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Is it possible for a substance to absorb a longer wavelength of EM wave and emit a shorter wavelength?
I know a fluorescent lamp works by emitting UV first and then the specific substance inside absorbs UV and finally emits visible light. An object may emit infrared under sunlight due to heating. But those examples are absorbing short wavelength of EM and then reemitting a longer one. Is there a substance that can do the opposite (e.g.: absorb visible light and then emit UV)?

Maybe not possible in theory (like claimed in the answers) but in practice many fluorophores (and perhaps some phosphorophores) will emit a small percentage of photons with a shorter wavelength than the absorbed photon. Normally some energy is lost in vibrational energy level crossings (see Pritt's answer), but a photon can also be absorbed by a molecule that's already vibrationally excited, giving it the energy to emit a photon with more energy (from the incoming photon + some vibrational energy). Of course it's not easy / common because energy flows downhill (to longer wavelenghts).

Take any green laser pointer and you've got an example. The core laser is in the IR, and it's frequency-doubled via second-harmonic generation.

It is possible, but impractical. You can "upconvert" light. This effect, as Wikipedia relays, is due to the sequential absorption of two photons per emitted photon.
This is certainly possible to make in a fluorescent / phosphorescent light scenario, but the engineering challenges (or, more specifically - the cost) involved in such a solution would be steep.
There is a chemical model of such a compound in the two-photon-absorbance wiki site, but I have not the nomenclature skills to name it for you. I even tried using online search for the structure (and it took me a while. I hate nested benzene groups - probably why I chose inorganic chemistry)

Does this happen in other fractions as well? 3 in 2 out, or even 5 in 3 out?

@tuskiomi There is no reason why not, it is just that the 2-into-1 version is rare enough

Is this the sane thing that's happening In a Nd:YAG doubling crystal?  https://www.rp-photonics.com/frequency_doubling.html

@tuskiomi No. Well. You could call frequency doubling a version of 2-absorption-1-emission, but there are different mechanisms at play. And you should think of them as separate phenomenon. I was always in a permanent state of confusion when learning about qm and lasers, but I see there are answers below that covers it.

Yes it is possible for molecules to absorb a single photon at a longer wavelengths than some of those that they emit at. If the molecules are thermally isolated from their surroundings then the result is that these molecules will be cooled. The effect is, however, small.
The effect is caused when the fluorescence and absorption spectra overlap to a significant extent. This can be seen at the longest absorption wavelengths and shortest fluorescence ones. This occurs in almost all molecules but particularly whose excited state geometry is rather similar to that of the ground state, e.g many type of dye molecules, porphyrins, chlorins etc. which means molecule with large planar aromatic rings.
The first figure gives two examples of spectral overlap, but the class of molecule in which this is most important are the chlorophylls and the overlap of absorption and emission is essential for efficient light harvesting in photosynthesis. In fact without it photosynthesis would be incredibly inefficient.

Fig 1. The figure shows absorption and fluorescence spectra for trans-stilbene and anthracene. The peaks in the anthracene spectrum are due to vibrational transitions in addition to the electronic transition forming the excited state.
The second figure shows, in schematic form only, the potential energy of a molecule in its ground and first excited state and a few vibrational energy levels. Because molecules have vibrational (and rotational) energy levels at room temperature it is possible by thermal means alone for a few vibrational levels to be excited.
If excitation is at long wavelength, shown in the figure by the red arrow, absorption occurs from the $v = 1$ ground state into the $v = 0$  level of the excited state. Subsequent emission to a lower level ($v=0$) in the ground state produces a more energetic photon than occurs in the absorption process. (Fluorescence emission can also occur to $v=1,2, 3$, in the ground state as determined by the Franck-Condon factors).
The black arrow indicates, additionally, that if the excited state is relatively long lived, a few nanoseconds, for example, in solution, that it is possible for thermal excitation in the excited state to populate $v=1$  and so also produce higher energy emission than absorption (green line).

Fig 2. Schematic of vibrational energy levels in the ground and excited state of a molecule. Only transitions to lower levels are illustrated.
(Note. If the molecules are in the gas phase then collisions with inert gases can promote energy to higher vibrational levels in the excited state; this effect was described decades ago in naphthalene, for example; see Chem. Phys. Letts v22, 235, 1973, & Proc. Roy Soc. (Lond). Ser A, v340, 519, 1974)

There are indeed: we use them in "Laser Viewing Cards" to be able to see IR Laser. An example of such a card is presented below:

When illuminated with IR light, it emits visible light. The principle is that the molecule are trapped in an excited state that requires an IR photon to de-excite. These cards need prior charging but it last several hours.

That's not really what he's talking about because of the prior charging requirement.  If they could  charge from the laser they are detecting it would be what he's asking for.

What about (pure) rotational Raman spectroscopy?
The specific selection rule is $\Delta J = \pm 2$. The anti-Stokes lines, with $\Delta J = -2$, arise from the emission of a photon with shorter wavelength (higher energy) than the incident photon.

Here's an idealised Raman spectrum of $\ce{H2}$, to illustrate the point (from Oxford lecture notes). The wavenumber scale at the bottom refers to the shift in wavenumber of the emitted photon relative to the incident photon. If this number is positive, the emitted photon has a higher wavenumber, i.e. higher energy/shorter wavelength.
Edit: As porphyrin has kindly pointed out - Raman spectroscopy is a scattering phenomenon. As such it is not technically an absorption + emission; although I guess (I hope) it is in the spirit of the question.

This is a nice example of energy being taken from or given to a molecule via the radiation field but it does not involve absorption or emission as it is a scattering process and need not be resonant with an absorption.

@porphyrin True that, it did cross my mind, but unfortunately only after posting my answer; it's technically not an excitation + decay.

Doesn't Raman scattering still involve absorption and desorption (emission) of a photon? I cannot think of another way how the photon ought to impart energy to the analyte. Even though the excitation is to a virtual state, the photon "before" should be different from the photon "after" (regardless of the change in frequency). Is this wrong?

+1: while being scattering, anti-Stokes Raman is an experiment that I've seen in labwork practica, so this is an experiment where an undergrad student has chances to personally do an experiment where (single) photons gain energy. (Would also be possible for @porphyrin's answer, but I don't remember any particular emphasis of the fluorescence experiments in that fashion. I do remember the exam question, though.)

It is possible and is applied in optical microscopy. This method is called two-photon microscopy. Two photons of longer wavelength are absorbed by a dye molecule which then emits one photon of shorter wavelength. As mentioned before, this process is highly unlikely, but this property can be used to reduce the size of the fluorescent spot in confocal mocroscopy. (Confocal microscopy is a techique where a fluorescent sample is scanned with a focused laser spot while fluorescence emitted by this spot is detected. The resolution of the acquired image depends on the spot size.) Especially the extension of the spot along the optical axis of the microscope is reduced. Furthermore, excitation with a longer wavelength is beneficial in thick biological samples (tissues) because red light is scattered less and is less harmful than, say, blue light.
I have used this technique with the fluorescent dyes Alexa Fluor 488 and Alexa Fluor 594, but I guess there are many more. We excited them with far red light at 800 nm and detected fluorescence in the green and orange range.

"Especially the extension of the spot along the optical axis of the microscope is reduced." Can you elaborate it? It would be helpful if you add a link to Confocal Microscopy.

For the sake of completeness: One way is the already explained chemical two-photon absorption, another option is the second harmonic generation of non-linear optics.
If you have a stereo and you increase the loudness, after a specific level the speakers begin to sound displeasing. The problem is that the speakers cannot amplify anymore without disturbances. These disturbances can be approximated by sums of frequency part, the base frequency part, the second harmonic, the third harmonic etc. The interesting thing is now if very strong light goes through material, the electrons do not react linearly anymore. So if you send infrared light with a wavelength of 1064 nm from a Neodym-YAG laser with enough intensity, you see the doubled frequency light, a green with 532 nm.

Picture by kkmurray, CC BY 3.0, 1997-04-11
To be fair, the conversion ratio is very, very low. CORRECTION: The double frequency crystals are quite efficient, the KTP crystal for Nd-YAG reaches 80%. The thing is the conversion rate is AFAIK low for  lasers in the range of hard UV or X-rays which cannot be produced directly.

It bears mentioning that this mechanism is in widespread use in just about all green laser pointers - they're IR lasers with a frequency-doubling SHG step at the end.

I don't think so.
The type of absorption and remission is called phosphorescence. The way phosphorescence works is, the shorter wavelength of light is absorbed by a suitable material and electrons are excited to many states higher above the ground state.

However, unlike most excitations, these do not de-excite immediatly. The decay back to regular states involve spin-flips, a forbidden transition. As a result, the decay can take over hours to decay step by step, with each step having a longer radiation due to lower energies.
For more information about Phosphorescence, check out the Wikipedia page.

Nitpick: it doesn't have to be phosphorescence. Fluorescence also (nearly always) occurs at longer wavelengths than excitation.

@Pritt Great answer. Can you suggest me a link to understand spin-flips?

@Mockingbird I'm no expert on that topic, but see if this helps you: http://astronomy.swin.edu.au/cosmos/S/Spin-flip+Transition

Quantum dots can be used to convert infrared light to visible light

Their approach was to use two films placed on top of a clear plate of glass. The film layer on the bottom was made using a type of quantum dot—an inorganic semiconductor lead sulphid which had been coated with a single layer of fatty acids to make the surface passive. The top film layer was crystalline and made of rubrene, an organic molecule.
In the two film approach, the top film absorbs infrared light, and the energy from it is then transferred to the bottom film. That energy, which exists in the form of excitons is then diffused as it passes through the rubrene—a process called triplet–triplet annihilation. They tested the films by shining an infrared laser at the finished product and found it glowed with visible light—it works, the team notes, because the collision of two low-energy excitons create high-energy excitons, i.e. singlets, which can emit visible light. The team reports that the films are quite efficient at converting the infrared light to visible light.


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Is enthalpy the same thing as potential energy?
From what I know in a closed system the potential energy + the kinetic energy is constant since no external work is done on the system. In my textbook it says that enthalpy is equal to the sum of the KE + PE meaning total energy. My question is how can the energy be lost to the environment after the collision of two particles. Since as the particle approach each other KE is turned into PE until activated complex is reached. Once it is reached they break up into smaller particles, repel each other and convert the PE back to KE. In conclusion all energy in the system is constant and no energy is lost. I don't quite understand why we have Enthalpy vs Time graphs. Since the energy in the system was never changed it simply converted between forms. For my homework when I pretended that enthalpy was simply PE and that endothermic = increase in PE and exothermic = decrease in PE I usually got it corrected(Such as question below). Is it not true that if you decrease PE by a amount KE would increase by that amount hence no change?
For instance a question from my homework "After a reaction, the product have less KE than original reactant molecules. Is the reaction endothermic or exothermic?"
Im thinking to myself that if KE has decreased PE must increase hence no change in energy so neither endothermic or exothermic. But the answer is endothermic.
In conclusion my main question is can I ignore the definition the textbook gave me and define enthalpy as total PE and delta enthalpy = change in PE rather than total energy.

No, potential energy is internal energy. Enthalpy is a different thing.

