Why rutile structure have primitive unit cell instead of body centered? In the literature (Wikipedia), I read that $\ce{MgH2}$ have a structure of rutile. Then, I looked at its space group which is $P4_2/mnm$. From this notation, I understood that it possesses a (tetragonal) primitive unit cell. However, when I looked at the structure of $\ce{MgH2, Mg^2+}$ ions are seemed to be ordered in body centered. Indeed, at each corner there are $\ce{8 Mg^2+}$ ions and in the middle, there is another one. Then, why is this called $P4_2/mnm$ given that it resembles more to body centered? Is there any smaller unit cell hidden here?

Thank you in advance.

You said you read that $\ce{MgH2}$ has a structure of rutile. Actually it is $\alpha$-$\ce{MgH2}$, which forms at room temperature and atmospheric pressure and have rutile type structure. The crystal structure of $\alpha$-$\ce{MgD2}$ was studied as early as 1963 (Ref.1). The abstract states that:

  A magnesium deuteride preparation of composition $\ce{Mg(D_{0.9}H_{0.1})2}$ is found to be tetragonal with $a = 4.5025, c = \pu{3.0123 \mathring A }$. The atomic positions are: $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$, $\ce{4 D + H}$ in $\pm (xx0)(\frac12 +x, \frac12 –x, \frac12)$ with $x = 0.306 \pm 0.003$. The bond lengths are $\ce{Mg-6D} = \pu{1.95 \pm 0.02 \mathring A }$. The structure is of rutile type.

The authors state that the choice of $\ce{MgD2}$ in place of $\ce{MgH2}$ because hydrogen atoms had no measurable effects on diffraction intensities (note that they were doing neutron diffraction studies). These studies were a followup to confirm their previous studied on $\ce{MgH2}$ (Ellinger, et al. in 1955; Ref.2). In that studies on $\ce{MgH2}$ with X-ray diffraction, it has been shown that $\ce{MgH2}$ crystallizes in the tetragonal system with $a_\circ = \pu{4.5168 \mathring A}$ and $c_\circ = \pu{3.0205 \mathring A}$. The measured density of the crystals was $\pu{1.45 \pm 0.03 g/cm-3}$, which has shown that there are two molecules in the unit cell (Ref.2). The calculated X-ray density is $\pu{1.419 g/cm-3}$.

The last statement answers your question. The rutile type structure provide two $\ce{MgH2}$ molecules in the unit cell. There are one $\ce{Mg}$ atom in the center (shares with one unit cell) and eight other $\ce{Mg}$ atoms at corners, each of which shares it with eight unit cells. Thus, number of $\ce{Mg}$ atoms in a unit cell is: $1 \times 1 + 8 \times \frac {1}{8}=2$. There are six $\ce{H}$ atoms in the unit cell. If you look closely, you'd realize two of them in inside the unit cell (shares with one unit cell) and rest are on two opposite surfaces, two in each surface (two $\ce{H}$ atoms in each surface share with two unit cells). Thus, number of $\ce{H}$ atoms in a unit cell is: $2 \times 1 + 4 \times \frac {1}{2}=4$. Hence, number of $\ce{MgH2}$ molecules in a unit cell is $\ce{Mg2H4 # 2 MgH2}$.

  The X-ray powder pattern shows that two $\ce{Mg}$ atoms in the unit cell are in a body-centered configuration, i.e. $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$. A consideration of all tetragonal space groups shows that there is one and only one possible choice of positions for the four $\ce{H}$ atoms leading to a plausible structure. This is,

  $$\text{space group }  P4/mnm \ \left(D^{14}_{4h} \right)\\
\ce{2Mg} \text { in } (000)(\frac12 \frac12 \frac12)\\
\ce{4H} \text { in } \pm (XX0)\left(X + \frac12, \frac12-X, \frac12\right) \ \text{with } X=0.306$$

  In this structure (rutile type), each magnesium is coordinated to six hydrogens at a distance of $\pu{1.95 \mathring A}$ and each hydrogen is coordinated to three magnesiums. One $\ce{H-H}$ distance is $\pu{2.49 \mathring A}$ and the others are $\pu{2.76 \mathring A}$. The distance of $\pu{2.76 \mathring A}$ compares favorably with the diameter of $\ce{H-}$ ion ($\pu{2.72 \mathring A}$) as found in $\ce{LiH}$. One short $\ce{H-H}$ distance is characteristic of one anion-anion distance of the rutile type structure.

By these statements, it can be concluded that the structure of $\alpha$-$\ce{MgH2}$ is indeed rutile type (not body-centered cube).

In another point of view, one would also suggest that the structure of $\alpha$-$\ce{MgH2}$ is face-centered cube, based on equal distances of $\ce{6 Mg-H}$ (see $\color{blue}{\text{blue box}}$  in following diagram; opposite face $\ce{H}$ atoms are highlighted in blue, red, and gray for visualization). However, again, calculations lead to $\ce{MgH3}$ in the unit cell, which would rule out the possibility.

The above diagram shows the crystal structures of other modifications of $\ce{MgH2}$ along with $\alpha$-$\ce{MgH2}$ (Ref.3 & 4). These other modifications are formed in high pressure situations. For example, when low-pressure phase $\alpha$-$\ce{MgH2}$ was heated at $\pu{1070 K}$ and in $\pu{2 GPa}$ pressure, the high-pressure phase $\gamma$-$\ce{MgH2}$ was formed in the orthorhombic (space group: $Pbcn$) structure (Ref.3):

  The high-pressure phase $\gamma$-$\ce{MgH2}$ was formed by heating the low-pressure phase $\alpha$-$\ce{MgH2}$ in a multianvil press at $\pu{2 GPa}$ pressure to $\pu{1070 K}$ for $\pu{120 min}$ and successive rapid quenching. Investigation by X-ray and neutron powder diffraction on the deuteride at ambient conditions revealed that it crystallises with the orthorhombic $\alpha$-$\ce{PbO2}$ type structure (space group $Pbcn, Z=4, a=4.5213(3), b=5.4382(3), c= \pu{4.9337(3) \mathring A }$ (hydride); $a=4.5056(3), b=5.4212(3), c= \pu{4.9183(3) \mathring A}$ (deuteride) at $T=\pu{295 K}$). The deuterium atoms surround magnesium in a distorted octahedral configuration with bond distances $\ce{Mg-D} = 1.915(3), 1.943(3)$ and $\pu{2.004(3)  \mathring A }$. The rutile structure of $\alpha$-$\ce{MgH2}$ was re-evaluated.

The bonding nature of $\ce{MgH2}$ is analyzed in recent studies with the help of charge-density, charge-transfer, electron-localization-function, and Mulliken-population analyses, which clearly have shown that all polymorphs of $\ce{MgH2}$  are to be classified as ionic materials with $\ce{Mg}$ and $\ce{H}$ in nearly 2+ and 1−  states, respectively (Ref.4).

References:

W. H. Zachariasen, C. E. Holley, Jr., J. F. Stamper, Jr., “Neutron diffraction study of magnesium deuteride ,” Acta Chrystallographica 1963, 16(5), 352-353 (https://doi.org/10.1107/S0365110X63000967).
F. H. Ellinger, C. E. Holley, Jr., B. B. McInteer, D. Pavone, R. M. Potter, E. Staritzky, W. H. Zachariasen, “The Preparation and Some Properties of Magnesium Hydride,” J. Am. Chem. Soc. 1955, 77(9), 2647–2648 (https://doi.org/10.1021/ja01614a094).
M. Bortz, B. Bertheville, G. Böttger, K. Yvon, “Structure of the high pressure phase $\gamma$-$\ce{MgH2}$ by neutron powder diffraction,” Journal of Alloys and Compounds 1999, 287(1-2), L4-L6 (https://doi.org/10.1016/S0925-8388(99)00028-6).
P. Vajeeston, P. Ravindran, B. C. Hauback, H. Fjellvåg, A. Kjekshus, S. Furuseth, M. Hanfland, “Structural stability and pressure-induced phase transitions in $\ce{MgH2}$,” Phys. Rev. 2006, B73, 224102 (https://doi.org/10.1103/PhysRevB.73.224102).


Is it possible to quantify the effects of a chelating agent? If one wants to see what compound is a more effective chelating agent, how does that get measured?

Generally, a good starting point is the Formation Constant (β) between the metal and the ligand. The greater the constant, the better the ligand is be at chelating the metal center (as evidenced by the free energy change).


What's the difference between a tetrahedron and a trigonal pyramid? The shapes respectively denoted by 'tetrahedron' and 'trigonal pyramid' seem to be the same. Is there a difference between the two? If not, why are the two presented as different concepts in introductory chemistry courses?

In a rigorous geometrical sense, there is no difference between tetrahedron and trigonal pyramid--the terms both mean the same thing. In colloquial and chemical use, however, 'tetrahedron' typically implies the 'regular tetrahedron', where all four faces are equilateral triangles.

Chemically speaking, when referring to these two shapes as descriptors of molecular geometries, there is (usually) a central atom in addition to the four atoms at the vertices.  In the symmetrically-proper regular-tetrahedral geometry, the central atom is inside the solid volume of the tetrahedron and equidistant from all of the four vertex atoms.  In the trigonal pyramidal geometry, the central atom can be located inside the solid volume, contained within a plane of the trigonal pyramid/tetrahedron, or I believe even outside the tetrahedral volume.


Can insect repellents kill insects? One day, there was a fly in my house. I really hate insects flying around me, but I don't want to catch it with my hands, so I sprayed some insect repellent directly onto the fly to try to get rid of it.
This is the insect repellent that I used.

Sticker reads:

"Wirkstoff basierend auf Zitronen-eucaluptusöl"

"Active ingredient based on lemon-eucalyptus oil"

When the insect repellent hits the fly, it slowed down and stopped flying, but it's still moving slowly on the window, so I applied the insect repellent again.
Eventually, the fly completely stopped moving, so I assume it's dead. Does the insect repellent actually killed the fly, or just stunned it?

This Wikipedia article on "Insect repellent" pretty much answers your question.

An insect repellent isn't necessarily the same as an insecticide.

The repellent, well, "repels" insects; i.e- it discourages them from approaching you (or whatever you sprayed the repellent on) for as long as the chemical(s) linger. These (particularly those based on essential oils) aren't much of a health hazard to other animals.

An insecticide is actually designed to kill insects ("No sh!t Sherlock") rather than simply keeping them at bay. The term "insecticide" refers to substances that kill insects/arthropods at very low concentrations ( "very low" is subjective... I don't think there's an authoritative, precise definition for "insecticides" anyway).

FYI, any substance (including common stuff like air and  water), supplied in excess (again, "excess" is also highly subjective) can kill. Ever heard of "Too much of anything can kill"?

The opposite effect is something called "Hormesis"; where just the right amount of anything (including radiation and toxins) can have a positive effect on you.

Back to question.

