Air- and light-sensitivity of phenol I've read that phenol should be stored under inert gas and protected from light.

What reactions does phenol undergo during long-term storage in the presence of air and light?
Is it really necessary to take such precautions when storing or handling phenol?
How does the stability of solid phenol compare to that of phenol dissolved in, say, cyclohexane?

Phenol slowly oxidizes on exposure to air and turns pink in color. At first, quinone (p-benzoquinone) is formed which again reacts with phenol to form a polymerized product phenoquinone which is red in color.

There has been research on the oxidation of phenol like wet-air oxidation and catalytic oxidation but those are out of scope.
Oxidation of phenol using light has been studied using UV/$\ce{H2O2}$/ozone system. From this paper:

The photolysis with 253.7 nm radiation practically does not degrade
phenol. But the combination UV/ozone gave better results than the
using only ozone. In the degradation of phenol using ozone there are
two main reactions: the direct reaction of ozone with phenol and its
products and the reaction of the hydroxyl radical generated. Some
intermediate products as benzoquinone, catechol and hydroquinone were
detected during the degradation of phenol.

So, yes. Necessary precaution are taken such as phenol are kept at brown bottles.

I think solid phenol is less prone to oxidation due to air/light than liquid/azeotropic phenol because of $\ce{\pi}$ stacking, efficient packing of crystals and hydrogen bonding leading to stability.


Nuclear spin multiplicity of radicals How do you calculate the nuclear spin multiplicity of radicals such as $\ce{N2H+}$? How does the net charge on the molecular species effect its overall nuclear spin?

The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).

So on a basic level, the net charge does not affect the overall nuclear spins.

The somewhat longer answer is that the nuclear spin states depend on the local chemical environment. This means that we can "see" different proton nuclear spins based on the molecular structure.

If there's are unpaired electrons, paramagnetic NMR requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).

Two main effects in paramagnetic NMR are pseudo-contact and contact shifts.

  The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).


Stoichiometric calculation of composition of a compound I have a question from a high school review package. I don't need the answer, but one thing that is bugging me is that, I feel as if I do not understand the question at all. I don't know what it wants me to find.
Can anyone please tell me what the question is about?
The question is:

Copper oxide is a black powder. It can be decomposed by heating it with an excess of charcoal, a form of carbon.
The charcoal reacts with the copper oxide to produce copper and carbon dioxide. Any excess charcoal that was used can be separated from the copper by adding water. The charcoal will float on the water while the more dense copper will sink to the bottom. The charcoal can then be skimmed off.
Mass of copper produced = $\pu{2.76 g}$
Mass of copper oxide used = $\pu{3.45 g}$
a) Use this data to determine the simplest formula of copper oxide.
b) Predict the valence/charge of the copper in the copper oxide formed.

I basically got:
$$\frac{2.76}{63.546} = \pu{0.043 mol}$$
The balanced chemical equation that I deduced was $$\ce{2CuO + C -> 2Cu + CO2}$$
However, I am not sure that my understanding of the question is correct. Is there a mistake in my understanding.
Note: I was able to solve the question.

This is an example of a classical method of determining chemical formulas by elemental analysis.
You are not supposed to know the chemical formula of copper oxide neither they are asking you to write a balanced equation. All they are telling you is that
(i) Mass of pure oxide of copper = 3.45 g
(ii) Mass of copper obtained after reduction (we do not need to know the chemical equation) = 2.76 g
Use mass balance, how many grams of oxygen atoms must be there in that oxide?
Now you know the masses of Copper and Oxygen.
Two options
(a) Use mol and find out molar ratios between the two elements
(b) Alternatively, if you do not need to invoke the mole concept ... old school way. Just for trying
Think about the fact that 16 (exactly 16) is the mass unit of 1 oxygen atom. What mass of copper atoms would be associated with 16 units of O if copper's atomic mass is 63.54 units. Use mass balance and some arithmetic.


Simple enantiomer in powder form? I want to do some experiments on enantiomers. As it involves some calculations of potential energy surfaces, I would like to do this experiment on a simple enantiomer (For example $\ce{CHBrClF}$). But this is in liquid form I think. I need an enantiomer sample in the powder form. Can anyone suggest a simple sample? Either one I can buy or can easily make in a chemical lab with the help of a chemist.

Enantiomorphs are non-superposable mirror images (all signs of one coordinate axis reversed).  What is the maximum enantiomeric divergence possible?  If you have a  quantitatively negligible chiral divergence, there is nearly no quantitative difference to be realized in calculation or measurement.  BTW, the maximum energy difference between enantiomers cannot be more than a part-per-trillion difference/average.  If it were larger, there would be easily detectable enthalpy of fusion anomalies for calorimetry secondary standard benzil.

Chirality is geometry.  Geometric chirality is easily observed (no ${S_n}$ symmetries[A]) and easily calculated[B].  alpha-Quartz and gamma-glycine single crystals have $\text{CHI} = 1 \text{, COR} = 1 \text{, DSI} = 0$, maximum mathematical chirality[C], but it cannot be measured.  Solution optical rotations ignore atomic mass distribution[D].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-${42d}$, rotates 522°/mm along [100] at 497.4 nm[E].  ${P3_121}$ quartz and ${P3_221}$ berlinite are both  levorotatory[F].  Achiral $\ce{PhCOCH3}$ impurity changes resolved $\ce{PhCH(OH)CH3}$ optical rotation[G].  CIP notation flips with composition at constant geometry (look at natural protein cysteine versus alanine, both L-amino acids but opposite CIP notations).

How do you obtain $100\%$ optically pure maximally divergent enantiomers to your specs?  No problem!  Benzil, $\ce{(C6H5CO)2}$, is achiral when molten, gas phase, or dissolved. Solid state crystal lattice forces twist and stack benzil molecules into homochiral helices, space groups ${P3_121}$ (right-handed)or ${P3_221}$ (left-handed)[H].

You obtain (racemic) benzil, modestly resolved (1 $\mathrm{S}$)-(-)-alpha pinene and (1 $\mathrm{R}$)-(+)-alpha pinene (both enantiomers), and mixed xylenes.  $20$ ml scintillation vial, foil-lined cap.  $100$ mg of benzil or so, $10$ ml resolved, alpha-pinene, heat to $100$ C. Drip in mixed xylenes until the molten benzil is just soluble, add another drop or two, very slowly cool undisturbed.  Only one space group crystallizes out.

Enantiomorphic space groups[I] are mathematically perfectly divergent geometric chiralities.  Here is the calculation of benzil crystal lattice's atomic mass distribution from Petitjean's software, CHI = $1$ (rapidly asymptotic) COR = $1$ DSI = $0$

If you want a discrete molecule as opposed to a crystal lattice, resolved camphor comes in around $\text{CHI} = 0.5$.  Resolved ${D_3}$-trishomocubane is a much better geometric example and is easily synthesized in large quantities starting with benzoquionone and cyclopentadiene (Diels-Alder).  If you have a methyl group sticking up from its ${D_3}$-axis, calculated $\text{CHI} = 0.996$.

Look at that methyl-bearing carbon!  It has no ${S_n}$ symmetries, for R must reflect or invert as S.  This chiral center's sense of chirality cannot be named.  Try doing it with CIP notation.

Uncle Al has been very naughty with geometric chirality, "[6.6]Chiralane: a remarkably symmetric chiral molecule," Symmetry: Culture and Science 19(4) 307-316 (2008).  Note that the core carbon atom has four rigorously identical substituents yet it is a maximally geometrically chiral center.  Chemical chiral nomenclature is fundamentally defective, for said chiral center (and the four to which it is attached) cannot be assigned a direction of handedness by any means, even in theory.

[A] http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
  pp. 11-13.
[B] J. Math. Phys. 40, 4587 (1999)
  http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM
[C] http://www.mazepath.com/uncleal/qzdense2.png for alpha-quartz
  http://www.mazepath.com/uncleal/glydense2.png for gamma-glycine
[D] http://www.mazepath.com/uncleal/norbors.gif
    http://www.mazepath.com/uncleal/norone.png
[E] Appl. Cryst. 33, 126 (2000)
[F] J. Appl. Crystallogr. 19, 108 (1986)
[G] J. Org Chem. 38(10), 1870 (1973)
[H] Acta Cryst. B43 398 (1987)
[I] http://elib.mi.sanu.ac.rs/files/journals/publ/69/7.pdf Section 2: eleven pairs of enantiomorphic space groups;
 http://www.math.ru.nl/~souvi/papers/acta03.pdf  Section 3ff.


Has QTPIE ever been applied to 1D metals such as polyacetylene? QTPIE is a fluctuating charge model that substantially improves modeling of polarization and charge transfer. I was curious whether it has ever been applied to 1D metals such as polyacetylene to provide new perspectives on topics such as Peierls' condensation and phase solitons?

How flattering that someone asked this question ;)

The short answer is yes, I've tried QTPIE on series of conjugated molecules like the oligoacetylenes and oligoacenes. The results, however, are fairly dismal, as can be seen in the preprint. The problem is that in these empirical charge models, the polarizability grows faster than the size of the molecule, which means that these models predict infinite polarizabilities and charge transport behavior for bulk materials, which is obviously nonsense.

The problem is quite intimately related to that of self-interaction error in approximate density functional theories, which also exhibit similar problems.


Chirality in coordination octahedral compounds Isn't this molecule achiral? I mean if I flip the mirror image on right by 180 degrees(like we do to determine chirality in tetrahedral carbons) it would be superimposeable? But apparently it is chiral. How am I flipping wrong?

EDIT
I finally got the hang of it (I think). Are the images below correct?

See following image:

Look at the purple loop (en) on upper left hand side structure. Now, if you rotate the whole molecule ${90^o}$ clockwise around an axis going through top $\ce {NH2}$- $\ce {Ni}$-bottom  $\ce {NH2}$, you will get that loop overlapping on similar loop on the mirror image. But, now look what happens to the other two loops? They are not overlapping on other two loops on mirror image. For example, the purple loop on bottom of left hand side structure will move to back side of the same side (it would be ${180^o}$ far from similar loop in the mirror image). Thus this molecule would not superimpose with its mirror image.


aluminum hydroxide as an electrolyte A friend and I recently produced some aluminum hydroxide, $\ce{Al(OH)3}$ for a project, and we wondered how it functions as an electrolyte. We know it has a low conductivity and solubility, but we wanted to know for knowledge's sake. We put some aluminum hydroxide in a bowl and used a power supply running at 31.9 volts (The resistance was high enough for the current to be immeasurable). We also used aluminum electrodes (aluminum foil).

The electrodes were bubbling and a white precipitate was forming on the top, and it seems we have less hydroxide than we started with...

So what exactly did we do?
$%edit$

When you are applying 31.9V across I am assuming cm-ish distances, you are definitely causing a lot of electrochemistry to happen in the solution.

You are oxidizing one of your aluminum electrodes to $\ce {Al^3+}$, therefore precipitating $\ce{Al(OH)_3}$ and probably reducing water on the other electrode to evolve $\ce{H_2}$.Along with these, you are probably doing a lot of other reactions with any and every impurity around, but these two should dominate.

