4.6043 O      0
    4.7213    0.8967    7.6943 O      0
    2.6223    1.3499    8.2786 O      0
    2.9354    1.3046    9.0347 H      0
    7.1953    9.8372    3.3639 N      0
    7.6456    5.9591    2.4701 N      0
    8.4385    5.8198    2.0682 H      0
    7.8134    6.2148    3.3161 H      0
    5.3238    7.9653    3.5835 C      0
    4.0381    8.0880    4.1077 C      0
    3.8953    8.5011    4.9288 H      0
    2.9772    7.5765    3.3740 C      0
    3.1781    6.9816    2.1438 C      0
    2.4563    6.6559    1.6556 H      0
    4.4680    6.8774    1.6459 C      0
    4.6013    6.4768    0.8168 H      0
    5.5699    7.3547    2.3503 C      0
    6.9378    7.0993    1.7546 C      0
    6.7575    6.7565    0.8547 H      0
    7.9197    8.2523    1.5451 C      0
    8.8050    8.0470    0.4790 C      0
    8.7594    7.2477    0.0049 H      0
    9.7458    8.9915    0.1090 C      0
   10.3266    8.8270   -0.5988 H      0
    9.8106   10.1839    0.8033 C      0
   10.4293   10.8326    0.5572 H      0
    8.9631   10.4081    1.8540 C      0
    9.0175   11.2135    2.3156 H      0
    8.0176    9.4560    2.2492 C      0
    6.3410   11.0331    3.1948 C      0
    6.8195   11.7005    2.6977 H      0
    6.1061   11.3837    4.0569 H      0
    5.5420   10.7916    2.7197 H      0
    6.8948    4.6534    2.5072 C      0
    7.5354    3.9263    2.4577 H      0
    6.3296    4.5991    1.7205 H      0
    6.0338    4.4553    3.7359 C      0
    5.2898    5.0768    3.7165 H      0
    6.5567    4.6337    4.5332 H      0
    5.5123    3.0333    3.7738 C      0
    5.0259    2.8601    2.9524 H      0
    6.2680    2.4268    3.8010 H      0
    4.6053    2.7307    4.9451 C      0
    3.8347    3.3179    4.9116 H      0
    5.0816    2.9111    5.7713 H      0
    4.1326    1.2872    4.9481 C      0
    4.8942    0.7100    5.1149 H      0
    3.7873    1.0713    4.0679 H      0
    3.0575    0.9929    5.9842 C      0
    2.3293    1.6247    5.8729 H      0
    2.7039    0.1030    5.8300 H      0
    3.5713    1.0798    7.3895 C      0
    6.3200    5.1537   -1.1307 Cl     0
  1 15  1
  2  3  1
  2  4  1
  2  8  1
  2 12  1
  5 55  1
  6  7  1
  6 55  1
  8 32  1
  8 33  1
  9 10  1
  9 11  1
  9 21  1
  9 37  1
 12 13  1
 12 20  1
 13 14  1
 13 15  1
 15 16  1
 16 17  1
 16 18  1
 18 19  1
 18 20  1
 20 21  1
 21 22  1
 21 23  1
 23 24  1
 23 32  1
 24 25  1
 24 26  1
 26 27  1
 26 28  1
 28 29  1
 28 30  1
 30 31  1
 30 32  1
 33 34  1
 33 35  1
 33 36  1
 37 38  1
 37 39  1
 37 40  1
 40 41  1
 40 42  1
 40 43  1
 43 44  1
 43 45  1
 43 46  1
 46 47  1
 46 48  1
 46 49  1
 49 50  1
 49 51  1
 49 52  1
 52 53  1
 52 54  1
 52 55  1
M  END
$$$$


text:
How to get Jmol to automatically determine and display bond order?
Some molecule file formats do not detail bond order and when I open these formats in Jmol implied (by the valency, of course) double bonds are not shown as double bonds but rather as single bonds. Is there a way to correct this behaviour? I tried to answer this question myself with a "calculate bond order jmol" DuckDuckGo search. It returned:

http://wiki.jmol.org/index.php/Jmol_as_editor
http://www.wikihow.com/Calculate-Bond-Order-in-Chemistry
http://wiki.jmol.org/index.php/Support_for_bond_orders

as the top three results. Not a single one mentioned something relevant. 3 seemed most relevant but it seems to talk about how to specify bond orders manually, not getting Jmol to automatically calculate them.
Bond lengths are shown in this screenshot:

The file opened is http://crystallography.net/cod/2236556.cif which is tianeptine hydrochloride.

I don't know if your problem is that Jmol incorrectly assigns a single bond or if it does take some (wrong) information from the input file. Could you try to use "set autoBond" to force Jmol to calculate bond orders? ( http://wiki.jmol.org/index.php/Jmol_as_editor )

Tried that, it was mentioned in 1. Did not do what I was looking for.

Can you measure the bond lengths of the bonds with the wrong assignment? Jmol uses bond lengths for automatic assignment so I suspect that the limits for each type are set differently from what you expect, we should see that if we look at the lengths. That way we should be able to figure out where the problem is. Is it also possible for you to post/link such an example?

Added some with the wrong assignment to my question.

You should be aware that you are not calculating a bond order in any case. You are assigning a single, double, triple rod to some (basically arbitrarily) chosen bond lengths. It is trying to force a Lewis-type structure onto a 3D model. It's tempting, and very suggesting, and experimentalists like this very much, but there is little to no justification to do that. Bond orders are not discrete; they are not natural numbers!

The current definition of the *.cif file does not care about bond order, rather than indicating connectivity of atoms within a molecule.  This is why representing organometallic complexes (like ferrocenes) drawn in sight of what is important for a crystallographer may differ from the one drawn by an organo(metallic) chemist. Among other data, *.cif include the position of the (hopefully correctly assigned) atoms, but nothing about hybridization ($sp^3$, $sp^2$, ...) -- which is one reason why you offer a drawing of the molecular structure of your product to your crystallographer when it comes to characterization by single crystal diffraction or powder diffraction analysis.  In the style of a the ball-and-stick representation, the above drawing is correct.
It is however possible to request Jmol to discriminate between single, double, triple bonds.  One approach to solve such a question is based on tables which relate typical bond lengths to single, double, triple bonds.  To document this, I used the data set indicated by you, as well as Jmol (version 14.13.1, 2017-04-09) under Xubuntu 16.04 LTS with openjdk version "1.8.0_131".  As far as I know, one has to proceed manually, individually for each bond in question.
As a preparation, be sure that Jmol may calculate the bonds, the corresponding menu is accessible via edit -> preferences:

If the option previously was disabled, restart Jmol now.  Subsequently, open your *.cif file and activate the model kit module

which will enable a little drop-down menu on the left hand side, marked by the ribbon in magenta. For now you do not need to access the options there, but hover the mouse over the first, subsequently over the second atom along the bond you are interested.  You will see that first the first atom, then the second atom is encircled.  If you now return hovering from the second atom in question to a position between the two atoms in question (and still on top of the bond in question), and left-click, the bond order may be recalculated, and its display will change if deemed suitable -- as shown in the example of the oxygen atoms of the sulfonamide.
This takes some practice, and frankly, the operator needs to stay vigilant.  In the case of aromatic cycles, the model kit module may correctly recognize the $sp^2$ character of the carbon atoms, but on the expense of hydrogen atoms attached to it.  Hence a result like the following may be expected, where I intentionally halted to "re-assign" the bond order on the last two carbon atoms of this cycle:

Be aware that if you left-click on one of the atoms in question, you may alter its designated element.  In the following, I did so while tapping on the carbonyl oxygen of the acid residue:

In such an instance, you should access the pull down menu of the model kit, and re-assign the atom type. Aberrant hydrogen atoms will be removed by this automatically, too:

The compute bonds option, for me (on Arch Linux), was set to automatically by default so didn't need to change it. Secondly this isn't really automatic is it? I have to manually select each bond and change it myself.

@BH2017 I will not exclude that there is a script option (as Jmol, similar to pymol, offers both GUI and CLI) to automate this.  In sight of "just display double bonds as such", you may load more than one cell in Jmol (type in the CLI |load "foo.cif" {1 2 1}|), write a pdb (|write "foo.pdb"|, the commands are within the pipes)  and proceed further in Avogadro (e.g. the elder branch, version 1.1.2) in ball-stick modus.  For your test data, it works just fine (carbonyl / sulfonamide C=O, alternating C=C/C-C in the arenes).