The insect repellent you're using contains (according to the sticker on the cap) lemon-eucalyptus oil, whose active ingredient is p-menthane-3,8-diol (and its geometrical isomers). Also, it's supposed to be used against mosquitoes (Mücken) and ticks (Zecken).

It serves as a deterrent, on account of how strong it smells.

Ectoparasitic insects, such as the said mosquitoes and ticks, rely on (to a large extent) their sense of "smell" to locate hosts. So spraying around strong odorants will do a good job of interfering with the bug's ability to track you down.

Remember my "anything can kill" rant from earlier? Well, the same goes for your repellent. If I were a mosquito or a tick (or whatever insect for that matter), I doubt I'd make it out alive if you decided to smother me with some godforsaken liquid/aerosol. For all I know, your fly could've died because it suffocated, rather than being poisoned.

Heck, you could achieve the same effect if you decided to drown the fly in deodorant or even perfume.

So there isn't really much behind the mysterious death of your fly; in all likelihood it suffocated (You murderer!). Take @Mithoron's advice; a fly swatter works better.

If you want to know what biochemical processes the repellent would tamper with in a bug, then that's a question for the Biology.SE. ;)


Photoionisation microscopy of hydrogen - where are the p orbitals? In this Physics Review Letters article, which was published in 2013, but I found recently, the authors report photoionisation microscopy images of hydrogen atoms in various electronic states.

  A. S. Stodolna, A. Rouzée, F. Lépine, S. Cohen, F. Robicheaux, A. Gijsbertsen, J. H. Jungmann, C. Bordas, and M. J. J. Vrakking, "Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States,"
  Phys. Rev. Lett., 2013, 110, 213001; url: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.213001

Here are some non-paywall summaries of this work:

physicsworld.com http://physicsworld.com/cws/article/news/2013/may/23/quantum-microscope-peers-into-the-hydrogen-atom
APS Physics http://physics.aps.org/articles/v6/58
io9 http://io9.gizmodo.com/the-first-image-ever-of-a-hydrogen-atoms-orbital-struc-509684901

The imagery is compelling:

I see images that look like $1s,\ 2s,\ 3s,\ \&\ 4s$ orbitals with appropriate number of nodes.

The solutions to the Schrodinger equation for hydrogen-like orbitals include wavefunctions corresponding to $p$ oribtals. Why did the authors not observe images that appear to look more like the $p$-orbitals?

These states aren't 1s, 2s, 3s, 4s, like you suggest. These are actually highly excited states of hydrogen. Note the negative energies in wavenumbers (cm$^{-1}$) in the rightmost column of the figure you provide. This is how much lower in energy the states are relative to the ionization limit, 13.6 eV, so we are talking about excitations ~13.57 eV in energy above the ground state. In fact, these energies are so high they refer to them as "quasibound" states.

The states the authors are looking at Stark states (the experiment is done in a field of 808 V/cm). So the analytical results they compare to are the solution of the hydrogen atom in parabolic, rather than spherical polar coordinates. This is why the quantum numbers given in the middle column are a bit funky. So the (0,29,0) does not mean $n=0$, $l=29$, $m_l=0$ (as I'm sure you can see why!) The quantum numbers define the state in parabolic coordinates, so the numbers are ($n_1$, $n_2$, $m$).

The reason they do this is that parabolic coordinates are still separable (so you can solve it analytically) when an external electric field is applied along some axis. Spherical coordinates, on the other hand, are not. So you don't see pure s, p, d orbitals (as nice as that would be!). My math gets a bit rusty here, so maybe some one else can add to this, but I believe the images you see on the left and the middle column are actually linear combinations of the spherical solutions we are used to as chemists. In fact, there is a line in the paper that in the experiment they prepare their states by exciting to a mixture of $s$ and $p$ states.

I believe you can connect the parabolic and spherical solutions in an electric field by a unitary transformation if the field is weak enough, since you can get the Stark effect in spherical coordinates by perturbation theory.

Hopefully that helps clear things up a little bit. I'm not an atomic physicist, so I'm often confused by some of the language and tools they use. There is a nice Wolfram Demo on the hydrogen atom in parabolic coordinates with and without an external electric field you may find interesting here:

http://demonstrations.wolfram.com/TheHydrogenAtomInParabolicCoordinates/


Why does STED microscopy cause photodamage? STED (stimulated emission depletion) microscopy is a technique which enables sub-diffraction limit of light imaging.
The depletion laser is at 775nm, and about 500 mW, concentrated for a short sequence of pulses on an area about the size of ~0.2 um^2
As far as I can tell very few organic molecules, nor water, absorb in this range of 775nm.
So why does it cause photodamage?
Thermal damage sure, but why bleaching fluorophores so fast?

Ballpark calculation: Wikipedia mentions the sun irradiation received at earth's sea level is up to about $\pu{300 W/m^2}$.  This is after the atmosphere filtered out already much.  You describe an experiment that where $\pu{0.5 W}$ set foot on about $\pu{0.2 µm^2}$ (or, equivalent to to this, $\pu{0.2 × 10^{-6} m^2}$).
This translates to a power density of
$$\frac{\pu{0.5 W}} {\pu{0.2 × 10^{-6} m^2}} = \pu{2.5 × 10^6 W/m^2}$$
which is more than 8000 times larger than above assumption.  This one of the reasons why samples are excited by pulsed laser radiation with repetition frequencies in the range of MHz.  And still degradation by photobleaching may occur, which one already tries to reduce, e.g., by excitation which do not hit the centre, but a tail of an absorption band.
(The mechanisms are different, but the results of exciting a small area with too much energy are similar to microscopic techniques based on second harmonic and complementary third harmonic generation (an example).)


How is plane polarized light rotated by chiral compounds? There are plenty of questions related to this topic on this site but no proper answer.
Can anyone please explain to me how plane polarized light is rotated by chiral compounds, and why it cannot be rotated by achiral compounds?

Circularly polarized light is like a helix that twists through space. The two components are mirror images of each other.
Now, every molecule interacts with both the left-handed twisting light and the right handed twisting light. The interactions differ. Every molecule, in different orientations, interact differently with the left-handed and the right-handed circularly polarized light.

Animation Reference: https://en.wikipedia.org/wiki/Circular_polarization
Now, if the molecules in solution have any mirror symmetry themselves, then if averaged over all the molecule interactions the left-handed and right-handed circularly polarized light interact in the same way, and so the polarization does not change.
But this does not hold for chiral molecules: for a particular interaction between a molecule in a certain orientation and the left-handed circularly polarized light there is no corresponding identical interaction with the right-handed circularly polarized light. There is no such molecule of that mirrored shape around.
This results in a net difference in interactions of the molecules with the left-handed and right-handed circularly polarized light, which can be described as a difference in refractive index for the two light waves. And this difference in refractive index can be detected as a change in the direction of the polarization for the sum of the two light waves.

Now, if we want to consider this using the concept of the speed of light varying in media, light goes slower in matter than in vacuum. This slowing down of light in matter is measured using the refractive index n. A higher refractive index means that the light goes slower and so keeps the light longer in the medium. Now if the medium is chiral, it gives the light two different speeds, one for the light that rotates its polarization clockwise and the other one for rotating polarization counter-clockwise. Any polarized light has only two parts (clockwise and counter-clockwise).
The two parts are combined and so the light shows a direction of polarization. When these two parts of light pass through chiral matter, one goes faster and the other slower. The result is that the polarization of light is rotated.

Response to @ Michael Seifert comment;
The way of splitting the light really depends on the matter, rather not decided by the light itself. . If matter has two directions (like a calcite crystal), light is split linearly (parallel and perpendicular). If matter is chiral, (like a solution of bio-substance) light is split circularly (clockwise and counter-clockwise). If matter is both, then all four polarizations are possible, which makes the out-going light hard to nail down.

References:
https://en.wikibooks.org/wiki/Organic_Chemistry/Chirality/Optical_activity#Why_Polarized_Light_Is_Affected
https://en.wikipedia.org/wiki/Optical_rotation
https://www.quora.com/How-do-chiral-molecules-rotate-the-plane-of-polarised-light


Why is NaBr only a reducing agent? Can you tell if something is a reducing/oxidizing/both agent? I came across a question today that read "Which of the following can only act as a reducing agent?" The answer was $\ce{NaBr}$, and I knew this from a previous question I had seen before. Is there any way to determine these properties without rote memorization?

(If it helps to use as an example, the other answer choices were $\ce{I2,\;BrCl,\;HIO4}$)

In the presence of something reasonably oxidizing, it could form $\ce{Br2}$ and reduce whatever oxidizes it. It could also oxidize to $\ce{NaBrO_{x}}$.

Those other things are not going to easily oxidize because that would put a formal positive charge on a highly electronegative atom. In the case of $\ce{HIO4}$, it would increase the charge on iodine from $+7$. This would be very hard.


How to optimize graphene geometry with Gaussian? I am new to Gaussian and I want to optimize graphene using DFT. After the the geometry optimization it has been severely changed, and some bonds were set as double and triple. Why am I unable to get the structural values similar to the input structure?

The decision for a bond to be visually represented as a single, double, or triple bond in a molecular visualization program (e.g. GaussView) is based on rather arbitrary distance rules that often do not represent the actual bond order. Unless you are investigating the bond order in other ways (e.g. Wiberg bond index), what you're seeing for the bond order is nothing more than a guess. If you switch visualization programs, you'll likely find the bond orders to change.

If, indeed, the actual geometry of the graphene is completely unphysical after the geometry optimization, here are some questions to ask yourself:

How good is the initial structure you gave the optimization algorithm? Are you using a template graphene structure that should be close to the final geometry? If you made it "from scratch", it would be worthwhile to apply symmetry to your structure prior to running the calculation in Gaussian.
How have you terminated your graphene? You say you're using Gaussian, so I doubt you are using periodic methods. Have you terminated all dangling bonds with hydrogens, for instance?
In line with Point #2, what is the size of your graphene "cluster" model? Perhaps the cluster you chose is too small.
Ensure you don't have various input file errors (e.g charge/multiplicity).


How to calculate the percentage of calcium carbonate in sea shells in an experiment We have a investigation in my chemistry class in which we must calculate the amount of calcium carbonate in sea shells. We can have as many sea shells as we wish, lab equipment, and hydrochloric acid of varying concentration from $0.5~\mathrm{mol~dm^{-3}}$ to $2~\mathrm{mol~dm^{-3}}$. How is it to be done?

$\ce{CaCO3}$ is basic so it will neutralize the acid.

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O +CO2}$$

In your experiment you need to devise a way of finding out what the end point of the reaction is (using a $\mathrm{pH}$ indicator to find when the mixture is neutral). You need to be able to measure the amount of hydrochloric acid used in moles  (you should measure the volume and you know the concentration. The number of moles of water is twice the amount of moles of the calcium carbonate (from the chemical equation above).