Measuring ionic conductivity properly requires doing an AC measurement since you cannot have a sustained DC current without doing electrochemistry of some sort. The electrochemistry on the electrode surface ends up the main factor influencing your measurement in any DC setup.


How to separate benzyl alcohol and meta-cresol Benzyl alcohol and meta-cresol boiling point's are not so different, the fractional distillation will be very difficult, what can i do for separate this two composites ?

Your observation on the boiling points is correct.

Take a look at the melting points instead: the m.p. of benzyl alcohol is -15 °C, while the m.p. of m-cresol is 11 °C. You might want to cool down the mixture and see whether you can see some enrichment.


dependence of Intensity on penetration of light I want to know about a transparent material,In which penetration of light decreases as intensity of light increases.

In general, when light penetrates a material (water for example) its intensity decreases exponentially with increasing depth. This process is known as attenuation or absorption. The severity of attenuation differs with the wavelength of electromagnetic radiation. In the region of visible light, the red part of the spectrum attenuates more rapidly than the shorter-wavelength blue part.

But what you are asking about is a non-linear optical property known under the name: Saturable absorption It is a property of materials where the absorption of light decreases with increasing light intensity. At sufficiently high incident light intensity, atoms in the ground state of a saturable absorber material become excited into an upper energy state at such a rate that there is insufficient time for them to decay back to the ground state before the ground state becomes depleted, and the absorption subsequently saturates. The material becomes transparent at that wavelength. Saturable absorbers are useful in laser cavities.

Another kind  of non-linear materials is known under the name "optical power limiting materials". In these materials the absorption of light increases with intensity such that beyond a certain input intensity the output intensity approaches a constant value. Such a material can be used to limit the amount of optical power entering a system. This can be used to protect expensive or sensitive equipment such as sensors, can be used in protective goggles, or can be used to control noise in laser beams.

For more details, please see: http://en.wikipedia.org/wiki/Saturable_absorption
and http://en.wikipedia.org/wiki/Two-photon_absorption


New York Times Crossword: Is "electrically balanced" in chemistry "apolar" (given), "neutral," or something else entirely? As a lifelong New York Times crossword puzzler, I was baffled by clue 38-Across in today's puzzle (Saturday, May 22, 2021). The solution is a 6-letter word, and the clue is:

Electrically balanced, in chemistry

The word that comes to my mind is "neutral." The accepted answer is "apolar," a word I've never heard or muttered in my life. I would have gone with "non-polar" long before I hit "apolar," and I'd be in trouble due to the fine print around "non-polar" and "electrically neutral."
The Webster's Third New International Dictionary has this definition for "apolar":

having no poles - used esp. of nerve cells formerly believed to lack processes

My go-to text for historical chemistry - M. P. Crosland's Historical Studies in the Language of Chemistry - yields no answers. From the definition above, it might be something one would encounter in biology, specifically in regards to cellular biology.
Is "apolar" an accepted or commonly-used word in chemistry as a synonym for "electrically neutral," and if so, in what context has this appeared?

Apolar is certainly a valid alternative to non-polar.
Either is natural in English as it is for, e.g. "asymptomatic", or its use in chemistry for non-polar has been introduced by non native English speakers quickly translating in their works basically the same word, "apolaire" in French and "apolare" in Italian, for instance.
Let me skip this discussion as matter for another SE site.
However, a quick Google chemical oriented search showed me that apolar is less diffused than what I expected, and seems to be more recurrent in biochemistry and astrochemistry, or as suggested in a comment, computational chemistry.
For the point more relevant to Chemistry SE,
"1) Is "apolar" an accepted or commonly-used word in chemistry as 2) a synonym for "electrically neutral," and if so, in what context has this appeared?"
For 1) see above. For 2) the answer is NOT. It just means, closer to the general definition that OP has found and reported, that a chemical entity has not (electrical) poles, i.e. no excess or partial electrical charges are present over its part.
While speaking of bonds, or simple biatomic molecules, apolar and non polar means that negligible or no electrical dipole moment is present. Still, an ensemble of polar bonds can nevertheless lead, as per vectors summation, to an apolar, or non-polar if you prefere, molecule.
While apolar chemical entities are indeed neutral, the same can be true for very polar uncharged molecules.
Another use of the terms is to classify solvents. In this case partial or overall electrical dipole moments are just one of the many factors influencing the interaction between molecules of different solvents or between them and those of a solute. The interested reader can refer to the section "Other polarity scales" of this Wikipedia article https://en.m.wikipedia.org/wiki/Solvent for more details.
(In the context of the puzzle, the given word definition is somewhat acceptable, because an apolar molecule won't present much of charge whatever finely examined, unless one discerns electrons and nuclei. And the fact that - as for the title rather than the OP text - they gave a somehow vague "electrically balanced" description, makes it quite accurate for apolar, in a crossword.)


What molecular features create the sensation of sweetness? There are many compounds that taste sweet. Many of the artificial ones taste much sweeter than natural sugars (which is why they contribute little to energy intake when use as sweeteners).

A range of sweeteners is shown in the figure below:

Structurally there are common features for some. We can see why sucralose (a non-metabolisable sweetener based on sucrose) might taste sweet. Rebaudioside a (a major ingredient of the sugar substitute Steria) shares some of those features. But the very sweet artificial sweeteners seem to have less in common.

Do we know what makes molecules generate the sensation of sweetness? Which features matter and why are some sweeter than others?

For humans, the "sweet taste buds" are the T1R2+T1R3 & T1R3+T1R3 G-protein coupled receptors.

Chemicals that are sweet are the ones that can bind to these receptors. We know some structures that tend to do this, namely ketones and aldehydes (carbonyl groups).

However, there are chemicals that are sweet that we do not know the mechanism for, such as Curculin, which is up to ~2000 times sweeter than sucrose by weight.


What is the convention on color of molecular orbitals in publication figures? I'm new to MO-visualization and I'm using the program Orca to generate visualizations of molecular orbitals of a system of interest. I've been able to produce some nice images in VMD but I'm not sure what they actually mean:

I've seen pictures like this before with blue/red blobs to represent molecular orbitals, but I don't know what (specifically) they are representing. My suspicion is blue=HOMO and red=LUMO, or blue = negative wavefunction and red = positive wavefunction, but I'd like to know if there is a universal standard.

For reference, this is the "Graphical Representations" panel in VMD for this system:

The colours represent the signal/phase of the molecular orbital, i.e. positive or negative.
Below is the HOMO (made with ChemCraft) of butadiene (PBE-D3BJ/def2-SVP with ORCA) as an example.

Figure 1: HOMO of butadiene (cutoff 0.03)

ChemCraft shows by default negative phase as blue and positive as red.
VMD seems to follow the same convention.

You can certainly change the colours if you want.
Below is how the controls for surfaces in ChemCraft look like:

Figure 2: surface controls in ChemCraft

Other programs may use different conventions, such as NBOView (yellow and blue).


Will an iron nail sunk in a highly concentrated salt solution rust? Assume I have a salt solution with the same concentration as the dead sea and I put a iron nail into the solution (assume it sinks). Will it rust? If yes, why? If no, why not?

Rust is iron oxide, $(\ce{Fe2O3.nH2O})$

It forms from iron(III)hydroxide made when iron reacts with oxygen in the presence of water. The chemical reaction can be summarized as follows:

$\ce{4Fe(s) + 6H2O +3O2 ->4Fe(OH)3(s)}$

Now, water 'naturally' contains some dissolved oxygen (8-14%), depending upon the source of the oxygen (photosynthesis from underwater plants, diffusion from the atmosphere) balanced against oxygen demands (such as fish breathing and decomposition of organic matter). The temperature of the water also affects the solubility of oxygen (gases are less soluble at higher temperatures).

The ocean (which is typically 3.5% salt) typically has 20% less dissolved oxygen in it than freshwater. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby driving dissolved oxygen out.

Now, the salinity of the Dead Sea is about 290 (that's 29% salt!), which is over 8 times the salinity of the ocean!  At such high levels of salinity, most of the oxygen is 'salted out' of the water, which means that the levels of dissolved oxygen are very low, especially below the surface, so an iron nail totally immersed in water with 29% sodium chloride (salt) will not rust.


Equilibrium for very small amounts of reactants Is there any difference in the concept of equilibrium when it comes down to dealing with extremely small amounts of reactants?

Say we have $$\ce{A + B <=> C + D}$$
 and $K_c$ is $100,000$. If we only have $10$ atoms of each $\ce{A}$ and $\ce{B}$ in a beaker, we would expect the container to be totally full of $\ce{C}$ and $\ce{D}$. But $K_c$ is not infinite, so there must be some unreacted $\ce{A}$ and $\ce{B}$. However, if there was even one atom of reactants, the reaction would quickly shift back to satisfy the equilibrium.

Does the concept of equilibrium only apply to macro-sized reactions? Is there an extendable concept for the micro scale?

Thermodynamics is only applicable to the systems of size in between astronomic and microscopic scales. Systems with a small number of particles can spontaneously abandon
equilibrium state due to fluctuations. Statistical physics tells us [1] that the relative magnitude of the fluctuations of thermodynamic quantities $x$ (e.g. pressure, temperature, volume, internal energy, entropy etc.) is inversely proportional to the root of the number of particles in the system:

$$
\frac{
\sqrt{\left\langle\left(\Delta x \right)^2\right\rangle}
}{
\left\langle x \right\rangle
}
\sim
\frac{1}{\sqrt{N}}
$$

Lets arbitrarily set the detection limit for experimental determination of the relative change of $x$ to optimistic value of $10^{-8}$. In this case we would need at least $N = (10^{-8})^{-2} = 10^{16}$ particles belonging to the thermodynamic system before we can speak of equilibrium at all.

References

Landau, L. D.; Lifshitz, E. M. Statistical Physics, 3rd Edition, Part 1: Vol. 5; Butterworth-Heinemann: Amsterdam u.a, 1980.


What is the origin of the colour of azo dyes? Is it the same reason as to why transition metal complexes have colour?

Yes, it is all about the absorption of light at specific wavelength.

Azobenzene, the parent compound has an absorption maximum around $\lambda$= 430 nm in the visible spectrum.

The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling.

Some examples are Allura Red (1), Chrysoine Resorcinol (2), Janus Green B (3) and Direct Blue 1 (4).

The colours of the molecules are chosen to resemble the colour in solution.

In order to make the dyes soluble in water, either sulfonate groups are attached or the dye molecule is just one fat cation.

As already mentioned by GM in another answer, +M substituents elongate the $\pi$ system (auxochromic effect).


Who came up with the box and arrow diagrams for electron configurations? In general chemistry texts, the electron configurations are sometimes shown in boxes with up and down arrows to show the concept of paired spins. My impression for years was that Hund used such diagrams. However, flipping through his book Linienspektren und Periodisches System der Elemente which used the s,p, d, f notation in a modern way, there is no such use of box diagrams with up and down arrows. Does anyone know who introduced this so-called box notation? Thanks.

