Wax and oil polish on wood Wood countertops are frequently sealed, when a non-hardening, food-safe finish is desired, with some magic cocktail of waxes and oils. Typical components might include mineral oil, beeswax, and carnauba wax, mixed in some unspecified ratio and then sold at premium prices.

Not being content paying for a bottle of magic goo, I'm seeking a scientific understanding of how these products actually work.

These products claim to (over time) build a coating of wax on the surface which is more durable than simply mineral oil would be. If the product is a mixture of oil and wax which is quite thin, how can it build a layer of wax on the surface of the wood with repeated applications? Does the wood selectively absorb the oils? When subsequent applications are made, are the thinner oils able to diffuse through whatever wax has built on the surface?

What are the relevant concepts in play here, and in what directions might I research to gain a deeper understanding?

Mineral oil serially penetrates the wood to render its surface hydrophobic.  Nothing eats mineral oil and it does not easily oxidize (go rancid).  It is also the solvent vehicle to transport beeswax (plasticizer, compatibilizer, internal soft sealant) and the hard wax surface sealant. Nothing polymerizes (re linseed or tung oil) or chemically bonds to the wood.  All three are FDA GRAS.

Like dissolves like.  While the surface will be sealed against polar materials and provide no food for bacteria and fungi of itself (and absence of water is a good biostat), it can imbibe fats and oils.  Butcher block surfaces and such are best made of dense oily woods that are naturally sealed against water.  Top them off.  Lignin is a fair antibacterial and antifungal agent.  Polyethylene cutting boards have higher sustained levels of viable bugs than do wood cutting boards.

http://faculty.vetmed.ucdavis.edu/faculty/docliver/Research/cuttingboard.htm


Why are vinylic and arylic carbocations highly unstable? I have checked the internet and read quite a few books, but I still am not able to understand why vinylic and arylic carbocations are highly unstable.

What I found while surfing the internet is: For arylic carbocation, the carbon bearing postive charge cannot participate in resonance with adjacent carbons because it's p orbital is perpendicular to the other orbitals. But, I didn't really understand what this means.

Background

S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital.  In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.

This is the basis of the inductive effect.  For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - the methyl group is inductively electron donating to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.

What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.

Answer

Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon.  For a positive center to be attached to a more electronegative group is destabilizing.  Hence the vinylic cation is less stable than a typical alkyl cation.

Things are even worse with the aryl carbocation on the right.  Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry.  A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$.  These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.


Synthesis of bromobutane What will be the cost efficient and high yielding method of synthesis for 1-bromobutane?

I can react butanol with $\ce{HBr}$ but I am worried butanol is not present at the moment in lab. But I need to hurry and make bromobutane.

Current prices are about 34 € (500 g) for 1-bromobutane and 32 € (500 mL) for 1-butanol in decent lab quality.

The most cost efficient method is ordering 1-bromobutane from a chemical supplier of your choice!

On a technical scale, one would probably react 1-butene with anhydrous hydrogen bromide under radical conditions, using peroxides or UV light to initiate homolytic cleavage of the $\ce{H-Br}$ bond.


Determining (approximately) the amount of a strong base needed to change a buffer's pH by a certain number of units I'm measuring buffer capacity by titrating $\ce{NaOH}$ into a buffer of $\ce {H3PO4 + NaH2PO4}$. I'm measuring buffer capacity by recording the amount of $\ce{NaOH}$ required to increase the $\mathrm{pH}$ of the buffer by 2 units, and then taking the ratio.

My problem is that I have no idea how much $\ce{NaOH}$ I'll need to change the $\mathrm{pH}$ by that much. If I've underestimated the strength of my buffer, I might have to reduce the drop in $\mathrm{pH}$ to just one unit. I need to have a rough idea so I can organize how much of each I'll need to prepare.

Basically, is there any way I can predict (roughly) how much $\ce{NaOH}$ I'll need?

Details:

$1\:\mathrm{M}$ $\ce{H_3PO_4}$ and $1\:\mathrm{M}$ $\ce{NaH_2PO_4}$
Buffer volume: $50\:\mathrm{mL}$
$1\:\mathrm{M}$ $\ce{NaOH}$

You can calculate the quantity of sodium hydroxide, required to change the $\mathrm{pH}$ of your system by two unites,  by calculating the concentration of sodium ion in the charge balance equation:
$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$
In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of $\ce{NaH2PO4}$), and you calculate the final concentration of sodium ion  after adding sodium hydroxide by the above equation. So you can deduce the required quantity  of sodium hydroxide.
As for the  other ions in the charge balance equation, they are calculated as follows:
At first the  $\mathrm{pH}$ of the buffer system $\mathrm{pH=pK}_a=2.15$.

If you want to change the $\mathrm{pH}$ of your system by two unites, then the final $\mathrm{pH}$: $$\mathrm{pH}= 4.15\,\,\,\,(2)$$
Now you can estimate the concentrations of ion hydronium and ion hydroxide.

The use of Henderson-Hasselbalch equation gives: $$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$

The equation of matter conservation gives: $$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$ As the buffer is $25\,\mathrm{mL}$ of $1\,\mathrm{M}$ of each of the components, so there's a total of $1\,\mathrm{M}$ in the $50\,\mathrm{mL}$ solution due to dilution (not as you mentioned in your comment).
Now, you can calculate the concentration of ion $\ce{H2PO4^{-}}$
from equation (3) and (4):$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$

If we substitute the numerical values in equation (1):
$$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$
$$n=\ce{  [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$
The volume of sodium hydroxide is $$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$
I suggest to use a  solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.


In a galvanic cell,do charges accumulate on electrodes 1st (then redox occur) which produce E in the wire when we connect between electrodes? We study that when we connect a wire between the terminals of a battery ,an electric field in established between terminals  due to potential difference .Is the potential difference result of accumulated charges on terminals. In other words   when an electrode is put in the solution like Zn in ZnSO4  will Zn electrode will become -vely charged ? If not then how does electric field established into wire at exact moment when we connect wire between terminals (electrodes) which essential for electric current .

You can imagine the cell electrodes as capacitors, that are charged by chemical way.

But the capacitance of these capacitors

$$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$

is very small, so does the accumulated charge $q$.

The charge loses due pushing electrons  through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox reactions.

If we consider the zinc electrode of the Daniell cell, the reaction :

$$\ce{ Zn -> Zn^2+ + 2 e-}$$

is charging the electrode negatively, while the opposite reaction:

$$\ce{  Zn^2+ + 2 e- -> Zn}$$

is charging the electrode positively.

The electrode gains such a potential, where both reactions are ongoing at the same rate.


Why does reducing pressure cause the equilibrium to shift towards the side with less moles? Take the example:

$$\ce{N2(g) + 3H2(g) <-> 2NH3(g)}$$

I understand that if I increase the pressure of the system, it'll shift towards the $\ce{NH3}$ side. This is because of the reaction rate increasing (increased collisions) more for the forward reaction than the reverse reaction. What I don't understand is why a decrease in pressure of the system would shift the equilibrium to $\ce{N2(g) + 3H2(g)}$. There is overall less collisions, but there's still more moles of reactants to collide and less molecules for $\ce{NH3}$ to collide and decompose.

Actually, the shift of reaction towards left on decreasing pressure (and towards right on increasing pressure) is due to Le Chatelier's Principle, which states that if a change is brought in the equilibrium conditions of a reaction, the reaction will proceed in such a manner that it counteracts the change.

In case of increasing pressure the reaction shifts to right due to lesser number of moles on right. And according to gas equation, lesser moles means lesser pressure. The opposite happens when the pressure is decreased.


Is this Boiling point apparatus? I have some chemistry equipment but I am not sure what it is for; I think it may be apparatus for measuring elevated boiling points. I have included a drawing of how I think it should go together. Basically, it consists of a conical flask connected to a graduated boiling tube (D) (for volume); the conical flask has a two-holed bung: one aperture contains a delivery tube (B) connected to the boiling tube (there are small holes in the delivery tube within the boiling tube, and the other has a glass cylinder passing vertically upwards, which is angled at the bottom (A; this might be to do with maintaining atmospheric pressure?). The boiling tube has a bulb on the upper portion. The boiling tube is contained within a larger glass container (C). There is an aperture in the boiling tube opening into the larger container (E). The larger container has a spout at the bottom.

So the liquid in the conical flask is heated and passes through E and condenses in the boiling tube (D). I am not sure if the delivery part, E, has to be lower, so that as the liquid condenses the delivery tube becomes submerged in the liquid. Then I am guessing the liquid that condenses in the boiling tube will at some point be hot enough to pass from the boiling tube, through aperture E into the larger container (C). Quite what is happening, I am not sure...

If anyone has seen this equipment and knows how it works, that would be much appreciated.

To me this looks like an apparatus to measure water in fat, butter or oil. A weighed sample of butter is placed in the conical flask. The tube A goes below the level of the melted butter or oil. It's purpose is to maintain atmospheric pressure or in case of a bump in the comical flask or back pressure or suction from the boiling tube D or the condensation flask C. Flask C should have a hole open to the atmosphere through the bung. The hole at the bottom of C is to remove water from the condensed ice. The conical flask is heated. The butter (oil) has a much higher boiling point than the water in it. So the water boils out. The tube from the conical flask will go well down into the boiling tube D. More than you have indicated in your diagram. Flask C should be full of an ice water salt mixture at −4 °C surrounding the boiling tube up to below the ventilation hole in D. It must not cover it or spill over into D as this would give you a wrong result. The thermometer measures the temperature in the region of D where the water condenses. If the thermometer indicates a rise in temperature above a set limit (I do not now have access to data where I could look up this temperature. However, I think it should be in the region of 5–10 °C) then the rate of heating show be reduced as you might overload the condenser and loose moisture. If the test has been running for some time it may be necessary to add ice as the ice melts to water the cooling efficiency of your condenser would be compromised. To do this some water should be run off from the tube at the bottom of flask C. This would be done Using the glass spout which you have shown. What you have not shown is the piece of rubber tubing (now long perished) and the clip attached to the spout at the bottom of C which would allow you to remove excess water from C. If you do not remove the excess water from C you could cause water to spill over into flask D as the ice is added from the top. If water were to get into D from C then your result would be destroyed and after cleaning everything out you would have to start over again. Remember it is necessary to have some water in C so that the ice can make better contact with D and act as a much more efficient condenser. The volume of water in the butter can be read on the side of flask D. It works well on low fat spreads which contain lots of water. If you want to use do not send for me. Good luck.