You should measure the mass of the amount of shell you use at the start. You will be able to figure out the mass of the calcium carbonate in that sample of shells that you use (since you now know the number of moles of calcium carbonate). Then:

$$(\frac{\mathrm{Mass_{\ce{CaCO3}}}}{\mathrm{Mass_{shells}}}) \times 100\%=\mathrm{Percent}~\ce{CaCO3}$$

You should be able to do it with just one concentration but your teacher probably wants to talk to you about reliability and experimental errors.


ICP-OES data treatment: correct for dilution and inweight, the units Let's say the ICP-OES software gives me a concentration of 186,49 ppm (mg/L) of Mn. I diluted the acid digested sample to 14.2 mL. I weighed in 10.17 mg of the powder I analyzed. I need to calculate the actual concentration.

It is quite logical in my head what I need to do: multiply by diluted volume and divide by inweight. However, I don't understand my units.

When I multiply 186.49 mg/L with 0.0142 L, I get 22,648 mg. When I then divide this by 10.17 mg, I end up with 0.260. But this is unitless. How do I know what unit I have? I thought I converted the w/V concentration to w/w by this procedure, but the units does not follow.

Is it so, that I end up with a ratio (not sure what kind of ratio)? Then I think I understand. 0.26 would then represent 26 %, meaning my sample contains 26 % Mn. But how do I know that this is in w/w, and not w/V as I started with. I just don't get the units to add up.

I think you have a typo in your calculation (also, using both dots and commas as the decimal point is rather confusing), but you're on the right track. The unfortunate thing about elemental analysis is that %, ppm, ppb, ppt, etc. get misused in a confusing way. These units are supposed to be used for only dimensionless ratios, the units you start with, ppm is actually mg/L which represents mass concentration, not mass ratio. Your calculations are correct:

$$\left(186.49 \frac{\mathrm{mg}}{\mathrm{L}}\right)\left(0.0142\mathrm{\ L}\right) = 2.648 \mathrm{\ mg}$$
gives the mass of the Mn in your sample, so when you divide by the sample mass:

$$\frac{2.648 \mathrm{\ mg}}{10.17\mathrm{\ mg}}=0.260$$
you get the mass fraction of Mn in your sample, and you know it's a mass fraction because you have mass units in both the numerator and the denominator, leaving you dimensionless. You can express it just like that, or you can convert it to 26.0% (and this is a dimensionless ratio, so it actually makes sense to use % here), or else express it as the mass of Mn in an arbitrary sample mass: 26.0 g/100 g.


How do I derive the Nernst equation in the case in which R is absent? Does anyone know what the intermediate calculations are to obtain equation 1.4.14? Taken from Fundamentals and Applications (Bard) page 31
To be more precise, how to obtain (1.4.14) from $\frac{nFAm_O}{nFAm_R}=\frac{i_l-i}{i}$

Thank you for the comment Philipp. With this clarification, the derivation is simple:

$E=E_0+\frac{RT}{nF}ln\frac{C_O(x=0)}{C_R(x=0)}$ (1.4.12) where

$C_O(x=0)=\frac{i_l-i}{nFAm_O}$ (1.4.11) and $C_R(x=0)=\frac{i}{nFAm_R}$ (1.4.13).

Than, combining equations (1.4.11) and (1.4.13) to (1.4.12), we obtain:

$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=

$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$


Are benzyl alcohol and o-cresol functional isomers or position isomers? My textbook categorizes them as functional isomers (phenol and aromatic alcohol) whereas

The above mentioned isomers are given as position isomers on 'chemguide.co.uk'

The chloride example is interesting because it highlights a fine line separating both.

The distinction between functional and position isomers isn't as simple as the typical examples make it to be - sometimes, what a substituent is attached to is critical to its functional group, so the same sequence of atoms with the same connectivity can behave, chemically speaking, very differently depending on what they are bonded to. And, conversely, position isomers in general don't have exactly the same chemistry either - you know, for instance, that n-butanol, 2-butanol and t-butanol will behave similarly but not identically.

So where do we draw the line? Technically, it's true that both are position isomers. In both cases the same substituent is attached to different positions (unlike, say, acetone and propenol, where the connectivity is clearly different). But it's also true that their chemistry is distinct - aryl compounds, in general, have different reactivity than their equivalent alkyl compounds, which is why we talk about the reactivity of "phenols" and "alcohols" as separate ideas.

So, from your examples, I'd say benzyl alcohol and o-cresol are clearly functional isomers, as their chemistry is quite different. For instance, compare pKa values:

o-cresol: $\mathrm pK_\mathrm a=10.3$

benzyl alcohol: $\mathrm pK_\mathrm a=15.4$

2-ethylphenol (similar to o-cresol): $\mathrm pK_\mathrm a=10.2$

2-phenylethanol (similar to benzyl alcohol): $\mathrm pK_\mathrm a=15.9$

Chloride, on the other hand, is less sensitive to the substrate; but I'd still classify chlorotoluenes and benzyl chloride as functional isomers, since I think they are different enough to merit that distinction, but it's in the twilight zone of vague definitions.


1H NMR of 4-Methylanisole I have been pondering the assignment for the 1H NMR peaks observed. I can't really make sense of the observations from what I know.
https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm

My question is the assignment of the signal between 6 and 7.5ppm. Unlike the way it is currently assigned, I would imagine Ha to be the most upshifted peak (near 7ppm) because methoxy group is slightly electron withdrawing and therefore de-shield the protons of Ha more than Hb (Hb is near methyl group which is slightly electron donating).

Any idea of how you would interpret the peak assignments?

Data retrieved from here: https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm

The methoxy group actually interacts with the aromatic ring via two competing mechanisms.

It is indeed electron withdrawing via the sigma bond interaction between the O and C of the aromatic ring. However the lone pair on the oxygen conjugates through the delocalised pi network and increases the electron density at both the ortho and para positions. If you draw out the conjugation pathway you will see the electrons can sit on the ortho and para carbons. This is the more significant contributor to the overall electronics and, therefore, chemical shift. So, the protons ortho to methoxy are far more shielded than those ortho to the weakly electron donating methyl group.

This effect can be generalised across all aromatic directing groups, and the size of chemical shift difference from benzene (7.3ppm) is directly correlated to the strength of the directing group. So, aniline has ortho chemical shift of about 6.5ppm (-NH2 is a strong EDG), while the ortho protons in nitro benzene are up at about 8.5ppm (-NO2 is a strong EWG).


Need to specify cyclic alkenes as cis or trans I'm studying organic chemistry from John McMurry's book. Double bond's substituents can be on cis or trans configuration. An example for cyclic molecules is trans-cyclooctene and cis-cyclooctene, where the cis compound is more stable. On the other hand, in a naming exercise this molecule is named simple as -cyclohexene. None of the other molecules  in the same exercise was named as cis or trans.

So, my question is: In small rings is not necessary to specify cis or trans orientation because the trans orientation is less stable (maybe impossible to obtain)?

Both @Waylander's and my Comment are true as well as your conclusion. From the IUPAC Blue Book 2013:


Can a spark ignite hydrogen? Can a spark ignite hydrogen? Also, what about the tip of a cigarette? I've heard that hydrogen is very explosive. Thanks

Can a spark ignite hydrogen? Also, what about the tip of a cigarette?

Yes, if the $\ce{H2(g)}$ concentration in oxygen or air is greater than about 4% not only will it burn, but most likely explode. You should avoid all heat sources when dealing with it, including direct sunlight. Just so you know, hydrogen safety datasheets include words such as "extremely flammable" and "DANGER" in red capital letters.
Besides its high flammability, another thing that makes it so dangerous is that you can barely see its flames when there's a high oxygen concentration, as seen below. It also reacts violently (A.K.A. explodes) with most oxidizing compounds, so you should be careful where you are storing it.
A rocket engine at full thrust. There's a barely visible bluish flame below it, that's the hydrogen-oxygen mix used as fuel.


Elucidating reaction mechanism and purpose Please help me understand synthesis sequence shown below:

Benzylbromide is used for protection, probably from action of $\ce{CuBr2}$, but it's reaction mechanism is unknown for me. The bromination of the methyl group is some kind of radical reaction? What would it do to the rest of the molecule weren't it for benzylbromide? Next, 18-crown-6 is clearly used as catalyst for halogen exchange — but how does it function? Does it enclose bromine making $\ce{Br-}$ a better leaving group or $\ce{F-}$ better nucleophile? Finally, how does $\ce{CF3COOH}$ manage to remove that protecting group? Maybe it protonate benzilic $\ce{O}$ leaving the stabilized carbocation behind? But this seems quite unlikely since this would probably attacked fluorine or carbonyl.

I have no idea what the addition of benzyl bromide accomplishes. There doesn't seem to be any need for a protecting group in any of the subsequent reactions. The rest of the reactions are relatively easy to explain.

This paper reports essentially 100% yield for bromination of 4-hydroxyacetophenone with $\ce{CuBr2}$, refluxed in chloroform-ethyl acetate for 15-20 mins. There is no mention of any need for a protecting group for the phenol. No mechanism is given but you are correct in your assumption that radicals will be involved.

The second reaction is a regular $S_N2$ reaction. The transition state of the reaction is highly stabilized by the adjacent carbonyl group and so the reactivity is dramatically increased compared to a regular primary halide. 18-crown-6 is used to solvate the $\ce{K+}$ cations, for which it has a very high affinity. This leaves the fluoride free to react, making it far more nucleophilic than it would otherwise be. The reaction will go to completion easily because bromide is a far more stable leaving group than fluoride.

The final reaction is a standard ether cleavage, catalysed by a strong acid. The oxygen is protonated by the acid and leaves to reform the phenol and the stabilized benzyl cation.


Calculating surface area from BET results I have been trying to work out the workings of the BET method and I have come to a standstill.

The surface area of a sample is given by

$$
S_{\text{BET}} = \frac{\nu_{\text{m}}Ns}{V}
$$

where $\nu_{\text{m}}$ is the amount in the monolayer (given in $\mathrm{cm^3/g}$), $N$ is Avogadro's constant, $s$ is the adsorption cross-section of the adsorbate ($\pu {0.162 nm^3}$ for $\ce{N2}$) and $V$ is the molar volume of the gas ($\pu {22.4 L/mol}$).

If I run this sample through a commercial software that comes with the adsorption equipment, for a given mesoporous sample, it gives me $\nu_{\text{m}} = 209.6160~\mathrm{cm^3/g}$, and a final result of $S_{\text{BET}} = 912.5 ~\mathrm{m^2/g}$.

If I try to calculate this myself (converting everything into SI units), however, I get that same result, but divided by a factor of $1\times 10^9$.

I think I'm making some stupid unit conversion mistake, but I can't see it.