This image is from Atomic Spectra and Atomic Structure by Gerhard Herzberg, John William Tranter Spinks - 1944. It does not really have boxes (just boxes of a table), but spin +1/2 and spin -1/2 are shown as arrows. Of course, I am not claiming this is the first usage, just that in 1944 it existed.


Appropriate container for storing/keeping chromic acid? I am planning to make chromic acid, however I the information for keeping chromic acid are conflicting. Can it be stored in a glass bottle even though it's used as a glass etchant? Also I saw some information that it's dangerous to keep it in plastic containers. What's the common practice?

Chromic acid can well be stored in glass bottles with stopper/lid but should be kept away from any organic compounds. Chromic acid is a dangerous oxidising agent and incompatible with various organic compounds. Also, it should be well marked with proper label and hazard symbol to distinguish it from other mineral acids. It has a decent shelf life and can be kept for long period without decomposition if kept well.

Also make sure it remains separated from mineral acid. Chromic acid is known to form solution with sulfuric acid forming chromosulfuric acid which is ironically used to clean laboratory glasswares.

Chromic acid is not a glass etchant.


Why is HCl not considered to have hydrogen bonding? A molecule that has hydrogen bonding usually follows these two premises.

1.) There is a hydrogen atom involved
2.) Hydrogen must be bonded to a highly electronegative element which are nitrogen ($\ce{N}$), fluorine ($\ce{F}$) and oxygen ($\ce{O}$).

Seeing that both oxygen and chlorine have a small difference in their electronegativity (oxygen being roughly 3.5 and chlorine being roughly 3.0), why does chlorine in a hydrogen chloride molecule ($\ce{HCl}$) have a dipole-dipole interaction, while the oxygen in a water molecule ($\ce{H2O}$) causes the water molecule to have a stronger form of dipole-dipole interaction called hydrogen bonding? I do not understand this since chlorine is just as electronegative as oxygen and nitrogen?
Edit: I should also added that nitrogen and chlorine have the same EN value (3.0)

Very, very related. Check it for a more in depth answer.

Well, first-of-all you need to understand that the $\ce{H}$ bonding isn't actually bonding. It is just a covalent attraction. Also since $\ce{Cl}$ is larger than $\ce{N}$, $\ce{F}$ and $\ce{O}$ it does not make a strong $\ce{H}$ bond. The size of the $\ce{Cl}$ makes the dipole-dipole attraction weaker. However $\ce{N}$, $\ce{F}$ and $\ce{O}$ are smaller and thus have an $\ce{H}$ bond.

Although in reality, compared to other covalently bonded structures, $\ce{HCl}$ has a very strong covalent bond.


How is the triferric tetroxide of iron possible? If oxygen has a valency of $-2$, $4$ of it would result in $-8$. But iron doesn't have a (variable) valency of anything that goes to $-8$. Then how is $\ce{Fe3O4}$ possible?

Iron, like many of the Transition Metals does have a variable valency, it can valencies of 2+ (ferrous iron), 3+ (ferric iron) and in some cases, 4+ (tetravalent iron).

In regards to $\ce{Fe3O4}$, according to the Wikipedia page Iron(II,III) oxide,

  contains both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions

A 2-dimensional diagram of $\ce{Fe3O4}$ is below, from the relevant ChemSpider page:

A 3-dimensional model is below (from Wikipedia):

red indicate oxygen


What's the reaction when bleaching hair with lemon juice (citric acid)? Is the acid working as an oxidizer or reducer with the melanin of the hair breaking down the bonds?

Hair is mainly keratin, the same protein found in skin and
  fingernails. The natural color of hair depends on the ratio and
  quantities of two other proteins, eumelanin, and phaeomelanin.
  Eumelanin is responsible for brown to black hair shades while
  phaeomelanin is responsible for golden blond, ginger, and red colors.
  The absence of either type of melanin produces white/gray hair.

I think the citric acid (in lemon) acts as a bleaching agent. The "bleach" reacts with the melanin in hair, removing the colour in an irreversible chemical reaction. The bleach oxidises the melanin molecule.

The illustration of the molecules perhaps might give a better picture on which groups are oxidised in the process.

The melanin is still present, but the oxidised molecule is colourless. However, bleached hair tends to have a pale yellow tint. It behaves as an oxidising agent at least according to my findings.

Acknowledgements

https://www.thoughtco.com/salon-hair-color-chemistry-602183
http://www.keratin.com/as/as002.shtml


Why are weak acids not considered good leaving groups? So today in my first year lecture my professor said that weak bases are good leaving groups. I understand why they are good leaving groups in comparison to unstable strong bases but why are weak acids not included? Acids are electron pair acceptors, shouldn't they be better able to take on the donated electron pair? Furthermore, when we say weak bases, weak in comparison to what? To be substrate the solvent or to other potential leaving groups?

Leaving group ability is tied to the stability of the leaving group. Acidity is directly related to the stability of the conjugate base. If you're a strong acid, the conjugate base is relatively stable in solution. Less so, if you're a weak acid. So the stability of the conjugate base is directly proportional to both acidity and leaving group ability.

Edit:

$$\ce{HA->H+ + A-}$$

$\ce{A-}$ is the species we're evaluating as the leaving group. Basically, it's a good leaving group if it's stabilized. If that's the case, the acid dissociation equilibrium also shifts toward products, making $\ce{HA}$ a stronger acid. You should be careful to differentiate between $\ce{HA}$ and $\ce{A-}$ when you're thinking about these concepts.


Why do some diagrams of some molecules show the bonds of only some of their submolecules? In the diagram of ATP below this paragraph there are no lines between the atoms in the $\ce{NH2}$ molecule or in the $\ce{OH}$ molecules; however, there are lines between all the other atoms. Does the omission of those lines represent some facts about  $\ce{NH2}$ and $\ce{OH}$ in the ATP molecule, or did the illustrator omit them in order to emphasize the rest of the molecule?

Your example is one of a variety of shorthands that exist because every chemist after the third year of studies knows what is meant so they are too lazy to draw.

In a nutshell, the $\ce{NH2}$ group will always be a nitrogen with a lone pair, two single bonds (one to each hydrogen) and one bond to the molecular skeleton. Much like $\ce{-OH}$ will always be $\ce{-O-H}$. It is also common practice to omit any insignificant hydrogen atoms in organic molecules; in the image above, a total of eight hydrogens has been omitted because every chemist will know where they are (and that they are there).

In principle, the phosphate chain could also be shortened to $\ce{HO3P-PO2H-PO2H-{}}$ rather than writing out all of it. Any chemist will mentally ‘expand’ the molecule in the same (correct) way.

As you continue to study, you will notice when which groups are generally shortened, which rules to observe while doing so and which hydrogens are so superfluous that they are not drawn — and, most importantly maybe, how atoms are labelled that actually are missing hydrogens. it is beyond the scope of a single answer to note them all.


What chemicals are present in burnt toast? I'm planning to do an experiment where I extract burnt toast and test how that affects the growth of mung beans. I know burnt food contains PAHs, but what are some other chemicals present and how would they affect plant growth?

First of all, one should have look a look on the ingredients of your toast. It mainly consist of starch, proteins (2% maybe), fat (a few %), water and some inorganic compounds such as NaCl.
Well Salt normally won´t interact with anything below 1400°C so no need to look at this.
Water just evaporates, i think we must not discuss that here.
Fat: that's more interesting because fat is known to degenerate into different compounds when heated.

But the definitely most interesting reactions will be present when one heats carbon hydrates together with protein: so called Maillard Reaction(s) those kind of reactions are still issues of ongoing research although this type of reactions have been discovered long ago. As a good starter on the various compound which can occur during Maillard Reaction you might read Mottram or Monnier. Especially acrylamide could be of interest when looking a toxicity. I don't have a certain paper in mind but i know that there has been research about toxicity of acrylamide in fish and other animal. So it's likely that there was some research about plants.

When you look at starch and protein separately you will encounter heat degeneration as well as caramelisation.


Why Phenyl Carbocation is unstable? Phenyl carbonation is said to be unstable because the + charge on it cannot be stabilized by Resonance or any other electronic effects, so my question is that if the resonance occurs in the direction shown in the below picture the + charge can be delocalized to some extent, so will it adapt that ?

Your resonance structure doesn't do anything to stabilize the cation. You've created a singlet carbene at the top most carbon, so it has both the cation and the anion on the same carbon. These do not cancel out. As drawn, you have two hydrogens on the top carbon. You need to replace it with a single lone pair and an empty orbital.


What is the intuition and motivation behind the definition of the thermodynamic activity? Stuck proving the relation $$\mu_i=\mu_i^\circ + RT\ln a_i,\tag{1}$$ I looked up the definiton of activity. It is

$$a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}},\tag{2}$$

but the the intuition/motivation for such a definition is unclear. Is it just because we wish for a relation like $(1)$ to exist by analogy with $(3)$? $$G_i=G_i^\circ+n_iRT\ln{\frac{p_i}{p^\circ}}\tag{3}$$

That is, if we define activity as given in $(2)$, we can take the natural logarithm of both sides, multiply with $RT$, add $\mu_i^\circ$ to both sides and arrive at $(1)$. And hence $(1)$ has a similar form to $(3)$ which holds for ideal gases.

Or am I mistaken and there are other reasons at play?

$\mu_i$ is chemical potential
The superscript $^\circ$ denotes the standard state
$a_i$ is activity
The subscript $_i$ denotes the $i$-th component of a mixture

So I'm taking Pchem right now so I'm no experts, but here is my two cents.

I'll be honest I essentially followed my books lead on the determination of $a_{i}$, and $\mu_{i}$. When looking at problems like this I try to go backwards. I see that $a_{i}$ is derived from $\mu_{i}$, and I try to find where $\mu_{i}$ comes from then see where that comes from until I get a definition or a well known relation like Gibbs free energy. I'm not savy enough to be like "Oh yea $\mu_{i}$, that's basically $\int dG(T, P_{2})$ I'll just show all that". But, I can slowly go backward. So, I followed the following path.
$$a_{i} \Rightarrow \mu_{i} \Rightarrow (\mu(T,P_{2})-\mu(T,P_{1}) ) \Rightarrow d \mu \Rightarrow d G$$
Then when you have this general path you reverse it to get the derivation.

Also remember that you have to keep track of all the assumptions you make. Its important to remember that. My answer is only applicable for a perfect gas at constant temperature, and a mixture of gas at low pressure.

From deferential of Gibbs free energy the chemical potential of a pure gas is,
$$dG = -SdT +VdP = d\mu$$
at constant temperature,
$$d \mu = V_{m}dP = \frac {RT}{P}dP$$
To get $\mu$ we integrate with constant temperature essentially  expanding the gas on an isotherm for $P_{1}$ to $P_{2}$.
$$\int _{1}^{2} d \mu = RT \int _{P_{1}}^{P_{2}} \frac{1}{P}dP$$
$$\mu(T,P_{2})-\mu(T,P_{1})=RT \ln \Big{(}\frac {P_{2}}{P_{1}}\Big{)}$$
Let $\mu(T,P_{1})=\mu^{\circ} $,  $P_{1}=P^{\circ}=1bar$, $P_{2}=P$, and $\mu(T,P_{2})=\mu_{i} $,

$$\mu = \mu_{i}+ RT \ln \Big{(}\frac {P}{P^{\circ}}\Big{)}$$

Side note your equation 3 comes from this equation. $\mu_{i}=G_{i}$.