Expected accuracy of experimental IR spectra compared to literature reference I have a table from a journal which indicated signals of an alpha chitin that I used as theoretical values. I did my own analysis to identify a chitin sample and it had signals that slightly varied from the theoretical (ex. $3106\ \mathrm{cm}^{-1}$ symmetrical N-H vibration and I got $3103\ \mathrm{cm}^{-1}$). Should I consider this? And what caused the variation?

In addition to @porphyrin's comment, the interaction of your sample with the surrounding may overall cause some changes in shape and place of the absorption bands; namely if your analysis

follows the classical sample preparation, grinding your compound altogether with excess of KCl to press a pellet (ionic environment) / mixed in nujol (apolar environment) that is then characterised in transmission

.or.

is an analysis by ATR with single or multiple reflection where your sample basically is a neat film on top of your optical window; either probe head or between crystal and anvil. (The ATR correction offered by the spectrometer's software, not every time active, basically will account for the penetration depth varying by the wavelength used, hence influencing rather the displayed %transmittance, then the position of the barycentre of the absorption band in question.)

If you aim for a qualitative IR analysis of a small organic molecule, I would not worry too much if the reported IR band was reported  at 3106 cm$^{-1}$ and is now spot at 3103 cm$^{-1}$.  The iS10 ATR-FT-IR spectrometer by Nicolet/ThermoScientific, for example, provides a data spacing of around 0.7 cm$^{-1}$.  Performing a library search in an electronic database to identify / narrow which compound the sample in question is hopefully would not be affected, either.

(A quantification, like determining the composition of gasoline by PCA, would of course need some calibration.)

The original question however may be read as if you aim to monitor the (partial) digestion of larger proteins and polysaccharides, to say something about their secondary structures.  A shift of 10...20 cm$^{-1}$ of an amide absorption band may be relevant; yet then more often the ones around 1600...1700 cm$^{-1}$ are taken into consideration.


Reaction mechanism for oxidation of primary alcohol to carboxylic acid Can someone help me identify the steps for the following reaction? I'd like to keep track of everything that happens. I can't figure out what happens with oxidation in this context, and thus can't get started.
It's 2-methylpropan-1-ol oxidized with potassium dichromate(VI) in aqueous sulfuric acid to yield 2-methylpropanoic acid (also known as isobutyric acid).

A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid in situ, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl π-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.


Nitrobenzene - Aromatic Substitution Reactions Which EAS (electrophilic aromatic substitution) reactions does nitrobenzene show?

We know that nitrobenzene is particularly deactivated towards most such reactions, as the nitro group is strongly electron withdrawing via resonance and inductive effects. However, I wish to know exactly where to draw the line - which reactions nitrobenzene undergoes, and which ones it doesn't.

Of course, there's no clear boundary as to whether or not a compound undergoes a particular EAS, but can't we at least comment on the yield obtained in such reactions?

I haven't been able to find much about EAS reactions of nitrobenzene online, hence I come to Chem SE, seeking data and sources.

The reactions of interest are:

1.Friedel Crafts Alkylation, Acylation

2.Nitration (concentrated sulphuric and nitric acid)

3.Sulphonation (fuming sulphuric acid)

4.Halogenation in presence of Lewis acids
etc.

Please help! Thanks a lot!

Of course, there must be some boundary as to whether or not nitrobenzene undergoes a particular electrophilic aromatic substitution (EAS) since nitro-group is very deactivating. I leave it to you to decide that fact on the yield obtained in following reactions, because there are some data you may fail to find but, nonetheless, exist in literature.

Chlorination of nitrobenzene at $\pu{35\!-\! 45 ^{\circ}C}$ in the presence of iron (III) chloride ($\ce{FeCl3}$) gives an isomer mixture, which contains 86% 3-chloronitrobenzene, 10% 2-chloronitrobenzene and 4% 4-chloronitrobenzene (Ref.1). Even if nitro group is $m$-directing, this reaction has given significant amount of $o$-prodouct. You may compare this reaction with nitration reaction of chlorobenzene at $\pu{40\!-\! 70 ^{\circ}C}$ with mixed acids (30% $\ce{HNO3}$ : 56% $\ce{H2SO4}$ : 14% $\ce{H2O}$), which typically gives a 98% yield of an isomer mixture consisting of 34-36% 2-chloronitrobenzene, 63-65% 4-chloronitrobenzene, and about 1% 3-chloronitrobenzene (Ref.1).

A simple and efficient method for the sulfonation of nitrobenzene using liquid $\ce{SO3}$ as the sulfonating agent in a microreactor at $\pu{40\!-\! 110 ^{\circ}C}$ was developed and conducted recently, and 88% yield of 3-nitrobenzenesulfonic acid (m-NBSA) has been obtained (Ref.2).  The main by-product is bis(3-nitrophenyl)sulfone (BNPS) and proposed mechanism was depicted in the following Figure:

On the other hand, the sulfonation of toluene has given $o:m:p$ ratio of  21.2% : 2.1% : 76.7% in 77.8% $\ce{H2SO4}$ (Ref.3). The $o:m:p$ ratio has become 38.8% : 2.6% : 58.6% when solvent has changed to 84.3% $\ce{H2SO4}$ (significant decrease of $p$-product in increasing acidity; Ref.3). Also, note that the sulfonation by sulfuric acid of nitrobenzene itself and of various aromatic compounds in nitrobenzene solution has been studied at a series of temperatures (Ref.4). Yet, no concomitant sulfonation of nitrobenzene in the presence of other activated substrates such as toluene and xylene has been reported.

Using $\ce{NO2+PF6-}$ as the nitrating reagent, Olah had studied nitration of nitrobenzene in two different solvents: He achieved $o:m:p$ ratio of  7.1% : 91.5% : 7.4% in 96% $\ce{H2SO4}$  and that of 10% : 88.5% : 1.5% in $\ce{CH3NO2}$ (Ref.3).

For comparison, same nitrating agent ($\ce{NO2+PF6-}$) has given $o:m:p$ ratio of  68.2% : 2% : 29.8% in $\ce{CH3NO2}$ when subjected to nitrate toluene (Ref.3).

To my knowledge, Friedel-Crafts reaction of nitrobenzene (alkylation or acylation) is yet to be known. I'd love to hear if somebody say otherwise.

References:

International Agency for Research on Cancer (World Health Organization), lARC Monographs on the Evaluation of the Carcinogenic Risk of Chemicals to Humans, Vol. 65, 1996, “2-Chloronitrobenzene, 3-chloronitrobenzene and 4-chloronitrobenzene,” pp. 263-296.
Y. Chen, Y. Su, F. Jiao, G. Chen, “A simple and efficient synthesis protocol for sulfonation of nitrobenzene under solvent-free conditions via a microreactor,” RSC Adv. 2012, 2, 5637–5644 (DOI: 10.1039/C2RA20406A).
George A. Olah, “Aromatic substitution. XXVIII. Mechanism of electrophilic aromatic substitutions,” Acc. Chem. Res. 1971, 4(7), 240–248 (DOI: 10.1021/ar50043a002).
F. J. Stubbs, C. D. Williams, C. N. Hinshelwood, “The Kinetics of Aromatic Sulphonation Reactions. Sulphonation by Sulphuric Acid in Nitrobenzene Solution,” J. Chem. Soc. 1948, 1065–1078 (DOI: 10.1039/JR9480001065).


A molecule absorbs light having a specific wavelength, why doesn't it absorb shorter wavelengths? I am currently studying computational chemistry and I am trying to understand the UV-VIS absorption of molecules. I know that UV-VIS absorption is electronic excitation and requires a specific energy.

What I don't understand is why a molecule can not absorb a shorter wavelength as it has more energy? A slightly higher energy causes vibrational excitations which is then released through vibrational relaxation (releasing heat). Isn't it the case when a very shorter wavelength is used so that the electrons in the molecule are excited from more lower MOs (lower energy) to the higher ones? Isn't it still absorption? It is supposed to be by definition?

Molecule can absorb a photon that has "red" amount of energy, it can also absorb "green" and "blue" amounts. But why can't it absorb anywhere between red and green or more basically isn't it supposed to absorb anything more than "red"? Just to be clear, by red I mean the energy difference between $n = 1$ and $n = 2$.

Also, I am already aware of the fact that, for example, any energy between $n = 2$ and $n = 3$ can not be emitted fully as some of them will be lost for rotational and vibrational transitions until the energy hits the $n = 2$ state but I am not expecting this to stop a molecule from absorbing that light.

Classical approach
Absorption occurs when there are "oscillators" that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies.
These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing their spins(NMR, ESR) orientation (microwave), configuration (infrared) nuclear configuration  ($\gamma$ rays) and finally electronic distribution (VIS, UV, X-Ray).
Of course, not all the materials have oscillators that absorb in all the spectrum. For wavelength between 10 nm and 100 pm the oscillators, that mainly involve electrons in these regions, have to absorb energy between $10^{7}$ and $10^{9}$ J/mol without being ionized, in this case only the inner electrons of some atoms (almost always metals) absorbs these wavelengths see XAS, so in our organic world seems that most compounds are X-Ray transparent but in fact there are a lot of inorganic compounds that absorb X-Ray.
When radiation is absorbed there are many ways to release his energy(see figure 1). We can divide them into two great categories radiative relaxation and non-radiative relaxation. In fact radiative relaxation is most common when we are dealing with shorter wavelength because with more energy is possible to excite an electron to a higher level and this can sometimes re-emit radiation when it goes from a higher level to the ground level, this as you know, is called fluorescence and is very common when you deal with UV, and X-Ray (see X-Ray Fluorescence spectroscopy). Eventually, there is also non-radiative relaxation process that are the main processes that occur from VIS (note that however is possible a VIS-NIR luminescence for some compounds) to the other spectral region with lower wavelength.

Quantum approach
Compound energy levels are quantized, so compounds can absorb only some discrete energy values. These values are determinate by transitions that respect strictly selection rules. If you know the energy of the transition, from Plank relation:
$$E=\frac{ch}{\lambda}$$
you can see that for every energy value there is a specific wavelength, so this means that compounds can absorb only at specific discrete wavelength. However if you see a UV-VIS spectrum you will see a continuous spectrum because roto-vibrational changes affect the energy levels of the compounds involved.


How are electrons in conjugated molecules excited? I have recently started reading about colour in organic molecules and come across conjugation of pi bonds. My question is pretty short... In transition metal ions I understand colour is caused by molecules jumping between the d sub orbitals as a result of ligand splitting them. The concept of excitation easily applies here. So what I want to know is. How do electrons get excited in a conjugated molecule?