If enthalpy is different what is the difference?

$$\underbrace{\hskip 0.75em H\hskip 0.75em}\text{enthalpy};=;\underbrace{\hskip 0.75em U\hskip 0.75em}\text{energy};+;\underbrace{\hskip 2em pV\hskip 2em}_\text{that's the difference}$$

@IvanNeretin Those some godly TeX skills, sir. Nice

I find this somewhat confusing as well.
It is often said in textbooks that change in enthalpy refers to heat, and NOT to pressure-volume work. Hence, the enthalpy would only be a part of the total energy. On the other hand H = U + pV suggests the other way around, that enthalpy includes pressure-volume-work. Is H a part of U or U a part of H? And why in Gibbs energy, is not the pressure-volume work exerted by a system part of what makes a reaction spontaneous or not?

@IvanNeretin  Internal energy is not potential energy, it is potential plus kinetic energy.  For instance, ideal gas particles have no potential energy, yet the internal energy of an ideal gas is not generally zero, since its internal energy also includes kinetic energy.  Thus the difference between potential energy and enthalpy is not pV, it's pV plus the kinetic energy: H = PE + (KE + pV)

@user83918  The enthalpy change equals the heat flow when the process is at constant pressure and the only type of work is $pV !$-work. Let's see why this follows from $H=U+pV$:  $H=U+pV \Rightarrow \Delta H = \Delta U + \Delta (pV) =  \Delta U + p\Delta V + V  \Delta p$.  At constant pressure: $ \Delta H =\Delta U + p\Delta V  =  \Delta U - w_{pV}$. And if the only type of work is $pV!$-work, $w_{pV} = w$, so: $\Delta H = \Delta U - w \Rightarrow \Delta U = \Delta H + w$.  But, since $\Delta U = q + w$, we have: $ \Delta H = q$.

In summary, at constant $p$, with $pV!$-work only,  $ q = \Delta H = \Delta U - w$.  We can see that, to get the heat flow, we need to subtract the $pV!$-work  from $\Delta U$, which means we need to add the $pV!$ term.   That's the way to understand why $H=U+pV$:  It's not that the enthalpy change includes the $pV!$-work; it's the opposite: it's obtained by subtracting  the $pV!$-work from $\Delta U$, which is why (again, under the specified conditions) it gives the heat flow.

In any way, ignoring the definitions given in a textbook is a bad idea. As for the question, the main problem in your thoughts is that you mixed too many things from different fields of science. Your example (two particles, kinetic energy, potential energy etc) looks like classical mechanics. So if you want to explain it, please, use tools offered by classical mechanics. Enthalpy is a measure of total energy in thermodynamics. It is used to explain processes taking place in large systems, which can be characterized with volume, pressure, temperature etc. It is not used to describe one particle.
Maybe you can take it into account and revise your question. Make it clear exactly what you want to understand: change in the kinetic and potential energy in interactions of two particles or change in enthalpy in exothermic and endothermic reactions.

No. H = U + pV, where H is the enthalpy of the system and U is its internal energy.  And the internal energy is the sum of the internal potential energy and the internal kinetic energy.
Thus:
enthalpy = internal PE + internal KE + pV
N.B.: The term "internal" refers to energy internal to the system.  For example, the internal kinetic energy is the energy the system has as a consequence of the thermal motion of its atoms and molecules, as contrasted with the external kinetic energy the system might have as a result of the motion of the system as a whole through space.

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Why is the 2-position of pyridine electron richer than the 4 position
I am wondering about the electronic structure of pyridine.
If we look at nucleophilic substitution, which favored on electron poor centers, for 2-chloro and 4-chloro pyridine we see that the substitution in 4 position is about 5 times faster (Ethanol, 20°C, taken from The Chemistry of Heterocycles, page 277

This SE question and answer explain why this might be the case and the explanation there is that the lone pair of the nitrogen is repelling the nucleophile. This might be the case to certain extend, but from geometry of the attack I doubt that it has much of an effect. In addition if we look at the electron distribution in the pi system we get this picture (also taken from The Chemistry of Heterocycles, page 271)

So the 4 position seems to be electron poorer than the 2 position. Usually we explain electron poor and rich centers in heterocycles by using mesomeric forms:

And the usual comment is that the 2 position is closer to the nitrogen so the electron withdrawing effect is stronger than in 4 position and it should be electron poorer, which isn't the case.
I also did some DFT calculations using 2- and 4-pyridyl substituents and 4-pyridyl is more electron withdrawing than 2-pyridyl in that case.
Is there an understandable explanation why the 4-position is electron poorer despite beeing furtehr away from the nitrogen?

I think when + charge is on the 4th position there is a type of cross conjugation which makes the cation lesser stable .. and when the + charge is on the second carbon there is extended conjugation which stabilizes the cation to more extent ..

@KushalTripathi besides mesomeric forms not being a real thing this explanation would lead to the exact opposite of what is observed. If the form with the positive charge on position 4 is less stable then it should contribute less to the overall electronic structure, which would mean it's not that electron poor in position 4.

Theres a related issue which is why 2-pyridyl boronic acids are so unstable, I'll try and dig those references up, but the essence is that the bond is significantly lengthened when its at the 2 position (i.e. the sm is just more reactive, rather than any argument about nucleophiles being repelled)

The relative energies of the transition state often are reflected in the relative rates (or vice-versa) so the electron distribution of the anionic intermediate is what should be calculated.

You described the first scheme as $\mathrm{S_NAr}$ being 5 times faster at C-4, but the relative rates given indicate that it is 5 times faster at C-2. Typo, maybe? (I didn't want to assume which was correct.)

Providing the electron density of pyridine is misleading since the chlorine atom will greatly perturb that density

When you deal with an aromatic ring, pairwise covalent bond structures miss a lot of key features.  You need to consider the entire conjugated ring as a unit and take account of symmetry.
Consider almost a benzene ring.  We may describe its $\pi$ orbitals as linear combinations of the out-of-plane atomic $p$ orbital on the carbon atoms.  Let $p_k$ represent the wave function of one such atomic, with the positive lobe on a specified side of the ring plane. The same orbital but with the positive lobe on the"wrong" side of the ring plane is rendered as $-p_k$.  The subscript $k$ refers to atom number $k$ as determined in the standard IUPAC scheme.
We then describe the canonical, occupied $\pi$ orbital as the familiar linear combinations:
$\psi_1=(1/\sqrt{6})\times (p_1+p_2+p_3+p_4+p_5+p_6)$
$\psi_2=(1/\sqrt{12})\times (2p_1-p_2-p_3+2p_4-p_5-p_6)$
$\psi_3=(1/2)\times(p_2+p_3-p_5-p_6)$
Now suppose that atom $1$ is changed from carbon to nitrogen to make pyridine.  Now that atom, being more electronegative, acts an an electron sink on occupied orbitals having an interaction with that atom -- meaning $\psi_1$ and $\psi_2$ are vulberable to thus $\pi$-electron withdrawal but $\psi_3$ which has a node at atom $1$ is "protected" by its symmetry.  Then positions $2$ and $3$, with some of their $\pi$- electron density on the protected $\psi_3$ orbital, are less susceptible than atom 4 whose $\pi$ electrons are all on the vulnerable $\psi_1$ and $\psi_2$ orbitals.
Now, going a little beyond the question, suppose we were to N-oxidize the pyridine, as with a peroxy acid.  In pyridine N-oxide the oxygen acts as a $\pi$ donor so that the nitrogen (or more accurately, the N-oxide functional group) is now effectively an electron source instead of a sink.  The "vulnerable" orbitals are now electron enriched and this characteristic is most strongly imparted, by conjugation and symmetry as argued above, to position 4.  Then electrophilic substitution, enhanced (or enabled?) by the N-oxidation, is most strongly so at position 4.  So, the primary electrophilic substitution product, after subsequently reducing the N-oxide, is the 4-substituted pyridine.  Which is what we see empirically with a wide variety of pyridine N-oxide electrophilic substitutions.

Oh wow, thanks. I think I got it. Just one follow up question: In benzene the orbitals look like this: http://www.chemtube3d.com/benzenemos.html . I could now put the nitrogen in two different position, for example the most left one or one right next to it. You described the case where it's in the left position. In the other case the 4 position would be part of all three orbitals. Is it the described case "just" because there's more symmetry to it?

@DVSA I might not be understanding your comment, but there shouldn't really be different positions to place the nitrogen in benzene. Whatever carbon you replace is effectively carbon $1$ due to the symmetry of benzene. Or, phrased another way, one of the linear combinations formed would not include the 4th carbon away from it.

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Narrow range of pH for the formation of polyoxometalate clusters

The molybdate ion in aqueous solutions, speciates as clusters known as polyoxometalates (POMs) when it is acidified, but these clusters only aggregate in a very narrow pH range "polymeric range" in the diagram. What causes this pH dependence of cluster formation?

I hope I expressed better this time. Please answer or share any reference where i can find the answer. Thanks all


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Can someone help me finish this reaction?
This question revolves around sodium percarbonate (https://en.m.wikipedia.org/wiki/Sodium_percarbonate) and its reaction with sulfuric acid.
I didn't find anything, so I tried assembling it myself.
Wikipedia says that $$\ce{2Na2CO3·3H2O2 -> 2Na2CO3 + 3H2O2}$$ in water.
Now I'm not sure if the $\ce{Na2CO3}$ does this: $$\ce{Na2CO3 + H2SO4 -> Na2SO4 + H2O +CO2}$$ or this: $$\ce{Na2CO3 + 2H2SO4 -> 2NaHSO4 + H2O + CO2}$$
Last, the $\ce{H2O2}$ reacts with $\ce{H2SO4}$ to peroxymonosulfuric acid (https://en.m.wikipedia.org/wiki/Peroxymonosulfuric_acid) $$\ce{H2SO4 + H2O2 <=> H2SO5 + H2O}$$ in high concentrations / to a small extend.
So, my final equation could be: $$\ce{2Na2CO3·3H2O2 + 2H2SO4 -> 2Na2SO4 + 2H2O + 2CO2 + 3H2O2}$$ or $$\ce{2Na2CO3·3H2O2 + 4H2SO4 -> 4NaHSO4 + 2H2O + 2CO2 + 3H2O2}$$ or $$\ce{2Na2CO3·3H2O2 + 5H2SO4 -> 2Na2SO4 + 5H2O + 2CO2 + 3H2SO5}$$ or $$\ce{2Na2CO3·3H2O2 + 7H2SO4 -> 4NaHSO4 + 5H2O + 2CO2 + 3H2SO5}$$
Which is correct or the most correct? Any help appreciated. Thanks!

We prefer not to use MathJax in titles, sine it looks terrible from Google.

Duplicate: https://chemistry.stackexchange.com/questions/51652/sodium-percarbonate-and-sulfuric-acid-reaction


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How does WD-40 remove rust?
One of the often-touted uses of WD-40 is that it "dissolves rust". The official website states that WD-40 "breaks down the bonds between metal and rust".
I can't understand how this would work, since there seem to be no rust-dissolving ingredients in WD-40 and it seems the product is more basic than acidic.
A (possibly not exhaustive) list of ingredients:

Mineral oil, I guess for lubrication and rust protection layer (but not a rust solvent?)
Decane, freeze-resisting
Nonane, water-displacing
Tetradecane, freeze-resisting and water-repelling
Undecane, freeze-resisting and water-repelling
Dimethyl Naphthalene (C12H12), likely used as a solvent
Cyclohexane, solvent
Carbon dioxide, likely used as a propellant

I can imagine that the low viscosity of WD-40 allows it to get in all nooks, making it easier to grind rust particles away because it acts as a short-term lubricant. Is that actually how it works, or am I missing something?