For a mix of gas at low pressure, the activity is equal to the ratio of the partial pressure of the gas over the standard pressure,
$$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}$$

$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$$

I think that's how you prove that relationship. I hope that helped.
$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i} \Rightarrow a_{i}= e^{\frac{\mu_{i}-\mu^{\circ}}{RT}} $$
This relation is trivial and just some algebra. When you are trying prove something its important to start as far back as possible.

Citations:

Chemical Activity

Levine, Ira N. Physical Chemistry. 6th ed. pg 175-176 . New York: McGraw-Hill, 1978. Print.

Notes from my professor (sorry).


Reductions by Na + EtOH According to my chemistry teacher, sodium in ethanol can reduce ketones, aldehydes, acid halides, esters, cyanides, isocyanides, nitro compounds and acid anhydrides.

Now surfing over the web, I could only verify that it reduces, aldehydes, ketones and esters. Can someone please confirm whether Na in EtOH reduces other mentioned compounds also?

$\ce{Na + ROH}$  reduces:

Aldehydes to primary alcohol
Ketones to secondary alcohol
Esters ($\ce{RCOOR'}$) to mixture of two alcohols ($\ce{RCH2OH + R'OH}$)
Cyanides to primary amine
Isocyanides to secondary amine
Alkyl nitro to primary amine
Aryl nitro to aryl amine, can $\ce{LiAlH4}$ do same?

Couldn't verify for Acid anhydrides and for acid halides, but my instinct says it must reduce acyl halides and anhydrides too. As @Waylander states "anhydrides and acyl halides are likely to be converted to ethyl esters by the EtOH solvent and then reduced."


Reduction of acetaldehyde from rectus and sinister faces I am asked to consider the reduction of acetaldehyde from the re and si faces. I am also asked whether the added hydrogen (labeled H* in my picture) is pro-R or pro-S in both cases. In my attached picture I replaced the added hydrogen (H*) with D to determine the absolute configuration.

If I did this correctly, addition of hydrogen from the si face results in a pro-S product with respect to the added hydrogen (H*). And addition of hydrogen from the re face results in a pro-R product with respect to the added hydrogen. Am I correct?

Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):

When we add from the si face, we get an (R) center, and if we add from the re face, it's the other way around (as it should be!) and we have (S) configuration.

Now, what if we reduce "normal" acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-R or pro-S. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.

From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.


How does chromium help stainless steel prevent rusting? Various sources (such as this one) say that stainless steel is stainless because a passivation layer of an oxide is formed, but I haven't found details on why chromium helps create this passivation layer. I can think of a couple possibilities:

Chromium changes the lattice constant of the iron to better match the spacing of the oxide layer.
Chromium migrates to the surface, and with enough Cr a Cr-rich layer forms on the surface which in turn creates the passivation layer.

Are either of these possibilities true, and if not, how does Cr help make the passivation layer?

TL;DR
    Note that the passive layer forms on the surface, there needn't be any change to lattice constant. Chromium needn't migrate , the Cr present on the surface will form the layer to protect it. The key point is how the layer develops from say a single-atom layer of oxide to the usual/maximum width by migration of electron and oxygen in the oxide layer. The process at molecular and atomic level is more complex than you think.

From wikipeddia article on Passivation

  There has been much interest in determining the mechanisms which describe how the thickness of the oxide layer on a material increases with time. Some of the important issues include: the relative volume of the oxide compared to the parent metal, the mechanism by which oxygen diffuses through the metal oxide to the metal - oxide interface and the relative chemical potential for the oxide to form. Boundaries between micro grains, if the oxide layer is crystalline, form an important pathway for oxygen to reach the unoxidized metal below. For this reason, vitreous oxide coatings – which lack grain boundaries – can retard oxidation. The conditions necessary (but not sufficient) for passivation are recorded in Pourbaix diagrams.

  Some corrosion inhibitors help the formation of a passivation layer on the surface of the metals to which they are applied. Some compounds, dissolving in solutions (chromates , molybdates) form non-reactive and low solubility films on metal surfaces.

The mechanism

  All passivating‐type inhibitors are oxidizing agents. However, not all oxidizing agents are inhibitors. A passivating ‐type inhibitor functions by producing local‐action current which anodically polarizes a metal into the passive potential region and thereby provides the means for obtaining a noble mixed potential. This mechanism is independent of whether passivity is caused primarily by oxide or adsorbed films.

  The major factors which determine whether a particular system will exhibit passivity are:
  (a) reversible potential of the redox system created by the oxidizing inhibitor;

  (b) electrochemical parameters for reduction of the inhibitor on the metal surface (exchange current, Tafel slope, and limiting diffusion current), and

  (c) critical anodic current and Flade potential(The potential of a passive metal immediately preceding a final step fall from the passive to the active region.) of the metal. If oxygen is present in the system, some chemicals may function by changing its reduction kinetics.

  It is further shown that the amount of inhibitor found associated with the surface is not necessarily related to adsorption. Data are presented for passivation of stainless steel and titanium. The mechanism is discussed in terms of various oxidizing agents including oxygen, chromate, molybdate, and pertechnetate.

Some theories I could find for passivation are Kinetic Theory of Inhibition and Passivation in Electrochemical Reactions and Scaling theory in a model of corrosion and passivation


Charge, multiplicity, level of theory and coordination coordinates of gold compleaxes Good day,

I need to learn a trick in answering the following questions:

How to come up with cartesian coordinates of a complex say, [Au(CH2)2(S)PPh2]2, to be ran on Gaussian?
The charge and its multiplicity?

Kindly please help.

There are various software to draw cartesian coordinates. Gaussian has its own GaussView, for example. I believe MolView or Avogadro are also popular. You can draw the molecule in any of these, then use the export option to create your coordinate file (export to Gaussian input file or xyz file). Take advantage of the fragment options for things like phenyl ligands. Also, depending on your goals, it's sometimes helpful to align your molecular axis with a principle axis (e.g. z-axis) for easier reading later, if you're using the calculation for more than just basic output parameters.

Furthermore, you can try to find the molecule in an x-ray database and adjust the geometry yourself to approximately match, which can pay dividends later by drastically simplifying the Gaussian calculation, particularly if you are using a larger basis set. You at least want to make sure you're drawing the correct stereoisomer, because the calculation may not correct this on its own.

The charge and multiplicity are manually defined in the input file for Gaussian. I believe you type in the values in the line just above the input coordinates if I recall correctly. You should check the official manual for sample inputs. There's much more syntax to be observed for an input file.


Why are Group 11 elements unreactive? Group 11 metals namely $\ce{Cu, Ag, Au}$ are known for their low reactivities. They are thus rightly called coinage metals and have been historically used to make currency owing to this very property of theirs. My question is, what is the reason behind this behaviour?

Is it because of the stability of fully filled and half filled orbitals? But then the counterexample of $\ce{Cu-Zn}$  presents itself. These are neighbours in the periodic table, and the differentiating electron in each goes completes a full or half-filled orbital, yet $\ce{Zn}$ is quite reactive.

This behaviour can be seen (sort of) between $\ce{Hg-Au}$ as well which differ in one electron and yet are completely different in terms of their properties.

EDIT: The other two questions referred to as duplicates of this one cite different properties of the elements that make them unreactive. Gold has strong $\ce{Au-Au}$ bonds (due to relativistic effects) and $\ce{Zn}$ is more reducing than copper due to factors like hydration energy and the enthalpy of atomization. But this does not answer my question because:

I was looking for the root cause of this behaviour. If gold is stabilized by relativistic effects, why don't these effects operate on other transition metals? If zinc has high hydration enthalpy, what is the reason?

Since all the coinage metals occur in a single group, I really wanted an explanation that could apply to all of them, such as a peculiarity of their electronic configuration that renders them inert to most available reagents.

The first step toward answering this question is understanding the extent to which the statement "Group 11 metals are unreactive" is or isn't true.

As stated in How do plants respond to copper deficiency? Plant Signaling and Behavior vol. 3, pages 231-232:

  The transition metal copper is essential for all organisms.

It is not just that copper is present in every living thing.  Copper is reacting in every living thing. Copper ions cycle between the +1 and +2 oxidation states in Plastocyanin,  Copper/zinc superoxide dismutase and cytochrome oxidase. So as you are saying copper is unreactive, it as reacting in cytochrome oxidase in every mitochondrion of your body, and in Cu/Zn superoxide dismutase in the cytosol of cells through your body.

One way to quantify the reactivity of metal ions in through reduction potential, here listed for reduction of the ion to the zero oxidation state metal.  (limiting to d-block metals at least as unreactive as Cu)

\begin{array}{c @{:} c} \\
\color{blue}{\text{Ion}} & \color{blue}{\text{Reduction potential}} \\\hline
\mathbf{\color{red}{Cu^{2+}}} & 0.3419  \\\hline
\ce{Tc^2+} & 0.400   \\\hline
\ce{Ru^2+} & 0.455  \\\hline
\ce{Cu+} & 0.521 \\ \hline
\ce{Rh+} & 0.600  \\
\ce{Rh^2+} & 0.600  \\
\ce{Rh^3+} & 0.758  \\ \hline
\mathbf{\color{red}{Ag^+}} & 0.7996   \\\hline
\ce{Os^2+} & 0.85  \\\hline
\ce{Hg^2+} & 0.851  \\\hline
\ce{Pd^2+} & 0.951   \\\hline
\ce{Ir^3+} & 1.156  \\\hline
\ce{Pt^2+} & 1.18   \\\hline
\mathbf{\color{red}{Au^{3+}}} & 1.498  \\
\ce{Au+} & 1.692  \\
\end{array}

Looking at this list, we see that there are a substantial number of d-block metals that are less reactive than copper.  A few are less reactive than silver, and gold is the least reactive on the basis standard reduction potential.

The overall trend in d-block metals, is down and to the right in the block is less reactive. The exception is that, for a given row, the group 12 metal is more reactive than the group 11 metal, but due to the formation of a 2+ ion, not a 1+ ion.

Next we should look at ionization energies of d-block metals to see how much of the reduction potential corresponds to properties of an individual atom, versus bulk or solvent effects.

Looking at first-ionization energies, for each row of the d-block, the highest ionization energy is always the group 12 element (Zn, Cd, Hg). This corresponds well with the fact the Zn, Cd and Hg do not form +1 ions.  The group 11 elements have significantly lower first-ionization energies, and do form +1 ions.

Then looking at second-ionization energies, group 11 has the highest second-ionization energies for every row of the d-block!