As Buttonwood stated, CIF is not the file format you would expect to preserve bond order within, and Jmol wiki suggests using other formats for that: MOL, MOL2, PDB, CML. I would like to propose a method involving a third-party tool that works out for me.

Open CIF in Olex2 (free, available on Win, MacOS, Linux).

Run compaq -c to assemble all atoms together with the largest molecular fragment; then grow and grow -w (not necessary for this structure, but is obligatory for those with an inversion center when molecular residues are split about special positions).

Due to the limitation of bond type that can be displayed in Jmol, we need to get rid of dotted bonds showing short interactions with deltai 0.

Run chemdraw to automatically assign bond orders.

Export CIF to MOL by running file 2236556.mol. MOL file is saved in the same directory where the CIF file is.
Open 2236556.mol in Jmol. All bond orders are preserved.

Booo! Double bonds from (tetrahedral) sulfur to oxygen! (This is not criticising your answer, just the general practice of using such display styles.)

@Martin-マーチン I'm desperately trying to understand what is wrong with this. How the $\ce{[=S(O)O]}$ fragment should be shown then?

More correct would be single bonds; the octett expansion is basically obsolete, meaning: there are no d-orbitals involved.

@Martin-マーチン Ahaaa, I see. Thank you, I've never thought of that before taking it for granted (pretty much all crystallographic software I ever touched assigns double bonds there). Answering this question was educational for me too:)


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Xanthoproteic test for phenylalanine
I read on the net that phenylalanine gives a negative result with xanthoproteic test and the reason that was given was that the benzene ring is deactivated and hence the extent of nitration is very less.
What I dont understand is that isn't the benzene ring in phenylalanine actually activated through hyperconjugation from the $\ce{CH2}$ group of $\beta$-carbon and hence shouldn't it give positive results with the xanthoproteic test?

Phenylalanine can react with HNO3, but is less reactive then aromatic aminoacids.

that's exactly my question... why is it less reactive than other amino acids  even though the ring is activated through hyperconjugation?


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Abundance of sulfur containing compounds in the air
I noticed a piece of jewelry has tarnished, so I looked up information about silver tarnishing. I read that this is due to the formation of silver sulfide on the surface of my necklace and that the sulfur is pulled from the air.
My question is, how abundant are sulfur containing compounds in the air? I know sulfur is by no means rare, but I guess I'm just confused as to why pretty much everything silver ends up tarnishing -- a box of silverware unopened for years can be tarnished.
How abundant are sulfur containing compounds in the air? Is the tendency of silverware to tarnish often/easily due to reactivity? If so, why doesn't carbon react to silver in the same manner and thus give way to a carbon-based tarnish?

The tarnishing of silver by sulfur is primarily due to hydrogen sulfide. Hydrogen sulfide is found throughout the Earth's lower atmosphere and has many sources including the human body$\pu{^1}$.
The amount of hydrogen sulfide naturally found in the air has been estimated as $\pu{0.11-0.33 ppb}$ ($\pu{0.15-0.46 ug/m3}$). Lower levels ($\pu{0.02-0.07 ppb}$; $\pu{0.03-0.1ug/m3}$) have been observed in some remote areas$\pu{^2}$.
Although these are relatively low concentrations compared to many other trace gases found in our atmosphere, silver is highly reactive towards hydrogen sulfide, and thus tarnishing of silver is a ubiquitous issue. Additionally, some tarnish results from the oxidation of silver by oxygen.
Regarding your question about silver tarnishing from carbon, it's a matter of reactivity. The overwhelming source of carbon in the air is carbon dioxide, and methane to a much lesser extent. Neither of these gases is reactive towards silver nearly to the degree of hydrogen sulfide and oxygen.

Toxicological Profile for Hydrogen Sulfide and Carbonyl Sulfide, U.S. Dept. of Health and Human Services, November, 2016
Ambient Air Guidelines for Hydrogen Sulfide, CAS Registry Number: 7783-06-4, March 27, 2006

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What are the effects of frenkel and schottky defects on electrical conductivity of a solid?
How do these two defects impact dielectric constant of solid? Please explain these two questions in detail.
In case of frenkel defect the the smaller ions are just dislocated from their initial positions. In that case the electrical conductivity must remain unaltered. But I have found out from few sources that electrical conductivity decreases due to increase in resistance so I am a bit confused.

Welcome to Chemistry StackExchange. You should now give sentence or two to explain what you have found out so far otherwise this looks like a homework question and no one will answer it for you, and then  its likely to be closed.

I have edited the question. Please look into it.

@Sriharsha Nice! Your question is going to be reopened very shortly and soon you will get an answer.


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Why can't we calculate the atomic radii of noble gases?
My book says that only the Van der Waal's radius of noble gases can be measured as these gases generally do not form compounds (except Xenon.) Then why can't we just calculate the atomic radii by computation, say solving the Schrödinger equation? Can't we make computer simulation models of isolated atoms of noble gases?

We can all right, but that's not what is called a measurement.

@IvanNeretin I have edited the question accordingly.

My comment still applies. Simulation is not a measurement.

@IvanNeretin Agreed, then can't we accept the simulated model results as excatly as a measurement?

What's to accept and what for? We won't be needing those radii anytime soon. And in any case, the book says nothing about that. It just says they can't be measured, which is true.

@IvanNeretin What if someone asks you a question about van der Waal's radius of these gases? Also, these calculations will make noble gases fit in the trend of atomic radii across a period.

Van der Waals radii are a different thing: they can be measured. I was referring to covalent radii only. As for fitting the trend, well, yeah, I guess that kinda can be done.

Let us continue this discussion in chat.

You can experimentally measure the atomic radii using x-ray crystallography. All form crystals except helium.

@porphyrin Bingo! That's exactly what I meant to say in the chat for this question.

@porphyrin  Solid helium has a hexagonal crystal structure. Also there is no such thing as an "atomic radius", that is a catch-all term for vdW radius, ionic radius, covalent radius, metallic radius. https://en.wikipedia.org/wiki/Atomic_radius

@Karl, thanks, didn't know He formed a crystal.

Covalent radius is measured as the distance ($r_\text{cov} = d/2$) between the nuclei of two bonded atoms (covalent). But, if you try to do the same for noble gases/inert gases (good luck!), as they have fully filled $np$ orbitals, they will repel each other, hence the closest distance between the two atoms is taken (high pressure, low temperature) as the van der Waals radius.

This does not answer my question. I was asking why we do not accept values produced by simulation models.

@ApoorvPotnis See my remark in the chat. Your argument is invalid.

@ApoorvPotnis Noble gases have no covalent radius, because they do not form bonds. You cannot calculate something that does not exist.

@Karl Yes. I initially was asking for covalent radius but edited the question. See [this]{https://chat.stackexchange.com/transcript/message/38808251#38808251}

@ApoorvPotnis It is pointless to compare the 95% electron propability radii of single, isolated atoms. What kind of scientific question would you want to answer with that information? You can't even prove that your calculation is correct! The values here https://en.wikipedia.org/wiki/Atomic_radius#Calculated_atomic_radii are interesting from a didactic point of view, but nothing more.

@Karl Who knows when this information will be needed? Also, if they are interesting from a didactic point of view, shouldn't they be given in textbooks?

They are scientifically useless, because they do not appear in measureable reality! You cannot use those values to prove or predict anything, you cannot even prove the values themselves by some independent means. So why bother? It's unscientific. A nice mathematical excercise, didactically helpful, most likely correct, but, by all relevant criteria, unscientific. https://en.wikipedia.org/wiki/Falsifiability

You could do a calculation for a noble gas and get the radial distribution of the wave function (or derived from this the electron density). But this will be a distribution function of the radius. There is no atomic radius where the electron density suddenly stops. You could ask for characteristic points (like maximum, expectation value, where the density is below a certain threshold, where the integrated density/probability exceeds 90%, ...), but those would be arbitrary decisions.


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Is it okay to heat a test tube for several minutes for evaporation of solvent inside?
I have a mixture of water and a small amount of undissolved solute in a test tube. I want to remove the water in it by evaporation method. Can the test tube carries the evaporation process with a Bunsen burner? I don't want to transfer into different glassware to avoid contamination and reduction of the solution when transferring.

"Put a test tube on fire" and "Heat a test tube" are two very different things, keep that in mind.

I think the better method would be to transfer the mixture in, say, a petri dish. You might face problems removing the undissolved stuff from the test tube later...

There are two issues here:

Heat from a Bunsen burner is pretty strong. You better make sure whatever is left after the evaporation will survive.
If the solvent is organic (and not super halogenated), this is a terrible idea because you run a serious risk of igniting your solvent vapors.