This first picture shows what happens to the molecular orbitals of ethylene if we bring two ethylene molecules (the right and left sides of the molecular orbital vs. energy part of the diagram) close together and form 1,3 butadiene (center of the figure)

The next figure shows the result of adding a third ethylenic group to the molecule and form 1,3,5-hexatriene. Note how in each case the molecular orbitals combine (split) and generate a new set of molecular orbitals; and in each case the separation between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO) decreases.  When light is absorbed an electron is promoted from the HOMO to the LUMO.  If we make the HOMO-LUMO separation small enough by extending conjugation, we can shift absorbtion into the visible region (longer wavelength, smaller HOMO-LUMO energy separation). We can add other groups too, like cyano, nitro, phenyl, carbonyl, etc. to extend the conjugation in our system.


Correlation energy in quantum chemistry In density functional theory (DFT) there is classic Coulomb energy and non-classic correlation energy. But what is difference between them? In books these energies are explained very similarly. Both contributions are explained by the fact that electrons are negatively charged particles.

All mentioned terms -- classical Coulomb, exchange and correlation energy -- have the very same physical origin: the electron-electron Coulomb repulsion.

The split into these contributions is only an artifact arising from Hartree-Fock theory.
Here the electronic wave function is approximated by a Slater determinant (the anti-symmetrized product of orbitals). This results in a mean-field description of the electron-electron interaction, and the error with respect to the exact solution is the correlation energy.

Plugging the Slater determinant into the Schrödinger equation eventually leads to an expression for the Hartree-Fock energy:

\begin{equation}
  E_\mathrm{HF} = \sum_i^{N_\mathrm{elec}} \langle i|\hat h|i\rangle + \sum_{i>j}^{N_\mathrm{elec}} [ii|jj] - [ij|ji]
\end{equation}

The three terms in that equation use a short hand notation. Explicitly written those are the one-electron integrals
\begin{equation}
  \langle i|\hat h|i\rangle = \int\chi_i^*(\boldsymbol x_1)\hat h\chi_i(\boldsymbol x_1) \mathrm{d}\boldsymbol x_1
\end{equation}
which cover the kinetic energy and electron-nuclear Coulomb interaction.

The other two terms originate from the electron-electron interaction.
The first one is called Coulomb integral, and gives the contribution of the classical Coulomb energy
\begin{equation}
    [ii|jj] = \int\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
The only reason for its name is, because with $\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) = \rho_i(\boldsymbol x_1)$ and $\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) = \rho_j(\boldsymbol x_2)$ one obtains the formula for the coulomb interaction of two classical charge distributions $\rho_i(\boldsymbol x_1)$ and $\rho_j(\boldsymbol x_2)$. However, we are doing quantum mechanics here, so do not over interpret that analogy.

The last term is the exchange integrals
\begin{equation}
    [ij|ji] = \int\chi_i^*(\boldsymbol x_1)\chi_j(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_i(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
and simply named like that because it is essentially the Coulomb term but two indices have been exchanged.

In DFT these three terms show up, because it borrows some concepts from HF and but tries to add the missing electron correlation.

Interpretation of Electron Correlation

I always struggle with the "correlated motion" thing to explain what electron correlation is. I don't think it helps anyone to get an intuitive understand what this means. I do like however the following pictures from the book by Helgaker, Jorgensen and Olsen.

They consider the electron density in $\ce{H2}$ in HF (mean-field) approximation in the top plots and the exact solution below.
To the left we have the 1-electron density which is obtained by integration (=averaging) of the 2-electron density the in middle and right plots.
The positions of the two Hydrogen atoms are obvious by the peaks in the 1-electron density.

The 2-electron density of the HF solution has 4 peaks:
The 2 at ($r_1\approx 1$, $r_2\approx -1$) and ($r_1\approx 1$, $r_2\approx -1$) are the probability of finding both electron at different atoms. The other 2 at ($r_1\approx 1$, $r_2\approx 1$) and ($r_1\approx -1$, $r_2\approx -1$) are the probability of finding both electrons at the same atom. The latter 2 peaks are missing in the exact solution, because the electrons try to avoid such situations.
HF however is unable to consider this correlation. Note that this effect is not visible in the 1-electron densities. HF only works based the 1-electron wave functions (orbitals) and constructs the 2-electron wave function by simple multiplication the 1-electron wave functions.


"Proof" that the chemical potential is less than zero for irreversible material changes From Levine's Physical Chemistry 6th edition:

In a closed system, $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP$ not only applies to constant composition but also reversible changes in composition (page 116).
For an irreversible process: $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN \lt -S\,\mathrm dT + V\,\mathrm dP$, thus $\mu\,\mathrm dN \lt 0$ (page 130).

For 1), why do we even bother with the chemical potential for closed systems if we can just find a reversible path in which 1) applies?

For 2), I thought that although $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN$ can apply to irreversible processes, you had to calculate a reversible path. If so, that would mean that the $-S\,\mathrm dT$ on the left was calculated for a reversible path and the $-S\,\mathrm dT$ on the right was calculated for an irreversible path and they wouldn't cancel.

Q1. You won't need the chemical potential terms $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha$ (because it equals zero) in using the differential relation $\text{d}G = -S\,\text{d}T+V\,\text{d}P$ in this case. Nevertheless, the chemical potentials are still relevant in establishing criteria for phase equilibrium: $\mu_i^\alpha = \mu_i^\beta$, etc.

  Why should $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha = 0$?

Reversible changes in composition imply that the process is in equilibrium, so that $\text{d}G = 0$. Assuming thermal and mechanical equilibrium and constant $T$ and $P$ we have $\text{d}T = \text{d}P = 0$, and the result follows by considering the (full) differential relation satisfied by $\text{d}G$.

Q2. Yes, you do need a reversible path. The expression on the right, if you follow its derivation carefully, does represent a reversible path. The inequality occurs because the chemical potential terms have not been accounted for. So the cancellation works out.

It seems, however, that we can derive the same result much more simply. Consider a system at thermal and mechanical equilibrium with constant $T$ and $P$, but not chemical equilibrium. Then $\text{d}G < 0$, since the Gibbs free energy is minimized at equilibrium. If $T$ and $P$ are constant $\text{d}T = \text{d}P = 0$, so all that's left in $\text{d}G$ are the chemical potential terms, which must thus be negative.


Fragmentation of phenyl radical cations in mass spectrometry What are the main fragmentation pathways and products of a phenyl radical cation in mass spectrometry ? I notice sometimes that there's a m/z = 51 peak after the m/z = 77 for phenyl radical cation; I assume due to loss of acetylene.

What about $\ce{phenyl^{*+} -> Dewar\ benzene^{*+} -> cyclobutadiene^{+*} + C2H2}$?


How much does oxygen enrichment improve some combustion engines? This question is about how much the combustion efficiency increases with oxygen enrichment in modern combustion engine designs. A desirable way to answer is to give the fuel efficiency dependency on oxygen percentage of air volume intake.
Some clues can maybe be had from

Engineering Science and Technology,
Volume 19, Issue 1, March 2016, Pages 438-443
Effects of oxygen enriched combustion on pollution and performance characteristics of a diesel engine  P.Baskar, A.Senthilkumar

and

THE EFFECT OF OXYGEN-ENRICHED AIR ON THE PERFORMANCE AND
EXHAUST EMISSIONS OF INTERNAL COMBUSTION ENGINES
,VARADARAJA SETTY, B.E.
MASTER THESIS
IN
MECHANICAL ENGINEERING
, Texas Tech University

Reports like the following are less interesting, since spark timing is the regulating mechanism with increased oxygen levels. Spark timing techniques are considered outmoded.

J.X. Zhou, B. Moreau, Christine Mounaïm-Rousselle, Fabrice Foucher. Combustion, Performance
and Emission Analysis of an Oxygen-Controlling Downsized SI Engine. Oil & Gas Science and
Technology - Revue d’IFP Energies nouvelles, Institut Français du Pétrole, 2016, 71 (4), pp.49.
ff10.2516/ogst/2015035ff. ffhal-01597165f

What does "energy efficienct" mean?
It depends on what you mean by "efficient".
For combustion reactions that involve engines, if "efficiency" means how much work (the "good" type of energy) that comes out of the engine per unit input of fuel, then yes oxygen-enriched mixtures would be more efficient than regular old air.
If instead "efficiency" means how much of the fuel is combusted all the way to $\ce{CO2}$, the answer is also probably yes, but for different reasons.
The efficiency of engines
To understand why, keep in mind that no engine can ever be more efficient than an ideal Carnot engine.  The efficiency of a Carnot engine is $\frac{T_H - T_C}{T_H} = 1 - \frac{T_C}{T_H}$, where $T_C$ is the temperature of a cold reservoir (e.g. our everyday environment)   in Kelvins and $T_H$ is the temperature in Kelvins of a hot reservoir.
Thus, the hotter you can get the "hot" side of Carnot engine, the better.  In a car engine or other Otto engines, this would correspond to the maximum temperature of the combusted air (or oxygen)-fuel mixture just before the expansion stroke.
Adiabatic flame temperature
With air instead of pure oxygen as the oxidizer, some of the energy of combustion goes into heating up inert components of the air, mostly nitrogen.  This lowers the maximum temperature of the air-fuel mixture and decreases theoretically attainable efficiency.
The adiabatic flame temperature is a quantitative number behind this idea.  Wikipedia lists the adiabatic flame temperature, the maximum temperature attainable in the combustion of certain fuels, and the values for fuels burning in air are a bit less than the same fuels burning in oxygen.
Limits of this analysis
Of course, there are many factors that make real engine performance different than ideal engine performance.  In fact, designing engines that can operate with at the extremely high temperatures that would maximize efficiency is tough because of materials problems.  (The adiabatic flame temperature of gasoline in pure oxygen is somewhere around 3800 K, far higher than the melting temperature of steel, for example.). So whether a given real engine designed to work under particular sets of conditions will necessarily do better with pure oxygen instead of air is less clear.  Probably yes?  And for sure an engine designed appropriately for use with pure oxygen would be more efficient than an equivalent one designed for us with air.