Your problem is not the unit conversion, but the units you have used. The adsorption cross-section of $\ce{N2}$ ($s$) has been found using BET method is $\pu{0.162 nm^2}$ (Langmuir). Yet, you have used $\pu{0.162 nm^3}$, more than a factor of $1 \times 10^9~\mathrm{nm/m}$.


Why so few fluid phases? Taking a look at the phase diagram of water

By Cmglee - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=14939155

I see ice phases as much as up to 'Ice XI', but only one 'liquid' and one 'vapor' phase. Why is this? Either there are many ice phases, or too few liquid phases relatively.

The latter seems to be the case as it seems to be consistent with other substances. No matter what number of solid phases you get, there is usually only one liquid and one vapor phase.

Why are there so few common$^1$ fluid phases? What is the physical explanation for this?

1: For the sake of not complicating the answer too much, please ignore more exotic phases such as mesophases etc. If, though, your answer benefits from relating to them without confusing the reader too much, by all means $\ldots$

The solid (crystalline) state is defined by structure, for which there can be many. Liquid means loss of long range order, unless you talk about liquid crystalline, where orientational, but not positional long range order, is partially retained. In the gas phase, also short range order is lost.

So, in the liquid or gaseous state, phases can only differ by the molecular structure of the constituents. For water, that leaves only one stable possibility. Sulfur has a few different molecular modifications, but they're hard to separate and they tend to interconvert, so they're not really phases.

There are a few liquids with two phases which differ not in structure but in quantum states. Helium I and II are an example, para and ortho hydrogen another.


Can a base like OH- attack aldehyde and ketone to create hydrate instead of enol? I don't know if this question is kind of vague. But i was wondering like in the reaction of saponification of ester, where you have a base like OH- directly attacking the carbonyl carbon, can you also have a base like OH- attack the carbonyl carbon atom and then protonate the O- atom to create like a hydrate? Or does the base prefer attack hydrogen of alpha carbon of ketone and aldehyde? I am kind of confused about when a enol is create and when a hydrate is created since both of them can be formed by a base. So, hence I ask this question.

Ester hydrolysis and the soap-making process

In aqueous solution, esters are subject to hydrolysis to the corresponding carboxylic acid and alcohol components:

Hydrolysis is,  the reverse of the esterification reaction. The addition of
base accelerates ester hydrolysis because the nucleophile is a hydroxide ion, which is much more reactive than a water molecule.

Enol formation

The a-protons of carbonyls, are somewhat acidic, because the negative
charge on the conjugate base is stabilized by resonance with the carbonyl oxygen. In the formation of an enol, a base abstracts an a-proton from a carbonyl compound, and that same proton (or a proton on a nearby acid group) is
transferred to the carbonyl oxygen. The deprotonated form of an enol is an enolate.

However in this scenario  I don't think an enolate will form in this case.

enol formation is favored in acidic environments rather than basic

The preffered route would be the first one above.

Edit:

To avoid a bit confusion:

To avoid generalisation, the R1 group given on the ester can be anything from -C(CH2)3,  in this case 2 alpha carbons and no alph hydrogen. (No enol possible)
an alpha hydrogen may be there but reaction conditions won't favour enol formation.

Hope this clears doubts


Confusion about atomic/molecular orbital terminology I have come across several different orbital terms: atomic orbitals, natural orbitals, split-localized orbitals, molecule-intrinsic orbitals, quasi-atomic orbitals.

I don't know what the difference between these are, or whether some of them describe the same thing. Could someone help clarify what these are?

The short answer is that orbitals are defined to within a unitary transformation.

Canonical Hartree-Fock equations

The application of variational principle to the Slater determinant leads to the system of Hartree-Fock equations
$$
    \newcommand{\op}[1]{\hat{#1}}
    \newcommand{\mat}[1]{\mathbf{#1}}
    \op{F} \psi_{i}(1)
    =
    \sum\limits_{j=1}^{n} \lambda_{ij} \psi_{j}(1)
    \quad \text{for} \quad i = 1, 2, \dotsc, n \, ,
$$
which can be simplified by choosing a unitary transformation $\mat{U}$, defined by the way it transforms spin orbitals,
$$
    \psi_{k}' = \sum\limits_{j=1}^{n} u_{kj} \psi_{j} \quad \text{for} \quad k = 1, 2, \dotsc, n \, ,
$$
that makes the matrix of Lagrange multipliers $\lambda_{ij}$ diagonal and leads to the system of the so-called canonical Hartree-Fock equations
$$
    \op{F} \psi_{k}'(1) = \lambda_{kk}' \psi_{k}'(1) \, , \quad k = 1, 2, \dotsc, n \, ,
$$
or, in a much more familiar form using symbol $\varepsilon_k$ for $λ_{kk}'$ and dropping the primes
$$
    \op{F} \psi_{k}(1) = \varepsilon_{k} \psi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .
$$
The canonical Hartree-Fock equations can be thought of the eigenvalue ones, for solving which there exist a number of well-developed algorithmic approaches, thus, they are so convenient to work with.

Slater determinant invariance to a linear transformation

It can be shown that Slater determinants composed of original and transformed spin orbitals are related as follows,
$$
    \Phi' = \det(\mat{U}) \Phi \, ,
$$
which by taking the absolute values and squaring both sides leads to
$$
    |\Phi'|^{2} = |\det(\mat{U})|^{2} |\Phi|^{2} \, .
$$
This means that the transformed Slater determinant $\Phi'$ describe exactly the same system as the original Slater determinant $\Phi$ in every observable way. Or, in other words, the Slater determinant is invariant to any linear transformation of spin orbitals (except for normalization). The only advantage of the unitary transformation from that perspective is that $|\det(\mat{U})|^{2} = 1$, so that there is no even need for renormalization.

Transformed canonical orbitals

Now, as it was mentioned earlier, it is convenient to do calculations with canonical orbitals, however, once the Hartree-Fock equations are solved, the resulting canonical spin orbitals can be transformed by any arbitrary transformation making no changes with respect to observable properties of the many electron system in question. And it advantageous to use unitary transformations since there is no need for renormalization of the Slater determinant.

So, other sets of orbitals can be constructed by transforming the canonical orbitals, i.e. by forming linear combinations of them. The way these linear combinations are formed (i.e. the choice of a unitary transformation) is dictated by what is the ultimate goal of the transformation. For instance, canonical orbitals are usually fairly delocalized over a molecule, rather than concentrated in limited spatial regions of it as anticipated from general chemistry models. Thus, of a particular interest for a chemists is a transformation which localizes the orbitals. The way localized orbitals are obtained depends on a particular choice of the unitary transformation and there are in fact few different approaches.

But just localizing the orbitals might not be the end of the story. For instance, one might want to not just localize them, but do so in a way that the maximum-occupancy character in localized 1-center and 2-center regions of the molecule. This procedure results in the most accurate possible natural Lewis structure corresponding to a given Slater determinant and is the basis of the natural bond orbital model.


Limitations of extremes for the exchange-correlation functional - DFT In DFT the exchange-correlation function is critical for the correct description of the system. I have a question about the applications and shortcoming of some functionals.
LDA delocalizes electrons so are good for conductors. And pure HF localizes electrons so it's good for molecules/atoms.

So, is it safe to say that if one doesn't include correlation and only uses HF exchange energy that crystals, or any material with long range ordering, couldn't be properly described?

And the same around, if one only uses correlation, conducting crystals are properly described but no molecules/atoms could be studied correctly?

I want to understand the applications and limitations of both extremes and I am having a hard time finding that kind of analysis online.

Thanks!

One way to begin to gain some insight in this is to consider the work done on benchmarking density functionals. I shamelessly plug the work done by my old group, which can be found here. The GMTKN30 data include results for Hartree-Fock and LDA functionals with large basis sets. The description of the several benchmark sets involved and the analysis in the actual paper will illustrate many of the shortcomings and strengths of these approaches (and why better, more pragmatical DFT approaches are more widely used). While I have no access to the GMTKN55 paper, I trust that it contains references to other benchmark papers that may actually deal with crystals.


salts affecting the salinity of water I know there are different types of salts like Himalayan salts table salts etc. I was wonder do different types of salts have a different level in salinity when dissolved in water.

Table salt is obtained from various sources from mines and seas. The world's second largest salt mine (Khewara salt mine) is in Pakistan and this is where this so-called Himalayan pink salt originates. All of it is mostly sodium chloride however it appears pinkish because of iron impurities. If you dissolve it in water, a very small amount of brown precipitate can be seen which is mostly iron oxides.

There is no dedicated study which shows which salt is better for health or not. Himalayan pink salt has recently become a craze in gourmet (cooking) world and among the pseudo-medical sciences which claim that this pink salt contains 99 minerals and it can cure everything.

  wonder do different types of salts have a different level in salinity when dissolved in water.

You cannot quantify your personal taste for salinity. As chemists, we only care how much sodium chloride is there by mass percentage. Whatever table salt ends up in grocery stores is highly refined and purified. Grocery store salts contain small amounts of additives such as potassium iodide, potassium iodate, silicates, prussiates, thiosulfate, and so on. So the taste may be slightly different from brand to brand.


How is exact racemization possible? A racemate consists of 50 % $d$ and 50 % $l$ forms of an optically active compound. But how can someone ensure exact 50% quantity comparing molecule by molecule? There will always be some difference in number of molecules in both forms of optically active compounds. Is this sufficient for the mixture to rotate the plane polarised light in clockwise or anticlockwise direction?

According to wikipedia "Racemization can be achieved by simply mixing equal quantities of two pure enantiomers."

Racemization isn't "exact," but rather very very close to equality. It is just simple probability.

Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory  the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations difference, which is roughly at the 95% confidence interval.

If you flip 10 pennies then a two standard deviation difference is $2\times \sqrt{0.5^2\times 10} \approx 3$ in the number of heads.

Now flip $6.022\times10^{23}$ dimes then a two standard deviation difference is $2\times \sqrt{0.5^2\times 6.022\times10^{23}} \approx 7.8\times10^{11} $ in the number of heads.

But now think of the % difference.

$3$ heads in 10 tries for the pennies is $30\%$.
$7.8\times10^{11}$ more heads when flipping $6.022\times10^{23}$ dimes is only a difference of $1.3\times 10^{-10}\%$ which is an insignificantly small difference.


Hybrid orbitals, photoionization and time dependence Wikipedia, in the article on Orbital Hybridisation, states the following:

  One misconception concerning orbital hybridisation is that it incorrectly predicts the ultraviolet photoelectron spectra of many molecules. While this is true if Koopmans' theorem is applied to localised hybrids, quantum mechanics requires that the (in this case ionised) wavefunction obey the symmetry of the molecule which implies resonance in valence bond theory.

(Source; emphasis mine)

I am wondering about the bolded part. I interpret it as meaning that the wavefunction is an eigenfunction to both the Hamiltonian and to the symmetry operators of the molecule's point group. To me, the bolded part then seems equivalent to stating that the molecule emerges from the ionization process in one of the eigenstates of the ionized system, and I can understand how the symmetry requirement would hold true in that case.