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Why does the fructose monomer in sucrose appear different from isolated fructose?
I understood that the representation of fructose is:

But during the formation of sucrose:

How did fructose, whose representation is given in the first picture, turn out to be different in the second. I tried to visualize myself rotating the molecule but the position of oxygen changes and it doesn't match with the one in sucrose.

Hint: Draw a vertical line/axis on your first diagram, connecting the oxygen to the middle of the C3-C4 bond. Then rotate the molecule 180 degrees counter clockwise. Now compare this with the fructose residue in sucrose ;)

Ah, a (fairly) common conundrum that assails us Chemistry students when we start Biochem. ;-)
At first glance, the fructose molecule in your first picture, and that in the second picture appear to have different conformations:

Believe me, they're the same. "How?" you ask? Stay with me here (it requires a little imagination):
Draw a vertical line/axis on your first diagram along the plane of the furanose ring, connecting the oxygen atom at the apex, right down to the middle of the C3-C4 bond. Then rotate the molecule 180 degrees counter-clockwise.
Now compare this with the fructose residue in sucrose.
(I lack the necessary Org. Chem. sketching tools to depict this... apologies)

If you weren't able to wrap your head around "rotation" bit, then, let's assign locant numbers to the carbon atoms in the two diagrams and then compare them:

 I've drawn them tiny, my bad

Must've clicked by now ;)


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How to calculate the vapour density of a mixture?
I have a gaseous mixture of $\ce{H2}$ and $\ce{CO2}$ which contains $66\%$ by mass of $\ce{CO2}$. I have to figure out the vapour density of the mixture (defined as mass of a certain volume of a substance divided by mass of same volume of hydrogen).
My approach: let's assume $\pu{100 g}$ of mixture contains $\pu{66 g}$ of $\ce{CO2}$ and $\pu{34 g}$ of $\ce{H2}$. Hence, the amount of substance of $\ce{CO2} = \pu{1.5 mol}$ and the amount of substance of $\ce{H2} = \pu{17 mol}$. The total amount of substance $= \pu{18.5 mol}$.
The mass of $\pu{1 mol}$ of mixture is 5.4 g (mass of 18.5 mol is 100 g), and the vapour density $\text{VD} = 2.7$.
I got my answer right, but now I am doubtful of my approach. Was it safe to assume that the amount of substance of mixture equals that of the amount of substance of both the gases?

And where does that value come from? Confusing.

as VD= M/2 M(= molecular mass

Vapour density is molar mass divided by two? That sounds strange. Shouldn't it have units of gramm per liter?

@Karl The OP is right (vapour density). I wasn't sure whether it's $2$ or $29$.

@andselisk Thanks. I have never heard of this measure before, and hope I will never again. ;-)

Vapour Density is kind of ratio between the mass of mixture and mass of unimolar Hydrogen gas. So it doesn't have any units.

Was it safe to assume that the amount of substance of mixture equals that of the amount of substance of both the gases?

Amount of substance is additive, just like mass. You have to be careful with volume, though. So yes, it is safe to make that assumption.

I have to figure out the vapour density of the mixture

I'm not sure you are starting with a safe question, i.e. if the vapour density quantity was intended to describe mixtures. Wikipedia has "vapour density = molar mass of gas / molar mass of H2", and I'm not sure whether a mixture has a defined molar mass. The same source states that vapour density is sometimes based on the density of air rather than dihydrogen gas (to assess whether certain gases collect on the floor or ceiling of a room when released).

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What's the explanation for the chelate effect?
I got that multidentate ligands form more stable coordination complexes than monodentate ligands, but why?

Related: Why do dianions (such as malonate) bind cations more strongly than anions? and What's so special about chelation?

The chelate effect is generally agreed to be a thermodynamic effect caused by the change in entropy upon binding of a bidentate ligand.

General description of the chelate effect:
Consider a coordination complex,  [M]L2, where L = a standard monodentate ligand. If we now add a bidentate ligand, two monodentate ligands are released.

The ligand substitution reaction in the forward direction is therefore generally favourable:  ∆S, the entropy term is positive due to two molecules reacting to become three, with this favourable entropy term having the effect of tending to make ∆G negative (negative ∆G is favourable, i.e. the reaction likely to proceed).[*]
Once a bidentate ligand is bound, it then becomes less favourable to go in the other direction (the entropy term would be negative due to entropy decreasing in the system, which in turn would tend to make ∆G positive, and hence the reaction unfavourable).
Example and thermodynamic data:
To give a more concrete example demonstrating the significance of the entropy term, consider the following reaction and associated thermodynamic data (taken from Physical Inorganic Chemistry, S. F. A. Kettle, 1996):

Clearly, the entropy term dominates, causing ∆G to be negative in both cases (hence causing the forward reaction to be favoured in both cases).
[*]: The caveat here is that we're assuming that enthalpy really isn't contributing a whole lot, this is clearly a vast over-simplification, and there are cases where enthalpy becomes important (i.e. when the complex is significantly more stable than the starting materials). We're also ignoring the fact that the act of binding a bidentate ligand itself has some unfavourable entropy, since we're containing the ligand and removing some conformational flexibility

This answer seems convincing thermodynamically, but is there a more to-the-point explanation? An explain that involves the binding of the chelate ligand?

Feel free to answer Prit - th thermodynamic argument is the one I've most commonly came across. The binding enthlpy is, to my knowledge, negligible in relative terms

I'd love to, too bad I'm not aware of such an answer anyway. I wouldn't ask you to answer if I could myself would I? :)


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Name for a spin state with a multiplicity of 13
If I want to name a spin state with 12 unpaired electrons,
what is the correct name?
Tredecim is the latin name for thirteen, but tredecimet sounds odd.

I have seen multiplicity 13 referred to as simply a 13-let. This #-let nomenclature is sometimes used for high multiplicities (e.g. higher than 10).

I would go for the Greek form and call it trisdecatet. Or should it be with a k, trisdekatet?

All other spin states use latin for their name. This would be very inconsistent.


text:
Homemade Fabreze mystery substance
My wife made a cleaning solution of isopropyl alchohol (1200mL), baking soda (100 mL) and distilled water(2400 mL). She first mixes baking soda with warm water till it dissolves.  Then adds the isopropyl.  Sometimes, a solid material will form at the bottom of the solution. What is it? It has us confused...

Is the IPA warm? First guess is a precipitate of baking soda due to cooling of the solution.

No, it is not warm. It does look crystalline though.  Thank you for your answer. I appreciate it.


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Great books in Experimental Organic Chemistry?
This summer I've quite a lot of free time, so I decided to study experimental Organic synthesis. I don't even have basic experimental chemistry knowledge, however my theory is pretty firm( Clayden level maybe ). What I need is good experimental organic chemistry books. Books with good illustrations will be welcome.

Search for books on Practical Organic Chemistry

Shopping question are frowned upon here. Online bookshops are full of books on organic synthesis.


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multi state or multistate or multi-state
I am not sure if this question belongs to English language StackExchange.
In the context of multiconfigurational methods, what is the correct way to write multi-state character:

multi-state character
multistate character
multi state character

Personally I would go with option 1.

I don't think it changes the meaning, i.e. doesn't really matter.

Of course, but there is still something called orthography. If there is one correct solution I want to use it. Wif jur argument ai kud justify eni riting. ;)

@mcocdawc - I prefer option (1), and cite Donald Knuth's reasoning as given here. However, Knuth suggests that the hyphenated versions of "newly coined" words morph to unhyphenated versions (e-mail becomes email, for example) as they gain acceptance and are more frequently used, which indicates to me that option (2) is not off the table, certainly because the word is not newly coined. I would argue that wide acceptance and usage of the wording in the context given here is not demonstrated, however, and fall back to option (1).

I'm voting to close this question as off-topic because it's rather about language then chemistry.

@ToddMinehardt thank you very much for this answer. If you write it as "real" answer, I will happily accept it.

@mcocdawc - I would happily do so, will vote to reopen. Ultimately the community will decide. You might want to make it more chemistry focused or include wording to overcome the issues relating to its closure.

While I think this is as much a chemistry question as it is a language question, and I think it is better answered here, I am a little bit reluctant to overrule community decision. || On the matter at hand: 1 and 2 are correct and common spellings. Which one you choose is up to you basically, but be consistent. So if you write multiconfigurational (as you did), you should write multistate, multireference, etc.. If you prefer the hyphenated versions (I'd choose them), you should use them for all.