Disproportionation of plutonium species - Latimer's diagram Why (according to provided answers) $\ce{Pu(IV)}$ disproportionates  into adjacent species whilst $\ce{Pu(V)}$ does not in acid aqueous solution.

$$\ce{\underset{(+6)}{Pu}O2^2+ ->[\pu{+1.02 V}] \underset{(+5)}{Pu}O2+ ->[\pu{+1.04 V}]\underset{(+4)}{Pu}^4+ ->[\pu{+1.01V}] \underset{(+3)}{Pu}^3+}$$

I thought it is the other way - the more positive potential of $\pu{+1.04V}$ on the right of $\ce{PuO2+}$ species will drive the disproportionation while in $\ce{Pu^4+}$ the higher value of $\pu{+1.04V}$ to the left renders the species by itself thermodynamically stable.

In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar.
Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.

In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states:
$$\ce{2Pu^4+ + 2H2O <=> Pu^3+ + PuO2+ + 4H+}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:
$$\ce{3Pu^4+ + 2H2O <=> 2Pu^3+ + PuO2^2+ + 4H+}$$
Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture.
Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.

Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states:
$$\ce{2PuO2+ + 4H+ <=> Pu^4+ + PuO2^2+ + 2H2O}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$

However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.


How does mixing lemon juice in hooka water affect the composition of the mixture? The other day I saw one of my friends, who was mixing lemon juice directly into the flavour chimney (or wind guard) and water pot of hooka. When I asked why are you doing as such, he said that it makes inhaling hooka smoke passive.  By passive, I mean easily inhalable without any hiccups or harshness.  I have tried it, it really makes the smoke passive. But how did this happen?
Here is a diagrammatic view of working principle of hooka:

Here is a list of ingredients of hooka flavour which he used, it was of about 200 grams:

glycerol 70 %
18 % mollases
5 % synthetic edible flavours
5 % sodium xanthate
0.05 % nicotine and 1.95 % water

the temperature at which glycerol boils (and hence the smoke appears) is 290 °C,
By adding lemon juice, we are actually adding citric acid to it, so what would be the products obtained heating that flavour at the given temperature which reduces harshness of hooka.
My friend says the water present at the bottom in the hooka pot, serves for the purification of nicotine present in the flavour (which I can't believe), and he further adds that adding lemon juice to it will act as a filter for nicotine, hence reducing its harshness. [I think there shouldn't be any harshness due to nicotine as it is in a very low concentration i.e. measurable in the order of milligrams per puff, see the comments beside the answer below]
So the harshness must be due to other substances present in the mixture, such as glycerol ( which wouldn't cause harshness as it is in the vapour state), mollases and other flavoured substances along with sodium xanthate, here is molecular structure of mollasses

is mollasses causing harshness, then what is the left over debris after all the flavour is exhausted, many individuals say that the left over & exhausted part of the flavour is nothing but the molasses itself.
and coming to the other substances present in the flavour.
the knowledge which i have gathered from online sources is, the reaction between glycerol and citric acid yields a polymer in a controlled and catalysed reaction, called PGC, which is generally obtained at the temperature of 150°C,  but here in this case the temperature is above 290°C i.e. boiling point of glycerol.
so the fact is that, no polymer is obtained while inhaling hooka, as the glycerol is in its vapour state and citic acid being, very acidic {1M citric acid has a pH value of 2}, so it is nearly in its dissociated state {the lemon juice was mixed in water present in the water pot},

Citric acid

Glycerol

The acidic water is probably protonating nicotine, causing it to carry a positive charge. Once charged, the resulting nicotinium cation will not be volatile, but rather, highly soluble. Based on this, I would expect that acidifying the water causes a much lower amount of nicotine  to be pulled through the exit hose (because it is trapped in the acidic water).


Derivation and visualization of cell potential dependency on temperature and entropy Working from the thermodynamic identity
$$\Delta{S} = -\left(\frac{\partial{\Delta{G}}}{\partial{T}}\right)_p,$$
is it allowed to integrate both sides to obtain an expression derived from the Nernst equation
$$E_\mathrm{cell} = \frac{T\,\Delta S}{zF}?$$
Can one plot the results and obtain $\Delta S$ from the gradient, assuming $\Delta S$ is constant for the reaction and $E_\mathrm{cell}$ values at some temperatures are given?

What you are suggesting in your equation is that $\Delta G_m=-T\Delta S_m$ but you know that in fact $\Delta G_m=\Delta H_m-T\Delta S_m$. Your math is incorrect because you are missing a constant of integration.
The expression $\Delta G_m=\Delta H_m-T\Delta S_m$ can be rewritten using the Nernst equation as
$$-nFE = \Delta H_m -nFT\left(\frac{\partial E}{\partial T} \right)_p$$
by using the definition you provide for $\Delta S_m$ as the temperature-derivative of $\Delta G_m$, and the Nernst equation $\Delta G_m = -nFE$. In fact
$$\Delta S_m = nF\left(\frac{\partial E}{\partial T} \right)_p$$
Therefore if you plot $nFE$ as a function of T the slope of the curve at a given temperature will provide the value of $\Delta S_m$ at that T.


Why do selenium , tellurium and polonium have more negative electron gain enthalpy than oxygen? In my book(NCERT Chemistry Part I, Textbook for Class XI[1]) a list of values of electron gain enthalpies of various elements is given. According to that list, oxygen has less negative electron gain enthalpy than selenium, tellurium and polonium. Shouldn't it be the opposite because electron gain enthalpy becomes less negative down a group?

Reference:
[1]: Rajput, Anup, Kumar. Classification of Elements and Periodicity in Properties. NCERT Chemistry Part I, Textbook for Class XI, First Edition(October 2019 Reprint); Uppal, Shveta, Banerjee, Binoy, Eds.; National Council of Educational Research and Training: New Delhi, India, 2006; pp 90.

Electron Gain Enthalpy becomes less negative down the group.

This is true, as a general trend. But the 2nd period is an exception to this trend.
In p block, it is observed that the EGE of some 2nd period elements is actually less negative then their 3rd period counterparts.
In the case of oxygen, it’s EGE is even less than selenium, tellurium and polonium. It’s because oxygen has very small radius and thus have high electron density. Due to this, the incoming electron faces a lot of repulsion from the already present electrons.
So, it takes more energy to add an electron to an oxygen atom i.e. oxygen has less negative EGE. On the other hand, selenium and tellurium have much larger radius and thus the electron density isn’t very high. Hence, their EGE is more negative as its easier to add an electron to them.


How to label atoms or groups "backwards" in ChemDraw I'm trying to draw L-tyrosine with the hydroxyl part of the phenol as HO instead of OH as it appears in the ChemDraw image. Everytime I type in HO, it automatically corrects to OH. I have the settings in ACS Document 1996 style. Is there anyway to make the format render my "backwards" type?

ChemDraw is a piece of software that attempts to be as intelligent as possible. In many cases, that is truly helpful, in others it can be annoying.

When labelling an atom in text mode, ChemDraw will automagically assume a label from left to right (i.e. in standard North American/European writing direction) and interpret the label as such. If it finds that the atom has bonds to the right but none to the left, as tyrosine’s hydroxy group has in your example, it will group the label into ‘sensible’ fragments[1] and reverse them. This makes sense if you think of rotating a molecule: If you had drawn tyrosine rotated by $180^\circ$, you would have wanted to add OH on the rightmost atom, not HO. If you then manually rotate the entire molecule within ChemDraw, you will notice the OH flipping into an HO as soon as it graphically makes sense to do so.

Note that if you just hover over the atom in lasso mode or bond adding mode and then hit o (lowercase!) it will automatically add OH (or HO, depending on the orientation) at that position. Hitting o again will remove the additional hydrogen (this holds true for all chemical elements that are attached to single letters regardless whether the atom is attached to a lowercase letter (i.e. o) or an uppercase one (B for boron)).

[1]: Consider OTBS: ChemDraw automatically recognises TBS as an abbreviation for the $\ce{TBS}$ group so when reversing, it becomes TBSO. However ChemDraw does not standardly recognise the triethylsilyl group $\ce{TES}$, so when reversing OTES it arrives at SETO. The preferred way would be to add an abbreviation to ChemDraw.


Where does the extra electron go in ionic compounds? In ionic compounds, where does the extra electron go? Or, in case of a cation, from which shell does it come?

Consider, for example, $\ce{OH^{-}}$, where did this extra electron go in the electronic configuration? And, what about $\ce{OH^{+}}$? There exists no such ion. Why?

In $\ce{OH-}$, that extra electron is on the Oxygen atom.  Oxygen has a very large electronegativity, so you can be sure that extra electron is with the $\ce{O}$ not the $\ce{H}$.  A lone oxygen has 6 electrons in its valence shell (2s2, 2p4), and wants to get to 8 (2s2, 2p6).  In the $\ce{OH-}$ ion, both atoms are relatively "happy", the $\ce{H}$ has 2 electrons in its valence shell (1s2) shared with Oxygen in a bond, and the Oxygen has 8 electrons in its valence shell (2s2, 2p6), 6 unpaired electrons and 2 shared with hydrogen.

In general, in an anion the extra electron will be going onto the most electronegative atom.  In a cation the positive charge will go to the lest electronegative atom.

Here's a helpful picture, your extra electron is in red:

picture reference: http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html

As for your question about $\ce{OH+}$, it would never be stable because that oxygen is going to tear some electrons off of the first thing it sees, because it REALLY wants that full octet.  If you somehow made some $\ce{OH+}$, my guess would be that the $\ce{H+}$ would just dissociate, and your $\ce{O}$ would find another $\ce{OH+}$'s $\ce{O}$ and make $\ce{O2}$ immediately.


How to understand the (non-)degeneracy of states of singlet Σ and Δ terms for a ππ configuration of two nonequivalent electrons? As instructed in Levine's Quantum Chemistry 7th edition, Chapter 13.8, page 377, for a $\pi \pi$ configuration of two nonequivalent electrons, four functions $\pi_{+1}(1) \pi_{-1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{-1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{+1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{+1}(1)$
 are used to construct the wave functions of the singlet $\Sigma$ terms
$$
\begin{array}{ll}{^{1} \Sigma^{+}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)+\pi_{-1}(1) \pi_{+1}^{\prime}(2)+\pi_{-1}(2) \pi_{+1}^{\prime}(1)} \\ {^{1} \Sigma^{-}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)-\pi_{-1}(1) \pi_{+1}^{\prime}(2)-\pi_{-1}(2) \pi_{+1}^{\prime}(1)}\end{array}
$$

The text following these equations says "Clearly, these two spatial functions have different energies." Could you suggest how to understand this point ?