Related: How does WD40 displace water?

I'd say it works by coming out as a liquid. Then it diffuses through the porous rust layer or travels along the steel edge and expands to gas- as the mixture has a highly suppressed vaporisation point to push the rust from the inside. Steel has a high surface affinity for hydrocarbons as demonstrated by high difficult it is to clean. Partial vaporisation is plenty powerful enough to knock off rust.


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Ammonia sigma bonds p character
Why does ammonia's sigma bonds to H have more p character than let's say ethane.
I am guessing it has something to do with the effect the lone pair has on bond lengths and angles, but not quite sure what the effect is exactly.

You'd rather have a higher $s$ character for the lone pair orbital. An ordinary bonding pair is already stabilized between two nuclei, so stabilizing the lone pair can help to further stabilize the system. Putting higher $s$ character into the lone pair orbital makes the orbital smaller and pulls the electrons closer to the nucleus.
This leaves more $p$ character in the other orbitals.

Thank you for your explanation. Could you elaborate more on the nature of stabilization between the nuclei in an ordinary bonding pair that sets it apart from the lone pair bond?

First, the lone pair is not a bond.

Electrons are stabilized by positive nuclei. A bond essentially puts a pair of electrons close to two nuclei where as a lone pair is just close to one. You can help stabilize the lone pair by putting it closer to its nucleus. This effect wins over the destabilization of the bonds.


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Etymology of "normality"
Why the concentration of a solution expressed in gram equivalent weights of solute per litre of solution is termed as "normality" ?
Does it have anything to do with the word "normal" ?

Interesting question, though it might be better off on another SE [ https://hsm.stackexchange.com/ ] ;)

For the etymology, https://en.wikipedia.org/wiki/Equivalent_concentration offers some, and https://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html some more background.  Glad that my work may use mol instead.

The term "normality" stems from the older German term for equivalent concentration (die Äquivalentkonzentration), namely die Normalität, which was coined by Karl Friedrich Mohr.
According to German Wikipedia which also refers to Straube's Der historische Weg der Chemie [1, p. 220], Mohr introduced the concept of normal solution and normality (as it's known in the modern world) in the first edition of his Lehrbuch der chemisch-analytischen Titrirmethode [2], e.g. the preface contained the following definitions among others:

Die Normaltemperatur der Flüssigkeiten ist 14° R. = 17,5° Cent.
[…]
Normalflüssigkeit bedeutet eine Lösung, welche 1 Atom Substanz, in Grammen ausgedrückt, in 1 Litre enthält.
Zehend-Normalflüssigkeit bedeutet eine Lösung, welche $1/10$ Atom Substanz, in Grammen ausgedrückt, auf 1 Litre enthält oder mit $1/10$ Atom eines darin enthaltenen Körpers (z. B. Sauerstoff) wirkt.

The book also lists numerous "normal" definitions not really in a sense of "adequacy", but rather in a a sense of "standard": Mohr basically invented standardized quantitative analysis and proposed numerous analytical tools and techniques.
Among these definitions there are:

Normalwägung — standard weighing [procedure];
Normalbürette — standard burette (a glass tube $\pu{500 mm}$ long, $\pu{13.5 mm}$ inner diameter, holds $\pu{60 cm3}$ of liquid divided into five parts, each $\pu{cm^3}$ is $\pu{7 mm}$ long);
Normale Stärke — strength of a normal solution of acid/base;
Normalsäure and Normalnatron — standard/normal solutions of acid and soda, which, interestingly enough, Mohr suggested to label with the pieces of red and blue paper, respectively;
Normalkleesäure, Normalsalpetersäure, Normalchlorbaryum, etc. — normal solutions of oxalic, nitric acids and barium(II) chloride and so on.

At the same time, normality was used to describe the normal ratio between stoichiometry of the elements (esp. C, O, N) and the number of found equivalents in complex compounds such as proteins or drugs [3].
References

Wilhelm Strube: Der historische Weg der Chemie. Aulis Verlag Deubner & Co KG, Köln 1989, ISBN 3-7614-1180-4. (In German)
Friedrich Mohr. Lehrbuch der chemisch-analytischen Titrirmethode; Vieweg und Sohn: Braunschweig, 1855. (In German. Available for free at Google Play (accessed 2019-05-25))
Zeising, A. Das Normalverhältnis der chemischen und morphologischen Proportionen; Weigel, 1856. (In German. Available for free at Archive.org (accessed 2019-05-25))


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Purification of polar NHS esters
What are the options that I might be missing for the purification of a molecule with the following characteristic:

Contains an activated ester that reacts with water (NHS)
Reacts readily with alcohols
Does not elute on silica gel (typical or spherical) with THF, Acetone, EtOAc, DCM, or Acetonitrile
Is not a solid that can be precipitated, rather some horrible gum
Decomposes rapidly in prep-LC on C18 0.1% TFA (10% back to acid after 20min run in best cases) typical loss of material is about 50% from initial mass.
Greasiest solvent that solubilises is DCM

Impurities are mainly side product from NHS coupling reagents or relatively uncharacterized impurities with properties similar to final product.
Honestly, I have not much hope left to find something there but if you know of conditions what could provide separation in reverse phase HPLC with aprotic solvent, I'm all ears.
Edit: I'm open also to hear about volatile buffer system that could be in a pH range that would allow greater stability of the NHS in water.

How does it behave on alumina?

Haven't checked. I always thought it was rather similar to silica gel but really inconvenient. Would you expect a different result? The core structure look like silica gel a lot. I'll try to find tlc for that...

Alumina can give different results so it is worth checking. Do you have access to the range of 'flavoured' silicas?   Those can also give useful differentiation between substrates.

I can get pretty much anything, but I would like to get tlc before committing my product.

You can get tlc plates of the various favoured silicas

This is a good idea, I'll definitely order a couple of these, but again which solvent will be the key question. I've never been convinced that they effectively cap all the silanols on there... Sigh! Friday is time for a beer

Try C18 prep LC but use a different buffer like ammonium formate at pH 4 or even, if you have to, ammonium bicarbonate at pH 6.5. If the compound is stable at this more neutral pH, it could be a way to go.

Crazier options might be to try DMSO in place of water for the "weak" C18 solvent.  Not sure if anyone is your lab would like you if you pumped liter quantities of DMSO through a prep LC column though.


text:
Using group theory to make molecular orbital diagrams
Because of Orthocresols' meta post I ask this question about application of group theory to molecular orbital theory.
In my undergraduate lecturers, we've seen many examples of making qualitative MO Diagrams by inspection of what orbitals could line up to form bonding / anti bonding orbitals.
This works fine for simple things, but I know it's possible to use symmetry to also work out mathematically which orbitals can mix with which other orbitals by using group theory.
Perhaps using example of water could somebody show how this could be achieved?

Can you be a bit clearer? Are you talking about the mixing of the orbitals in Oxygen to the Orbitals in Hydrogen in water?

Group theory really just formalises the process that you're going through when you construct molecular orbital diagrams by inspection (though as the molecules get bigger it gets significantly harder to intuitively guess at what will mix and in what combinations).

In order to construct an MO diagram for water, we'll take a stepwise approach:

first, determine the way in which the hydrogen atoms can combine (in phase or out of phase, as per the standard dihydrogen MO diagram) which will give us a set of symmetry adapted combinations
second, we mix these symmetry adapted orbitals with the atomic orbitals on the oxygen, bearing in mind the 'rules' of the game

only orbitals of similar energy are able to combine
only orbitals of correct geometry/orientation are able to combine (i.e. head on or side to side)
only orbital of correct symmetry are able to combine. This is where group theory becomes useful

1. Symmetry of the central oxygen's orbitals
The symmetry of the oxygen orbitals (2s and 2p, the 1s isn't relevant to the bonding) can easily be read off from a character table. In this case, water has C2v symmetry, the character table for which is given below:
$$\begin{array}{c|cccc|cc} \hline
C_\mathrm{2v} & E & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_1} & 1 & -1 & 1 & -1 & x, R_y & xz \\
\mathrm{B_2} & 1 & -1 & -1 & 1 & y, R_x & yz \\ \hline
\end{array}$$
To summarise the information obtained:

the s-orbitals have a1 symmetry
the px orbital has b1 symmetry
the py orbital has b2 symmetry
the pz orbital has a1 symmetry

(as @Orthocresol noted in the comments, there is no convention on axis in the C2v point group,meaning x and y can switch depending on what book you get the character table from. The one given is the same as the commonly used set from Atkins' Physical chemistry, but overall it doesn't really matter as long as you're consistent)
It's worth noting that you can derive these symmetries above by considering how the atomic orbitals moves under the various elements within that point group, this is shown below for the hydrogens, and not really needed to be repeated here.
2. Symmetry of the hydrogen group orbitals
To construct the group orbitals for hydrogen (which will go on to be mixed with the oxygen atomic orbitals), we simply have to consider the possible combinations and then work out the symmetries of those combinations.
For the case of two hydrogens interacting, only two cases are possible: in phase, and out of phase.
If we consider these two cases pictorially, we can then determine what happens to them under the various symmetry elements within the C2v group. If the orbitals stay the same, we assign a character of 1, if they switch, we assign -1.

Once we have the full set of characters for both possible combinations, we can assign symmetries, again, using the character table for the relevant point group:

the in-phase combination has a1 symmetry
the out of phase combination has b2 symmetry

3. Mixing
Mixing is the part of this process that you're likely more comfortable with, except now rather than considering whether interaction (on symmetry grounds) is likely by inspection, we can do it by mixing orbitals with the same symmetry labels:

The diagram above is the one that you would likely have arrived at without any consideration of group theory (you can actually derive a more complex diagram in which the molecular orbitals also mix), some key points to note:

the oxygens orbitals have all (qualitatively) been placed lower than those of hydrogen. Electronegative atoms have more contracted orbitals and are lower in energy
two of the p-orbitals end up non-bonding, these are the lone pairs on oxygen
the s-orbital on oxygen interacts with the in-phase combination of hydrogen, this is the lowest energy combination rather than using the p-orbital (also with a1 symmetry)

Further reading
The general approach taken here was a mix between that taken in Group Theory for Chemists by Molloy, and Cottons' Chemical Applications of Group Theory. Most (if not all) books on group theory will provide examples of how to construct MO diagrams based on formal symmetry arguments, and many approaches exist.
I've followed a more qualitative method, relying on the character table to gain the required information, but there is a more formal approach in which irreducible representations are used... possibly someone else will provide an answer with this more formal method (the amount of MathJax required put me off doing this, especially given the ease at which the information can be accessed and the fact that given you're asking the question, you're probably not at the point at which this has been formally taught).

Explained it better than I could. I guess it is better not to mention perturbation theory in this explanation.