However, my intuition wants to think about the ionization process in such a way that the departing electron leaves a "hole" somewhere in the wavefunction, which then evolves over time as a superposition state; i.e., the system does not emerge in an eigenstate, and the "hole" left in the wavefunction will "travel" through the molecule in some way. I am under the impression that the time-dependent wavefunction at any point in time could in principle break the symmetry imposed by the molecular point group.

So this leads me to two questions:

Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry? And
is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

Update: I am of course aware of the argument that we can make unitary transformations of the overall wavefunction as we please, as it leaves the total electron density distribution and energy unchanged; this is also how hybrid orbitals are formed when starting from an MO picture. However, because the orbitals involved in hybridization belong to different energy eigenvalues, this creates a time-dependent superposition state with no defined energy. In extension, hybridization silently implies that the MO energies themselves are not observables that would describe the initial states in the ionization process. And, in addition, the hybrid orbitals do not necessarily form irreducible representations of the underlying molecular point group anymore, and do thus not obey the bolded symmetry requirement in the quote above.

So here are two questions which are quite closely related to the two I had originally formulated above:

What is the justification for the assumption that the MO energies are not observables? And
how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

There are really two parts to your question - one is to interpret the wikipedia entry, the other is the more general question about the transition of methane from the parent molecule to the ionized $\ce{CH4+}$.

For the first question, I believe the author has muddled the explanation a bit (though it's of course just as likely that I'm the one who is muddled). It sounds to me like they are explaining the reconciliation of molecular orbital (MO) theory and valence bond (VB) theory for $\ce{CH4}$ rather than for the cation or the ionization process itself. Here is what I think they are trying to say:

Electronic structure of $\ce{CH4}$

MO theory
If we carry out an analysis of the delocalized molecular orbitals of methane, we find that the four highest energy occupied orbitals are spread between two energy levels, with one orbital at the lower energy level and the other three at the higher level. The orbitals meet the quantum mechanical requirements of being orthonormal and matching the symmetry operations of the molecule. The energy levels are eigenvalues of the wave function energy operator, so we'll call them eigenstates.
This representation is supported by the experimental data from PES, which shows two ionization energies, with a relative intensity of 3:1.
VB theory
In contrast, the valence bond view of the molecule has four orbitals of equal energy, each one corresponding to a specific C-H bond. These orbitals can be constructed as linear combinations of the MO theory orbitals and are orthonormal to each other. As such, they form an alternate basis set for the same space as described by the MO theory orbitals (that is, any linear combination of the MO theory orbitals can be described by a linear combination of the VB theory orbitals). However, they do not match the symmetry of the molecule, and their energy levels are not eigenvalues for the energy operator.
This representation seems to be in disagreement with the experimental evidence, since it seems to suggest that only one ionization energy should be observed. However, it is not incorrect, as I will explain.
Reconciliation
The reconciliation between the VB description and the experimental result comes from realizing that "a linear combination of MO theory orbitals" is just a different way of describing a superposition of states. All quantum systems exist as a superposition of states. In the mathematical edifice that is quantum mechanics, we require that when we make a measurement of an observable of a superposition of states, the result must still be an eigenvalue for one of the states in the superposition. The probability of a specific eigenvalue appearing is a function of the contribution of the corresponding state to the superposition.

The simplest example of this is measuring the spin of an electron. Unless a constraint is imposed, we expect to measure ($+\frac12$) 50 per cent of the time and ($-\frac12$) 50 per cent of the time. In neither case do we conclude that the electron was fixed in that spin state prior to the measurement. It was in a superposition of both spins until we made the measurement.

Getting back to the methane molecule, we say that all 8 electrons in the valence orbitals can be viewed as existing in superpositions of states, with the only requirement being that within each individual molecule, the electrons are in states that are orthonormal to each other. The VB and MO theory representations are just two of an infinite number of possibilities. When we do the PES experiment, the results must always be eigenvalues, and the intensities will always have a 3:1 ratio, because the average contribution of the eigenstates to the population of orbitals across the population of methane molecules must remain 3:1.

Ionization of $\ce{CH4}$
With the above in mind, we can move on to the harder general question about time evolution of states during and after ionization. Since the measurement in the PES experiment returns an eigenvalue, it would seem that it must force the molecule into an eigenstate, just as measuring the spin of a free electron forces it into a single spin state.

We know that initially the molecule retains the tetrahedral geometry of $\ce{CH4}$, but within a short time it relaxes to a $C_{2v}$ symmetry. This change in symmetry must be accompanied by a change in the wave function and the energies (eigenvalues) of the states. I would assume that this time-evolution of the wave function also allows for scrambling of the eigenstates back to a superposition of states, but I don't know that process well enough to give a clear description. Perhaps someone else can jump in with that.

Conclusion
Referring to your specific questions -

  1.Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry?

As I described, I would say you are correct. Superpositions do not need to meet the symmetry requirement. The only requirement is that they are linear combinations of the symmetry-consistent basis set and that as a group they form an equivalent orthonormal basis set.

  And is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

I think so, but I'm not certain.

  What is the justification for the assumption that the MO energies are not observables?

I don't understand this question. Energy is definitely an observable, and the PES experiment measures it. An observable is anything that can be measured.

UPDATE: Rereading your updated question, I think the confusion is between "not having a fixed value" and "not being an observable". Just because the electrons do not have defined energies does not mean that their energies are not observables. In chemistry, we use language that refers to fixed values of observables like energy, but a fundamental concept of QM is that observables do not have fixed values until they are measured. The measurement process itself collapses the wave function to a specific eigenstate. I tried to clarify that with the example of the electron spin. Position is another common example. Ideally, we would preface lots of statements with "if you were to measure x. . " but that would be really cumbersome.

  And how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

As I tried to explain, I believe the Wikipedia article was mixing up the parent molecule and the ionized state. The immediate effect of the ionization (ie measurement) is to force the molecule into an energy eigenstate. However, that is an eigenstate of the immediate product, which still resembles the methane parent in symmetry and orbitals. There is a time evolution to the ground state of the cation, which has a different symmetry ($C_{2v}$) and different eigenstates. Since we have performed no measurements on this molecule, it is not fixed in any of those eigenstates. Since it has evolved away from the initial $T_d$ symmetry, it is also no longer an eigenstate of the parent.

To summarize, the only point in the process in which the electrons are in a fixed eigenstate is immediately after ionization (before relaxation), and this eigenstate is described by the set of symmetry-allowed MO theory orbitals that correspond to the $T_d$ symmetry of $\ce{CH4}$, not the $C_{2v}$ symmetry of $\ce{CH4+}$.


Comparing the basicity of aniline and ortho anisidine In  my worksheet there is a question that asks for the comparison of basicities of the three anisidine isomers and aniline:

After my analysis using:

$\text{Basicity} \propto \text{+M effect (except for meta annisidine)} \propto \dfrac{1}{\text{-I effect}}$,

I could make the following conclusions:

m-anisidine should be the least basic
p-anisidine should be more basic than aniline.

But the main problem is comparing the basicity of aniline and o-anisidine and comparing them with p-anisidine. I doubt the presence of ortho effect in ortho isomer after seeing the ball and stick model which makes the interaction of lone pair of nitrogen with hydrogen atoms of $\ce{-OCH3}$ improbable.

The answer given is:

   p-anisidine> aniline > o-anisidine > m-anisidine (I have verified it with $pKa$ values from PubChem)

Why you considering the methyl group on $\ce{-OCH3}$ to be "locked" in position, facing away from the $\ce{-NH2}$ group? It is free to rotate in all directions, and it will come at a position next to the $\ce{-NH2}$ group, where the ortho effect will apply.

For a visualization, here is a similar image:

(taken from Martin's excellent post)

PS: I believe that Steric Inhibition of Resonance should not be invoked here, because both sides of the $\ce{-NH2}$ group are not blocked by the $\ce{-OCH3}$ group, and $\ce{-OCH3}$ is certainly not as bulky a group as t-butyl.


Is there any bad side effects for using calcium chloride to replace a running dehumidifier? My dehumidifier costs to much to run, so I’m thinking about using calcium chloride to keep my basement dry(er) ALSO, this may keep my mouse problem to a minimum... my questions are -Will this be harmful to me breathing this when I’m in the basement?
-Will any gases be omitted to me in my main floor?
-Will this remove to much moisture that my block foundation will start to degrade?
-Are there any other side effects I’m not aware of?
The only person that goes in basement is me so I’m not concerned about anyone consuming it.

As for safety, calcium chloride, $\ce{CaCl2}$, is in sea water and added to foods, such as beans and pickles. It's about as dangerous as table salt, sodium chloride, $\ce{NaCl}$ (actually, $\ce{CaCl2}$ is safer for concrete).

However, it absorbs only so much water until it dissolves and stops being effective. What would you do with the puddle of $\ce{CaCl2}$ solution? It can be dried with heat and reused, but then your just moving energy usage from a continually running electric dehumidifier to a batch process using $\ce{CaCl2}$... though if you can get heat for free, it would be cost effective! If you don't have a solar concentrator or access to geothermal energy, there would probably be no energy savings, though.


HPLC method for salicylic acid Salicylic acid is a relatively polar compound hence I thought that normal phase would be preferred as the stationary phase of normal phase is polar. However, it is said that reverse phase would be the preferred option instead.

Can somebody explain to me why reverse phase HPLC is preferred for salicylic acid? And would reverse phase produce better graphs/results than normal phase?

Reverse phase is the standard in HPLC so, if you can use reverse phase do not complicate matters.
Normal phase usually gives more problems than reverse phase and the stationary phases degrade faster.
Reverse phase works fine and in this case you would typically use a C18 column and water and acetonitrile.

In the case of salicylic acid and, in general, all the acids and phenols, the peaks in reverse phase are normally broad and the retention times vary.
That is because there is an equilibrium between the acid and the ionized form, the carboxylate (or phenolate).

To avoid that, you have to force the equilibrium to the acid form, which is also more apolar, so the reverse phase makes even more sense.
For this, it is sufficient to add a small amount of acid to the water used in the mobile phase.
Very often, 0.1% formic acid is enough.
Sometimes it is required to use higher concentrations up to 1%.
If you need a higher $\mathrm pK_\mathrm a$, you can use trifluoroacetic acid instead.


Is there any actually observed excited state of a "hypervalent" main-group molecule that actually involves d-orbitals? It is well known that, even in periods 3 and higher, the bonding in so-called "hypervalent" main-group molecules nearly always do not involve d-orbitals, and I've heard that it's because the d-orbitals are just too high in energy.

I now have a question: if the energetic difference is the only problem, then can bonds involving d-orbitals exist ("exist" as in "actually be observed") in excited states of "hypervalent" molecules?