The discussion further below suggests that when four functions $\pi_{+1}(1) \pi_{+1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{+1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{-1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{-1}(1)$
 are used to construct the ${}^1 \Delta^{+}$ and ${}^1 \Delta^{-}$ wave functions of the singlet $\Delta$ terms, "Since they have the same energy, there is no
point in using the $+$ and $-$ superscripts". Could you suggest how to understand this point and the reason of the difference from the above point ?

Many thanks for your suggestions !

The setup here is a diatomic molecule with two $\pi$ electrons in different subshells. A typical scenario would be one in a $\pi$ orbital and the other in a higher energy $\pi^*$ orbital.

In a diatomic molecule aligned with its molecular axis on the z-axis, the $\pi$ subshell consists of two orbitals, one with a node in the x-z plane and one with a node in the y-z plane. The $\pi^*$ subshell likewise consists of two orbitals, which are the antibonding counterparts to the two $\pi$ orbitals.

In the cited text, "+1" and "-1" indicate the two orientations, so it might be simpler to replace them with x and y. Likewise, the "$\pi'$" can be changed to $\pi^*$ for clarity.

Now consider the first $^1\Sigma$ term:
$$^1\Sigma^+: \pi_{+1}(1)\pi'_{-1}(2) + \pi_{+1}(2)\pi'_{-1}(1)+\pi_{-1}(1)\pi'_{+1}(2)+\pi_{-1}(2)\pi'_{+1}(1)$$

We can rewrite this in the friendlier notation and use only every other term, since the others are just exchange within the same orbitals. (What follows here is conceptual, not mathematically rigorous.):
$$^1\Sigma^+: \pi_x\pi^*_y + \pi_y\pi^*_x$$

The second $^1\Sigma$ term likewise simplifies to:
$$^1\Sigma^-: \pi_x\pi^*_y - \pi_y\pi^*_x$$

Now we further separate the contributions from each axis:
$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y $$

We see that what we have done is to "undo" the formation of the $\pi$ and $\pi^*$ orbitals and essentially restored the original atomic $p$ orbitals:

$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y=p_{Ax} + p_{Ay} $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y=p_{Bx} + p_{By}$$

where A and B are the two atoms in the molecule. If A and B are different elements, these terms will have different energy.

If you do the same exercise with the $^1\Delta$ terms, you'll get something like:
$$^1\Delta^+: p_{Ax} + p_{By}$$
$$^1\Delta^-: p_{Bx} + p_{Ay}$$

Since both terms have equal contributions from A and B, they will have the same energies whether or not A and B are the same element. And since the orbitals on the two axes are orthogonal to each other, the change in sign does not affect the energy.

Conceptually, the sigma terms are diradicals where either both electrons are on A or both electrons are on B in orthogonal p orbitals. The delta terms are diradicals in which one electron is on each atom.


Boiling aluminum in high-alkalinity water This question ultimately relates to brewing beer, but more specifically to water treatment before brewing.

Though I can't get a water quality report, I suspect I have high levels of bicarbonate alkalinity in my tap water because boiling it causes a white precipitate (which I think is calcium carbonate) to form. I base this assumption on this formula, in the book Brewing:

$$\ce{Ca(HCO3)2 ->[heat] CaCO3 + CO2 + H2O}$$

I wanted to sanitize some aluminum foil so I boiled it in fresh tap water for about 15 minutes. I noticed a few things:

The precipitate formed much faster than it normally would without foil in it.
The foil began to blacken in certain parts.
I'm not positive, but the precipitate seems a bit finer in texture than it normally is.

I thought at first maybe the foil was just providing nucleation points for the carbonate to form around. But then I thought I remembered reading somewhere that aluminum competes with calcium in certain situations (I think it was regarding osteoporosis, maybe). The question I have, then, is whether or not it's feasible that aluminum could be affecting a different reaction than what I'm assuming normally occurs in this situation?

It appears your creating calcium aluminate by reacting aluminum with alkaline water. Since that is probably not wanted in the brew, you might use glass or anodized aluminum, which is more resistant to chemical attack, and a water softener, if the calcium carbonate isn't wanted, either.


Can ozone have a triangular structure? The structure of ozone is traditionally depicted using two resonance forms:

However, what I am wondering is instead of using such an idea to explain the structure why can't we show the structure of ozone like this?

Some people have told me that such a structure is not possible but they fail to mention the reason. What I have in mind is maybe electron repulsion. But I would like to know exactly what mechanism would prevent this from happening?

Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\pi_\mathrm u$ orbitals are both singly occupied.

Note: 6a$_1$ and 2b$_1$ have the incorrect sign on the outer p-orbitals.

If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\mathrm{6a_1}$ and $\mathrm{2b_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.

So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\mathrm{1a_2}$ orbital and the $\mathrm{4b_2}$ orbital start rising in energy, eventually rising above the $\mathrm{6a_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\circ$.

This same diagram can be used for other molecules, such as $\ce{CO2}$. Try using it to figure out why $\ce{CO2}$ is linear.


How can you calculate the tendency of one compound to become another? Excuse my inexperience with chemistry here. I've been poring over papers with terms like "Standard Gibbs free energy of formation", "Hess's Law", "standard enthalpy of formation", "entropy", and several others, and I haven't been able to find an answer to one simple question:
Given I have a system with compounds A, B, and C, and there is an alternative set of compounds X, Y, and Z involving the same amounts of the same elements, how do I calculate how much the ABC system wants* to become XYZ?
As an example, if I have a system with propane and oxygen, with an excess of oxygen, this system wants to combust to become carbon dioxide and water vapor. A quick Wikipedia check reveals the balanced equation for this reaction:
$$\ce{C3H8 + 5O2 \rightleftharpoons 3CO2 + 4H2O}$$
Anyone who's played around with a camp stove knows that propane and oxygen like to burn to become carbon dioxide and water vapor, and that carbon dioxide and water vapor really don't like to "unburn" to become propane and oxygen. How would I measure this tendency for the propane-oxygen system to become the carbon-dioxide-water-vapor system? Similarly, how would I measure the distate for the carbon-dioxide-water-vapor system to become the propane-oxygen system?
If I remember correctly, the tendency of a reaction to happen and the speed of the reaction are entirely unrelated, one being a thermodynamic property and the other a kinetic property, but if you wouldn't mind throwing in just enough info about reaction rate in the above case that I can look up and learn myself on that topic as well, I'd really appreciate it.
*I'm falling victim to the oh-so-elegant abstraction of molecular anthropomorphism here. Yes, I understand that chemicals don't want anything, but given I'm looking for a unit most easily expressed as the degree of desire for chemicals A, B, and C to become X, Y, and Z, these human terms play out quite well. Please do let me know if there's better terminology I could be using

Depending on the constraints you impose on the system, there are different criteria for spontaneity, however, the commonly running theme is that it is related to minimizing some state function. For example, as a simple example from classical mechanics, the time evolution of the system is such that the potential energy of the mechanical bodies is minimized.
When we do a chemical reaction, we usually do it at constant temperature and pressure. In this case, the evolution of the system is determined by a potential function known as Gibb's free energy. If there is a state in which the system which reduces Gibb's free energy, then the system will evolve to reach that state. When I say evolve I mean to have processes such as chemical reactions.
I had actually written a very detailed answer about this on physics stackexchange, check it out for more information (here)

Refer these notes by MIT for the criteria for spontaneity for different constraints :
MIT ocw notes
Discussion about constant P and T chemical reactions:
Here


Aldehyde and ketone nomenclature when aldehyde/ketone group is bonded to an alkyl group The longest alkane chain is decane and, attached to the fifth carbon, is an ethyl group the end of which is double bonded to an oxygen atom. My best guess at naming it would be:
5-(ethyl-2-al)decane.

Here, the longest chain is again decane, but there is a propyl group and an oxygen atom double bonded not to the ends of the alkyl, but to carbon #2, which leads me to:
5-(propyl-2-one)decane
Now I have no clue as to whether I am right or not because I have never come across such molecules and I have just done what I found to be logical. Have I come to the correct conclusions?

The characteristic group $\ce{=O}$ corresponds to the substitutive prefix ‘oxo’.
Thus, the substitutive names that you are looking for are ‘5-(2-oxoethyl)decane’ and ‘5-(2-oxopropyl)decane’.

However, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred names for aldehydes and ketones derived from alkanes are constructed using the suffixes ‘al’ and ‘one’, respectively.

Furthermore, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present.

Therefore, the preferred IUPAC names for the given compounds are ‘3-butyloctanal’ and ‘4-butylnonan-2-one’.


Can molten sugar also rotate polarized light as sugar water does? It is well know that the sugar water can rotate the plane of linearly polarized light, so light bends when travelling in sugar water. However, do the liquid state of sugar but not the sugar solution also have this property?

Molecules in every state - liquid, solid and gaseous - rotate the plane of linearly polarized light - it doesn't matter, because this rotation arises from inherent asymmetry of molecules, which is causing their optical activity

More to the point, not only water solutions of sugars, but also their crystals, molten sugars and even vapor, rotate the plane of linearly polarized light. If you're asking specifically about sucrose checking this may be problematic as it's decomposing during melting, as you can read in this question and here, but I think it still can be done, under vacuum.

You can also check these related questions about chiral gases and rotation of the plane of polarization by molecules


Is there radioactivity at absolute zero? Theoretically, will a radioactive material still be radioactive at absolute zero? What would happen at the lowest realistic temperatures we have ever achieved?

Will radioactivity stop at absolute zero, since it is a nuclear phenomenon and nuclear motion slows down as we approach absolute zero (and theoretically stopping entirely at absolute zero)?

Theoretically, a radioactive material will still be radioactive at absolute zero, and its rate of decay will be $100.00\%$ of that at room temperature. Practically, at the lowest achievable temperatures we observe the same thing: radioactivity is still there, not affected the slightest bit.

Nuclear motion does not slow down as we approach absolute zero, because there is no such thing as nuclear motion in the first place. In a way, all nuclear motion has stopped already at room temperature. Each nucleus just sits there in the ground state and does not know what happens in the chemical world above. From its point of view, the room temperature is the same as absolute zero. To reach its first excited state, it would need energies a great deal greater than that.

Say, you heat your radioactive sample until it melts. Then you heat it a few more thousand degrees, until all materials, including tungsten, melt and then evaporate. Then you heat it some more, until even the strongest chemical bonds are broken and there are no more molecules, just atoms. Then you heat it about ten times more, until atoms lose much of their valent electrons and you have a highly ionized plasma. Then you heat it about a hundred times more, until all atoms lose all their electrons and you have something like a stellar plasma. Then you heat it some more, just in case. Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort.