Much of the reactive/unreactivity of d-block is explained by the increase of nuclear charge to the right and down the d-block.  The complete d-shell of neutral group 11 atoms shields the valence s electron well enough that it may be removed to react.  However, to lose 2 electrons, the group 11 elements would need to lose a d-electron from a completed d-shell, which is difficult.  For group 12, two s valance electrons can be lost, and the complete d-shell maintained.


Finding ions in water samples So for a school chemistry assignment I was given 5 water samples and a bunch of different reagents to help me find out what was in the solutions. I'm  stuck on two.

The first one had a pH of 7 and only reacted with silver nitrate and formed a white precipitate. the second had a pH of 14 and turned brown when silver nitrate was added to it. Does anyone know of any chemicals that would do this?

It's Silver(I) Hydroxide

The solution was very alkaline, as evident by it's pH value of 14. Although  $\ce{AgOH}$ is sparingly soluble, with a $K_{\text{sp}}$ of $\pu{6.8×10^{-9}}$, under alkaline conditions it precipitates as a brown solid.

This is a picture of precipitating silver hydroxide. Is it what you saw as well?


Why do ligands approach along the axes? According to the Crystal Field Theory (CFT), it is given that ligands (their lone pairs) approach the central metal along the axes. When they approach, the degeneracy of the $\mathrm{d}$ orbitals is lost. The explanation:

The $\mathrm{d}_{x^2-y^2}$ and the $\mathrm{d}_{z^2}$ orbitals(the $\mathrm{e_g}$ orbitals) are oriented along the axes while others are oriented in between the axes. In octahedral geometry having complexes, when the ligands approach, they approach along the axes and thus the $\mathrm{e_g}$ orbitals which are along the axes experience more repulsion. Hence, these orbitals increase in energy more than the $\mathrm{t_{2g}}$ orbitals. Thus, the orbitals split in energy due to the approach of ligands.

My question is  why do the  ligands (their lone pairs) have to approach along the axes. They can approach from anywhere.

You basically said it.

  They can approach from anywhere.

Well, I can define the axes to be anywhere.

An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. Hence, the $\mathrm{d}_{z^2}$ orbital does not have the ‘orientation’ you know from pictures — it only gets that once in some way axes are defined. You could now go ahead and define that one of your octahedral ligand approaches the central metal (placed at $(0,0,0)$) from the coordinates $(1,1,1)$. Get out a large sheet of paper and a new pencil and do the maths, calculating what the final orbital distribution will be. The maths will be a lot more complicated, but the result will be the same. Hence, chemists decide to not do complicated maths and just have octahedral ligands approach along the coordinate axes.

Note that in tetrahedral complexes, the ligands would indeed approach from $(1,1,1)$ and the symmetry-equivalent positions of a tetrahedron.


Nonpolar vs Polar Covelant Bonding So I'm having trouble understanding the difference between the two. Looking specifically at water and carbon dioxide, CO2 is non-polar, because it is symmetrical and carbon shares two electrons with each oxygen and vice versa. But doesn't oxygen also share 1 electron with each hydrogen and vice versa? I know water isn't symmetrical, but for a molecule which I haven't seen the shape before how would I figure it out?

Look at each bond separately. In $CO_2$, in each $C=O$ bond, electrons are shifted towards oxygen atom as is has more pulling tendency(electron affinity). Now there is a formation of $\delta+$ charge on carbon and $\delta-$ on oxygen. This forms a dipole and you write it vectorially. In carbon dioxide, the vectors are equal and at $180$ degrees, so they cancel out and net dipole moment is 0, thus nonpolar molecule. Similarly, for the 2 $O-H$ bonds you get a net nonzero dipole moment, leading to a polar molecule.


How can I find the smaller symmetric structure from big crystal unit cell? I have Pyrope (Mg3Al2(SiO4)3) crystal structure, downloaded from materialsproject:
https://materialsproject.org/materials/mp-6073/#

It appeared as triclinic cell, but if you download CIF file and open it from softwares like Avogadro or VESTA, it appeared as a orthogonal simple cubic (a=b=c, alpha = beta = gamma = 90 degree). I also checked that this geometry is the same with what I downloaded from American Mineralogy Material Database.

Now, I wish to run DFT or quantum calculations for them. So, I excluded atoms in periodic boundary, unit cell contains total 160 atoms. I'm afraid this might be too big, might takes too long time to calculate via DFT or any other quantum simulations. All I need to do is just geometry optimizations, but still, 160 atoms seems too big.

Using VESTA or Avogadro or any other software, is this possible to find the symmetric structure inside this or other "big" unit cell, and make a smaller / simpler version of unit cell for faster calculation?

Or, can I manually cut this unit cell in 1/4 size, 40 atom system with 1/2 size for x, y, and z direction, and run the minimization? It seems that the Pyrope unit cell have some sort of repeating structure inside the big unit cell. But I'm not sure if it is "safe" to find symmetry and divide the unit cell manually.

Thank you

The structure contains 4 independent atoms, Mg, Al, Si and O. The space group is a cubic one containing 96 asymmetric units. If you place an atom on a general position, it will appear 96 times in the unit cell. Mg, Al, and Si are located in various special positions, i.e. lie on symmetry elements. Placing the Mg and the Si atom will result in 24 atoms in unit cell, while placing the Al atom in its position (on the origin) will result in 16 atoms in the unit cell (on a three-fold and on the center of inversion).

So you could say that the asymmetric unit contains one oxygen atom, one quarter Mg and Si atom respectively (24/96), and one sixth Al atom (16/96). One of the sites you cited in a previous question gives the fractional coordinates as:

Si: 0.3750 0.0000 0.2500

Al: Origin

Mg: 0.1250 0.0000 0.2500

O: 0.0341 0.0490 0.6530

If you look at the space group, http://img.chem.ucl.ac.uk/sgp/large/230az1.gif,

you can see that Al, Mg, and Si are on special positions (you can either find the locations on the diagram or apply the 96 operations (48 combined with the I-centering) listed on the right to the coordinates and see how many unique locations you get), while oxygen is not.

  can I manually cut this unit cell in 1/4 size, 40 atom system with 1/2 size for x, y, and z direction, and run the minimization?

If you software is aware of crystallographic symmetry, it is a very simple system with 4 degrees of freedom (the oxygen position and the lattice parameter a). Everything else is fixed if you impose the symmetry of garnet.

  I have Pyrope (Mg3Al2(SiO4)3) crystal structure, downloaded from materialsproject: https://materialsproject.org/materials/mp-6073/#. It appeared as triclinic cell...

There are some inconsistencies in the way the site shows the data. The space group is (correctly) given as Ia3d. The 3D image does not show a body-centered cubic cell, but instead the primitive rhombohedral cell (the smallest unit that recreates the entire crystal through three translations). In one iteration ("primitive"), the coordinate file lists the symmetry as 1, applied the Ia3d operators to the atomic positions (maybe to "not to confuse" users unaware of crystal symmetry), and switched to the rhombohedral cell.


Why allene cannot be described with an allyl system? Why is it impossible in allene to have two pi-bonds in the same orientation?
=> Why isn't allene planar? (4 $p$ electrons are in two 2x2 $p$-orbitals)

From my chemical intuition I would guess that there are 4 electrons in a $p$-system with three orbitals, but frankly I don't know how to allocate the lone pair electrons of $\ce{CS2}$ into the $sp$ and $p$ orbitals.
=> Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?

I've tried to rationalize this by allocating electrons to different hybrid orbitals (especially in $\ce{CS2}$ I was not sure how many electrons do we have in the 2x3 $p$-orbitals):

Why isn't allene planar?

Let's start by looking at the structure of allene and seeing what happens if we make it planar.

(image source)

As the above image shows, the 3 carbons in allene are linear (form a straight line) and the terminal hydrogens lie in orthogonal planes (e.g. the planes containing the hydrogens are rotated 90° one from the other as shown in the bottom half of the drawing).  Allene has two orthogonal pi bonds (e.g. the planes of the two pi bonds are also rotated 90° one from the other).  Each one of those pi bonds contains 2 electrons.  The terminal carbons are $\ce{sp^2}$ hybridized, while the central carbon is $\ce{sp}$ hybridized.

Now let's twist one of the terminal methylenes by 90°.  It doesn't matter which one, so let's just say we twist the one on the right, and twist it so that its red p-orbital lobe lines up with the red lobes on the other double bond.  Now our molecule still has the 3 carbons arranged in a line, but all of the atoms - including the hydrogens - now lie in the same plane; we've made our molecule planar.

To accomplish this we broke one pi bond.  In our planar molecule we have 3 pi electrons (2 from the original pi bond on the left and 1 from the terminal methylene p-orbital that we rotated) in the 3 aligned p orbitals that exist in the plane of the screen.  We are also left with 1 electron in the p-orbital on the central carbon that is perpendicular to the screen (the p-orbital that was part of the double bond on the right side of the molecule).  So we took a molecule with two stable double bonds, rotated a p-orbital 90° and wind up with a planar molecule that has 3 aligned p-orbitals with 3 pi electrons and one electron by itself in a single p-orbital.  This is a biradical, an allyl-like (it's just allyl-like because the 3 carbons in our molecule are still linear, whereas the 3 carbons in the allyl system are bent) radical and a radical in a p-orbital.  Converting 2 double bonds to a biradical is energetically uphill, it does not happen at room temperature. Allene prefers to keep its 2 stable double bond structure rather than be a planar biradical molecule because the former geometry is much more stable.

  Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?

It doesn't. Carbon disulfide and carbon dioxide have the same structure and hybridization as allene.  It is just a little harder to notice because we don't have hydrogens on the ends of those molecules to "tip us off" that the two double bonds are orthogonal.

Here is a picture of the orbitals in carbon dioxide (it's exactly the same in carbon disulfide)

(image source)

Just like in allene, we have 3 atoms arranged in a linear fashion and the central carbon is $\ce{sp}$ hybridized; there are two pi bonds that are orthogonal to one another.


How dependent is the molal depression constant on the nature of solute? I've been reading about the molal depression constant (freezing point depression constant) for binary solutions in class.

I encountered a problem (see below) in which the constant $K_f$ was to be evaluated first for a sucrose-water solution ($\Delta T$ and concentration of the solution was given).

Then this value of $K_f$ was used to determine the depression $\Delta T$ of the freezing point of a glucose-water solution when given it's concentration.

This is the problem:

  A $5\%$ solution (by mass) of cane sugar in water has freezing point of $271.00\  K$. Calculate the freezing point of a $5\%$ glucose water if freezing point of pure water is $273.15\ K$.

My confusion: Is $K_f$ solute dependent? If no, then why not?

I have this confusion because I'm used to solving problems in which if the solute is changed, then most of the constants related to various properties of the solution also changes.