For a more concrete example- do the bondings in sulfonate groups in Brilliant Blue FCF and indigo carmine get "elevated" from S(-O)3 to S(=O)2(-O) (charges omitted and S=O bonds involving sulfur's 3d orbitals) as they absorb red-light (opposite of blue) photons, and, if they do, to what extent?

One candidate involves calcium, of all things. This paper by Krieck et al. [1] reports the synthesis of a dimeric calcium(I) complex $\ce{[(thf)3Ca\{μ-C6H3-1,3,5-Ph3\}Ca(thf)3]}$ in which the calcium atoms are bonded to each other and there are two unpaired electrons in antibonding orbitals of the aromatic ring system.

DFT calculations show that the orbitals containing the unpaired electrons have a significant contribution from calcium 3d orbitals which have the same symmetry as the ring antibonding orbitals.  A model of the compound is given below.

References

Krieck, S.; Görls, H.; Yu, L.; Reiher, M.; Westerhausen, M. Stable “Inverse” Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(Thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(Thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(Thf)3]. J. Am. Chem. Soc. 2009, 131 (8), 2977–2985. DOI: 10.1021/ja808524y.


If polyethylene glycol (PEG) is hygroscopic, how does it protect wood? Archaeologists and conservators use polyethylene glycol (PEG) to replace water inside wood to protect it from shrinking and expanding with changes in humidity. But it is hygroscopic—even more so than wood, it sounds like. So how does it work? Is it like the polymer in a diaper that is able to hold on to moisture and therefore not allow any additional moisture to expand the wood further? Or is it simply filling in all the holes and its hygroscopic character is inconsequential?

The reason why PEG is suitable for this purpose has many reasons including the wood chemistry and degradation of wood in water.

Degraded wood usually very characteristic chemical losses. The cellulose of the wood is usually lost due to bacterial action and/or due to the action of the water itself. Over long period of time wood looses water-soluble components like starch and complex sugar like molecules to water. What remains in there is the water-insoluble component in the cell walls and wood tissues primarily consisting of lignin.

If the wood is in water, lignin provides structural stability to the wood. But wood becomes spongy and easy to break. If this wooden piece is exposed to air, water evaporated and wood becomes dry and shrinks. Most of the spaces between lignin layers is nor air, which makes it easy to break.

Here, comes in PEG. PEG is remarkable because it can dissolve in water.
PEG solution has very high osmotic pressure and this is one of the reason why PEG solution treatment works. PEG seeps in the tissues and provides the required osmotic pressure to keep lignin (and wood tissue) stable. This prevents the wood from expanding and shrinkage. While treatment, it is ensured that PEG penetrates the wood tissue which takes considerable times, and water is removed from the wood. In the end a large part of the original water in wood is replaced by PEG in the tissues providing structural stability.

Coming to your point on the hygroscopic property (also known as deliquescence). Higher molecular weight PEG has lower hygroscopic property because of more alkyl chained part. So it depends on the type of PEG used. But a balance has to be made since, lower MW chains are easier to penetrate tissue.

In case the relative humidity changes, for example increases, the effect due to absorption due to PEG inside tissue is negligible. Only the PEG which is outside tissues (excess, which was not removed during PEG treatment) might cause additional increase in weight. The removal of excess PEG is usually taken care of by experts.

Still however, it is the norm that >80 % relative humidity is not good for PEG treated wood.

Reference

D.W. Grattan, "A Practical Comparative Study of Treatments for
Waterlogged Wood, Part II, The Effect of Humidity on Treated Wood",
(Ottawa: ICOM Waterlogged Wood Working Group, 1981), pp. 243-252.
Taduteru Nishiura and Setsuo Imazu, "Experimental study of the
dimensional change of highly degraded waterloged wood according to
ambient humidity after preservation treatement"Bremerhaven: ICOM
Waterlogged Wood Working Group, 1990) pp. 107-118.


Is Delta H = Delta U for bomb calorimeter or Delta H = Delta U + Delta n(g)RT? As we know volume change in bomb calorimeter is zero. So if we use $\Delta H = \Delta U + p\Delta V$ then we get $\Delta H = \Delta U$. But in some reaction there is change in number of moles like $$\ce{NH2CN(g) + 3/2O2(g) -> N2(g) + CO2(g)+ H2O(l)}$$ so here $\Delta H = \Delta U + (-0.5)RT$.

First and foremost your $-0.5$ should have units of moles (i.e. should be $-0.5~\mathrm{mol}$). Don't leave out units. They're important.

  $$\Delta H = \Delta U + p\Delta V$$

If you were not taught what conditions this equation holds true, shame on your teacher. If you were taught it, you should probably try to remember that this holds for constant pressure.

In general

$$\begin{align}
H &= U + pV \\
H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
&= (U_2 - U_1) + (p_2V_2 - p_1V_1) \\
\Delta H &= \Delta U + \Delta (pV)
\end{align}$$

Only under constant pressure $p_2 = p_1 = p$ does this simplify to

$$\begin{align}
\Delta H &= (U_2 - U_1) + (pV_2 - pV_1) \\
&= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V \\
\end{align}$$

If this is not true (and think about it, if your volume is constant, then what do you think is happening to the pressure inside the calorimeter?), then you may use the ideal gas approximation

$$p_2V_2 = n_2RT_2;\quad  p_1V_1 = n_1RT_1$$

and if you further assume constant temperature $T_2 = T_1 = T$ (which is probably not true, but it is close enough to the truth since the heat capacity of a calorimeter is usually large leading to a very small difference between $T_2$ and $T_1$), then

$$\begin{align}
\Delta H &= (U_2 - U_1) + (n_2RT - n_1RT) \\
&= \Delta U + (n_2 - n_1)RT \\
&= \Delta U + RT\Delta n
\end{align}$$

The further approximation is that the non-gaseous species do not contribute significantly to the $\Delta(pV)$ term, so $\Delta n$ here can be identified with $\Delta n_g$, the difference in the number of moles of gas.

TL;DR

$$\color{red}{\textbf{Learn when the equations you are using are appropriate to use.}} \\
\large\color{red}{\textbf{Don't just blindly use them!}}$$


Why is acidic hydrogen peroxide a potent oxidizer? When hydrogen peroxide acts as an oxidizer it itself is reduced. Broadly this means that the peroxide bond is broken and that water is formed.

However, why is there a tendency for hydrogen peroxide to be such a potent oxidizer in acidic media?

  In acidic solutions, H2O2 is one of the most powerful oxidizers known—stronger than chlorine, chlorine dioxide, and potassium permanganate.

Why is it not as potent without added acid, i.e. 3% hydrogen peroxide, the stuff you find in medicine cabinets, isn't a very good oxidizer?

The "duh" answer is that hydrogen peroxide is especially unstable in acidic media. But why can't acidic media stabilize hydrogen peroxide, and specifically the peroxide bond?

Simple electrostatics tells me that having a bunch of positively charged hydronium ions around the positively polarized oxygens in the peroxide bond isn't a "good" situation. Of course there are anions from the dissociation of the strong acid but these anions necessarily have diffuse negative charge, which is probably not enough to stabilize the polarized peroxide bond to significant extent. Is there anything else?

Just consider the Nernst equation for the half-reaction
$$\ce{H2O2 + 2e- -> 2OH-}$$

You see the potential is clearly pH-dependent. Increased acidity will decrease the concentration of the reduced form, thus shifting the equilibrium to the right and consequently making peroxide a more potent oxidizer.

(BTW, even pH-neutral hydrogen peroxide is far from harmless. It can leave white marks on your hands and gnaw a hole in your jeans, albeit not instantly.)


What to do when there is an odd number of valence electrons left on the the central atom? For example, I'm trying to write the formula for $\ce{NO}$, and $\ce{ClO2}$.
If you count the total number of electrons in these molecules, it's odd. This means that no matter how you arrange the electrons, some atom is not going to have a full octet, and some atom is going to have an unpaired electron somewhere - it's inevitable. So my advice is: don't worry about the central atom not having a full octet. Just try and get as many full octets as you can. Of course, don't ever stuff more than 8 valence electrons into a Period 2 element. That's a big no-no. Once you have a structure that you think is reasonable, just check Google for the structures.
The steps my teacher gave me always assume that the central atom has a full valence, or more than full, after the outer atoms have been bonded with it. I'm running into trouble here because my central atom in these cases does not have enough electrons!
I know I could Google what they look like, but it doesn't help me understand why.

There are some molecules where the full octet isn't possible (like you've found with $\text{NO}$ and $\text{ClO}_2$). In these cases, the important thing to consider is trying to keep the formal charge (Wikipedia) on each of the atoms as close to zero as possible.

For example, for the $\text{NO}$, we know that $\text{N}$ contributes 5 electrons and $\text{O}$ contributes 6 electrons for a total of 11 electrons. We essentially have to decide between the following Lewis structures:

$$
(1) \qquad :\ddot{\text{N}}::\ddot{\text{O}}. \quad \text{Single electron on O} \\
\qquad \text{vs.} \\
(2) \qquad .\ddot{\text{N}}::\ddot{\text{O}}: \quad \text{Single electron on N}
$$

Here is where calculating the formal charge is important. For $(1)$ above the formal charge on the N is (5 - 4 - (4$\div$2) = -1) and for O is (6 - 3 - (4$\div$2) = +1, whereas for $(2)$ the formal charge on the N is (5 - 3 - (4$\div$2) = 0) and for O is (6 - 4 - (4$\div$2) = 0.

Therefore the second Lewis structure is more favorable.


How to write an entire trajectory into a single PDB file? I have a trajectory (few frames) that I want to stack together and build a single PDB. I thought either to use VMD or MDAnalysis library.

With VMD if I do atomselect it takes a frame identifier as an argument. With MDAnalysis I saw this:

with MDAnalysis.Writer("all.pdb", multiframe=True) as pdb:
    for ts in u.trajectory:
        pdb.write(u)

However, I am not sure how to use it in a complete script.

#!/usr/bin/env python2

import MDAnalysis

# file formats are automatically deduced; this is a standard topology and binary
# (NetCDF) trajectory from AMBER
topology = 'peptide_1.prmtop'
trajectory = '02_peptide_1_equil_NVT.nc'
u = MDAnalysis.Universe(topology, trajectory)

with MDAnalysis.Writer("all.pdb", multiframe=True) as pdb:
    for ts in u.trajectory:
        pdb.write(u)

One thing that may be important: I see lots of warnings when trying to use Python 3, so 2.7 is probably better at this point.


Measuring Cu2O Thickness in Oxidized Copper Sheet I have Cu2O fabricated using thermal oxidation of Cu sheets in air at  high temperatures. The thickness of Cu2O is expected to be at least 100 microns. There is still plenty of unoxidized Cu metal inside the Cu2O, so cleaving it to perform cross-sectional SEM is not trival. It is also too thick for spectroscopic ellipsometry and XRR.

What are some other methods of measuring the thickness of this Cu2O thermal oxide layer?