A caveat with $C_\mathrm{2v}$ is that every book will have a different convention on the $x$ and $y$ axes... and hence the mirror planes $\sigma_\mathrm{v}$ and $\sigma_\mathrm{v}'$, and the irreps $\mathrm{B_1}$ and $\mathrm{B_2}$, get swapped in different books.  $D_\mathrm{2h}$ is even worse, it's a complete nightmare. But as long as one is consistent throughout, the final result will always be the same.

Thanks for answering the question. If you say everything is from point group table does that mean all molecules with c2v symmetry would have the same MO diagram? For example if we draw formaldehyde which also has a c2v group

Perhaps it is worth mentioning that this approach, while valuable, only produces the orbital symmetry and not the actual shape or size of any orbital. You use nice circles and p-orbitals because of your chemical intuition but any suitable symbols could be used as far as symmetry is concerned. Also the orbital energy is not known, but the ordering may be 'guestimated' based on the number of nodes produced, zero nodes being lowest.

@JamesDealon - not really, formaldehyde has more 4 atoms (not 3 like water), which are arranged differently. Whilst both molecules have the same symmetry (and symmetry elements) their bonding is vastly different (formaldehyde has pi-interactions between C=O).


text:
Can the summation of activation energy and reactant's initial internal energy change with temperature?

Look at the graph.
The orange arrow represents the summation of activation energy and reactant's initial internal energy. Can it change with temperature?
Is there any specific name for this summation?

There's certainly a name, it's the energy of the transition state.

Is it fixed for a specific reaction irrespective of temperature? @orthocresol

The average internal energy increases with temperature as more reactant rotational and vibrational energy levels become populated as described by the Boltzmann distribution. So does the population in the transition state (even though it exists only fleetingly, fs to ps). Thus reactions get faster as the temperature increases because a larger fraction of molecules have enough energy to cross the potential energy barrier assumed constant, independent of temperature as this is determined by electronic configuration of atoms in a molecule.

Activation energy= Energy of transition state - Energy of Reactants. Of which temperature energy of reactants in considered here?

The activation energy is determined experimentally and over a small temperature range is often independent of temperature. ($d\ln(k_{obs})/dt = E_a/(RT^2)$) From a stat. mech. approach the activation energy is the difference in zero point energies between reactants and transition state.

//the activation energy is the difference in zero point energies between reactants and transition state.// I was wondering this for a while. Can you please, provide any reference for this? It would be much satisfactory for me and enough to close this topic In Shaa Allah. And I also didn't understand 'stat. mech. approach'. What's that?

No. As the label at the y-axis indicates, we are talking about potential energy. Potential energy itself does not mean anything, only energy differences can be meaningfully interpreted. (It is the same with heights, to quantify how tall a mountain is you give its height with respect to the sea level.)
The activation energy is defined as the difference between the transition state's energy and the reactant's initial internal energy, and has therefore a physical meaning.
The reactant's initial internal energy itself is just some arbitrary value. Unless you define some reference point, which is not the case in your figure.
Therefore the orange arrow ("the summation of activation energy and reactant's initial internal energy") is conceptually wrong.
Furthermore, a change in temperature does not change your potential curve. This curve is a minimum energy path. It represents the ground state of the system at each point along the reaction coordinate.
But when we have a ground state, then there are excited states as well. And thermal energy makes the system partially occupy these states (either statistically in the classical sense or by probability distributions in the quantum mechanical sense). Effectively the reactant will be higher in energy, but potential remains the same! This is actually the way you can reach the transition state.
Think of it like there is a ladder on the floor. The floor will be the ground state, and on the top of the ladder there is the transition state. Now the higher the temperature, the higher you can step on the ladder. With high enough temperature you can reach the transition state and descend on the other side towards the products.

//With high enough temperature you can reach the transition state// doesn't that mean that this state must have a definite value of potential energy? Maybe I can't measure it, but it does have a definite value, doesn't it?

The transition state has a definite value, just in the same sense as the reactants and product have a definite value. You can't measure this value directly since it is an potential. But you can very well measure the activation energy.

If I try to start the reaction at a high temperature, the reactants will already have a high internal energy and they will need less energy to reach the transition state, meaning the activation energy will be less at a higher temperature. But again, I have read that activation energy doesn't change with temperature. So, there's something wrong with my understanding. Please, help me.

The activation energy is defined with respect to the reactants ground state at 0K and does not change!
With the energy of the reactants we have to be careful. There are two different quantities here (I'm not sure about their correct names): The first one the ground state at 0K and is given by the potential energy curve in your figure. This one is a fixed property of the system and does not change on external conditions. The second one includes the thermal energy of the reactants, therefore it depends on temperature. A reactions is only possible if this energy is above the activation barrier.

//A reactions is only possible if this energy is above the activation barrier.// Which energy needs to be above the activation barrier? The potential energy of the reactants or the thermal energy of the reactants or the summation of both of them?

The thermal energy needs to be larger than the activation barrier. If the potential energy of the reactants would be larger than the activation barrier, there would be no activation barrier. It would just be a descending energy path, and the "reactants" wouldn't be stable chemical compound(s).

Let us continue this discussion in chat.


text:
Effects of Ionizing a Radioactive Isotope
If you took your favorite radioactive isotope and ionized it using a strong electric field (or any other ionization method), thereby stripping it of an electron, what would happen when it decayed? For simplicity, assume this process takes place in a vacuum.
Would the resulting material also be ionized? (Main Question)
If the isotope decayed via beta decay, is there any chance the produced electron could take the place of the stripped electron?

An intriguing question, especially as (https://en.wikipedia.org/wiki/Radioactive_decay#Types_of_decay) hints that there is more than just $\alpha, \beta^+, \beta^-$-decay.  Not only in terms of "what does it mean for a single, isolated atom", but even for radiopharmaceutical compounds like 18F-FDG (https://en.wikipedia.org/wiki/Fludeoxyglucose_(18F)), where the $\ce{^{18}F}$ turns into $\ce{^{18}O}$ ...

I don't really think so. Energy levels in the nucleus are really large compared to electronic energy levels. A beta electron would have a lot of energy, it would overcome all the electronic energy states and simply fly away.

@PrittBalagopal - That is what I was thinking as well. If I'm not mistaken, the electrons produced through beta decay move at a significant fraction of the speed of light. Is that correct?
Do you have any thoughts on the first question, though? Would the resulting material also be ionized?

There is no effect. Imagine you are demolishing a building with high explosives; does it matter whether the small window on the third floor was open or closed prior to the explosion? No, it doesn't. The building will go crashing down all the same.
I believe this is a correct metaphor of the relation between chemistry and radioactivity. Of course the product will be ionized. Quite probably, it will be multiply ionized via Auger events, and also will certainly ionize quite a few other atoms around. Whether it was ionized before is about as unimportant as that small window on the third floor.


text:
The definition of density
Could density be defined as the measure of matter per unit volume instead of mass per unit volume because mass is a measurement of matter?

It depends, there are different kinds of densities: mass density, particle density, charge density, probability density ...
Depending on how your measure your matter, mass, number of particles, charge (e.g. for electrons), you could use the corresponding density.
"Matter per unit volume" would only work in a broader sense, since it does not specifically state what kind of measure you want to use. Most of the time one simply defaults on mass, but this depends on the specific field.

I'd be more heavy handed. I don't see how matter per unit volume could ever work. "Matter" is ambiguous as you point out. Does "matter" mean mass, number of particles, charge, ect?

You could still state that you have more (or less) of something. And the different kinds of densities may correlate, e.g. more particles are usually more mass (unless you replace heavier particles by one lighter ones). But if you need to put numbers on it, then yes, you need to state what you are talking about.


text:
Why is pi bond in N=N stronger than sigma bond?
Bond energy for N-N is 163Kj/mol
whereas bond energy for N=N is 418kj/mol even though it should be lesser than twice bond energy for N-N. Why is that?
For carbon it is:
          Bond energy for C-C is 348KJ/mol
          Bond energy for C=C is 614Kj/mol

My book explains the reason for carbon but it contradicts with the case of nitrogen.

=======================The reason can be described both qualitatively and quantitatively. Qualitatively, you can think of it in the following fashion: There are four sigma levels when one looks at the MOs formed from the 2s2s and 2p2p orbitals of N2N2 : σs,σ∗s,σp,σ∗pσs,σs∗,σp,σp∗. These levels interact as they have same symmetry and the σpσp level, being third highest in energy, is pushed up more (think level repulsion from second order perturbation theory), making that interaction weaker. ππ levels do not have others of the same symmetry to mix with, making them stabler (especially because overlap is massive as the system is small).
Quantitatively, within the Atomic Orbital basis of Nitrogen, the Hamiltonian matrix is pretty big (considering only the 2s2s and 2p2p levels, we have 8 orbitals and thus an 8 by 8 matrix of elements ⟨m|H|n⟩⟨m|H|n⟩ where m,nm,n are two of the 8 orbitals mixing (and they can be the same-like 2s2s on same atom). I am not writing the 64 element matrix out for obvious reasons, but symmetry makes a host of elements zero, simplifying our work. We are left with block diagonalized bits, two 2×22×2s for the 2 pi levels (which work out simply) and one 4×44×4 for the sigma level.
The σσ Hamiltonian matrix is still too big for me to fully write out, but it contains terms like interaction between 2pz2pz on one atom and 2s2s on the other. Even 2p2p on one atom and 2s2s on the same are not perfectly non-interacting now (they were previously orthogonal when there was no second atom, but the perturbation to the Hamiltonian from the presence of the second atom throws it off)-although this is truly speaking the smallest interaction to consider.
Now for some numbers. Please note that I making lots of approximations and so this is not really a reflection of real N2N2 but is simply a crude estimate. Ionization of 2p2p for Nitrogen is 14.53eV. I looked at some NIST data (Energy Levels of Neutral Nitrogen ( N I )) and seems like ss about 8eV below in energy. Bond-strength of N2N2 is about 941 kJ/mol and so we can guess the coupling to be ~3eV3eV ish (roughly 1/3rd of that). As an approximation, let us claim that individual orbital energies themselves are not greatly perturbed by the presence of the other atom (coupling is though-which is why we have bonding). Lets claim that the p−pp−p coupling is V,s−pV,s−pcoupling is 0.75V0.75V and s−ss−s coupling is 0.50V0.50V (p has greater coupling due to directionality). Note that I made the numbers up (although they 'seem' reasonable) and thus they need not reflect reality.
The dashed curve represent how energies will look if the σsσs and σpσp do not communicate with each other. As you can observe, the 2p bonding (cyan ) line is distinctly less bonding than before and the 2s antibonding orbital is quite a bit antibonding. Although my numbers are all guesses, overall qualitative behavior ought to be the same.
Found this but coud not understand.

What molecules are you asking about when you use n-n and n=n? I assume its not $\ce{N2}$ as this has a dissociation energy of $945, \pu{kJmol^{-1}}$. But the nature of sigma and pi bonding is different so why should they be equal, other than by accident?