Some solutes form nearly ideal solutions up to moderate molalities, examples being glucose and sucrose in water. Such solutions allow fitting of freezing point temperature data to the following equation:

$$\mathrm{log}(1-x_s)=\frac{\Delta_{fus} H_m}{R}\left(\frac{1}{T_{m}}-\frac{1}{T}\right)$$

Taking various approximations which includes assuming that the heat of fusion is constant from $T$ to $T_m$ (the melting point of the pure solvent), that $x_s<<1$, and that $T_m\approx T$ you obtain the expression for the freezing point depression, $\Delta T = T_m - T$, in terms of the cryoscopic constant $K_f$ and solute molality $m_s$:

$$\Delta T = K_f m_s$$

where

$$K_f = \frac{M_wRT_m^2}{\Delta_{fus}H_m}$$

Since $K_f$ contains parameters that depend only  on the solvent (not on the particular solute) the equation can (to within the limitations imposed by the above approximations) be applied to any solutes with which the solvent forms ideal solutions. Which is why you can determine $K_f$ with one solute only to use that same constant to later determine the concentration of another solute.


Why is hydrochloric acid often sold as 33% concentration? What is the historical perspective for the choice of 33% concentration that is often found in commercial sales of Hydrochloric acid?

Is this the same reason we find sulfuric acid sold at 37% and Nitric at 68%?

I am guessing these are some limit at which further concentration, synthesis or purification reaches a chemical or economic limit but am curious to find out about the specific numbers selected.  Obviously, all of these are also sold in higher purities/concentrations as analytical reagents but that obviously takes more effort to produce and will cost more.

The user ssavec prompted me to research further on my own and post an answer.

It seems the limits are practical ones.

For hydrochloric acid — $\ce{HCl}$ a practical limit is 38% with absolute
limit around $40~\%$ and commercial concentrations ranging from 30 to 35%
for convenience in transportation infrastructure.
For nitric acid — $\ce{HNO3}$ the $68~\%$ limit is the maximum that can be
reached by direct distillation, higher concentration up to $98~\%$
(fuming nitric acid) needs alternative dehydration or synthesis techniques.
For sulfuric acid — $\ce{H2SO4}$, which can be distilled to $98.3~\%$ (and can be used
to concentrate nitric acid) it is sold at concentrations that are
directly suited for the intended use ranging from $20~\%$ to $80~\%$. With concentrations around $30~\%$ common for lead-acid battery electrolyte preparation.

So it seems the limits are all practical but for different reasons, one was close to the absolute limit for efficient transport.  The second was at the direct distillation limits and the last one was to suit the most popular consumer applications.

The spread in the numbers had caused me to wonder on their relation but it seems there was no direct relation to the concentrations we see for consumer sale.


How do I predict the solubility of a compound in a mixture of water and an organic co-solvent? I'm trying to predict the solubility of long alcohols (e.g. 1-octanol, 1-decanol or 1-dodecanol) in a mixture of water and an organic co-solvent such as DMSO, acetonitrile or ethyl acetate.

For example, decanol in water with 25% DMSO or saturated with ethyl acetate (8.3%). I of course know the solubility of alcohols in water, but not in organic solvents or in mixtures.

I found a server to predict the solubility of any compound in any solvent, but I'm not sure how accurate it is and it only does predictions for pure solvents, not mixtures: http://showme.physics.drexel.edu/onsc/models/multisolvent.php

Any input on this would be greatly appreciated!

I think this is very much an open problem. As you may see from the Drexel pages, even solubility prediction in one solvent is still fairly hard, partly because of the lack of experimental data.

The only software that I know capable of making predictions about solubility in mixed solvents is COSMOtherm, based on the COSMO solvation model.

I have never used it, but the capability is mentioned on the website.


Why do people often capitalize element names? According to IUPAC rules, names of chemical elements should not be capitalized. See Wikipedia’s take on the issue:

  According to IUPAC, chemical elements are not proper nouns in English; consequently, the full name of an element is not routinely capitalized in English

However, I commonly see people (even in research articles) capitalize the element names, and I wonder why. Is there a historical trend of doing so? Did the rule change at some point? Are element names capitalized in other languages?

This is a Google ngram of the usage of the words: iodine, carbon, nitrogen, zinc and Iodine, Carbon, Nitrogen, Zinc (not at the beginning of the phrase).
There is no historical trend, at least not an usage of an archaic form (we can see that the maximum  usage of the capitalized element name is from 1900 to 1960 but normalized to the overall usage is steady).

We can make some hypothesis about why someone could choose to write capitalized element names:

If we must write element abbreviation capitalized (Zn, Cu, C, N) I  can think I have to write the element capitalized even if is not abbreviated.
Someone (me too) may think that if I write Zinc it is more unambiguous and means I'm talking about the concept of element "all atoms with the same number of protons in the atomic nucleus" and not about the metal zinc.  If you write report with some elemental techniques (XRF, AES) sometimes you feel the urge to specify to the reader (that maybe is not a chemist): "when I write that I've found zinc I mean that in the sample could be present $\ce{Zn}$ but also $\ce{ZnO}$, $\ce{Zn+}$, $\ce{Zn^{2+}}$".
The fact the atoms are identical and the definition of a proper noun is: "a noun that in its primary application refers to a unique entity" could induce to think that elements may fit the definition, and so gain the privilege to be capitalized.
In alchemy elements such water and fire were capitalized.

However up to my knowledge the form not capitalized is preferred even in the others languages, in Italian I'm sure.


Why is sand that insoluble? Why is sand that insoluble?

Sand is a polar molecule but does not dissolve in water.
Is it because of the big structure that sand makes?

What is Sand?

Scope of this Answer

In order to give a good answer, we first have to decide on the type of sand we want to talk about, because its composition is highly dependent on that. Is it beach sand? If so, where from? All these questions should have been addressed in the question. I will limit the scope of this answer to pure silicium dioxide $\ce{SiO2}$ sand.

Is Sand a Molecule?

Sand is, strictly speaking, not a molecule. It's a crystal. As such it has properties such as the lattice energy. Maybe have a look at that link, it might clarify it all for you already.

So now we have talked about what we want to talk about and clarified the basics, so let's dive right in, shall we?

Why does $\ce{SiO2}$ not Dissolve in Water?

Basically this is an energy calculation. On one hand you have the crystal, with its lattice energy, and on the other hand you have the solvated $\ce{SiO2}$ units. A tentative reaction equation would look like this:
$$ \ce{Si_{n} O_{2n} <<=>[E_\text{lat}] SiO2 <=>[E_\text{solv}] SiO2(aq)} $$

The energies I've indicated are the lattice energy $E_\text{lat} = 12535~\mathrm{kJ\,mol^{-1}}$ (Source) and the solvation energy $E_\text{solv}$ (see figure below, where I've also included sodium chloride as a substance to compare with).

Data from here (PDF) ($\ce{SiO2}$) and here ($\ce{NaCl}$). It is obvious that the solvation of $\ce{SiO2}$ monomers is highly unfavored (if I got all the sign conventions right, that is...)

Basically, for ease of understanding, imagine that the $\ce{SiO2}$ unit in the middle has to decide on which side it wants to go. Either it forms a crystal with other units (to the left) or it gets solvated by (surrounded by molecules of) water. Which way will it go? We already know that it will take the route to the left. This means that the whole system can release more energy this way. That is why $\ce{SiO2}$ crystals are stable in water.

Or somewhat. Some theoretical investigations point towards the possibility that the surface of quartz crystals undergo chemisorption of water, thereby generating $\ce{SiOH}$ terminal groups, which would then generate more favorable interactions between the crystal surface and the surrounding water.

Coming back to (a part of) your question:

  Is it because of the big structure that sand makes?

You are kind of on the right track here. It is because of the lower energetical state as a crystal that these big structures (crystals) are formed, vs. the units just being dissolved (solvated by water).


Analysis of post-HF wavefunctions Hartree-Fock method introduces electron (spin)orbitals and they are commonly used for qualitative rationalization of many molecular properties. However, MOs have meaning only if we ignore electron correlation. Post-HF non-DFT methods build correlated wave-function, but due to correlation extracting independent orbitals from them is not possible.
Electron density still is a somewhat informative concept even with no orbitals involved, but we like to talk about $\pi$-systems, sigma-bonds and the like and relate electron transitions to molecular features. So, I'm interested in existing approaches to analysis of post-HF wavefunctions (for example, obtained from Coupled Cluster method).
Are there any well-known approaches to analysis of correlated wavefunctions that do not rely on concept of orbitals that allow to relate some molecular properties to local structural features?

There are! First off, you're correct that Hartree-Fock and DFT produce effective one-particle states $\phi_i$ which are the molecular orbitals (MOs) in the case of HF or the Kohn-Sham (KS) orbitals in the case of DFT (I'll focus on Hartree-Fock since I am more familiar with it than DFT). The MOs are defined via expansion coefficients ($C_{\mu i}$) in the basis set $\{\chi_\mu\}$ such that $\phi_i = \sum_\mu C_{\mu i} \chi_{\mu}$. The more physically meaningful quantity one works with is the one-particle reduced density matrix (1PRDM) defined as $\Gamma_{pq} = \sum_{\mu \nu} C^*_{\mu p} P_{\mu \nu} C_{\nu q}$ where $P_{\mu \nu} = \sum_i C_{\mu i}^*C_{\nu i}$. Diagonalization of the 1PRDM produces what are known as natural atomic orbitals (NAOs) $\phi^{\rm{NAO}}_i = \sum_p a_{p i} \phi_p$ where the coefficients $a_{p i}$ are defined by the eigenvalue relation $\Gamma_{p q} a_{q i} = \rho_i a_{p i}$ where $\rho_i$ is the occupation number corresponding to th $i^{\rm{th}}$ NAO. Since $\Gamma$ is Hermitian, the NAOs are orthonormal and form a perfectly suitable basis to perform any quantum chemistry calculation, however, their value lies in the fact that NAOs actually show you where the electron density is so it's a very useful visualization tool. Once you have NAOs, all of the analysis you want is available. There are things called natural bond orbitals (NBOs) which rotate NAOs in such a way that they maximize the electron density between 2 atoms, thus simulating the notion of a chemical bond. Here, you find the orbitals describing your $\sigma$- and $\pi$-bonds. The point I'm trying to make is that what's important for this kind of analysis is having $\Gamma_{pq}$.
So the answer to your question depends on whether or not you calculate $\Gamma_{pq}$ for a variety of post-HF methods, such as configuration interaction (CI), coupled-cluster (CC), and others. If we denote the Hartree-Fock determinant as $|\Phi\rangle$, the 1PRDM from before is simply $\Gamma_{pq} = \langle \Phi | a^+_p a_q | \Phi \rangle$. The wavefunction originating from post-HF calculations is $|\Psi_\mu\rangle = (1 + C_\mu)|\Phi\rangle$ in the case of CI and $|\Psi_\mu\rangle = R_\mu e^T |\Phi\rangle$ in the case of equation-of-motion (EOM) CC. The analogous expression of the 1PRDM for the $\mu^{\rm{th}}$ excited state ($\mu = 0$ is the ground state) is
$$\Gamma^{\mu}_{pq} = \langle \Psi_\mu | a^+_p a_q | \Psi_\mu \rangle$$
You can calculate this object directly for the various types of wavefunction that come out of correlated electronic structure calculations. It's not light work for sure, but if you know your way around many-body algebra (e.g. Slater's Rules, diagrammatics, Wick's Theorem), you can calculate this object and diagonalize it to find correlated bonding orbitals just as you would in Hartree-Fock. Also note that in the case of CC, the bra state $\langle \Psi_\mu | \neq [|\Psi_\mu\rangle]^+$ and instead we use the biorthogonal left-CC parameterization $\langle \Psi_\mu | = \langle \Phi | L_\mu e^{-T}$ where $L_\mu$ is a linear de-excitation operator complementary to $R_\mu$. These and other technicalities are of course well-documented in the CC literature. The key observation here is that we never explicitly build a correlated $|\Psi_\mu\rangle$ defined via Slater determinants directly. In the case of CC wavefunctions, you'd blow up your computer since $e^T$ is an object as large as the full CI wavefunction for any level of truncation. So the way things are done is you obtain $|\Psi_\mu\rangle$ indirectly though a set of CI coefficients or CC cluster amplitudes and calculate the 1PRDM, again, completely defined through this sequence of finite coefficients. It may still be a single-particle orbital, but it has many-body information embedded within it.
That being said, these calculations are available in most electronic structure programs.


why do we consider CuO as a base? According to Arrhenius acid/base theory, a base is a substance which when dissolved in water furnishes $\ce{OH-}$.