I like your idea to follow up the oxidation of metallic copper by analysing a by means of X rays. If you find a way to obtain a nice cross-section of your plates, I would suggest to analyse them however by means of EDX, where electron beams are used to excite the material locally, which relaxes by the emission of X rays of characteristic energy.  This will offer you both a qualitative, as well as a quantitative characterisation of your sample.  Monitoring corrosion is a recurrent task (cf. here) for this technique, recording concentration profiles, too.
In the easiest case imaginable, your scan trace would simply recognise "high level" (for pure copper), "low level" (for the adlayer of $\ce{Cu2O}$, lowered because of the presence of oxygen) and "background level" (around your plate, practically equal to zero), as schematically depicted in the following drawing.  Assuming you follow the concentration of copper as indicator, the higher the concentration of copper, the higher the read-out signal (the intensity of the $K_\alpha{}$ X-ray generated):

You mentioned the layer of $\ce{Cu2O}$ is not homogeneous, likely contains grains of unreacted copper.  Tentatively I tried to depict this situation in the following drawing, where the darker polka dots are meant to be representative for remaining starting material:

You may, depending on the scan trace chosen, encounter earlier or later such a grain.  Depending on the relative size of this grain, the beam diameter, the distance towards the pure copper matrix, the read-out signal will be influenced more, or less; hence the attenuation of the signal likely will not be as smooth as idealised here. Yet while scanning along the $x$-direction, it is possible to record $i$) $x_1$ when the high-level read-out decreases, and $ii$) $x_2$ when you reach the surface of the mixture of $\ce{Cu2O}$ and $\ce{Cu}$ towards the vacuum chamber, and hence to measure the thickness of the oxide layer.


DNA's physical properties If we took a bunch of DNA molecules and put them together what properties would it have?

I learned that long molecules tend to make a solid substance because they get tangled. A single DNA molecule is very very long so I guess it has to be solid as a substance. I also guess it would be resistant to pulling it apart. Is it true? Has anyone actually tried doing that?

Regarding first of two questions:
Assuming duplex DNA (dsDNA), the molecules should repel each other and thus stay in solution at low ionic strength, such as in Tris/EDTA buffer. This is due to the abundance of negative - repulsive - charges. Increasing the ionic strength of the solution, such as high concentration of sodium acetate should neutralize the charges and allow the dsDNA fibers to sediment. In fact, a controlled sedimentation condition may help crystallize the nucleic acids.


Hydroxide formation: reaction of group 13 elements with water I was studying about 13th group elements and I found that in this group Al and Tl react water to form a hydroxide and oxide respectively, but the rest of the elements don't react to form hydroxides.

Can anyone tell me why this is the case?

Can anyone tell me why this is the case?

Though the elements reside in same group, they react with water and form different products. This is mainly due to inert pair effect.

Boron

$$\ce{2B + 3H2O ->[700-800 C] B2O3 + 3H2}$$

  Boron react with water to produce boron trioxide(III) and hydrogen.
  This reaction takes place at a temperature of 700-800°C.(source)

Aluminium

$$\ce{2Al + 6H2O → 2Al(OH)3 + 3H2}$$

Gallium

$$\ce{2Ga + 6H2O → 2Ga(OH)3 + 3H2 ~~~~~~(1)}$$

$$\ce{2Ga + 4H2O ->[350 C] 2GaO(OH) + 3H2O~~~~~(2)}$$

  Gallium react with water to produce gallium oxohydroxide and water.
  This reaction takes place at a temperature near 350°C. (source)

Thallium

$$\ce{2Tl + H2O ->[atm. O2 ~~~50-70 C] 2TlOH}$$

  Thallium react with oxygen and water to produce thallium(I) hydroxide.
  This reaction takes place at a temperature of 50-70°C. (source)


why is potassium the most free element in most of plants Potassium is known to be the most freely available mobile element.
I am wondering why that is the case?  Is the answer hidden in its atomic structure or its monovalent cationic existance?

There are many functions related to potassium some of which are shown below

Potassium regulates the opening and closing of the stomata by a potassium ion pump. Since stomata are important in water regulation, potassium reduces water loss from the leaves and increases drought tolerance. Potassium deficiency may cause necrosis or interveinal chlorosis. K+ is highly mobile and can aid in balancing the anion charges within the plant. Potassium helps in fruit colouration, shape and also increases its brix. Hence, quality fruits are produced in Potassium rich soils. It also has high solubility in water and leaches out of rocky or sandy soils. This water solubility can result in potassium deficiency. Potassium serves as an activator of enzymes used in photosynthesis and respiration[4] Potassium is used to build cellulose and aids in photosynthesis by the formation of a chlorophyll precursor. Potassium deficiency may result in higher risk of pathogens, wilting, chlorosis, brown spotting, and higher chances of damage from frost and heat.

and the answer to your question is the small size of potassium when compared to all the other cations in plants

another reason for this is its association with the chloride ion .
chloride and potassium ions will be playing follow the master game in which if any cation takes chloride ion near it then potassium will also follow chlorine and the same is the case with potassium whenever any other anion attracts it this makes potassium highly mobile.  source wikipedia

hope you got your answer. if any doubts left feel free to comment


Retrosynthesis of a lactone We have been asked to do a retrosynthesis of (6S)-6-[(2S)-2-hydroxy-4-phenylbutyl]oxan-2-one J from 3-phenylpropan-1-ol K:

My first instinct would be to do a functional group interconversion of the alcohol to a ketone, to then do a Baeyer-Villiger oxidation.

The key to this synthesis should be — in my opinion — how to generate the two hydroxy functions with anti stereoselectivity starting from at least one achiral reagent. Most of the time you would attempt building up chirality de novo using chiral auxiliaries, catalysts or reagents as these methods are often more cost-efficient than acquiring a specific chiral reactant — the obvious exception being if a compound exists in the natural chiral pool that you can easily use (which is not really the case here). Having identified our most important problem, we can turn to the other ones:

we need to generate a δ-lactone
we need to form a $\ce{C-C}$ bond, preferably in a chiral manner, to a $\ce{C6}$ body (the bit of the final product that is missing in the reactant).

Scheme 1: General retrosynthesis idea.

In my opinion, the δ-lactone is a non-issue. If you have an open-chain ester such as 3, you can probably form the six-membered lactone simply by $\ce{K2CO3}$-mediated transesterification. Practically all literature methods to synthesise esters should also predominantly give the δ-lactone especially since the unfavourable ζ-lactone is the alternative. (In the unlikely case that it is formed, it can easily be transesterificated as noted.) I would put this step last.

Scheme 2: Transesterification step.

We now have 3, an anti-configured 1,3-diol. When I see this, my mind immediately jumps to the Evans-Tishchenko asymmetric reduction. Thus, we can arrive there from either of the two possible β-hydroxyketones 4 or 5. For this to work best, I would want the carboxy group to be masked as a protected alcohol here, cf. 6.

Scheme 3: Evans-Tishchenko step. $\ce{X}$ would be $\ce{H}$ for the newly-formed hydroxy group and $\ce{CH3CH2CO-{}}$ for the inducting one.

Both of these β-hydroxyketones could be generated from aldol reactions; however, only ketone 4 suggests that it can easily be reached from the desired reactant 2. If we wanted to synthesise 5, we would need an extra carbon atom on 2 which we would need to introduce in an additional reaction. However, the route to 4 only requires selective oxidation to the aldehyde — my personal favourite is Dess-Martin–oxidation, but you should probably check all the others that suggest themselves (Swern and relatives, TEMPO, Ley-Griffith, etc.) as one could be superior over the others due to your lab spirits liking it.

Finally, this also tells us that the aldol addition to generate 4 from 7 and 8 should proceed in a stereoselective manner giving the (S)-configured product predominantly. Again, many methods are known and I would suggest a Paterson aldol reaction. Analysis of the relevant literature (unfortunately, I did not find a website, so a citation it is) reveals that $\ce{(-){-}Ipc2BOTf}$ and Hünig base should give the (S) product preferentially.[1]

Scheme 4: Final aldol disconnection.

This leads us to the following forward synthesis:

$$\ce{\mathbf{2} ->[DMP] \mathbf{7} ->[(-){-}Ipc2BOTf][ (iPr)2NEt, \mathbf{8}] \mathbf{4} ->[SmI2, EtCHO] \mathbf{6} ->[1) TBSOTf; 2) PPTS][3) DMP; 4) {Pinnick}] \mathbf{3} (R $=$ H) ->[DCC, DMAP][K2CO3] \mathbf{1}}$$

Reference:

[1]: A. S. Franklin, I. Paterson, Contemp. Org. Synth. 1994, 1, 317–338. DOI: 10.1039/CO9940100317.


Are there ways to fully dissociate a complex salt? A complex salt is a salt where the metal and ligands do not dissociate in water.

For eg : $\space\ce{K_4[Fe(CN)_6]}$ when put into $\space\ce{H_2O}$ dissociates as : $\space\ce{4K^+} + \ce{[Fe(CN)_6]^{4-}}$

where $\ce{[Fe(CN)_6]^{4-}}$ is a complex ion which cannot be further dissociated.

So, is there any way to dissociate them, either by heating or by adding something? Or will they dissociate in some other solvent? Or is there no way to dissociate them?

Let's consider the formation constant of this ion complex at $25^\circ\mathrm{C}$:
$$\ce{Fe^{2+} + 6CN- <=> [Fe(CN)6]^4-}\qquad K_\mathrm{f}=10^{+24}$$
This means that, the equilibrium is highly displaced towards the formation of the  complex ion. The ion complex is very stable.

So, if we want to displace this equilibrium towards the dissociation of the complex ion, we have to add an adequate quantities of  a metallic ion or a ligand that form with $\ce{CN^{-}}$ or $\ce{Fe^{2+}}$, respectively more stable complexes.


How to determine what organic molecule has the highest boiling point? Compare and answer which one has the highest boiling point:

$\ce{CH3CH2CH2CH3}$ (butane) ... [$\pu{−1}$ to $\pu{1^\circ C};\ 30$ to $\pu{34^\circ F};\ 272$ to $274\ \pu K$]
$\ce{CH3NH2}$ (methylamine) ... [$−6.6$ to $−6.0\ ^\circ\pu C;\ 20.0$ to $21.1\ ^\circ\pu F;\ 266.5$ to $267.1\ \pu K$]
$\ce{CH3OH}$ (methanol) ... [$64.7\ ^\circ\pu C\ (148.5\ ^\circ\pu F;\ 337.8 \pu K)$]
$\ce{CH2F2}$ (difluoromethane) ... [$−52\ ^\circ\pu C\ (−62\ ^\circ\pu F;\ 221\ \pu K)$]

My effort:
At first, without referring to factual data (but the factual data has been provided in parenthesis, see above-options, Source: wikipedia.org), I reached the following conclusion:

Butane appears to have the lowest boiling point because there is only Van-der-Waals forces, which are weaker than dipole-dipole interactions, present in the rest of molecules.