Short of that, you could just as well have asked if the radioactivity in a sample stops when you paint it blue.


Diastereomers, enantiomers, and meso compounds of molecules with lots of chiral centres For a molecule with 2 chiral centres, there are four possible combinations, (R,R), (R,S), (S,R) and (S,S). For example, 3-bromo-2-butanol has four stereoisomers: (2R,3R)-3-bromo-2-butanol, (2R,3S)-3-bromo-2-butanol,
(2S,3R)-3-bromo-2-butanol and (2S,3S)-3-bromo-2-butanol.

Of these combinations, the following pairs are diastereomers:

S,R and S,S

S,R and R,R

R,S and S,S

R,S and R,R

I notice that only one chiral centre has been flipped to form these pairs.

The following pairs are either enantiomers or meso compounds (e.g. for (2R,3S)-tartaric acid):

S,R and R,S

S,S and R,R

I notice that both chiral centres have been flipped to form these pairs.

Could we  generalise that flipping an even number of chiral centres always gives either an enantiomer or the same molecule if it's a meso compound, and flipping an odd number of chiral centres always gives a diastereomer?

Could we generalise that flipping an even number of chiral centres always gives either an enantiomer or the same molecule if it's a meso compound, and flipping an odd number of chiral centres always gives a diastereomer?

No. The generalization is slightly different: Comparing chiral centers in two stereoisomers with more than two chiral centres, if

all chiral centres are the same, it is the same molecule.
some but not all are switched, it is a diastereomer.
all are switched, it is an enantiomer (or meso compound).

Open-chain carbohydrates are a good example (compare D-glucose, L-glucose, D-galactose, D-talose etc.). For this class of molecules, there is one more term, epimer (when exactly one chiral centre is switched).

This generalization excludes the more exotic cases of stereoisomers, see e.g. Why are allenes chiral? and Why atropisomers are called conformers?


How can I tell which parts of this example drug are necessary parts of its pharmacophore? $\hspace{10mm}$
$\hspace{10mm}$(Click image for larger version)

This picture identifies a generic drug-enzyme complex taken from an exam script. There is no name given.  One part of the exam question states:

  Propose a structure based pharmacophore for the family of drugs to which this one belongs.

I understand that a pharmacophore is the minimum structure required to exert a certain biological activity. Therefore, my answer would go something like the following:

  We would need a drug with similar steric conformities to fill the entire active site (competitive inhibition) and that has some form of interaction at either end which keeps it in place, e.g., $\pi-\pi$ stacking with Phe groups or the H-bond and electrostatic interactions with Arg.

But would that paragraph above be a viable answer? Would this drug work just as well if it didn't have the $\pi-\pi$ stacking with the Phe groups, for instance? Are the interactions between the drug and the Lys and Glu residues dispensable - maybe that region of the drug is only there for its length?

From this? You can't. You need more information. Examples:

Do you know other drugs of this family? Which chemical features do they have in common? The interactions mediated by these features are likely important then.
Do you know how the drug is supposed to function (you mention competitive inhibition)? What does the natural substrate look like, and what are the similarities and dissimilarities to your example drug?

If you know only the figure shown above, you're really asked to identify electronic and steric features of the drug, that can possibly interact with the amino acid residues in the active site shown. In this case, all amino acid residues are capable of favourable interaction with a part of the example drug, so I would mention which interactions are present.

There are many good ionic interactions/H-bond interactions, which are qualitatively the strongest.

One possibility is that it is sterically unfavourable to jam something into the hydrophobic pocket, but the payoff is, that this is what stabilises a certain (active or inactive) conformation of the protein, and the ionic/H-bond interactions simply serve to increase binding affinity / to keep the drug in place.

Another possibility is that the interaction with the hydrophobic pocket only increases binding affinity. The drug then might share the same ionic interactions as the real substrate, but placed differently in space, so they don't induce the same conformation change as the natural substrate.

The point is, without knowing what the action of the drug is, and some knowledge about other drugs of the same family or the natural ligand/substrate, anything beyond a listing of the interactions you can spot between the amino acid residues shown, and the chemical features of the drug will really just be guess work :-).


Is reduction of copper oxide a combustion reaction? Is the following reaction a combustion reaction or just a single displacement? Could be both?
$$\ce{H2 + CuO -> Cu + H2O}$$

A friend of mine claims it is a combustion reaction because it is exothermic and a redox reaction.

By the question asked, I assume OP is in high school of freshman in college. Thus, I'd use this general description of combustion as a starting point:

  Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't $\ce{O2}$ or $\ce{F2}$) is called the fuel. Combustion reactions are a type of redox reaction.

As  M. Farooq pointed out a combustion reaction happens quickly, producing heat, and usually light and fire. For example, lets look at combustion reaction of an alkene (a hydrocarbon). If it is a complete combustion, the fire have a blue flame:

$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$

If it is a partial combustion, it can have a multiple $\ce{C}$ compounds as products, and have a yellow flame due to presence of elemental $\ce{C}$:

$$\ce{C_nH_{2n} + x O2 -> m C + p CO + $(n-m-p)$CO2 + n H2O}$$

where $x = \frac{2(n-p-m) +p}{2} = \frac{2n-2p-2m +p}{2} = \frac{2n-p-2m)}{2}$. In your reaction would not produce fire and it didn't use either oxygen or other oxidants ($\ce{CuO}$ is not that type of oxidant). It is true that the reaction is a redox reaction.


Chemical elements or compounds with yellowish white color Are there chemical elements or compounds that have a yellowish white (cream) color in their natural state?

Aluminium(III) chloride - often described as white but samples is sometimes contaminated with iron(III) chloride giving it a light yellow color
Antimony(V) oxide - pale yellow solid
Bismuth(III) oxide - light yellow solid
Rhodium(III) oxide - lemon yellow
Cerium(IV) oxide - pale yellow-white compound
Dysprosium(III) nitrate - white to pale yellow
Holmium(III) oxide- Color depends on lighting condition. In daylight, it is tannish light yellow; Under trichromatic light, it is fiery orange red (Thanks @EdV)
Samarium(III) oxide - light yellow
Silver(I) iodide - samples can have a range of color(white to pale yellow to yellow) due to impurity and photosensitization
Selenium(IV) chloride - white to faint yellow volatile solid
Gold(III) chloride - pale yellow to yellow
Potassium Telluride - pale yellow
Thallium(III) bromide - pale yellow
Some vanadate salts are light yellow

Bromium nitrate is pale yellow but not confirmed. Silver(I) bromide and lead(II) bromide are light yellow and white cream colored respectively.
Above data source: http://lanthanumkchemistry.over-blog.com/article-the-colors-of-chemicals-complexes-104821449.html


Would standing next to a pool of mercury do you any harm? I was watching this clip from the 2017 film The Mummy in which the characters stand next to a pool of mercury which was being used as part of an elaborate prison.

Several comments under this video imagine that standing next to a pool of mercury would be very dangerous, but I can't see any reason to think that a significant amount of vapour would be coming from the pool so it seems like the mercury just being there in liquid form would be quite harmless. But maybe I shouldn't go robbing any graves any time soon.

The automatic warning dialogue told me the question appears quite subjective. So, let's say you're standing next to the pool for no more than ten minutes and we'll define harm as requiring medical intervention or (if chronic) as having a demonstrable, long-term health effect, such as erethism. I have no idea how large the pool is in the video, but for the purposes of the question, we'll say it's a circular pool of 10 metres in diameter.

It depends. But there is little short term risk.

Mercury metal isn't very toxic. You can even ingest it without significant short or long term effects. But mercury vapour is nasty and you don't want to be continuously exposed to any significant amount of that (see this answer for more detail). And some mercury compounds are pretty nasty too.

Of course mercury metal emits mercury vapour so there will always be some in any room with exposed mercury. But the reason why this isn't an immediate worry is that it takes time to absorb enough vapour for it to become notably toxic. Chemists are used to handling liquid mercury in the open lab but are a lot less comfortable if it has been spilled and not cleaned up. This didn't used to be so which is one of the reasons we know the dangers: chemists who were exposed to vapour over years did suffer toxic effects. Mercury miners suffered similar long term toxic effects from the vapour. In modern labs we make sure the vapour doesn't build up by good ventilation and making sure any spills are cleaned up quickly so exposed mercury doesn't hang around emitting vapour.

So would you be in danger standing next to a pool of mercury? No, if you didn't do it for long. Or, if the ventilation in the room were good and the vapour wasn't allowed to build up. Even then, though, modern safety standards wouldn't be comfortable having pools of mercury exposed to the atmosphere: you wouldn't want to regularly work in that room with the pool.

In short, standing next to a pool of mercury even in a badly ventilated room wouldn't cause immediate harm; working next to that pool for months or years would certainly cause you harm.


Accuracy benchmarks for atomic multipole moments Thermochemical accuracy is regularly quoted as 1 kcal mol$^{-1}$, or 4.2 kJ mol$^{-1}$. I'm expressing some data and I want to use benchmarks like this to validate them, however, they are not all in units of energy. My analyses involve various ranks of a multipole expansion.

I have found a reference when benchmarks dipole moments$^{1}$, but I can't find anything in the literature which similarly benchmarks higher rank multipole moments. Any pointers to the literature or comments from experience would be greatly appreciated.

[1]: J. Phys. Chem. A, 2014, 118 (20), pp 3678–3687.

To the best of my knowledge, there is no definition of "chemical accuracy" for higher-order (electric) multipole moments, in part because they are difficult to measure experimentally, with relatively large uncertainties. I found one recent paper that has a small comparison, prefaced with:

  There are several techniques to determine experimentally the dipole moments, [4,5] but it is still very difficult to obtain precise experimental values of higher multipole moments such as quadrupole or octupole moments, [1,6,7] independently of the experimental conditions. Theoretical calculations are therefore essential but challenging for quantum chemistry methods. The accurate calculation of these properties is highly dependent on the method employed, [8] either regarding approximate density functionals [9] or methods based on wavefunctions. [10,11] Consequently, calculating the multipole moments is a way to assess any electronic structure method.

The authors use this work as a method development showcase for the Piris natural orbital functional 6 (PNOF6), the details of which I will generally leave out, but they provide valuable data for Hartree-Fock (HF), coupled cluster with single and double excitations (CCSD), and multireference configuration interaction with single and double excitations (MR-CISD, they abbreviate as MRSD-CI after the old abbreviation SDCI). I will leave in the PNOF data, since it (to my understanding) is a natural orbital-based almost black-box correlated method when the number of occ-virt correlated pairs can be controlled. For the molecules studied (small, organic, closed-shell), CCSD in a triple-$\zeta$ basis set is most likely a reasonable choice.