@ToddMinehardt Q is reopened.

My rule of thumb: Leeloo Dallas had Multi Pass, so all other arguments are invalid:)

I prefer option (1), and cite Donald Knuth's reasoning as given here.
Knuth asserts that hyphenated versions of "newly coined" words morph to unhyphenated versions (e-mail becomes email, for example) as they gain acceptance and are more frequently used, which indicates to me that option (2) is not off the table, certainly because the word is not newly coined.
However, I argue that wide acceptance and usage of the wording in the context given here is not demonstrated and fall back to option (1). Finally, as pointed out by Martin in the comment thread, be consistent and use (or don't use) hyphenation regardless of which variation you choose.


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Gibbs Free Energy Paradox
From my experience so far with chemistry, it seems like every reaction that happens is to attain stability. So shouldn't a lower entropy be good? Since less disorder means more stability.
But by Gibbs Helmholtz equation, a negative entropy change value favors a positive $\Delta G$.... But shouldn't a reaction which leads to stability be spontaneous?

Second Law of Thermodynamics: the entropy of the universe always increases.

Why? Any proof. I don't mean to disrespect any laws... But im curious. I have a hard time just accepting facts without knowing why it is so

Less disorder doesn't mean more stability.

@ULTIMATEGAMER07 You are asking for a full introductory lecture on thermodynamics. That's not quite what SE is about, I think. Pick up your chemistry textbook, and come back when you have an actual question.

The laws of thermodynamics are fundamental axioms, and cannot be derived. We postulate them based on physical experience.

But if you accept a correspondence between entropy and disorder, then this might help: say we have particles zooming around in a box with total energy E. States that look orderly are few and far between; all other states look disorderly. So the disorderly states are far more probable, especially in the thermodynamic limit of infinite particles. Orderly states thus, on average, evolve toward disorderly states, and entropy increases.

Possible duplicate: https://chemistry.stackexchange.com/questions/69478/spontaneity-of-reactions-gibbs-energy-or-entropy-change/76981#76981

There exists a "Gibbs paradox" (cf. Wikipedia article, section 3), but it has always been explained as gas mixing paradox. It is a strange thing that the problem that disorder (or better: spreading) leads to stabilization of energy is exactly this "Gibbs paradox": If an ideal gas is expanded into a vacuum, temperature does not increase by 180°C per mole, because it is stabilized by entropy. This paradox has been solved by quantum mechanics (particle in a box/Schrödinger equation).

@a-cyclohexane-molecule Some of the fundamentals in thermodynamics are observeable - notably the conservation laws - they are observeable as symmetries.

In chemistry it has often been explained that entropy increase had something to do with increasing disorder. The word disorder stems from Helmholtz (1882). But this is outdated, it was only a first attempt at theoretical interpretation.
Entropy = Spreading of energy or of particles of molecular (or sub-molecular) scale against constraints on their free motion. These constraints may be outer (volume) or inner (aggregative) constraints. This spreading leads to quantum mechanical stabilization effects (which may be explained with the “particle in a box”).[1] — This is the reason why entropy increase leads to more stability.

K thank you very much.... This makes a lot more sense


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Can the Smirnov-Zamkow reaction be described as a pericyclic process?
The Smirnov-Zamkow reaction involves the reaction of 2-butyne and chlorine in the presence of a Lewis acid catalyst to afford a dichlorocyclobutene:

At first glance, it appears to be an example of a thermal [2+2]-cycloaddition, which, by the Woodward Hoffman rules is forbidden.[*] This either means that (1) the reaction is stepwise, or (2) the intermediates are able to arrange themselves such that both components are not reacting suprafacially.
Very few mentions of the Smirnov-Zamkow reaction appear in the literature. The only real description is found in Fleming [1] where the reaction is proposed to involve initial formation of an allene. All other references on the internet simply link back to Fleming, with no mention made of this reaction in March, Carey, or any of the other usual suspects.
Clearly the reaction isn't synthetically that useful, and its initial reports occurred before pericyclic reactions had been widely studied/rules existed governing what was 'allowed',[**] but it is a curious example, to which I can't (yet) convince myself is an allowed pericyclic reaction on the grounds of orbitals.

Can the Smirnov-Zamkow be satisfactorily described as a pericyclic process (i.e. can it be drawn in such a way as to obey the thermal Woodward Hoffman rules), or is it best described as a stepwise process?

References:
[*]: The reaction was so strange that allegedly R.B. Woodward was left 'open-mouthed, with nothing to say' when he first read the report of the reaction taking place, under such 'mild' conditions, if a set of second year undergraduate notes are to be believed.
[**]: The reaction is actually referenced in Conservation of Orbital Symmetry by Woodward and Hoffman, they actually give a slightly different explanation to Fleming/the lecture notes I first encountered this reaction in.
[1]: Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference Ed.), Wiley: 2010
[2]: I. V. Smirnov-Zamkov, Doklady Akad. Nauk SSSR, 1952, 83, 869 (Chem. Abs., 1953, 47, 2711g); I. V. Smirnov-Zamkow and N. A. Kostromina, Ukrai. Khim. Zh., 1955, 21, 233 (Chem. Abs., 1956, 50, 9302e); R. Criegee and A. Moschel, Chem. Ber., 1959, 92, 2181.   (I also don't speak Russian to read some of these)

I was going to write a comment, but I think this is good enough for an answer.
The issue that symmetry rules forbid a [2+2] thermal cycloaddition if the addition is suprafacial-suprafacial.
However, you can achieve a suprafacial-antarafacial [2+2] thermal cycloaddition. This is symmetry allowed because the antarafacial symmetry is exactly opposite suprafacial symmetry.
I'm not familiar with your exact example, but I am familiar with [2+2] thermal cycloadditions of ketene. It is thought to proceed via suprafacial-antarafacial additions.
See for example: Burke, L. A. J. Org. Chem., 1985, 50, 3149.

+1 for the general description, but I was really asking specifically about this weird named reaction. I alluded to the non-all-supra condition in the q.


text:
Why would the pH of a solution not be 7 at the equivalence point of an acid-base titration?
Is it because the new substance formed might be acidic/basic ?
The internet defines this point as when the number of moles of the acid equals the number of moles of the base. But if we have for example 1 mole of NaOH and 1 mole of H2SO4 (both in dilute solutions) mixed together, the resulting solution will still be acidic because 1 mole of H2SO4 releases 2 moles of H+ right ? So we will have an extra mole of H+ which will make the resulting mixture acidic. So is this still the equivalence point even though it isnt really equal (bcz we have 1 extra mole of H+) or is it 1 mole of NaOH + half a mole of H2SO4 ?

The equivalence point is defined as when you have an equal amount of acid and base. It you have a strong acid and a strong base, the pH at the equivalence point should be pretty close to 7, but if you have a weak acid or a weak base involved, there is not reason why the equivalence point should be at pH 7. You would have a conjugate base or acid that is reasonably basic or acidic at the equivalence point.


text:
Chemical function of Mg and Conjugated bonds in Chlorophyll
This question has 3 parts, each related to the title.
In biochem, we were shown the chemical structure of chlorophyll.  The light-absorbing head has many carbon-rings with alternating single and double bonds (conjugation).  We were taught that this arrangement of bonds is what allows the molecule to absorb light energy.
Question 1: I understand that pi-bonds are generally better at absorbing photons than single bonds, but why do conjugated bonds increase this ability?
Question 2: Also, this molecule does not seem to be have any resonance structures needed for delocalized bonds and thus does not have the light absorbing capabilities like that graphene/graphite or benzene rings.  (If we could engineer an artificial chlorophyll molecule would this increase its efficiency?)
The head of the molecule also contains a Mg ions that is coordinate bonded to four surrounding nitrogen atoms.
Question 3:  If the alternating double and single bonds are responsible for the absorption of light, what role does this metal complex serve in the molecule?  Other pigments like carotenoid don't have this metal complex and they're still able to absorb light.

but why do conjugated bonds increase this ability

Adsorption of light results in excitation of an electron, moving from low-level (usually bonding or non-bonding) to a higher (usually antibonding) orbital. (for understanding of what bonding, non-bonding and antiboding orbitals are, see here)
Generally, when two orbitals of same energy partially overlap, two more are formed: one with slightly higher energy and one with slightly lower. When orbitals of conjugated bonds interact, energy of one bonding orbital goes up and of one antibonding orbital goes does, narrowing the gap. When the gap is withing or closely near ~1.6-3.2 eV it can participate in visible light adsorption.

this molecule does not seem to be have any resonance structures

It does, see carefully.

what role does this metal complex serve in the molecule

The relevant metal-free molecule do exist in nature. It has, however, different color. $Mg$ ion seems to adjust the gap mentioned above. For a more clear understanding of the matter you might look for data on phtalocianin pigments, that may be easily obtained with different central atom in a very similar structure.

this MO thoery seems to be really useful as it keeps popping up wherever I go.  Do you know any sources that would allow me to learn this concept?  Thanks !