Welcome to ChemSE!  To avoid errors in asking and answering questions on ChemSE, please adhere to proper use of dimensions, which includes "k" (lowercase) as prefix for $\pu{1E3}$, "m" (lowercase) for $\pu{1E-3}$, but "M" (capital) for $\pu{1E6}$. Use element symbols as IUPAC does without alteration; e.g. $\ce{PU}$ refers to phosphor and uranium (if there is such a compound?), $\ce{Pu}$ to plutonium. Learn mathjax works, allowing  you to type single, double, triple bonds, too: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88

Guys the question is very clear, (though I was befuddled myself and close voted it) and Oscar Lanzi has answered it satisfactorily.

The $\pi$ bond in $\ce{N=N}$ is actually not stronger than the $\sigma$ bond.  Rather the $\sigma$ bond is stronger in $\ce{N=N}$ than in $\ce{N-N}$, and this plus the $\pi$ bond gives the double-bonded linkage more than twice the binding energy.
So what makes the $\ce{N=N}$ $\sigma$ bond stronger?  Fewer valence electrons are there when you have a double bond, that's why you form double bonds to begin with.  You have $12$ valence electrons around a pair of double-bonded nitrogens versus $14$ with a single bond.  Having fewer valence electrons means less repulsion between them to interfere with forming a stronger, shorter $\sigma$ bond.


text:
Gradient of integrals with Obara Saika recurrence scheme
Update I made a mistake in the first formula, now everything is correct as far as I can tell.
I am calculating integrals with the Obara–Saika recurrence scheme[1]. Now I want to calculate also the gradients of each integral. I suppose it is calculated like in the equation below which I derived with the product rule.
$$
\hat\partial_i ( a || b ) =
- 2\zeta_a ( a +  {\bf 1}_i || b )
- 2\zeta_b (a ||b + {\bf 1}_i)
+ \sum^{l_a}_{j=1} \delta_{ij} ( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik} ( a || b - {\bf 1}_k )
$$
I use the same nomenclature as Obara and Saika did, since the paper is somewhat loaded with math I will explain those here again. $\hat\partial_i$ is the derviative in $i$-direction, $a$ and $b$ are cartesian gaussian functions, $\zeta_a$ is the orbital exponent of the gaussian function $a$, $\zeta$ is a shortcut for $\zeta_a + \zeta_b$, $\delta_{ij}$ is the kronecker delta and ${\bf 1}_i$ is an increment or decrement of the polynom of the gaussian function in $i$ direction—that basicly means you get an $d_{ij}$ from $p_i+{\bf 1}_j$. Oh, I almost forgot that one, $(a||b)$ denotes an integral over two gaussian functions.
Update or What I have done so far:
I've done some calculations with the Obara–Saika recurrsion so far, that suggest that the expression I derived for the derivative is correct. If you calculate the derivative explicitly as Erik Kjellgren had done in his answer you get the expression I have shown above. It is a bit cryptic and but if you take a closer look you will see it. Please note that we both have tackled a different problem, I have calculated the derivative in x-direction, while he has calculated the derivative for the movement of the nucleus A in x direction. So mabye I am dealing with the wrong problem.
Let's solve both starting with my problem. The gauss integral as used in the paper from Obara and Saika is given as
$$
\varphi({\bf r}; \zeta, {\bf n}, {\bf R}) =
(x-R_x)^{n_x}(y-R_y)^{n_y}(z-R_z)^{n_z}\times\exp(-\zeta({\bf r-R})^2)
$$
this is nearly the same as used by Erik, but I want to stay in this whole post in the same nomenclature. The next step is to calculate the derivative
$$
\hat\partial_i ( a || b ) =
\hat\partial_i \int{\rm d}{\bf r}
N(\zeta_a,{\bf a})\varphi({\bf r}; \zeta_a, {\bf a}, {\bf A})
N(\zeta_b,{\bf b})\varphi({\bf r}; \zeta_b, {\bf b}, {\bf B})
\\
= N(\zeta_a,{\bf a})N(\zeta_b,{\bf b})
\hat\partial_i \int{\rm d}{\bf r}
(x-A_x)^{a_x}(y-A_y)^{a_y}(z-A_z)^{a_z}\times\exp(-\zeta_a({\bf r-A})^2)
\\\times
(x-B_x)^{b_x}(y-B_y)^{b_y}(z-B_z)^{b_z}\times\exp(-\zeta_b({\bf r-B})^2)
\\
= N(\zeta_a,{\bf a})N(\zeta_b,{\bf b})
\int{\rm d}{\bf r}\biggl(
\hat\partial_i
(x-A_x)^{a_x}(y-A_y)^{a_y}(z-A_z)^{a_z}\times\exp(-\zeta_a({\bf r-A})^2)
\\+
(x-A_x)^{a_x}(y-A_y)^{a_y}(z-A_z)^{a_z}\times\hat\partial_i \exp(-\zeta_a({\bf r-A})^2)
\\+
\hat\partial_i
(x-B_x)^{b_x}(y-B_y)^{b_y}(z-B_z)^{b_z}\times\exp(-\zeta_b({\bf r-B})^2)
\\+
(x-B_x)^{b_x}(y-B_y)^{b_y}(z-B_z)^{b_z}\times\hat\partial_i \exp(-\zeta_b({\bf r-B})^2)\biggr)
\\
= - 2\zeta_a ( a +  {\bf 1}_i || b )
- 2\zeta_b (a ||b + {\bf 1}_i)
+ \sum^{l_a}_{j=1} \delta_{ij} ( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik} ( a || b - {\bf 1}_k )
$$
Okay this was straigth forward. Because I don't know anything about the derivative I applied kronecker-deltas. This should be okay because you
can also write something like that for the above expression:
$$
\hat\partial_i ( a || b ) =
- 2\zeta_a ( a +  {\bf 1}_i || b )
- 2\zeta_b (a ||b + {\bf 1}_i)
+ \delta_{ix} a_x ( a - {\bf 1}_x || b )
+ \delta_{iy} a_y ( a - {\bf 1}_y || b )
+ \delta_{iz} a_z ( a - {\bf 1}_z || b )
+ \delta_{ix} b_x ( a || b - {\bf 1}_x )
+ \delta_{iy} b_y ( a || b - {\bf 1}_y )
+ \delta_{iz} b_z ( a || b - {\bf 1}_z )
$$
I am now at the same point as Erik Kjellgren, nothing new so far, but maybe
a bit more general. Now I am going to get rid of the integrals with the Obara–Saika recurrsion, because that is what it is all about.
$$
\hat\partial_i ( a || b ) =
- 2\zeta_a ( a +  {\bf 1}_i || b )
- 2\zeta_b (a ||b + {\bf 1}_i)
+ \sum^{l_a}_{j=1} \delta_{ij} ( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik} ( a || b - {\bf 1}_k )
\\
=
-2\zeta_a \Bigl(
(P_i-A_i)(a||b)
+ \sum^{l_a}_{j=1} \delta_{ij}\frac 1{2\zeta} ( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik}\frac 1{2\zeta} ( a || b - {\bf 1}_k )
\Bigr)\\
-2\zeta_b \Bigl(
(P_i-B_i)(a||b)
+ \sum^{l_a}_{j=1} \delta_{ij}\frac 1{2\zeta} ( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik}\frac 1{2\zeta} ( a || b - {\bf 1}_k )
\Bigr)\\
+ \sum^{l_a}_{j=1} \delta_{ij}( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik}( a || b - {\bf 1}_k )
$$
After applying the basic recurrence scheme I got something that looks interesting and promising
$$
\hat\partial_i ( a || b ) =
-2\zeta_a (P_i-A_i)(a||b)-2\zeta_b (P_i-B_i)(a||b)
\\
- \sum^{l_a}_{j=1} \delta_{ij} ( a - {\bf 1}_j || b )
- \sum^{l_b}_{k=1} \delta_{ik} ( a || b - {\bf 1}_k )
\\
+ \sum^{l_a}_{j=1} \delta_{ij}( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik}( a || b - {\bf 1}_k )
$$
The rest is simple math and so I derived the following expression
$$
\hat\partial_i (a||b) = -\bigl((2\zeta_a({\rm P}_i - {\rm A}_i) + 2\zeta_b({\rm P}_i - {\rm B}_i)\bigr) (a||b)
$$
where $\bf A$ and $\bf B$ are the centers of the gaussian functions $a$ and $b$ and $\bf P$ is the center of the product of both gaussian functions.
Orginal question
Have any of you experience with gradients of molecular integrals and can give me a hint on that issue? Is this the right way to calculate the gradient?
[1] S. Obara and A. Saika, J. Chem. Phys. 84 (7), 1986

If you could put your derivation from (a||b) all the way to the final equation, into your question that would be awesome

I believe eqs. 41 and 42 as well as chapter 14 of this review have the answer to your question. It seems like a lot of work to actually understand everything that's going on and translate notations so maybe someone braver than me will do it.

@ErikKjellgren I added my derivation for both problems, now my question is a wall of math. I tried to get away from the Obara–Saika notation, but it would be more formulas. Hope it is helpful.

@awvwgk I naivly thought that derivative in the x direction implied with the derivaitve with respect to the geometry in the x direction, but can see I missinterpreted that. When you write "the basic recurrence scheme", what relations are that excactly? When I think about the OS relations for overlap integrals I think of these

@ErikKjellgren I get the same shape in the final formula either way, so it doesn’t matter for the question which derivative we are looking at. It does matter in case I going to code it, so I have to reconsider on this. That are exactly the recurrence schemes I used, I made some modificatons which seemed quite resonable to get a more general form.

@awvwgk When I look at your final expression I got a feeling something is not right. $2\zeta (P_i - A_i)$ is "just" a constant. So it can be written as: d/dA(a|b)=C*(a|b).  Since a and b are Gaussian type functions, this implies that d/dA(a) = C*(a), which we know is not true.

@awvwgk I think that the OS scheme is used to  calculate $S_{i+1,j}$ and $S_{i-1,j}$ for the derivatives, just like "normal" overlap integrals. Ofc, I see a great point in manipulating the equations to something more useful, before implementation. And I hope we work out something good.  P.S. I got a code that can calculate geometrical derivatives, so if you need some numbers to test against, I can send you some.

@ErikKjellgren There was an error, now it should be correct.