  Table III. $\Theta_{zz}$ component of the quadrupole moments, in atomic units, computed with the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.

  \begin{array}{lllll}
\hline
                     & \text{HF} & \text{CCSD} & \text{Expt.}             & \text{PNOF6} \\
\hline
\ce{H2}              & 0.4381    & 0.3935      & 0.39 \pm 0.01            & 0.3935 \\
\ce{HF}              & 1.7422    & 1.7156      & 1.75 \pm 0.02            & 1.6939 \\
\ce{BH}^{a}          & 2.6772    & 2.3388      & 2.3293~\text{(FCI/aVTZ)}^{b} & 2.3706 \\
\ce{HCl}             & 2.8572    & 2.7233      & 2.78 \pm 0.09            & 2.7753 \\
\ce{HCCF}            & 3.3530    & 2.9335      & 2.94 \pm 0.10            & 3.2482 \\
\ce{CO}              & 1.5366    & 1.4889      & 1.44 \pm 0.30            & 1.4562 \\
\ce{N2}              & 0.9397    & 1.1712      & 1.09 \pm 0.07            & 1.0530 \\
\ce{NH3}             & 2.1258    & 2.1661      & 2.45 \pm 0.30            & 2.1080 \\
\ce{PH3}             & 1.7217    & 1.5695      & 1.56 \pm 0.70            & 1.6507 \\
\ce{ClF}             & 0.9413    & 1.0514      & 1.14 \pm 0.05            & 1.1122 \\
\ce{CH3F}            & 0.3482    & 0.3002      & 0.30 \pm 0.02            & 0.3269 \\
\ce{C2H2}            & 5.3655    & 4.6850      & 4.71 \pm 0.14            & 5.1531 \\
\ce{C2H6}            & 0.6329    & 0.6234      & 0.59 \pm 0.07            & 0.6275 \\
\ce{C6H6}            & 5.5418    & 5.6653      & 6.30 \pm 0.27            & 6.3571 \\
\ce{CH3CCH}          & 4.2913    & 3.6939      & 3.58 \pm 0.01            & 4.1146 \\
\ce{CO2}             & 3.8087    & 3.1966      & 3.19 \pm 0.13            & 3.6012 \\
\hline
\text{MAE}           & 0.2646    & 0.0902     &                           & 0.1517 \\
\text{MAE}~(\mu_{z}) & 0.0843    & 0.0177     &                           & 0.0309 \\
\hline
\end{array}

  a. Calculations performed with the aug-cc-pVTZ basis set.
  b. Full CI calculation reported by Halkier [29].

I included the mean absolute error (MAE) for the dipole results from Table II to give an idea of the relative difficulty compared to calculating quadrupole moments. The quadrupole MAEs are about 3, 5, and 5 times larger than the dipole MAEs for each method.

Plotted, it looks like

which is admittedly not very useful other than to show pictorially that CCSD is generally in at least qualitative agreement with the experiment. I personally would consider HF in qualitative agreement to the level that HF usually gets geometries qualitatively correct, though bond lengths are usually too short due to lack of electron correlation. Mean-field electrostatics are the dominant contributor.

For percent errors,

\begin{array}{lrrr}
\hline
&  \delta_{\text{HF to CCSD}} &  \delta_{\text{CCSD to Expt.}} &  100 * \text{err}/\text{Expt.} \\
\hline
\ce{H2}        &                       11.33 &                           0.90 &                           2.56 \\
\ce{HF}        &                        1.55 &                           1.97 &                           1.14 \\
\ce{BH}        &                       14.47 &                           0.41 &                           0.00 \\
\ce{HCl}       &                        4.92 &                           2.04 &                           3.24 \\
\ce{HCCF}      &                       14.30 &                           0.22 &                           3.40 \\
\ce{CO}        &                        3.20 &                           3.40 &                          20.83 \\
\ce{N2}        &                       19.77 &                           7.45 &                           6.42 \\
\ce{NH3}       &                        1.86 &                          11.59 &                          12.24 \\
\ce{PH3}       &                        9.70 &                           0.61 &                          44.87 \\
\ce{ClF}       &                       10.47 &                           7.77 &                           4.39 \\
\ce{CH3F}      &                       15.99 &                           0.07 &                           6.67 \\
\ce{C2H2}      &                       14.53 &                           0.53 &                           2.97 \\
\ce{C2H6}      &                        1.52 &                           5.66 &                          11.86 \\
\ce{C6H6}      &                        2.18 &                          10.07 &                           4.29 \\
\ce{CH3CCH}    &                       16.17 &                           3.18 &                           0.28 \\
\ce{CO2}       &                       19.15 &                           0.21 &                           4.08 \\
\hline
\end{array}

this is further confirmation that CCSD is probably a good qualitative benchmark.

  Table IV. $\Theta_{zz}$ and $\Theta_{xx}$ components of molecular quadrupole moments, in atomic units, computed using the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.

  \begin{array}{lrrr}
\hline
                & \text{HF}  & \text{MR-CISD} & \text{Expt.}  & \text{PNOF6} \\
\hline
\ce{H2O}~(xx)   &  1.7966    & 1.8050         & 1.86 \pm 0.02 & 1.7808 \\
\ce{H2O}~(zz)   &  0.0981    & 0.0950         & 0.10 \pm 0.02 & 0.0869 \\
\ce{H2CO}~(xx)  &  0.1019    & 0.1100         & 0.04 \pm 0.12 & 0.0516 \\
\ce{H2CO}~(zz)  &  0.0921    & 0.2230         & 0.20 \pm 0.15 & 0.1255 \\
\ce{C2H4}~(xx)  &  2.7819    & 2.3700         & 2.45 \pm 0.12 & 2.5892 \\
\ce{C2H4}~(zz)  &  1.4942    & 1.1700         & 1.49 \pm 0.11 & 1.3266 \\
\ce{O3}~(xx)    &  1.1175    & 1.2830         & 1.03 \pm 0.12 & 1.2426 \\
\ce{O3}~(zz)    & -0.2387    & 0.1680         & 0.52 \pm 0.08 & 0.3606 \\
\hline
\text{MAE}      &  0.1772    & 0.1448         &               & 0.1066 \\
\hline
\end{array}

The ozone result for Hartree-Fock most likely has the wrong sign due to its multiconfigurational ground state, with the MRCI result still in relative error of ~67%.

For octupole moments, they only present results for methane ($\ce{CH4}$), for which the first non-vanishing multipole moment is the octupole moment, due to its $T_{d}$ symmetry.

  \begin{array}{ll}
\hline
                 & \Omega_{xyz}~(\text{a.u.}) \\
\hline
\text{CCSD}      & 2.0595                     \\
\text{PNOF6(14)} & 2.1142                     \\
\text{Expt.}     & 2.95 \pm 0.17              \\
\hline
\end{array}

Considering the lower bound of the experiment, the PNOF6 result is in 23% error, and considering the upper bound of the experiment, the CCSD result is in 36% error.

I won't draw any more conclusions about the data other than to say that in general, there may be large errors in the experimental data that are hard to predict (non-systematic). In that case, if you were to perform a study looking at convergence of a result with respect to some other parameter (say, cluster size), I would first look at convergence of said result with respect to method (HF, MP2, CCSD) and basis (aVDZ, aVTZ, ...) at the largest cluster size, just to check that CCSD/aVTZ is a reasonable benchmark if you're after absolute accuracy. Since experimental error is large, it may be more reasonable to just compare against higher-level, full cluster calculations.

References

0: Mitxelena, Ion; Piris, Mario. Molecular electric moments calculated by using natural orbital functional theory. J. Chem. Phys. 2016 144, 204108, DOI: 10.1063/1.4951685
1: A. D. Buckingham. Direct Method of Measuring Molecular Quadrupole Moments. J. Chem. Phys. 1959 30, 1580,  DOI: 10.1063/1.1730242
6: W. H. Flygare and R. C. Benson. The molecular Zeeman effect in diamagnetic molecules and the determination of molecular magnetic moments (g values), magnetic susceptibilities, and molecular quadrupole moments. Mol. Phys. 1971 20, 225, DOI: 10.1080/00268977100100221
7: E. R. Cohen and G. Birnbaum. Far infrared collision‐induced absorption in gaseous methane. II. Determination of the octupole and hexadecapole moments. J. Chem. Phys. 1975 62, 3807, DOI: 10.1063/1.430932
8: R. Glaser, Z. Wu, and M. Lewis. A higher level ab initio quantum-mechanical study of the quadrupole moment tensor components of carbon dioxide. J. Mol. Struct. 2000 556, 131, DOI: 10.1016/S0022-2860(00)00658-X
9: A. J. Cohen and Y. Tantirungrotechai. Molecular electric properties: an assessment of recently developed functionals. Chem. Phys. Lett. 1999 299, 465, DOI: 10.1016/S0009-2614(98)01317-7
10: P. Bundgen, F. Grein, and A. J. Thakkar. Dipole and quadrupole moments of small molecules. An ab initio study using perturbatively corrected, multi-reference, configuration interaction wave functions. J. Mol. Struct. 1994 334, 7, DOI: 10.1016/0166-1280(94)03974-P
11: J. M. Junquera-Hernández, J. Sánchez-Marín, and D. Maynau. Molecular electric quadrupole moments calculated with matrix dressed SDCI. Chem. Phys. Lett. 2002 359, 343, DOI: 10.1016/S0009-2614(02)00722-4
29: A. Halkier, H. Larsen, J. Olsen, P. Jørgensen, and J. Gauss. Full configuration interaction benchmark calculations of first-order one-electron properties of BH and HF. J. Chem. Phys. 1999 110, 734, DOI: 10.1063/1.478042


How do you draw nitrogen in a newman projection? I have to draw a Newman projection for a molecule, sighting along the bond between a carbon and nitrogen. Given that the nitrogen is behind the carbon from this viewing angle, how would I depict the nitrogen? I know that I cannot draw it as a circle, as that represents carbon, so I am assuming that it is not possible to draw a Newman projection of this molecule from this particular angle. Is that a valid assumption?

And, if that is true, what would be considered an acceptable answer to that question? Should I just draw a standard 3D structure from the proposed viewing angle?