@AniekanUmoren Modern university-level inorganic-chemistry textbooks usually provide a qualitative introduction.  Quantum-chemistry books that are not focused on DFT provide a more detailed introduction.

@ AniekanUmoren Chlorophylls play a major role in light harvesting and in electron transfer in photosynthetic organisms, plants, algae  and bacteria. The Mg seems to adjusts the redox potential wrt. electron transfer  in the protein of photosynthetic organisms. It may also play a role in binding to the protein. (The metal free chlorophylls are called pheophytins and have slightly different spectra. Generally chlorophylls are derivatives of chlorins and are closely related to porphyrins)


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How does hybridisation affect an otherwise chiral centre?
In basic theory, a carbon atom with four nonidentical substituents attached, makes a chiral centre. Thus any molecule is chiral as long as it has a chiral centre (except meso compounds). I thought identifying a chiral centre was straight forward until I saw the following video.
Is Carbon (with the ability to bond 4 covalent bonds) the only atom qualified as a chiral centre (stereogenic centre)? Wikipedia link states, other atoms like Nitrogen or Sulfur could also be a chiral centre.
The above video shows that other atoms too can be chiral as well... However a comment in the same video, replies that a hybridised Nitrogen atom is not qualified to be a chiral centre. Original comment : That Nitrogen is sp2 hybridized, the lone pair of electron are in a unhybridized p-orbital; it would not be considered a chiral center.
With reference to this question 1-, it has number of comments, but not quite the answer I am after. On the other hand this question 2- has more emphasis on the lone pair for chirality - an atom with lone pairs can't be a chiral centre.
So given these, why can't a hybridised atom be a chiral centre?

I'm afraid you misunderstood it. A lot of chiral centers can be considered sp3 carbon atoms so I don't see much sense here.

@Mithoron I am learning. And at most times I feel it's just better to stick to text book and not look outside the box - because when I try to reason beyond the oversimplified model, the density of information can be confusing. I would have even accepted the above video as it is, if I didn't see that comment. Up until I saw this video - my textbooks only talk about a Carbon as a chiral centre and had no clue that hybridisation has such effect on chirality. I am not sure why the question is invalid.

Phosphorus and sulfur can both be chiral centers with 3 bonds and a long pair. It's also true of nitrogen, but with phosphorus and sulfur, the rate of inversion is slow enough that the chiral center is stable.

@Zhe before this questions gets closed, could you please clarify these two statements? Qualifying an atom as chiral : 1. 4 non-identical substituents (including a lone pair being one). 2. Hybridisation has no impact on a chiral centre

As is noted in the answer you accepted, hybridization is a relevant factor.

I really wouldn't put too much stock into YouTube comments. Often, it's just the blind leading the blind. The comments on this video have all sorts of incorrect statements, it's really not where you want to be learning from...

Firstly, chiral centres are not restricted to only carbon. Source: IUPAC Gold Book. With that out of the way, we can enumerate the three main possibilities for nitrogen.

$\mathrm{sp^3}$ nitrogen with a lone pair is usually not considered to be a chiral centre, as the rate of inversion is very rapid (google "nitrogen inversion").
There are exceptions, such as if the amine geometry is constrained by a bicyclic system (something similar to quinuclidine), but that's a discussion for another day.
$\mathrm{sp^3}$ nitrogen without a lone pair e.g. in ammonium cations, can be chiral if there are four different groups attached. So these two molecules are enantiomers of each other.

In fact, it is this context in which Wikipedia says that nitrogen can be a chiral centre.

Racemization by Walden inversion may be restricted (such as ammonium cations), or slow, which allows the existence of chirality.

The implication is that nitrogen inversion is fast in the case where there is a lone pair. However, it is very unclear, so I don't blame you.
$\mathrm{sp^2}$ nitrogen, e.g. amide nitrogen, is planar so it cannot be chiral. So, this video you watched is incorrect in stating that the amide nitrogen is a chiral centre.
And, that also means the comment you quoted is correct. But, I would caution that it is not a direct cause and effect relationship between "$\mathrm{sp^2}$ hybridisation" and "not a chiral centre". It is more like: because it is $\mathrm{sp^2}$, therefore the geometry around it is planar; and because it is planar, it cannot be chiral.

As was mentioned in the comments, phosphorus and sulfur behave differently. This has been covered multiple times here; here are some links. 1 2 3 4

That comment in youtube was the mother of all confusions I had - I knew for sure the original statement had a valid reason to be there. I just lacked knowledge to reason it and perhaps articulate my question to receive the answer for clarification. But I see what you mean now. Muchas gracias! At basic level I learn things in isolation, and rather ignoring actual concepts such as inversion rates, resonance and stability etc. But put together, it makes absolute sense.

No problem, glad I got the question right! Don't worry about all the other things - you'll come to them in due time (if you want to, of course).

On a side note, chirality also emphasises much on the non-superimposability of the molecule in terms of the plane of symmetry. In that case, can I sum up that, for a molecule to be chiral and an atom to be chiral centre (in extremely basic/beginner level), 1.molecule must be non-superimposable with its mirror image, where there's no plane of symmetry (here it links to hybridisation discussed) 2. it has 4 non identical substituents something to note: chiral centre is not limited to carbon.

Yes, that's probably a good starting point.


text:
How to determine conformation using IR
In my advisor's Ph.D. research, she used IR spectroscopy to (help) determine the conformation of a compound of interest (its structure is known.)
In organic chemistry, I learned about using IR as a way to identify functional groups (and in conjunction with HNMR, CNMR, & mass spec, I was able to determine the structure of a compound.)
How can one determine conformation using IR? What are the limitations of this? I feel like it could get messy really quickly.

You maybe need to specify the type of compounds you're interested in. I've seen this kind of thing for large macromolecules, but not in small molecule elucidation

Assuming your advisor performed gas-phase IR spectroscopy, she could have determined the conformers based on predictions (e.g. DFT) for a known chemical structure. The intensities of the spectral lines are very sensitive to the dipole moments along the 3 principal axes of the molecule and they are typically different for different conformers and relatively easy to calculate.

Another option is possible when she performed rotationally resolved spectroscopy (this is less likely). When you selectively replace each element in a molecule by a heavier isotope (e.g. replace H by D and $^{12}$C by $^{13}$C, the rotational constants change and by determining the constants from the spectrum you can work out the coordinates in the center-of-mass frame of the substituted atom (using the so-called Kraitchman Equations).

Also I would mention that IR can help identify conformers in all cases in which different conformer distribution can be altered prior of or during the measurement. As Paul said, generally one already knows the chemical structures.

I believe you're referring to this thesis. She uses vibrationally and rotationally resolved electronic spectroscopy. I don't know much about this so I'm not gonna answer, but the information is there if anyone wants to provide an answer. Feels kind of creepy I could find that using just your name and the question details. Haha :)

@jheindel I have to admit, that's pretty impressive. Blast my unusual name!

@Alchimista Perhaps that is what was done. I'm pretty sure, in this case, that the chemical structure was known. The only unknown structure thing I could think of would be degradation. I don't know to what extent that would be an issue.

This is perhaps my biased view as a theorist, but I've collaborated quite a bit with a condensed-phase 2D-IR group and a little with a UV resonance Raman group, so I have some understanding of the experimental perspective as well.
For anything outside of the simplest models, like a 2- or 3-state system you can scribble on an envelope, or the simplest molecules that would display configurational dynamics, like short peptides or single amino acids, experiment strongly benefits from computation. Have fun searching the literature for tables of IR peak frequencies, looking at the assigned normal modes, (which are sometimes compound motions), and then seeing if any of those match your hopefully very similar structure. If the molecular structures are known, then for simple cases a few of the conformers can be made by hand using your favorite computer program or a partially-automated conformer search, have their geometries optimized, and harmonic vibrational frequencies calculated to at least qualitative accuracy in an afternoon. Then it's a matter of clicking around in a visualization tool to see which interesting peaks correspond to which normal modes.
If populated conformers are harder to find, such as in systems that are larger with many rotatable bonds or have substantial librational motion, an exhaustive conformer search may be too expensive in terms of computer time, and running molecular dynamics trajectories with some sort of post-trajectory clustering analysis is an alternative, since IR and Raman spectra can be generated from trajectories via velocity, dipole, and polarizability autocorrelation functions.
Example
Here is an example where a lithium cation is interacting with L-proline in two ways: 1. with the backbone-bound carboxylic acid, and 2. with the ring. Perhaps this is cheating, because it isn't a single molecule, it's two different coordination modes, but you would see something similar with an intramolecular conformational change or rearrangement such as keto-enol tautomerization.