I see you have not gotten any answers for some time. I have not myself become familiar enough with the notation used in S. Obara and A. Saika, J. Chem. Phys. 84 (7), 1986. But I have a suggestion for how you can help your self proceed. If you use your derived formula to calculate some derivative, and you try to calculate the same derivatives with finite difference, then you can check and see if you get the same result. If the results are identical to some nummerical error, your equation are proberly correct.
UPDATE:
I have tried to make a derivation of the derivative myself. Here is what I have come up with:
The overlap integral is given as:
$$S_{\mu,\nu}=\left(\mu|\nu\right)$$
Here $\mu$ and $\nu$ represent Gaussian functions of the type:
$$\phi_{l,m,n}=N_{l,m,n}x_{a}^{l}y_{a}^{m}z_{a}^{n}\exp\left(-\alpha\left|r-R_{A}\right|^{2}\right)$$
Now the derivative with repsect to the movement of nucleus $A$ can
be calculated by using the chain rule:
$$\left.\frac{\partial S_{\mu,\nu}}{\partial X_{A}}\right|_{X_{\mu}\neq X_{\nu}}=\left.\left(\left.\frac{\partial\mu}{\partial X_{A}}\right|\nu\right)\right|_{X_{\mu}=X_{A}}+\left.\left(\mu\left|\frac{\partial\nu}{\partial X_{A}}\right.\right)\right|_{X_{\nu}=X_{A}}$$
Note that $X_{\mu}\neq X_{\nu}$, this is imposed by the invariance
of moving all centers [1]:
$$\frac{\partial S_{\mu,\nu}}{\partial X_{A}}+\frac{\partial S_{\mu,\nu}}{\partial X_{B}}=0$$
Here $A=B$. Also a wavefunction must have a depends on $A$ for a
derivative to give a value, thus for atom centered wavefunctions,
the wavefunction must be placed on the atom for which the derivative
is taken. Now the the derivative of the Gaussian functions is given
as [2]:
$$\frac{\partial\phi_{l,m,n}}{\partial X_{A}}=\sqrt{(2l+1)\alpha}\phi_{l+1,m,n}-\left.2l\sqrt{\left(\frac{\alpha}{2l-1}\right)}\phi_{l-1,m,n}\right|_{l>0}$$
It should be noted that the factors in front of the Gaussians is because of normalization. Thus now giving:
$$\left.\frac{\partial S_{\mu,\nu}}{\partial X_{A}}\right|_{X_{A}=X_{\mu}\neq X_{\nu}}=\left(\left.\frac{\partial\mu_{l,m,n}}{\partial X_{A}}\right|\nu_{l,m,n}\right)$$
$$=\left(\left.\sqrt{(2l+1)\alpha}\mu_{l+1,m,n}-\left.2l\sqrt{\left(\frac{\alpha}{2l-1}\right)}\mu_{l-1,m,n}\right|_{l>0}\right|\nu_{l,m,n}\right)$$
$$=\sqrt{(2l+1)\alpha}\left(\left.\mu_{l+1,m,n}\right|\nu_{l,m,n}\right)-2l\sqrt{\left(\frac{\alpha}{2l-1}\right)}\left(\left.\left.\mu_{l-1,m,n}\right|_{l>0}\right|\nu_{l,m,n}\right)$$
Now $S_{i,,j}=\left(\mu_{l=i,m,n}|\nu_{l=j,m,n}\right)$and by insertion
into the found expression for the derivative:
$$\left.\frac{\partial S_{\mu,\nu}}{\partial X_{A}}\right|_{X_{A}=X_{\mu}\neq X_{\nu}}=\sqrt{(2l+1)\alpha}S_{i+1,j}-\left.2l\sqrt{\left(\frac{\alpha}{2l-1}\right)}S_{i-1.j}\right|_{i>0}$$
Now from here on I was unable to find the equation you found. Also my notation is abit different. I thought you might want it as an alternative way to find the derivative (or maybe you will be able see from this if your equation is right or wrong). Again if you do the calculations with numbers you can see if the two different equations give the same results. Now the Obara-Saika recurrence relation can be considered [3]:
$$S_{i+1,j}=X_{PA}S_{i,j}+\frac{1}{2p}(iS_{i-1,j}+jS_{i,j-1})$$
This can be insterted into the above derivative to eliminate the $i+1$ term:
$$\left.\frac{\partial S_{\mu,\nu}}{\partial X_{A}}\right|_{X_{A}=X_{\mu}\neq X_{\nu}}=\sqrt{(2l+1)\alpha}\left( X_{PA}S_{i,j}+\frac{1}{2p}(iS_{i-1,j}+jS_{i,j-1}) \right)-\left.2l\sqrt{\left(\frac{\alpha}{2l-1}\right)}S_{i-1.j}\right|_{i>0}$$
[1] Trygve Helgaker, Poul Jorgensen, Jeppe Olsen; Molecular Electronic-Structure Theory. Equation: 9.3.4
[2] Attila Szabo, Neil S. Ostlund;Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory. page. 442
[3] Trygve Helgaker, Poul Jorgensen, Jeppe Olsen; Molecular Electronic-Structure Theory. Equation: 9.3.8

That is clear, this is almost the first formula in my question, now in the correct form. I’m more interested in the application of the Obara–Saika-recursion on the $S_{i+1,j}$ part.

Thanks for your input. I was looking at the wrong problem all the time.

@awvwgk I have tried to insert the OS relations into the derivative. Now my derivation, was not very general at all. Also because of my ignorence (imfamilarity) with your notation, it is a bit hard to compare the results. To complicate even more, I have normalized the expression.  But I think our two expressions look alike very much.

They are the same.

I moved my extended explanation from my question to a real answer. The given problem was to calculate the gradient of overlap integrals. Thanks to Erik Kjellgren I noticed that I am looking at the false problem.
So far I have calculated derivatives for cartesian coordinates. This is possible but not very helpful in case I want to have gradient for geometry optimizations. I can use a bit from my previous work with
$$
\frac{\partial}{\partial R_i}\varphi ({\bf r}; \zeta, {\bf n}, {\bf R}) =
- \frac{\partial}{\partial r_i}\varphi ({\bf r}; \zeta, {\bf n}, {\bf R})
$$
and get
$$
\frac{\partial}{\partial A_i} (a || b) =
2\zeta_a ( a +  {\bf 1}_i || b )
- \sum^{l_a}_{j=1} \delta_{ij}( a - {\bf 1}_j || b ).
$$
This is the exactly same as in Erik Kjellgren’s answer.
If I use the Obara–Saika recurrsion formula on the expression above I get the following formula.
$$
\frac{\partial}{\partial A_i} (a || b) =
2\zeta_a \Bigl(
(P_i-A_i)(a||b)
+ \sum^{l_a}_{j=1} \delta_{ij}\frac 1{2\zeta}( a - {\bf 1}_j || b )
+ \sum^{l_b}_{k=1} \delta_{ik}\frac 1{2\zeta}( a || b - {\bf 1}_k )
\Bigr)
- \sum^{l_a}_{j=1} \delta_{ij}( a - {\bf 1}_j || b )
$$
From here on I cannot go any further.
The equation I have derived now does not look easy anymore, but I think it is fine, all intermediates are available from the calculation of the overlap integral itself, so I don’t need any additional expensive calculations.


text:
What is the reason for isotropy in amorphous solids?
In my book, it's been mentioned that crystalline solids are anisotropic whereas amorphous solids are isotropic in nature. The reason for these has been explained (with a diagram) as:

Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystals. This arises from different arrangement of particles in different directions. Since the arrangement of particles is different along different directions, the value of the same physical property is found to be different along each direction.
Amorphous solids on the other hand are isotropic in nature. It is because there is no long range order in them and arrangement is irregular along all directions. Therefore, value of any physical property would be same along any direction.

My question is: similar arrangement of particles in different directions in crystalline solids gives rise to different physical properties. Then, how can disarrangement in amorphous solids in all directions give rise to same physical properties?
In other words, why do crystalline solids have properties (like electric conductivity, thermal conductivity, mechanical strength, and refractive index) different in different directions, and the amorphous solids have these properties same in all directions?

Imagine that your amorphous solid is made up of many small crystals, each anisotropic in properties, but in a random arrangement both in position and orientation.

Because it's disarranged in all directions equally, "degree of disorder" is same in any direction. It's not disarranged in any direction more than in another. It's isotropic because it's uniformly disarranged in all orientations.

@Jamil Ahmed Keep in mind that the mentioned anisotropy electrical resistance, refractive indeces, etc. affects them because they are (at minimum) vectorial properties.  It will not affect properties of purely scalar in nature, like density, for example.

Assume that the spheres in the drawing are atoms.
In the second picture atoms are not orderly arranged. No matter where you go in crystal of a amorphous solid, the disarrangement will be same. Since in every direction, disarrangement is same, physical properties along every direction is also same. Therefore, amorphous solids are isotropic.
In the first picture, the arrangement is orderly. So, when you go in any direction, the arrangement will be different. It is also clear in the rectangles shown in the picture. Try doing it yourself on a piece of paper. Try finding any four different directions along which the arrangement is same.
Ps: Ignore my poor drawing skills.

Read the following article by C.V. Raman which explains how structural anisotropy exists in amorphous solids and also gives rise to the phenomenon "Structural Birefringence". This is the first and the only article that speaks about birefringence and anisotropy in amorphous solids.
Structural birefringence in amorphous solids
Abstract:
The result reported in an earlier investigation with vitreous silica is now shown to be true also for other amorphous solids, including especially inorganic glasses; besides the well-known photo-elastic effect, another kind of birefringence may be observed differing from the former both in its origins and in its observable characters. This “structural birefringence” arises from anisotropy of structure present in the solid by reason of the circumstances of its formation. It is conspieuously seen with plate glass whose optical behaviour shows it to have a highly laminated structure, while in moulded glass, it exhibits itself as luminous streaks or sheets of variously curved forms.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
If I drink water without sipping, is there a chance of saliva diffusing into my bottle?
I always prefer to drink water without sipping, like this:

I do this so that my saliva doesn't contaminate the water contents, which would make the water impossible to share with my friends.
But lately, I have been thinking: Can the saliva diffuse through the water stream that pours into my mouth? Saliva is mainly composed of water, some salts, and amylase. Amylase is a rather bulky protein, so I don't expect it to diffuse so quickly, but I'm a little concerned about the electrolytes. Ions like $\ce{Na+, K+, Ca^2+, Mg^2+, Cl-, HCO3-, PO4^3-}$ are quite small and might diffuse into the bottle.
I don't have access to the values of the ionic speeds, so I'd like to ask a chemist for confirmation whether these ions can diffuse into the bottle.
To make the question a bit more informative, lets assume I keep the bottle about $\pu{3cm}$ away from my mouth, so the water would be flowing at a speed of approximately $\pu{77cm s^-1}$.

If there is no physical contact between your mouth and the bottle, then you can't contaminate the water. Bacteria from your mouth certainly wouldn't swim "upstream" to get into the bottle. This is also much faster than the diffusion rate of any ions. from your mouth. (I guess you could actually cough into the container...)

What brand of water are you talking about? Electrolyte concentrations vary depending on the brand; Aquafina has next to zero sodium content, while some bottled water brands can have sodium contents as high as 20mg/L

Diffusion coefficient of $Na^+$ in water at 25°C is $1.33×10^-5$ $cm^2/s$. I think it says pretty much enough. And you might be intrested in Can molecules diffuse upstream when emptying a cup of tea?.

Why would you even care if the ions you mention diffused into the bottle? First, they're sterile; second, the $\ce{Na+}$ etc ions in my saliva are literally identical to the ones in your saliva.

To be more cautious: You should not breathe into the bottle while drinking, you should hold your breath. If you are suffering from any diseases like common cold, pneumonia, etc the droplets from your nose could travel into the bottle.

This question reminds me of a germaphobic classmate I had back in Middle-school. He would always stay clear of public lavatories/urinals because he was under the impression that urinating at one would result in bacteria from the urinal climbing up the stream of urine and into his urethra...needless to say, he was a very lonely boy.