Next I would compare between methylamine and methanol because they both have dipole-dipole interactions and most importantly that these molecules associate themselves with hydrogen bonds, but as oxygen is more electronegative than nitrogen, hydrogen bonds formed within methanol molecules would be stronger than those formed between methylamine molecules. So methylamine takes the 3rd position.

Lastly, it becomes easy for me to differentiate between methanol and difluoromethane, because according to my knowledge the latter has got fluorine which is more electronegative than oxygen, present in former. Moreover difluoromethane has got 2 fluorine, which add weight to my above argument in the way that there will be even more extensive hydrogen bonding within the molecules.

Inferences and comparison with the source's answer
Answer: Methanol
I am full of doubts as to why difluoromethane is not the answer. Revisiting my arguments again and again, I always ended up with the same conclusion i.e., mine answer would always contrast with the source's answer.
Afterthoughts & Deductive Conclusions
A few days after extensive thinking and arguing myself, I ended with the following reason, which seemed plausible to my brain. It may be possible that the because of: the presence of two fluorine atoms, small and compact nature of carbon-fluorine bond (according to bond length - bond strength - bond dissociation enthalpy relation/ theory), and the overall small and virtually spherically shape (But, in fact, it is tetrahedral geometry) of difluoromethane; the two fluorine atoms might be finding themselves in a position, in which they cannot simultaneously form hydrogen bonds with the hydrogens of other molecule. My above explanation can also imply that they might be hindering into each other's hydrogen bonds (reason can be same as mentioned in list in the previous sentence).
This reasons and explanation thus explain why the boiling point of difluoromethane falls below that of methanol. Remember that in methanol, there is less compact nature of carbon-oxygen (similar size of orbitals of carbon and oxygen as well as bigger size of oxygen atom as compared to that of fluorine), a single carbon-oxygen bond is weaker than a single carbon-fluorine bond (based on electronegativity difference and concept of dipole-development & interactions), and the last that is a molecule of methanol is less symmetrical (3-H and 1-O) as compared to difluoromethane, which is more symmetrical (2-H and 2-F).
Final conclusion and additional Info.

Solution1: $\ce{C4H10 < CH3NH2 < CH3OH < CH2F2}$ (in increasing order from left to right)
Solution2: $\ce{C4H10 < CH3NH2 < CH2F2 < CH3OH}$ (in increasing order from left to right)
Solution3: None (it's should be totally based on your own conclusions and knowledge, backed up with info. from  appreciable sources)

There are commonly three types of intermolecular forces quoted when it comes to melting and boiling points: dipole-dipole interactions, London interactions and hydrogen bonding. Hydrogen bonding is by far the strongest force of them all. In all cases except for very rare ones, for a hydrogen to take part in hydrogen bonding, it must be bonded to $\ce{O, N, F}$ — one of the three very electronegative atoms.

In difluoromethane, there are no hydrogens bound to $\ce{O, N, F}$ — they are both bound to carbon. It is mainly carbon that experiences fluorine’s electronegativity and by the time the inductive effect reaches the hydrogens it has been substantially lowered. On the other hand, methanol and methyl amine both have hydrogens bonded to the electronegative atom — nitrogen or oxygen — so these compounds actually take part in hydrogen bonding. Since we have two compounds that participate in hydrogen bond networks it is safe to assume that these will be the top runners.

As you mentioned, methanol’s hydrogen bonds should be slightly stronger since oxygen has a higher electronegativity. I think your reasoning as to why methanol is stronger than methyl amine is sound enough.

Difluoromethane’s principal intermolecular forces will be dipole-dipole interactions. However, you should also note that the presence of fluorine atoms strongly reduces the underlying London interactions, since $\ce{C-F}$ bonds are more localised and fluorine’s lone pairs are much less disperse than most other carbon-heteroatom bonds — this is due to fluorine’s very high electronegativity. Thus, while we have significant dipole-dipole interactions, we are somewhat lacking London interactions in spite of the rather large molecular weight of $52$.

This is important when comparing it to both methyl amine $M_\mathrm{r} = 31$ and butane $M_\mathrm{r} = 58$. Butane, given its larger electron count ($26$ bonding electrons versus $14$ bonding and lone pair electrons in methyl amine) and its very non-spherical shape can give rise to much stronger London interactions, especially since all its bonds are almost or entirely unpolar $\ce{C-H}$ or $\ce{C-C}$ bonds. A general rule of thumb is the heavier a molecule, the stronger the London interactions and they quickly catch up with what dipoles or hydrogen bonds can do.

So putting it all together:

Difluoromethane has very weak London interactions but some dipole-dipole interactions. It is expected to have a rather low boiling point.
Methyl amine can form hydrogen bonds but is also rather small and thus lacking London interactions. It is less polar than methanol giving rise to lesser dipole-dipole interactions. Put it somewhere in the middle.
Butane is significantly larger in volume than all the others and also has much more electrons that can take part in London interactions. Thus, London interactions are predicted to be rather strong but nothing else is there.
Finally, methanol of course forms strong hydrogen bonds, is a good dipole and can take part in London interactions well. A clear winner.

I wouldn’t know how exactly to rank butane and methyl amine but I would put difluoromethane at lowest and methanol at clearly highest boiling point. This is consistent with the experimental data.


What splitting pattern is this? I am trying to find the structure of an unknown molecule. What is the splitting pattern of these sets of lines?

I think my unknown molecule is butanol and the hydrogens this splitting pattern is referring to are the middle ones, i.e. $\ce{H3C-CH2-C\color{\red}{\mathbf{H}}_2-CH2-OH}$.
They would couple with the hydrogens on the right as a triplet and the ones on the left as a triplet, but would they also maybe do a W-coupling and couple with the methyl on the left.

So what would the coupling pattern be?

This is the a peak from one of the vinyl protons on an allyl group. This splitting pattern is an ABX pattern from the bold hydrogen below. It is technically a triplet of doublets of doublets.

$$\ce{R-CH2-C}\textbf{H}\ce{=CH2}$$

Check out the following simulated NMR spectrum of 3-chloro-1-propene (allyl chloride) from nmrdb.org

Now image if the coupling constants are a little different and the structure collapses into what you have.


How is resonance energy measured? Is it the difference of energies of formation of the resonance hybrid and the average of the energy of formation of all the contributing structures? How would we even calculate the energy of a resonance contributer?

Resonance energy is a fictitious value, it cannot be measured, it is always only estimated. The definition in the gold book[1] clearly states this:

The difference in potential energy between the actual molecular entity and the contributing structure of lowest potential energy. The resonance energy cannot be measured, but only estimated, since contributing structures are not observable molecular entities.

The arising question is how to obtain the energy of the hypothetical contributing structure. One way to do it is to use tabulated (averaged) values for 'normal' bonds to estimate the total energy of the hypothetical molecule.[2]
Another way to do it is to extrapolate from measured hydrogenation energies. This was famously done for benzene. For example, the hydrogenation energy of cyclohexene to cyclohexane is about $\pu{110 kJ mol-1}$. This is the energy to hydrogenate one double bond. In the hypothetical Kekulé structure of benzene you'd expect three of them, giving you a theoretical hydrogenation energy of about $\pu{330 kJ mol-1}$. The measured hydrogenation energy of benzene is about $\pu{200 kJ mol-1}$. This difference of $\pu{130 kJ mol-1}$ can now be taken as the stabilisation due to delocalisation, i.e. the resonance energy.[3]
There is no universal way of doing this, because none of the ways can be proven to be right; it's not real after all. Therefore you may find vastly different values for the resonance energies. It is better to think about resonance as a phenomenological, qualitative argument of mostly pedagogical value. If you are comparing values of hydrogenation (something that can be measured) against extrapolated values (or expectation values) then it is unnecessary to obscure this comparison by giving it another label.

Resonance Energy in: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. DOI 10.1351/goldbook.R05333

You can find such values for example in this Chemistry LibreTexts article on Bond Energies.

Values are loosely converted to SI units from the article on Chemistry LibreTexts Structure and Resonance Energy of Benzene: A First Look at Aromaticity.


Measuring Colloidal copper PPM & how to ignore the salt PPM in soluton I made some colloidal copper with 9V battery. To make this water I had to add small amount of salt. However, the salt itself shows up in PPM count of my PPM reader. So how can I discount the salt PPM and come up with correct copper PPM in distilled water.

First method is to look for Faraday's law of electrolysis and put in the equation your data.

Second method is to add to TDS reading the initial reading of the salt solution before the electrolysis.


Can we observe the shapes of MOs? Condensed question formulation:

Is there an experimental method to directly visualise the 3D form of a MO wave function, or at least the electron density associated with it?

Full statement:
Molecular orbital theory is a crowning achievement of modern Chemistry: it is by far the most accurate description of chemical bonding we possess and accounts for the vast majority of related phenomena.
An array of evidence exists to confirm predictions from MO theory. Ab initio computational methods such as Hartree-Fock are able to produce a very accurate description of orbital coefficients and energies, which match experimental data for the energies (via photoelectron spectroscopy etc.) or predicted reaction mechanisms/FMO interactions (in kinetic runs).
But do we have any experimental evidence for the shape of MOs?
I'd presume the most feasible way of observing some measure related to the MO wave function $\Psi$ would be to follow a probabilistic approach and try imaging the electron density assumed to be given by $|\Psi|^2$.
Of course, X-ray diffraction first springs to mind; the notorious issue with that being the phase problem. In my superficial understanding: due to the loss of phase information, in going from the XRD pattern to the electron density isosurface (via ab initio methods), essentially we model some initial guess phase and try to fit it to the intensity data.
Since usually the location of nuclei is of far greater practical interest, the formalisms developed to model the phases (e.g. Hansen-Coppens) tend to focus on charge densities localised on each atom (though they are not necessarily of spherical symmetry). This is useful in actually getting the form of the molecule, but produces a picture different to the expected electronic occupation of delocalised MOs.
Is it possible to instead produce the delocalised picture if our ab initio phase guess is based on a RHF calculation? But even if possible, does this have any more legitimacy than the localised picture? I.e. it seems, as far as XRD is concerned, these are equally possible interpretations of the same intensity data where we just forced some phase information as convenient.
Sorry for the long description, this is an in depth presentation of my thoughts on the issue thus far. Any experimental technique/fundamental aspect of MO behaviour I'm missing would be appreciated. But the bottom line is, as stated in the beginning, can we experimentally visualise the shape of MOs or at least their modulus squared?

The very question you pose is addressed thoroughly in this open access work:
https://www.nature.com/articles/ncomms9287 .
The short answer to your question is that the electron density can be mapped using a technique akin to diffraction as described in their work.  You mentioned a distinction between the electron density and the orbitals, and the orbital is not itself observable; but as you suggested, the electron density can be mapped.