Here are their respective vibrational spectra within the harmonic approximation.

Already without any assistance, we can see huge changes outside of the fingerprint region. This is similar to how hydrogen bonding in water changes the IR spectra when moving from molecules to the bulk. The three most notable changes going from backbone- to ring- bound are

a blueshift (increase in energy) of the carbonyl peak, moving it from 1760 to 1852 wavenumbers,
the intensity increase of the ring $\ce{C-H}$ stretches below 3000 wavenumbers, and
a few changes above 3500 wavenumbers.

1 is probably due to charge transfer into an antibonding orbital localized on the carbonyl group. However, it's also strongly coupled to the $\ce{-OH}$, so interactions there will have an indirect effect. I'm less sure about the $\ce{C-H}$ stretches, but they belong to the three hydrogens closest to the $\ce{Li+}$ (7, 11, and 14), and Mulliken charges don't support a CT idea, which isn't surprising and doesn't rule it out. For number 3, the $\ce{-OH}$ stretch moves from 3686 to 3439. This redshift is probably a result of the blueshift to the coupled mode; assuming that the total vibrational energy between conformers is similar ($E_{\text{ZPE}} = 90.100$ vs. $\pu{90.182 kcal/mol}$ here), vibrational energy tends to be redistributed evenly and in ways that make physical sense; see here for a discussion about how this simplified two-mode coupling works which might also be applicable here.
Frequencies are in wavenumbers (cm$^{-1}$) and intensities are in units of km/mol. Modes below 1200 wavenumbers are commented out.
$$
\begin{array}{rrrr}%
\hline%
\text{backbone frequencies} & \text{backbone intensities} & \text{ring frequencies} & \text{ring intensities} \\
\hline%
%% 30.91 & 2.37 & 62.75 & 4.92 \\
%% 88.36 & 6.77 & 91.13 & 2.54 \\
%% 168.98 & 24.56 & 146.67 & 15.42 \\
%% 200.22 & 35.17 & 239.15 & 53.54 \\
%% 301.33 & 5.59 & 299.49 & 10.23 \\
%% 323.29 & 67.62 & 399.53 & 47.70 \\
%% 393.10 & 18.41 & 436.61 & 77.33 \\
%% 423.78 & 49.21 & 454.79 & 10.29 \\
%% 485.16 & 69.90 & 517.44 & 4.46 \\
%% 534.48 & 56.37 & 541.24 & 57.13 \\
%% 565.71 & 45.28 & 582.68 & 14.18 \\
%% 625.82 & 27.87 & 622.71 & 9.03 \\
%% 663.98 & 12.16 & 703.37 & 26.78 \\
%% 754.48 & 22.99 & 758.60 & 12.72 \\
%% 802.57 & 16.27 & 808.16 & 50.74 \\
%% 842.01 & 5.53 & 823.92 & 48.36 \\
%% 858.14 & 14.85 & 863.84 & 7.69 \\
%% 875.87 & 27.62 & 875.21 & 13.13 \\
%% 897.13 & 13.53 & 912.31 & 2.55 \\
%% 919.58 & 0.12 & 928.34 & 12.11 \\
%% 966.10 & 12.84 & 952.79 & 7.90 \\
%% 1007.42 & 0.12 & 1017.17 & 0.53 \\
%% 1034.74 & 23.77 & 1056.15 & 1.11 \\
%% 1070.65 & 26.45 & 1116.38 & 40.56 \\
%% 1154.24 & 13.04 & 1119.11 & 9.28 \\
%% 1161.61 & 3.40 & 1148.10 & 7.38 \\
%% 1181.00 & 74.31 & 1173.28 & 7.56 \\
1213.61 & 4.92 & 1203.94 & 6.81 \\
1222.70 & 2.73 & 1228.83 & 4.07 \\
1263.58 & 152.20 & 1261.21 & 39.44 \\
1279.48 & 0.73 & 1272.66 & 12.99 \\
1314.79 & 1.01 & 1321.29 & 57.22 \\
1328.39 & 58.17 & 1342.94 & 314.05 \\
1366.92 & 47.01 & 1365.77 & 74.02 \\
1390.24 & 127.09 & 1402.33 & 19.65 \\
1424.08 & 13.57 & 1522.07 & 16.68 \\
1432.59 & 5.88 & 1561.70 & 0.40 \\
1460.15 & 3.30 & 1572.71 & 24.88 \\
1760.04 & 284.35 & 1851.86 & 270.80 \\
2976.69 & 25.42 & 2607.42 & 324.10 \\
2983.82 & 7.16 & 2806.11 & 147.60 \\
2993.71 & 4.29 & 2900.28 & 54.24 \\
3000.08 & 8.94 & 3024.05 & 5.13 \\
3039.46 & 13.36 & 3027.04 & 2.76 \\
3063.00 & 5.22 & 3061.92 & 2.04 \\
3075.69 & 4.87 & 3074.89 & 3.80 \\
3477.77 & 29.07 & 3438.98 & 235.23 \\
3685.91 & 93.01 & 3527.08 & 36.70 \\%
\hline%
\end{array}%
$$
All calculations were performed using B97-D/def2-SVP. See source for XYZ structures.


text:
Halides in s-block elements
What is the reason behind the following statement:

The $Δ_fH^\circ$ for the fluorides become less negative as we go down the group while the reverse is true for chlorides, bromides and iodides.


text:
What determines how much water of crystallisation a salt has?
Is there any reason behind the number of water of crystallisation?
For example, $\ce{LiCl.2H2O}$ has two waters of crystallisation, and $\ce{MgCl2.8H2O}$ has 8 waters of crystallisation. Why does $\ce{LiCl}$ have only 2 waters of crystallisation and not 8?

It has to do with the coordination abilities of the metal ions and crystal lattice voids I believe, we'll wait for a more advanced chemist to come along.

There is no simple answer; if you think you have one, you are wrong.

Something like the topics discussed in this paper would be needed to get at why the particular number you end up with is the actual number. What you're asking is related to the paper I reference because the number of waters in a crystal structure will be related to the number of waters which solvate the ions present in a solution. But yes, the answer is exceedingly complex because it depends on the stabilization of multiple electronic states of ions in the presence of water, and is thus also not general.

The number of hydrate water molecules per formula unit is not fixed for a certain salt. Calcium sulfate for example exists as dihydrate and hemihydrate, sodium sulfate as hepta- and decahydrate. Pressure and temperature during the precipitation are important. MgSO4 has at least three different hydrates.

The water of crystallization in a crystal of any ionic salt can be associated with the size of the cation, magnitude of charge on it, tendency for co-ordination and hydrogen bonds, its lattice structure and other factors involving the anion. But we cant actually theoretically determine why a particular salt has these man water of crystallization. We determine this experimentally in the labs.Then after finding this data for a set of salts we try and link them with the above mentioned factors and try to give a suitable explanation which justifies real data to the greatest possible extent. A theoretical explanation would be very complex and definitely involve way more parameters than what I mentioned above to be accurate for max cases.Even there isn't any 100% full-proof theory or mathematical formula that would allow us to calculate the water of crystallization in any random salt.