The situation you described ("Drinking without sipping") is a modification (albeit, a more reasonable one) of the "Bacteria-climbing-up-a-stream-of-urine" theory.
I'm not a chemist myself, and like you, I am faced with... ahem, "issues", while searching for speeds of common ions in water (at a given temperature, and in the absence of an external electric field). But I feel this question can still be satisfactorily approached even without these values.
From the wording of your question, I suppose you assumed that potassium, sodium, magnesium and other ions would diffuse from a region of higher concentration (the saliva in your mouth) to a region of lower concentration of those ions (the stream of water entering your mouth). Now this would seem perfectly plausible if it weren't for a few things:
1) You're dealing with flowing water
This isn't a simple case of spitting into a glass of water and waiting for the contents of your saliva to (passively) diffuse into the water (and eventually form a dilute saliva solution). By drinking it the way you do, the passive diffusion of electrolytes in your saliva won't be able to match the speed of the water flowing into your mouth, much less exceed it in order to enter the bottle.
2) All the electrolytes in your mouth are trapped in a viscous protein-soup
Human saliva contains a variety of proteins (among other things). Common experience tells us that saliva is (a little more) viscous than water, and doesn't "instantly" dissolve in it (<----- Coming from a guy who had to spit in test tubes for his Biology lab assignments). Now not only do the electrolytes have to actively diffuse upstream to get into the bottle, but they also have to "free" themselves from the saliva first.
3) What do you mean by "saliva"?
This is the philosophically inclined conclusion to my answer. You're worried that the entry of electrolytes from your mouth into the bottle (won't happen) counts as contaminating it with your saliva. At first glance, even if the electrolytes seem to be able to make it to the bottle, you're pretty sure that larger molecules (such as proteins) won't be able to do the same.
If you don't count the proteins, epithelial cells and bacteria that normally constitute human saliva...how would this make saliva any different from a simple laboratory-prepared cocktail of sodium, potassium, magnesium, calcium, etc? Besides, ordinary (un-branded) bottled water would already contain these electrolytes; so by that logic, the water's already contaminated with your "saliva" even before you opened it!

@PrittBalagopal Diffusion is slow slow slow compared to water pouring out of a container from any practically meaningful height. See here.

I had to read this whole comment thread to see if there was anything salvageable in there. Use [chat] for general discussion.

With respect to the exact question in a highly literal sense, it is highly unlikely and next to impossible for ions to diffuse upstream in this manner; the linked question deals with this very nicely.
With respect to the implied question, however, as to whether you should trust a bottle of water, if someone with a highly contagious disease, known to spread though saliva droplets, drank from it from a distance of 3cm, then the answer is no. The problem wouldn't be one of upstream diffusion per se, but all sorts of splashes and aerosolised particulates dancing from one's mouth and onto the bottle opening itself.
Then again, if it's less about surviving a "Resident Evil" style incident and more to do with the "ew" factor, I wouldn't worry too much about it.


text:
Is it possible to have Zero order Elementary Reaction?
I was going through a question, which got me confused. Is decomposition of NH3, an elementary reaction?

I have studied that  for elementary reactions, Order = Molecularity,
if so, then why they are different in this reaction.
If this is not an elementary reaction, Then molecularity has no significance. Each step should have its own molecularity. isn't it?
So, Is it possible to have zero order elementary reaction ?

Yes, zero order reactions are possible. See -> https://en.wikipedia.org/wiki/Order_of_reaction#Zero_order

While zero order reactions are certainly possible and have considerable technological use, for example, in breathalysers used to determine alcohol content in ones breath, the question is what do you consider an 'elementary' reaction to be ?
In a pure sense the only 'elementary' reactions that occur are unimolecular and bimolecular ones (possibly also termolecular but exceptionally rare). A unimolecular reaction could be a bond dissociation or cis-trans isomerisation which can occur in an isolated molecule. Bimolecular reactions such as atom/molecule & molecule/molecule reactions are very common and strictly involve only two species.
So in this sense zero-order reactions have to have a more complex nature, such as involving an enzyme as in alcohol decomposition.
(Note also that the order of a reaction is measured experimentally as the sum of the powers to which reactants are raised in the rate expression, $rate= kA^aB^bC^c$ the order is $a+b+c$. Often it is the case that an apparently simple reaction such as $\ce{H2 + I2 }$ is only fully understood by a complex series of reactions, in this case the rate expression has fractional and reciprocal orders. The molecularity m refers to the coefficients in an elementary reaction, for example $m = a + b$ if the reaction actually occurs in a step $aA+bB \rightarrow$ so that m is a small positive number usually $1$ or $2$, for example $ m=1;\, A\rightarrow product$ or $ m=2;\, A+B\rightarrow product$)

Can you please reply all my questions w.r.t screenshot example.


text:
Why doesnt the overall rate of a reaction increase if you increase the concentration of a reactant in a non rate determining step?
We say that the overall rate doesnt increase if we increase the concentration of a reactant in a non rate determining step. Does that mean that the speed of that non rate determining step increase but at the same time something happens to counter the increase in speed? Because it doesnt make sense to say that the speed doesnt increase with increase in concentration.

Let's think about this reaction:
$\ce{ A + B->[k_1][] C + D ->[k_2][] E }$
let's say $k_1 << k_2$ so the first reaction is much, much slower than the second step (at equal concentrations)
Now it get's a little bit tricky since the reaction rates depend on the concentration. Let's assume for now that the concentrations of $A$, $B$ and $D$ are in a way that the reaction rate of the first step is really much faster than the second step.
What happens is that $A$ and $B$ react to $C$ quite slowly, but once you got any $C$ it reacts very fast with $D$ to $E$. This means your concentration of $C$ will always be quite low, since everything that is produced will react to $E$. But this also means that the concentration of $E$ is basically increasing with the rate of the first step. The slow reaction part is limiting the overall rate.
What happens if we increase the concentration of $D$? Well, the second step will be even faster, so all the $C$ that comes out of the first step will be converted to $E$ even faster. But the first step is still slow, so your $E$ concentration will still increse approximately with the rate of the first step.
Only if we would reduce the concentration of $D$ to very low levels (much lower than $A$ and $B$) we would see a difference, in that case the second reaction will get slower than the first one (reaction rate depends on concentration) and will be the limiting step.


text:
Synthesis Golf III: Chloramphenicol
A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This third round of golf concerns the synthesis of chloramphenicol 2,2-dichloro-*N*-[(1*R*,2*R*)-1,3-dihydroxy-1-(4-nitrophenyl)propan-2-yl]acetamide, an antibiotic on the World Health Organisations list of essential medicines:
InChI=1S/C11H12Cl2N2O5/c12-10(13)11(18)14-8(5-16)9(17)6-1-3-7(4-2-6)15(19)20/h1-4,8-10,16-17H,5H2,(H,14,18)/t8-,9-/m1/s1

In addition to being a useful target, chloramphenicol was chosen as
during previous rounds of synthesis-golf, several people commented that they'd like to see targets amenable to biosynthetic transformations, and scaleable process routes. This target very much fits both of those criteria (the drug is manufactured on a huge scale, and several enzymatic routes have been developed).
The challenge is to propose a route to chloramphenicol, including a method for setting the absolute and relative stereochemistry of the 1,2-hydroxyamine functionality. You may start from anything commercially available in the Sigma Aldrich catalogue, but to keep things interesting, no starting material may contain more than 8 carbon atoms.

Approach:
The approach taken in this proposed synthesis of chloramphenicol is largely based on Greene's synthesis of paclitaxel, using dihydroxylation methodology proposed by Sharpless.
Whilst not particularly elegant (I was hoping for some kind of asymmetric hydrogenation to set the stereocentres), it does get to the product in a respectable 7 steps (including setting both stereocentres, as required), and avoids having to introduce protecting/directing groups at any stage.
Forward synthesis:

Step 1: Horner Wadsworth Emmons olefination

The first step in this proposed synthesis is a Horner Wadsworth Emmons olefination to introduce the (E)-alkene required for the planned dihydroxylation. Without the nitro-group, the exact reaction has been done many times and with good selectivity, so no reason to assume that this HWE would be particularly challenging (the nitro group clearly makes the aldehyde electron poor).

Step 2/3: Sharpless dihydroxylation, epoxidation sequence

With the (E)-alkene formed, a Sharpless asymmetric dihydroxylation can take place using AD-MIX-B (taking the aryl group as 'large' and the ester as 'medium'). This step (without the nitro) is present in the Greene paclitaxel synthesis (above) and initially gives 82% e.e, though a single recrystallisation of the product brings this up to >95%.
Once recrystallised, treatment with mesyl chloride and subsequent heating (same pot) affords the necessary epoxide.

Step 4/5: Epoxide opening and reduction

Now comes the slightly dubious step....
In the paclitaxel synthesis, the epoxide is opened with sodium azide at the benzylic position (expected, the aromatic ring can stabilise the developing partial positive charge here), however in our case, we have an incredibly electron withdrawing nitro group in the 4-position, completely shutting down any stabilisation at the benzylic position.
Based on this lack of benzylic stabilisation, we could reasonable assume that it will take place with the opposite regioselectivity to the paclitaxel case. There is some evidence for this in the literature, (including where there is an ester present on the other side, but sadly not using azide as the nucleophile), and a cursory flick through an organic textbook will tell you that SN2 adjacent to carbonyls is generally fast.
Assuming it works fine, a Staudinger reduction converts the azide to the secondary amine with retention of stereochemistry.

Step 6: Ester reduction

Lithium borohydride is a mild and selective reagent, able to reduce esters in the present of various other functionality, including nitro groups. One draw-back of this is that lithium borohydride is significantly slower than its counterparts (not to mention more expensive), but, using LAH or other highly reactive hydride species would result in side reactions involving the nitro group.

Step 7: Amide bond formation

The required acyl chloride is commercially available, and should do the required reaction in place of the possibly competitive ester formations (nitrogen LP higher in energy).
Conclusion:
7 steps overall starting from the cheap, commercially available 4-nitrobenzaldehyde. The majority of steps have good precedence on similar systems (ignoring the epoxide opening, which may/may not be fatal), and all of the proposed reactions can (and have been) demonstrated to be possible on large scale (again, ignoring the epoxide opening, but TMS-azide is often substituted in the case of process syntheses involving azide).

Even if you form the ester in the last step, you might get transfer of the acyl group onto the nitrogen since that's an intramolecular process. I've had this cause problems when I wanted the amino ester instead...

Could you use LiBH4 to reduce the azide along with the ester? Saves a step and you don't have to get rid of Ph3PO which is always a pain.

Regarding the epoxide opening, Clayden gives typical $k_\mathrm{rel}$ for SN2 at (1) benzylic positions (2) alpha to C=O; the latter is faster by several orders of magnitude.

I believe you mean the Staudinger reduction will give you the primary amine, and not a secondary amine.

Some literature searching led me to a different linear synthesis with 8 steps.

Overall discussion of strategy
A different, and perhaps more direct, route to aminoalcohols is provided by the Sharpless asymmetric aminohydroxylation:

Of course, it's never that simple. Styrenes tend to undergo this reaction with the wrong regioselectivity, i.e. the nitrogen tends to add to the benzylic position. In particular, Kurti and Czako write:1

cinnamate esters react to give preferentially the β-amino ester product...