But $\ce{CuO}$ is insoluble in water and does not produce $\ce{OH-}$, so why do we consider it a base?

There are two other widely used acid/base theories that apply here: Brønsted–Lowry theory and Lewis theory.

A Brønsted base is defined as a proton acceptor. Technically, $\ce{CuO}$ is capable of reacting with $\ce{H+}$ to form hydroxide ions and eventually $\ce{H2O}$. Although $\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\ce{CuO}$ is present as an undissolved solid).

A Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\ce{CuO}$ has lone pairs of electrons available.

$\ce{CuO}$ is considered a base because it fits these two descriptions.


Assay over 100% (L-Cysteine Hydrochloride Monohydrate) In reading a COA (Certificate of Analysis), I came across an Assay line:

The Assay is 100.5% where "Lower Limit" is 98.5 and "Upper Limit" is 101.0

This product is L-Cysteine Hydrochloride Monohydrate.  How can the Assay be over 100%?

I have found this statement: "If the substance under test is purer than the standard, the assay comes more than 100 pct."  If that is the case, then the question becomes, where can I find out about this standard?

There is a simple reason to have the purity greater than 100% for this compound.

If the substance was exposed to a dry environment for several hours, a small amount of the  water of hydration could be lost, causing the calculation to have a higher purity.

Also, I believe that the purity is calculated from the perchlorate titration of the substance in acetic acid (I will need to check the USP monograph). There are many sources for error in the calculations: the weight of the L-Cysteine Hydrochloride Monohydrate, the volume of the perchlorate solution, the error from the standardization of the perchlorate solution, etc.

Under the right conditions, you can easily get 101% for the purity.


Chemoselectivity in synthesis of paracetamol In the reaction below, why does acetic anhydride not react with the -OH group to form an ester instead?

The reaction is likely with one equivalent of acetic anhydride and under thermodynamic control.

Kinetically, the nucleophilic attack of phenol on acetic anhydride should be a lot faster than that of aniline. Aniline’s lone pair is taking part in resonance across the phenyl ring so it is barely able to attack. Phenol has two lone pairs on oxygen, though, of which only one is able to resonate with the aromatic system. Under kinetic control I would thus expect acylation on oxygen.

However, there is a second process which renders us thermodynamic control: The initial product can react with a second molecule in a transacylation reaction.

$$\ce{RHN-C6H4-OAc + H2N-C6H4-OR <=>> RHN-C6H4-OH + HAcN-C6H4-OR}$$

Both the oxygen of a free phenol and the nitrogen of a free aniline (the latter slower, but still) can attack a corresponding acylated atom. In overall thermodynamic stability, the acetylamine is more stable than the acetate (and is also kinetically more inert) so after equilibration has taken place the acetamide will prevail.


"Nitrogen (gas) is physisorbed over iron (metal) at room temperature": Is this for real? My copy (PDF) of Levine's Physical Chemistry (6th ed.), page 570-571, section: "Adsorption of Gases on Solids", states:

  Physical adsorption is nonspecific. For example, $\ce{N2}$ will be physically adsorbed on any solid provided the temperature is low enough.

and

  ... N2 is chemisorbed at room temperature on $\ce{Fe, W, Ca,}$ and $\ce{Ti}$ but not on $\ce{Ni, Ag, Cu,}$ or $\ce{Pb}$.

Alright, fair enough, thanks for the examples.

But

A bunch of (notoriously error-prone-citation-deficient) course books issued by my school, all unambiguously concur on the contrary:

  At $\ce{83K}$, $\ce{N2}$ is physically adsorbed over iron metal. Raising the temperature to $\ce{773K}$ then results in the nitrogen being chemisorbed over the metal as atomic nitrogen.

Which in other words is: "At room temperature, nitrogen is physically adsorbed. The nitrogen is chemisorbed after the temperature's raised to 500 degrees Celsius".

But this contradicts the information in Levine's book.

Who's right here: Levine or my book? (References are welcome) O:)

EDIT-

LinearChristmas has correctly pointed out that the temperatures at which the physical adsorption --> chemisorption transition takes place depends on pressure as well as the nature of the iron surface.

In light of this, I'll modify my question:

Q- My coursebook and Levine's lists (implicitly in Levine's case) the transition from physical adsorption --> chemisorption at different temperatures. Between the two, which one lists the more improbable temperatures?

I think you are misinterpreting the course book issued by your school since it never mentions what happens at room temperature.
For me, none of the books is (seems) wrong, because they do not contradict each other. I will mention each affirmation including the textbook which said it (Levine, or Your book):

Levine says: N2 is physically adsorbed on any solid at low temperature. This agrees with your text saying N2 is physically adsorbed at $\pu{83 K}$ on Fe.
Levine says: N2 is chemisorbed at room temperature on Fe. Your book does not mention about room temperature. No contradiction here.
Your book says N2 is chemisorbed at $\pu{773 K}$. Levine does not mention this temperature, but if it already occurred at room temperature it will also occur at higher.

The only 'gray area' is at which temperature does N2 get chemisorbed (room temperature or not?). In the article of de Boer et al. mentioned by @Linear Christmas above, they measure chemisorption of N2 even at $\pu{23.9 ^\circ C}$.
They clearly state:

Hitherto it has been reported in literature$^{1,2}$ that an iron catalyst starts to chemisorb nitrogen at a measurable rate at about $\pu{200 ^\circ C}$. The chemisorption activity of our intensely reduced sample was considerably higher; already at $\pu{0 ^\circ C}$, nitrogen was chemisorbed at a measurable rate.

Reference:
J. J. F. Scholten, P. Zwietering, J. A. Konvalinka, J. H. de Boer, "Chemisorption of nitrogen on iron catalysts in connection with ammonia synthesis. Part 1.—The kinetics of the adsorption and desorption of nitrogen," Trans. Faraday Soc. 1959, 55, 2166-2179 (https://doi.org/10.1039/TF9595502166).


Methylammonium lead(II) iodide perovskites in bulk form I am trying to obtain $\ce{(MA)PbI3}$ as a bulk powder using lead(II) iodide and methylammonium iodide as precursors.

I have not yet tried simply mixing the two in methanol/ethanol and removing the solvent in a rotavap, but this will simply result in a mixed salt with very little actual perovskite crystals if I am not mistaken.

So, how could I obtain $\ce{(MA)PbI3}$ in the highest purity possible using lead iodide and methylammonium iodide?

Procedure from the paper by Xie et al. [1] (Powder XRD didn't reveal any other impurity phases in the bulk samples):

Synthesis of $\ce{CH3NH3PbI3}$. The as-prepared $\ce{CH3NH3I}$ and $\ce{PbI2}$ (Sigma-Aldrich, 5N) were mixed in stoichiometric ratio and dissolved in γ-butyrolactone (Sigma-Aldrich), and then a 40wt% solution was formed. The solvent of as-prepared solution was evaporated in an oven at 60 °C for 2 days and $\ce{CH3NH3PbI3}$ precipitate was collected.

Similar approach has been utilized by Yamada et al. [2] for the synthesis of monocrystalline samples. From the SI:

Sample Preparation
$\ce{CH3NH3PbI3}$ $(\ce{MAPbI3})$ single crystals were prepared from γ-butyrolactone solution at 100 °C. Purified lead iodide (2.73 g, 5.92 mmol) and methylammonium iodide (0.93 g, 5.84 mmol) were dissolved in dry
γ-butyrolactone (4.8 mL) at 60 °C by stirring. The resulting solution was poured into a screw-cap vial which was placed on a hot plate at 100 °C for three days to produce a single crystal with dimensions of around 5 mm x 5 mm x 5 mm. We obtained a clean surface by cleaving the single crystal with thin knife under an Ar atmosphere before being used for the measurements. The experiments were conducted in an Ar atmosphere. We confirmed that the same results could be obtained with multiple samples that were cleaved and kept in Ar atmosphere.

Alternative method by Hoke et al. [3]; from the SI:

Crystals of $\ce{(MA)PbBr3}$ (orange, rhombic dodecahedra) were first
grown by combining stoichiometric ratios of $\ce{PbBr2}$ and $\ce{(MA)Br}$ in concentrated (9 M) hydrobromic acid. Diffusion of acetone into this solution afforded crystals within 2 days, which were filtered and dried under reduced pressure.

It appears that methods described in [1] and [2] are based on the preparation of the solution of $\ce{MAPbI3}$ for the thin film deposition on $\ce{TiO2}$ layer, which was published earlier by Noh et al. [4] (SI).
References

Xie, J.; Liu, Y.; Liu, J.; Lei, L.; Gao, Q.; Li, J.; Yang, S. Study on the Correlations between the Structure and Photoelectric Properties of $\ce{CH3NH3PbI3}$ Perovskite Light-Harvesting Material. Journal of Power Sources 2015, 285, 349–353. https://doi.org/10/f7c87m.
Yamada, Y.; Yamada, T.; Phuong, L. Q.; Maruyama, N.; Nishimura, H.; Wakamiya, A.; Murata, Y.; Kanemitsu, Y. Dynamic Optical Properties of $\ce{CH3NH3PbI3}$ Single Crystals As Revealed by One- and Two-Photon Excited Photoluminescence Measurements. J. Am. Chem. Soc. 2015, 137 (33), 10456–10459. https://doi.org/10/f7p5s9.
Hoke, E. T.; Slotcavage, D. J.; Dohner, E. R.; Bowring, A. R.; Karunadasa, H. I.; McGehee, M. D. Reversible Photo-Induced Trap Formation in Mixed-Halide Hybrid Perovskites for Photovoltaics. Chem. Sci. 2015, 6 (1), 613–617. https://doi.org/10/gc3g49.
Noh, J. H.; Im, S. H.; Heo, J. H.; Mandal, T. N.; Seok, S. I. Chemical Management for Colorful, Efficient, and Stable Inorganic–Organic Hybrid Nanostructured Solar Cells. Nano Lett. 2013, 13 (4), 1764–1769. https://doi.org/10/gc3g8w.