In the solution then crystal, in the first example Li becomes Li+1 cation with one bonding positions, and Cl become Cl-1 ion, both having only one position so they can bond any way needed in 3D.  However, a direct bond between them will not create a crystal, as a crystal needs an continuous chain (one must be open to the next set) to continue.  You get Li-Cl single bond, and not crystallization.
Yet with H2O, the Cl-1 will bond width one H2O (hydrogen bond), and the another Hydrogen (in H2O) is available for crystal chain bonding.  That is why H2O are needed to create a crystal.  You get a +H-O-Cl.
The Li+ would bond to an outer electrons of the H2O, and again end the chain.  In that way, we only have the H2O Hydrogen bonds, and get a chain +H-O-H+ with Li attached in the other hemisphere.
With regular H2O in crystal as +H-0-H+ with two bonds. That means that you get a crystals with chains at the corners of the cube (8 positions).
The H2O in 3D has two Contributing Hydrogen bonds (proton outward), and six receiving outer electrons.  That leads to your conclusion that it would be 8H2O.
Yet, the Cl- ends the chain, Li+ does not.  That means you get a chainable structure with just 2 H2O where two Hydrogen are ready for other Hydrogen bonds:
                    H+
          +H-O-H+   O-H+Cl-
             Li+

There is two open H+ for hydrogen bonding along the crystal chain in a clean 2H2O:LiCl structure.  That makes this crystal possible with that low ratio (if build in that environment) as achievable.
However, Mg becomes Mg+2 with those bonding positions at endcaps.  That means the Mg+2 always creates bonds to two H2O atoms at the H+ positions.  The two contributing positions do not line up relative to the H2O H+ positions (or calc very well as attractive) as a double bond.
But, you cannot create a crystal without an H+ in H2O remaining open, so you get two H2O for every Mg, which then leads to 4 more H2O by hydrogen bonds.
                +H      H+
                  O+Mg+O
                +H      H+

The positions for Cl- still end the chain, so those can insert that pair anywhere; you can have a crystal ending one H+ in the Cl-1, and still build in the other three directions.
                           H+
                +H      H+ O-H+ Cl-
                  O+Mg+O
                +H      H+ O-H+ Cl-
                           H+

However, the +H bond to other H2O, so you get 2 direction from the Mg x 4 hydrogen H2O bonds = 8 H2O required for the second crystal.  It cannot work with just 2H2O in the 3D structure.
That is why the MgCl2 requires a ratio of at least 8 H2O, so MgCl2:8H2O, not :2H2O.
The main quality of water for creating crystals is its creation of Hydrogen bonds at two diagonal corners of the Scrunched (104.5 bonding from ~66 nucleomagnetics inclination angle) Cube in the same hemisphere.  In AVSC, the Hydrogen electron fills the Oxygen settling position.  That leaves the Hydrogen as settling Proton outward, which can create the location for Hydrogen bonds at specific angles.

Now that's a remarkable bunch of nonsense.


text:
Why energy of all d orbitals is same?
This question first popped up in my head when I learnt that $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and  $d_{z^2}$  orbitals are used in $sp^3d^2$ hybridization. If all the $d$ orbitals are said to be degenerated orbitals (i.e. they are said to have equal energy) then why these two orbitals are used in $sp^3d$ and $sp^3d^2$ hybridization?
Some said that $d_{z^2}$  is used because it is closer to the nucleus because of the ring of electron density surrounding the nucleus, but if that is the reason then why do we call the $d$ orbitals as degenerated in the first place when clearly two orbitals are closer to nucleus.
Here's what I want to know:
A) Why $d$ orbitals have same energy when its clearly apparent from their shape that two orbitals have lower energy than other three?
B) Why $d_{z^2}$ orbital is used in $sp^3d$ hybridization and $d_{x^2-y^2}$ and  $d_{z^2}$  orbitals are used in $sp^3d^2$ hybridization?

(1) in the atom with spherical symmetry, all five d-orbitals are degenerate, there is no dispute about this, and the argument of "clearly apparent from their shape" doesn't hold up, as our eyes aren't a good measurement tool for energy of an orbital. (2) in octahedral geometry the degeneracy is partially broken. The ${x^2-y^2, z^2}$ set is actually higher in energy than ${xy, yz, xz}$, so maybe that's something else that should have been added to your previous question on why hybridisation isn't looked upon very favourably anymore.

and the difference in energy isn't because it's "closer to the nucleus" or anything like that - it's better explained by MOT or at the very least, CFT. Anyway, for your other qn, the reason why those orbitals are chosen is because of symmetry properties - for example if you try to make an octahedral molecule using ${xz,yz}$ then you will end up with asymmetric hybrid orbitals (roughly speaking, they point more along the z-axis than along the x-/y-axes). Again, MOT explains it much better;  the use of the ${x^2-y^2,z^2}$ set falls out naturally from the fact that they transform as $e_g$.

@Orthocresol"Again, MOT explains it much better; the use of the ${x^2−y^2,z^2}$ set falls out naturally from the fact that they transform as $e_g$" I don't understand this. Can you provide me some resources about this phenomenon?

@Mockingbird It means that as a ligand approaches the central atom(in case of octahedral complex) along the axes, energy of these two orbitals increase more than the remaining three orbitals; these two orbitals are now called $e_g$ orbital. orthocresol means to say that the use of these two orbitals can simply be explained by this splitting of d orbitals as the ligand approaches the central atom.

@orthocresol by reading your comment, I clearly understood why the $x^2-y^2$ and $z^2$ set is used in $sp^3d^2$ hybridization but how do we explain the use of $d_z^2$ orbital in $sp^3d$ complex?

@Mockingbird This phenomenon is at the heart of crystal field theory. You might want to read up on that.

You asked https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes but still asking about " sp3d "? Really?

@Mithoron Much of what we are taught in class is based upon these less advanced theories of hybridization and that question was anyway specifically asking about coordination complexes.  "by still asking about " sp3d"" , are you referring to the not so favourable use of hybridization or do you mean that I am missing something trivial here?

I mean, why does $d_{z^2}$ exist in the first place? Why aren't there $d_{x^2}$ and $d_{y^2}$ orbitals?

A. The orbital shapes originate from solutions to the Schrodinger equation for hydrogen like atoms. The derivation of the orbitals come from the section for the solution to Schrodinger equation for Hydrogen-like atoms on Wikipedia: https://en.m.wikipedia.org/wiki/Hydrogen-like_atom
The solutions have the same energy because you solve the equation for specific energies. They really are the same, which can proven by doing an integral across the volume.
The different hybrid orbitals depend on what quantum numbers are possible. This question has been answered here: Why an asymmetric geometry with sp3d and sp3d3 hybridization?


text:
Charge Balance and power law?
I've come across a statement in a paper* that's poorly explained, and I can't work out the reasoning behind it. I suspect this is embarrassingly basic, but could anyone please explain the theory behind the following (from the bottom of the second paragraph of the discussion)?

For a reaction involving B(OH)4- and CO32-, a B(OH)4- can balance the charge of only 1/2 a CO32-, thus a dependence [of the B/Ca ratio in synthetic aragonite] on [B(OH)4-]/[CO32-]0.5 would be expected.

I'm unsure why a power law applies here.
* Holcomb, M.; DeCarlo, T. M.; Gaetani, G. A.; McCulloch, M. Factors affecting B/Ca ratios in synthetic aragonite. Chem. Geol. 2016, 437, 67–76. DOI: 10.1016/j.chemgeo.2016.05.007.

Your quoted statement would make sense in discussing the dependence of the equilibrium constant of a reaction involving $2\ce{B(OH)_4^-}$ on one side and $\ce{CO_3^{2-}}$ on the other, but I'm unsure if this is relevant.

Thanks @a-cyclohexane-molecule. Could you point me towards some reading so I can understand why this is this case?

I don't have a good resource on hand, but a Google search regarding equilibrium constant should be productive. Essentially, we define the equilibrium constant for the reaction $\ce{aA + bB \to cC + dD}$ as $$K = \frac{[A]^a[B]^b}{[C]^c[D]^d}.$$ Specializing to this reaction indeed shows the desired proportionality.
Again, do note that I'm not certain that this is what the authors of the paper are getting at.

I think you're exactly right. They discuss the stoichiometry of the reaction as $B(OH)_4 + 0.5 CO_3^{2-}$, so this makes sense. Thanks for the pointer!


text:
How to obtain hydrophobic Doxorubicin from its hydrochloride salt?
I have read literatures on dissolving Dox hydrochloride in DMSO and adding Triethylamine to the resulting solution. The purification is done using dialysis (incase of polymers). But I cannot do dialysis.
Another method I tried was neutralizing the salt solution of Doxorubicin with PBS (pH 8.04), and centrifugation of the resulting mixture. But centrifugation yielded no pellet.

I would imagine this compound is soluble in diethylether or dichloromethane. You can probably do a standard acidification with 1M HCL and then extract into such an organic solvent. Hard to know how much detail to provide here because I don't know your experimental skill level or experience.