A solution to this is to use a different chiral ligand, instead of the usual DHQ–PHAL and DHQD–PHAL ligands. In 1998, Sharpless reported the "reversal of regioselection in the asymmetric aminohydroxylation of cinnamates" , using an anthraquinone (AQN) core instead of the phthalazine (PHAL) core.2

There is another catch, in that electron-poor aryl groups don't work well in this reaction, for whatever reason:

Electron poor cinnamates proved less suitable with the AQN iigand system. For example, the AA of methyl 3-nitrocinnamate gave a 1:1 mixture of regioisomers and significant amounts of the diol by-product...

I don't really view that as a problem per se, though. The removal of the N-Cbz group is usually done with a hydrogenation, which is already incompatible with the nitro group.3 Instead, the entire problem can be circumvented by using the 4-chloro derivative, which was reported to give a 77:23 ratio of regioisomers B:A with an accompanying ee of 92%. The chloro group can then be converted to the nitro group near the end of the synthesis, using palladium chemistry developed by the Buchwald group.4
Okay - enough talking, now for the synthesis...

Forward synthesis
Methyl 4-chlorocinnamate is commercially available, but the restriction on starting materials to 8 carbons or fewer means that we need to make it. Of course, there are lots of valid possibilities. The most common way seems to be a Heck reaction, but olefin metathesis and the Horner–Wadsworth–Emmons have also been used. This cinnamate can then undergo the Sharpless aminohydroxylation as described previously, which creates the key C–O and C–N bonds with the desired stereochemistry.

The next step is DIBAL reduction of the ester to the alcohol, followed by protection of the 1,3-alcohol as the acetonide. I inserted this protection step as I'm afraid that the free hydroxyl/amino groups adjacent to each other would poison the Pd catalyst in the nitration step.

The synthesis is completed by:

hydrogenolysis of the Cbz group (the use of ammonia as a solvent inhibits benzyl ether cleavage5, eliminating any possible risk of destroying the acetonide).
acylation on N (the use of the acetonide also nicely sidesteps any potential acylation on O). I chose to do this before the nitration, just in case the free amine interferes with the Pd.
Pd-catalysed nitration, which seems to have quite a broad substrate scope, and
removal of the acetal with aqueous acid, preserving the amide.

Notes and references

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; pp 404–405.
Tao, B.; Schlingloff, G.; Sharpless, K. B. Reversal of regioselection in the asymmetric aminohydroxylation of cinnamates. Tetrahedron Lett. 1998, 38 (17), 2507–2510. DOI: 10.1016/S0040-4039(98)00350-5.
The use of a Boc carbamate (instead of Cbz) is not reported in the paper, but I wonder if it could be done, as the deprotection with TFA hopefully wouldn't mess up anything in the compound. However, the poor reactivity of the nitro cinnamate would then become a problem, necessitating some other workaround, so I didn't pursue this line of thought any further.
Fors, B. P.; Buchwald, S. L. Pd-Catalyzed Conversion of Aryl Chlorides, Triflates, and Nonaflates to Nitroaromatics. J. Am. Chem. Soc. 2009, 131 (36), 12898–12899. DOI: 10.1021/ja905768k.
Sajiki, H. Selective inhibition of benzyl ether hydrogenolysis with Pd/C due to the presence of ammonia, pyridine or ammonium acetate. Tetrahedron Lett. 1995, 36 (20), 3465–3468. DOI: 10.1016/0040-4039(95)00527-J.

I started my route from D-serine. Not as available as biological L-serine, but still fairly cheap.
(a) TBS protect the alcohol. Otherwise, I'm going to get annoyed later trying to remove a benzyl ether.
(b) Exhaustive benzylation leads to dibenzylamine.
(c) Reduction with LAH
(d) Swern oxidation with dimethylsulfoxide and oxalyl chloride.
(e) I found this nice paper from Lüdtke (JOC, 2017, 82, 3334). The conditions are to use an aryl boronic acid (here, 4-nitrophenylboronic acid) and diethyl zinc. Unfortunately, I don't have journal access, so I can't check the substrate table to see if this particular combination works and gives good selectivity. As a general note, Reetz reports that these types of dibenzyl aminoaldehydes generally afford the Felkin-Ahn product, which is not what we want. This method with diethyl zinc affords the chelation controlled product, which provides the desired syn selectivity.
(f, 2 steps) Time to remove protecting groups. 10% palladium on carbon with ammonium formate should remove both benzyl rings. $\ce{TBAF}$ should pop off the silyl group. This product has a good chance of being crystalline, so we might consider shoring up purity with a recrystallization.
EDIT: The hydrogenation is probably the most questionable step in this synthesis since I have to do it in the presence of a nitrobenzene. In the worst case, we might be able to oxidize the aniline back to a nitrobenzene.
(g) Finally, dichloroacetyl chloride with $\ce{DMAP}$ and triethylamine should provide the desired product.
Eight steps in the longest linear sequence. Not my best work, but I think it works.

I found the nitro group incredibly irritating too. Ugh. I had everything planned out and that nitro group was the only thing stopping a hydrogenation of Cbz + benzyl ether...

Fingers crossed that the benzyl deprotection doesn't just wreck the nitro group... @orthocresol

Intermediate 5 is really intriguing. (1) it looks tantalisingly like it might be possible by some desymmetrisation of a 1,3-propane diol derivative, Krische style maybe, and (2) because it looks worryingly like it would epimerise

@NotEvans. That's a really good idea. I'm sure there are plenty of good routes to desymmetrize serinol. Quick search found: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2529399/

If instead of benzyl you use dimethoxy benzyl for your N protection then it will come off more easily. However one should remember Murphy's Law applies.

@Waylander Good idea. The references I found for that selective addition to the aldehyde are from the dibenzyl amine, but it's unlikely that substitution on the benzyl rings would kill the selectivity...

I solemnly swear I came up with this on my own...
Retrosynthesis
The aminoalcohol motif (to me) strongly suggests the addition of an azide to an epoxide. The requisite chirality can be introduced with a Sharpless asymmetric epoxidation; then, the second stereocentre is easily set because the epoxide opening is stereospecific.
This target is relatively simple, so the retrosynthesis can probably fit in one scheme. I think it's also (mostly) pretty self-explanatory.

Forward synthesis
The choice of protecting group P in the starting material 5 is largely dictated by the presence of the nitro group in 6. Quite a lot of protected 2-bromoethanols are available in the Sigma–Aldrich catalogue; the obvious choice in terms of cost is probably the methyl ether (CAS 6482-24-2, 5 g for £20).
A standard Wittig reaction with 4-nitrobenzaldehyde 6 (CAS 555-16-8, 10 g for £19) should give the (Z)-alkene. Demethylation with $\ce{BBr3}$ and the appropriate Sharpless epoxidation should hopefully get us to 3:

and now I run into the same problem as described in NotEvans's answer: it would seem that the preferred regioselectivity of epoxide opening is at the benzylic position, where the $\mathrm{S_N2}$ process is accelerated.
My workaround would be to Swern oxidise the alcohol; open the epoxide (now the position α to the carbonyl group should definitely be favoured); and re-reduce the aldehyde (sodium borohydride):

The endgame is of course a Staudinger reduction and amide formation. As has been mentioned many times, dichloroacetyl chloride is commercially available (CAS 79-36-7, 5 g for £8).

If acylation on oxygen occurs I'd guess it can be reverted with $\ce{K2CO3/MeOH}$, and the amide should hopefully stay put.
All in all it's 6 steps if the epoxide is cooperative, 8 steps if it isn't.

$\ce{BBr3}$ on a non-aromatic methyl ether... you like to live dangerously!  Nice route though, 6 steps

@NotEvans. That's true... and it's also allylic... probably not the best idea. :( Any suggestions for a replacement? I was also thinking of using an MOM ether (sm also commercially available) and removing it with PPTS... (at least Greene said it was good for allylic alcohols!)

Good old THP ether would be OK here

Ok, so I decided to try this synthesis golf this time.
I decided to use benzene, 3-hydroxypropanoyl chloride, and dichloroacetyl chloride as my raw materials.
I'm starting off by performing a Friedel-Crafts acylation of benzene.

I will then do a Wolff-Kishner to get rid of the carbonyl.

Performing allylic substitution,

Dehydrohalogenation,

Diol addition using $\ce{OsO4}$,

Nitrate the benzene ring,

Replace $\ce{-OH}$ with $\ce{-Cl}$,

Replace $\ce{-Cl}$ with $\ce{-NH2}$

Add, dichloroacetyl chloride,

And, that's it. Hopefully, this synthesis goes well.

Your starting acyl chloride probably doesn't exist, it would react with itself. At the very least, I'm unable to find a reliable source selling me that compound.

@orthocresol Hmm, so I have to find a way to add it without the $\ce{-OH}$

There are a handful of reports of making that compound, probably intended for immediate use. Even so, it seems improbable that the acyl chloride will react with the benzene ring first before the hydroxyl group (much more nucleophilic).

Hey @orthocresol I got a way to add it. I'll use 3-chloropropionyl chloride and hydrolyze it after acylating it into the benzene ring.

Are you sure about the outcome of your HCl reaction with the triol? I'm not.

In addition to the above, you might also possibly replace the $\ce{OsO4}$ step with an enantioselective variant in order to make a single enantiomer, as required.  In any case, nice attempt!

@NotEvans Thanks. I don't know much reactions anyway, I only knew this reaction. I guess I'll get better at Synthesis Golf as time passes, and I learn more reactions.

Starting materials: acrylic acid, benzene, protected ammonia, dichloroethanoyl chloride.
6 total steps.

Conditions:
(1) - I2, NaHCO3
(2) - Benzene, Lewis Acid (AlCl3)
(3) - NH2Ts, base
(4) - DIBAL-H
(5) - H2SO4, HNO3
(6) - Dichloroethanoyl chloride, base
Reactions:
(1) - Iodolactonization, the 3-membered ring forms because if the 4-membered ring formed, Baldwin's rules would be violated.
(2) - L.A. addition to the carbonyl, benzene attacks the electrophilic carbon.  This creates a deprotonated alcohol alpha to an alkyl halide, therefore epoxide formation occurs with predetermined stereochemistry.
(3) - Opening of epoxide using protected ammonia, inversion of stereochemistry occurs.
(4) - Reduction of the carbonyl.  Occurs with this stereochemistry because the -Tosyl group is bulky and is hindering an attack from above.  Therefore the attack proceeds from below and the alcohol is formed facing upwards.
(5) - Nitration of this compound occurs using a nitrating mixture, due to the strong acids present in the mixture, the amine is deprotected (tosyl group is removed).
(6) - Amide formation, self-explanatory.

I'm surprised the free OH groups are untouched by the nitration conditions; there seems to me to be plenty of scope for this going wrong

I tried searching the literature for any precedent of the first step and couldn't find anything. It strikes me as being a little odd (at least, to my knowledge), as the alkene conjugated to C=O is more electrophilic than nucleophilic (and hence iodonium formation, which involves nucleophilic attack of C=C on iodine, is harder).

Polymerization might be the problem, I know alpha propiolactones are generated from the closing of deprotonated halopropanoic acids, so the same product may be able to form from deprotonation of 2,3-dibromopropanoic acid.  Also, regarding the first comment, the -OH groups could be protected, resolving the nitration issue that may arise.

A synthesis starting from a basic starting material. Asymmetric induction was through the Sharpless epoxidation.