If Beer's Law appears to apply to a high concentration (>0.01M) of a solute, is it valid to use for concentration calculation? Internet consensus seems to define "high concentration" for Beer's Law as >0.01M. If one has a solution at higher concentrations than that, but the absorbance vs. concentration is still linear, is it valid to use Beer's Law to calculate concentration from absorbance data (assuming that the concentration is less than the highest point for which one has experimental data)?

Internet consensus seems to define "high concentration" for Beer's Law
  as >0.01M

Keep in mind that Beer's law is an approximation. Look at the more advanced version. Beer's law-advanced version. The statement that a concentration of 0.01 M is the upper limit of Beer's law is incorrect due to several reasons. It is the absorbance value which we should worry about rather than the concentration. One can have a solution which is 0.01 M yet its extinction coefficient is not that large at that particular wavelength, we would still be fine.

This trick is used in atomic absorption spectroscopy, where you choose wavelength, which is not a resonance line of that particular atom. As a result, one can use relatively high concentration without getting a non-linear curve.

Secondly, chemists tend to infatuate little bit with linear calibrations. It is okay to have a non-linear curve. Quadratic equations can be used to fit the empirical curve and one can get accurate analytical results.


Why do H₂ and O₂ create H₂O? Both $\ce{H2}$ or $\ce{O2}$ are stable molecules. However when given energy they heat up and form $\ce{H2O}$. My teacher explained that the increased energy breaks the bonds, allowing new ones to form. However, why don't the old forms reform instead? Why are new bonds created?

Are you familiar with activation energy?  Consider a hole at the top of a large hill.  If a ball or some water was in that hole it would seek the lowest point it could find.  Clearly, the bottom of the hill would be preferable to the bottom of the hole at the top of the hill, however, in order to get there it must first be brought up over the top of the hole.

In this picture, the $H_2$ and $O_2$ are on the right side of the curve.  Just like water in the above analogy, they'd like to move to the lower area, but first must be brought up over the bump.  So some energy must be added to accomplish this, and then they end up in the much lower energy position on the left.

Once over the bump, they stay as $H_2O$ because that is the state they'd like to be in at any time, but were prevented from being in before by the activation energy bump.


Soil-solution cations in competition for exchange sites How could one determine the likelihood of one cation displacing another on a negatively charged exchange site of a clay particle?

For example, in a soil solution containing equal parts (by number of atoms) $\ce{Ca^2+}$, $\ce{Mg^2+}$, $\ce{K+}$, $\ce{Na+}$, $\ce{Al^3+}$, which cations will find the most exchange sites and which will ultimately be washed deeper into the soil? (Assuming there are not enough sites for all of them.)

Could two $\ce{Na+}$ cations displace an adsorbed $\ce{Ca^2+}$ cation simply because they "bump" into it? Or is there some mechanism where cations are dislodged? Is it easier for two $\ce{Na+}$ to replace $\ce{Ca^2+}$ or for $\ce{Ca^2+}$ to replace two $\ce{Na+}$?

Only half as many Ca2+ are needed as Na+ to neutralize the charge of the ion exchanger (here: clay).

The affinity of clay for bivalent ions is larger than that for monovalent ions:
Ca2+ > Na+

For most clays, the affinity for equally charged ions increases with the nonhydrated radius of the ions:
Cs+ > K+ > Na+ > Li+
Ba2+ > Sr2+ > Ca2+ > Mg2+


Why is there a subtractive volume correction in the van der Waals' equation for real gases? Why there is a subtractive volume correction in van der Waals' equation for real gases, both real and ideal gases occupy volume of container in which they are kept, so there shouldn't be any correction?

The van der Waals equation is an attempt to explain the pressure-volume relationship for real gases. For example, what will the pressure be of a given amount of gas in a vessel of a fixed volume.

In ideal gases the size and interactions between the atoms or molecules of the gas don't matter at all. We think of them as point-like entities that don't interact except to bounce elastically off each other and off the walls of the vessel. Their behaviour is described by the ideal gas equation and doesn't depend at all on the nature of the particles making up the gas.

Real molecules and atoms deviate from this ideal behaviour because the particles of the gas both have finite volume and attractive interactions which mean they don't behave exactly as the ideal gas equation describes. The van der Waals equation is a simple attempt to explain those deviations from ideality. The volume correction is one of those factors. It takes into account the fact that real gases are made from particles that actually have a finite size and are not mathematical points. The factor is about the volume of the particles of the gas not about the vessel the gas is in. The result of the correction is that the pressure of a given amount of gas in a vessel of fixed size is slightly different than would be expected from the ideal gas equation.


Does the rate constant depend on pressure? I have read in some books that when pressure increases, the number of effective collisions increases. Hence, $k$ depends on pressure. Is it true? If so, then why doesn't $k$ depends on concentration? The number of effective collisions should increase here also.

According to hard-sphere collision theory of gas-phase reactions, the rate constant ($k$) for the elementary bimolecular reaction $\ce{B + C -> products}$ is
$$ k = N_\text{A} \pi (r_\text{B} + r_\text{C})^2 \left[ \frac{8RT}{\pi}\left( \frac{1}{M_\text{B}} + \frac{1}{M_\text{C}}\right)\right]^{1/2}e^{-E_\text{thr}/RT}$$
where $N_\text{A}$ is the Avogadro constant, $r_\text{B}$ and $r_\text{C}$ are the radii of the hard-spheres, $R$ is the ideal gas constant, $M_\text{B}$ and $M_\text{C}$ are the molar masses of B and C, and $E_\text{thr}$ is the threshold energy on a per-mole basis. Therefore, no pressure dependence.

For nonideal systems, the rate law for the same elementary reaction can be written as
$$ r = -\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = k_r [\text{B}][\text{C}]$$
The Brønsted–Bjerrum equation relates the apparent rate constant, $k_r$ with the infinite-dilution rate constant, $k_r^\infty$, and the activity coefficients of the envolved species:
$$  k_r = \left( \frac{\gamma_\text{B}\gamma_\text{C}}{\gamma^\ddagger}\right)k_r^\infty$$
where $\gamma_\text{B}$, $\gamma_\text{C}$, and $\gamma^\ddagger$ are the concentration-scale activity coefficients of B, C, and the transition state respectively. These quantities depend on temperature, pressure, and on the concentration of all species on the same phase. For this reason, the rate constant of nonideal elementary reactions can have a dependence on the pressure.


Does one atom of an element show the properties of a bulk of that element? For example, does one atom of gold have a golden color?

Molecular properties such as melting point, conductivity, elasticity all require a multiplicity of atoms or molecules to be observed since each of those properties are defined interactions between the same particles. Color is different - when a photon hits an atom the atom's electrons are excited, the electron will fall back to it's unexcited (ground) state and when it does it emits a photon.  This photon will not always have the same energy as the photon absorbed, when this is the case the rest of the energy was converted into thermodynamic energy (heat). Since the color of a wave of light can be defined by its energy then atoms can have "colors". This is a bit of an oversimplification but I hope it helps!

Edit: as @alchimista pointed out the color of a single atom will not likely represent the color of a sample of that element large enough to see. Also the light reflected from most atoms will likely cover the entire visible spectrum which will result in the same color of light being reflected as was used to illuminate it.


Why don't alkaline earth metals lose only one electron when they are ionized? Why doesn't alkaline earth metals lose only one electron when they are ionized ? I know that magnesium atoms like to have the electronic configuration of neon but I don't understand the reason that we can't find $\ce{Mg+}$ atoms .

Magnesium can definitely exist as $\ce{Mg^{+}}$. Hence the availability of first ionization energy data for magnesium.

The same goes for other metals; there are only uncommon valences, such as $\ce{Na^{3+}}$.

If you had enough energy, you could make $\ce{Na^{4+}}$ if you wanted to. It would just cost, according to the table below, 9543 + 6912 + 4562.4 + 495.8 kJ/mol.

I think your question is why isn't $\ce{Mg^{+}}$ more commonly encountered. That has to do with stability. Losing two electrons and becoming isoelectronic with a noble gas is more stable than having an unpaired electron. An unpaired electron is available for bond formation, and bond formation tends to be exothermic.

If you want to read up on the +1 valence of magnesium, such as in $\ce{MgH}$, you can read more here:

http://bernath.uwaterloo.ca/media/24.pdf


Why do we need to incubate this system for 18 hours at 60 degrees Celsius? I am conducting a bio-chemistry related experiment and I have been unable to understand a step which is commonly performed. The procedure that is confusing me is outlined in the supporting information, beginning on page S2.

My aim in this step is to apply a PEG (Polyethylene glycol) silane layer.

After immersing ITO (Indium Tin Oxide) slides in a PEG-concentration, the slides are incubated for 18 hours at a temperature of 60 degrees Celsius.

Can you tell me why the incubation is performed? And why is it for 18 hours at 60 degrees Celsius?

Until you post the original source procedure, my answer is going to be generic.

  When I am repeating an experiment I found in the literature, why do I need to do [insert step that seems unnecessary, difficult, or just plain long]?

The short answer is, because it works.

You need to incubate the slides at 60 oC for 18 hours because that's what the author of the paper found to be necessary to guarantee that the slides were properly PEGylated. It's possible that the author of your procedure is also using a procedure that he or she found in the literature, and the the rationale for the specific incubation procedure will require some digging.

The longer answer is, the reaction probably needs some incubation to run to completion. The reaction between the PEG and the ITO surface is not instantaneous. If you want the coating to be good, you need to give it time to cure, and mild heating can accelerate the process. While 18 hours seems like a long time, it would likely take interminably longer at room temperature. Could you heat at a higher temperature for less time? Probably, but I would wait until you are certain you can make the existing procedure work before messing with it. That way you have some successfully coated slides to compare to.

The 18 hour period has one of two origins. The second is more likely.

The developer of the procedure maximized the efficiency of the process: He/she carefully experimented with the conditions until he/she discovered the exact time/temperature combination that maximized the efficiency of the reaction - getting the best coating in the shortest time.
The developer of the procedure maximized the convenience of the process: He/she determined 1) that the coated slide needed to cure for several hours (3 or 4) at 60 oC,  2) that higher temperatures were detrimental to the quality and integrity of the coating, and 3) prolonged heating at 60 oC is not detrimental. In this case, the researcher puts the slides into the oven in the afternoon to cure overnight and picks up the cured slides the next morning. 4 PM on Monday afternoon to 10 AM on Tuesday morning is 18 hours. As a synthetic chemist, I often wrote down '18 hours', when I really meant 'overnight'. The reaction may have been done at 6 hours or 8 hours, but 18 hours is more convenient. I set up the procedure one afternoon a few hours before I left for the day so that I could make sure it was stable. I worked it up the next morning.