Add a base to an aqueous solution of the compound.  The base could be ammonia water or ammonium hydroxide.  Make sure you add enough base to raise the pH to 12 or higher.
Add a water-immiscible organic solvent with a low boiling point to the solution and mix well.  A good solvent might be dichloromethane.
Let the phases separate and carefully remove the organic phase.  (If you use dichloromethane, the organic phase will be the bottom phase, but this is not true of all solvents.)
Repeat step 2 several times, each time combining the newest organic extract phase with the extract from the previous steps.
Evaporate the solvent under air, or, if you have it, an inert gas such as nitrogen.


text:
why bond energy of hydrogen molecule is twice the chlorine molecule?
why Bond energy of Hydrogen molecule is greater than chlorine and HCl molecule?
Here's a link

This could be relevant to you.

Bond energy of chlorine is less due to lone pair-lone pair repulsions.

Can you please explain in detail.?
My book claims that in the bonding molecular orbitals, there is more electron density between the two nucleus as compared to the atomic orbitals. And this more electron density acts a glue to bring two nuclues close to each other. That is why bonding molecular orbital is stronger?

Because there are shielding electrons, meaning that the atoms can't come as close to each other. The outermost electron is the one being shared to fill a p orbital in chlorine where as it fills the s orbital in the hydrogen

Can you please explain in detail.?
My book claims that in the bonding molecular orbitals, there is more electron density between the two nucleus as compared to the atomic orbitals. And this more electron density acts a glue to bring two nuclues close to each other. That is why bonding molecular orbital is stronger.


text:
What does the Pi stand for in the reaction for biological nitrogen fixation?
In the reaction for biological nitrogen fixation, what does the Pi stand for in the products?
$$\ce{N2 + 8H + 8e- + 16 ATP -> 2NH3 + H2 + 16ADP + 16 P_{i}}$$

It stands for inorganic phosphate (Pi). When ATP is broken down into ADP, energy is released along with a phosphate. You can visualise this if you look at the structure of ATP. ADP is then also able to be broken down into AMP and more Pi


text:
Which group is more suitable for metal nanoparticles functionalization?
I have synthesized bimetallic nanoparticles (gold core with a silver shell). I have two choices for functionalizing them, through thiol group or amine group.
Which of them is more attracted to this kind of nanoparticles?

Sulfur containing groups, such as thiols, are in general more strongly attached to silver nanoobjects than amine groups. However, the problem with sulfur and silver is that the strength and structure of sulfur-silver bonds depends critically on the local topology of the silver surface, and in-layer lateral interactions between stabiliser molecules can have a strong impact on this, so there isn't a clear pattern of reactivity of sulfur-containing groups and silver nanoobjects. You can easily get irregular, disimilar or patched coverage.
Amine is much more straightforward, links itself more weakly and is routinely used in the synthesis of silver nanoparticles as a stabiliser. However, if you want permanent functionalisation of the nanoparticle (vía covalent bonding), probably amine-silver bonding will be too labile.

Do you think thiol group is the best choice for permanent functionalisation of this kind of nanoparticles or there is another choice too?

Well, what do you want to functionalise them with?

I want to functionalize my nanoparticles with a kind of Gd ligand. I would like to know what is the best and most permanent bond with silver?


text:
Broken object bonds
This may be a stupid question but I was wondering why, if I break a stone or any other object in two parts, and then I try to put these two parts together they don't stick to each other.
I mean, the bonds that were keeping those atoms together have been broken so whenever they are put next to each other they should theoretically attract each other back..?
Thank you

@IvanNeretin Hmm. Well, I am not even sure what actually happens to the bonds when the object gets broken. do they rearrange themselves or they stay exactly as they were? if they don't move then your could be a valid answer :-)

Paraphrasing what  Feynman once said: Suppose you snap a block of pure copper. If you were to somehow prevent any impurity/oxide layer from developing at the newly formed surfaces, and if you were to bring the blocks together again... they would stick. It's because there is no way for the copper atoms  to "know" that they belong to different blocks. : This was an interesting piece of insight, so I thought I'd share it with you ;)

Depending on the chemical nature of the stone, the bonds may rearrange themselves to be quite content with their new situation, or they may be left dangling for a while.  In the last case you should be able to put the halves back together, given the following conditions:

The dangling bonds do not react with the components of surrounding air;
The break-up surfaces remain very clean (no debris from the fracture, no dust in the air);
You can put the parts together with atomic precision.

Each of these conditions is unlikely enough in itself. And if any of them fails, then the stone will behave as usual, that is, remain broken.


text:
Mechanism of Hofmann Rearrangement Reaction
In the Hofmann Rearrangement Reaction (in basic medium), when $\ce{RNH-}$ ion is produced, why does it not attack $\ce{Br2}$  or take $\ce{H+}$from the amide itself?
Also, why does it take $\ce{H+}$ from water only?

How much water is there and how much of everything else is there?

@jerepierre, we have taken bromine in excess and 1 equivalent of water.

It's very unlikely that there's only one equivalent of water...

@jerepierre, the reaction conditions state that there are 3 eq. of NaOH,   Br2 and 1 eq. water.

@jerepierre, it is also given that, sometimes we don't require water as it gets produced in the reaction itself.

@Ayushmaan Wait, wait, wait. There are 3 equivalents of sodium hydroxide and no water? Is it just a solid sitting at the bottom of the flask?

Most of the steps in the mechanism are reversible,  especially the proton transfers. So, the exact source of a proton doesn't matter, as that proton might have been exchanged with other protons many times before. Except for the actual rearrangement (i.e. the isocyanate forming step), most are reversible steps.

@Zhe, I meant to say that the solvent doesn't matter, as long as it does not hinder OH- attack.

@EashaanGodbole but won't it help reduce the amount of NaOH required in the reaction?

Review the mechanism of the reaction. The NaOH is required stoichiometrically. You need that much...

In Hoffmann Rearrangement water is used as solvent, so much more than 1 equivalent is added.

Water is present in large excess when compared to the amide, so it is statistically not likely that the proton would be removed from the amide rather then water. Even if it happens, proton exchange is reversible so it would be formed/unformed many times that eventually the removal of the proton by water would happen and proceed with the mechanism.
Regarding the attack to $\ce{Br2}$, I believe the reason it doesn't happen is that the ion would be surrounded by many many water molecules because: 1) the amide forms H-bonds with water, 2) the ion is solvated efficiently by water, so more molecules would be around to have their protons removed.


text:
Ketone/aldehyde synthesis from N-acylazetidines or aziridines
A common principle in organic chemistry is to make sure the product is less reactive than the starting material, to prevent overreaction. So, for example, the treatment of an ester with a Grignard reagent will afford not the corresponding ketone, but the alcohol.
However there are some cases where the contrary is true. In particular, when the tetrahedral intermediate is stable, then the reaction can be worked up to 1) produce the desired ketone and 2) destroy any unreacted organometallic reagent, thus preventing overreaction. The classic example is a Weinreb amide, where the tetrahedral intermediate is stabilised by chelation, AFAIK.
Somewhat recently the synthesis of ketones from N-acylazetidines was described:1

and also the reduction of N-acylaziridines is known to produce aldehydes:2

Presumably the idea is the same: the tetrahedral intermediate is stable under the conditions used. Why is this so? Is it connected to the small size of the ring?

References

Liu, C.; Achtenhagen, M.; Szostak, M. Chemoselective Ketone Synthesis by the Addition of Organometallics to N-Acylazetidines. Org. Lett. 2016, 18 (10), 2375–2378. DOI: 10.1021/acs.orglett.6b00842.
Brown, H. C.; Tsukamoto, A. Selective Reductions. I. The Partial Reduction of Tertiary Amides with Lithium Aluminum Hydride. A New Aldehyde Synthesis via the 1-Acylaziridines. J. Am. Chem. Soc. 1961, 83 (22), 4549–4552. DOI: 10.1021/ja01483a013.

I think in this case its a leaving group/pKa thing. In the Weinreb amide, the intermediate is stabilised by chelation, in the case of the azetidine, its likely more that the azetidine is just such a bad leaving group that the pathway is too high in energy at low T. Would be interested to see if anyone digs up a reference/proper explanation though :)

I think your suggestion about the stability of the tetrahedral intermediate is correct, but it's not that the tetrahedral intermediate itself has stability. I think it's because the collapse of the tetrahedral intermediate leads to a less stable intermediate.
My key insight for rationalizing this is that lithium hydride reduction of amides generally affords amines. We generally prefer to eliminate the lithium-oxygen adduct from the tetrahedral intermediate, forming an iminium ion.

With the two substrates you've provided, the iminium ion would be very highly strained because of the geometry of the small ring. I think collapse of the intermediate is thus disfavored. On work up, we actually kick out the amine to collapse the intermediate, providing an aldehyde or ketone.