If it were up to me, I would:

Draw the nitrogen atom as if it were carbon but label it as "N".
Assuming normal valence  (e.g. an amine but not a nitro conpound), include the formally nonbonded electron pair as one of the substituents on nitrogen.


What does m± represent in the equation for mean activity of a salt solution? I need to find the activity $a_\pm$ for a $\pu{0.0120 M}$ solution of $\ce{Na3PO4}$, and have calculated the activity coefficient, $\gamma_\pm = 0.389$.

I know that $$a_\pm = \frac{m_\pm}{m_0}\gamma_\pm,$$

but I must be plugging in the wrong numbers. What does the top $m_\pm$ mean? And isn't the bottom $m_0$ simply $1$?

The top $m$ is the mean ionic molality of the solution you have.  You must know the concentration since you need it to compute the activity coefficient.  The bottom $m_0$ is the standard molality which in general is $\pu{1 M}$.  It is there to make the activity dimensionless.

The required formula for $m_\pm$ is $$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$ where $m_-$ is the molality of the negative ions and $m_+$ is the molality of the positive ions. Also $\nu_-$ is the stoichiometric coefficient of the negative ions and $\nu_+$ is the same for the positive ions. Also $\nu = \nu_- + \nu_+$.

Thus in your specific case, where you have $\pu{0.0120 M}$ of $\ce{Na3PO4}$, we have

\begin{align}
\nu_+ &= 3 & \nu_- &= 1 & \nu &= 4 \\
m_+ &= \pu{0.0360 M} & m_- &= \pu{0.0120 M}
\end{align}

and hence

\begin{align}
m_\pm &= \left[(\pu{0.0360 M})^3(\pu{0.0120 M})^1\right]^{1/4} \\
&= \pu{0.0274 M}
\end{align}


Proton-decoupled phosphorus spectrum of a ruthenium complex Regarding the complex $\ce{[RuH2(CO)(PPh3)3]}$, there are three possible octahedral stereochemical configurations,
a) Draw these isomers indicating which, if any, are chiral and identify the spin systems in each case
b) Sketch the expected appearance of the $\ce{^{31}P\{^1H\}}$ ($\ce{^1H}$ decoupled) spectra
c) Sketch the expected appearance of the $\ce{^1H}$ spectra

My attempt:
I think I have completed (a) but would appreciate on any advice for (b) and (c)

For the isomer on the far left I think the decoupled phosphorus NMR would show a doublet and a triplet.

The middle one I think it might too be a doublet and a triplet.

The isomer on the right I'm leaning towards doublet of doublets but I'm unsure.

Let’s call the isomers mer-trans, mer-cis and fac from left to right for easier addressing.

What you didn’t indicate yet is which symmetry elements the different ones have and if they are chiral; this is helpful in determining the look of the NMR spectra because it helps us identify enantiotopic and homotopic atoms which give identical spins. The mer-trans isomer is clearly the one with the highest degree of symmetry: I judge its point group to be $C_\mathrm{2v}$. The other two should be $C_\mathrm{s}$, if I did the looking correctly.

Once we have identified the symmetry, we need to find out which atoms are chemically and/or magnetically equivalent. For the two $C_\mathrm{s}$ compounds it is pretty easy: If the plane of symmetry transforms one atom onto the other, they are chemically equivalent but enantiotopic (so magnetically non-equivalent). If the plane of symmetry contains the atoms in question, they must be chemically non-equivalent, thus have different spins. For the mer-trans isomer it may look slightly more tricky, but note that spinning around the $C_2$ axis will always place one atom of the same kind in question on top of the other thus again making them chemically equivalent and homotopic. (Also note that the three phosphorus atoms can never all be chemically equivalent to each other since one always occupies a special position by symmetry.)

With our sets of chemically equivalent or non-equivalent atoms in hand — to check: I arrived at sets of 1, 1 and 2 hydrogens and always two phosphorus; in one phosphorus set we have a pair chemically equivalent atoms, the other two contain a pair of enantiotopic ones — we can predict the look of the spectra and what will couple to what. Note that the $\ce{^{31}P\{^1H\}}$ spectrum will not contain any couplings to hydrogen atoms.

For mer-trans, your reasoning is sound and I also expect the result to be a doublet and a triplet; the coupling constants must be identical. If I had to guess, I would say that those closer to $\ce{CO}$ should have a higher chemical shift due to $\ce{CO}$’s electron withdrawing nature; but it could be due to a push-pull mechanism that its influence is greater in the para-position.
mer-cis should also be a doublet and a triplet for more or less the same reasons. The chemical shifts must be non-identical but could be closer to each other due to every phosphorus being a neighbour of $\ce{CO}$ (which is still the single most differentiating species in the compound).
fac cannot be a doublet of doublets since that would be a single signal for all three; the three are, however, chemically non-equivalent. The single special phosphorus will again display a triplet because it is coupling to two distinct phosphorus atoms. The other two should show up as a doublet. I’ll let you do the reasoning.

When proceeding to the hydrogen spectra, remember that $\ce{^{31}P}$ is also a magnetically active, spin ½, $100~\%$ abundance nucleus. Thus, you must consider the coupling of the protons with each other (if applicable) plus the coupling of the protons with the phosphorus atoms (the magnetically non-equivalent ones). In no particular order you should arrive at:

One spectrum containing two distinct proton shifts, each with two doublet couplings and a triplet coupling. I cannot say a priori which one should be larger or smaller.
One spectrum containing a single proton shift which is split up into a doublet and a triplet. Again, I cannot say which coupling is stronger.
One spectrum containing a single proton shift, split into a doublet of doublets of doublets (ddd). At least here I don’t have to say which coupling is stronger because they are all distinct.

Retroactively assinging proton spin systems, I get $\ce{AX, A2}$ and $\ce{AA'}$; for phosphorus I arrive at $\ce{AX2, AA'X}$ and $\ce{AX2}$, again in no particular order for all six spin systems (i.e. you would still need to assign them to the correct structures). If you wish to combine the spin systems into one entire one per structure, use $\ce{M}$ rather than $\ce{A}$ in the hydrogen ones (or $\ce{K}$ and $\ce{R}$ instead of the hydrogen $\ce{AX}$). Note however that I am terrible at spin systems and that you should take those with a grain of salt.


Maximum force that hydrogen-hydrogen bond can withstand For $\ce{H-H}$:

bond length is $74~\mathrm{pm}$
bond energy is $436~\mathrm{kJ~mol^{-1}}$

(Source: article chemical bond on Wikipedia.)

The University of Waterloo page that the article cites says:

  The amount of energy required to break a bond is called bond
  dissociation energy or simply bond energy.

and

  The bond energy is essentially the average enthalpy change for a gas
  reaction to break all the similar bonds.

So I'm assuming that as the average enthalpy change increases, so does the force that stresses the bond (and thus the forces on each hydrogen atom).

For a single molecule of $\ce{H-H}$, what is the maximum force that the bond can withstand before breaking?

The force and the bond enthalpy are not necessarily directly related.

The bond dissociation energy is the amount of energy released when the bond is formed, or the energy required to break the bond.

The force constant of the bond derives from the curvature of the potential energy curve. For example, if we assume a harmonic oscillator (spring) for a bond, then:

$$ V(x) \approx \frac{1}{2}k_f \; x^2$$

Real bonds aren't purely parabolic, but clearly the second derivative of the potential energy curve near the equilibrium bond length will give the needed force constant.

Here I've plotted two fictional bonds with the same length. The blue curve gives a bond with a higher bond enthalpy and force constant (i.e., the curvature is higher than the red curve).

It should be clear that this isn't necessarily true. We can easily imagine some bond that has a shallow curvature but a deep enthalpy.

I should indicate that in many cases, these are related - stronger bonds typically have higher force constants and are harder to distort, but the point here is that the bond dissociation energy does not directly tell us anything about the mechanical strength of a bond.


Groups of the periodic table The groups go from 1 to 18 in the periodic table. When looking at the transition metals in the periodic table, You see roman numerals labelled with 'B,' in the order of III, IV, V, VI, VII, VIII, VIII, VIII, I, II.
Why does III - VIII come before I and II? And why does VIII get repeated three times?

Also, what is the difference in the groups labelled A and B, besides the fact that the B groups are the transition metals?

Even though your question get downvoted, I think it is an legitimate question for people to understand the trend. I believe I-VIII Roman numeral nomenclature belong to CAS system while 1-18 Arabic numeral groups are recommended by IUPAC (Ref.1). The IUPAC recommended Periodic Table is given below (ignore the Roman numerals, which I put in there to explain):

According to Ref.1, the confusion of Periodic Table of Elements has been addressed as follows:

H.G. Deming used the long periodic table in his textbook General Chemistry (See following diagram with Roman numerals only; Ref.2), which appeared in the USA for the first time in 1923, and designated the first two and the last five Main Groups with the notation "A", and the intervening Transition Groups with the notation "B". The numeration was chosen so that the characteristic oxides of the B groups would correspond to those of the A groups. The iron, cobalt, and nickel groups were designated neither A nor B. The Noble Gas Group was originally attached by Ueming to the left side of the periodic table. The group was later switched to the right side and usually labeled as Group VlllA. This version of the periodic table was distributed for many years by the Sargent-Welch Scientific Company, Skokie, Illinois, USA. $[\cdot\cdot\cdot]$
The designations A and B have been extensively and rather arbitrarily used in the meantime in textbooks and in publications. An investigation of the application of the subgroup designations A and B in all articles, which appeared between 1972 and 1981 and covered by Chemical Abstracts, revealed a completely arbitrary use of the designation. Moreover, more than 10% of the articles it was nearly impossible, from the wording of the text, to recognize which elements were being discussed without further information (Ref.3).

The IUPAC Commission on the Nomenclature of Inorganic Chemistry (CNIC) proposed the designations of groups by Arabic numerals in 1984 (Ref.4), which was approved by the Nomenclature Commission of the American Chemical Society (ACS) and rest is history.
Thus, it is safe to say that this nomenclature is now outdated and not worth discussing.

References:

E. Fluck, “New notations in the periodic table,” Pure and Applied Chemistry 1988, 60(3), 431-436 (https://dx.doi.org/10.1351/pac198860030431).
Horace Grove Deming, General Chemistry; J. Wiley & Sons, Inc.: New York, NY, 1923.
W. C. Fernelius, W. H. Powell, “Confusion in the periodic table of the elements,” J. Chem. Educ. 1982, 59(6), 504-504 (https://doi.org/10.1021/ed059p504).
K. L. Loening, “Recommended Format for the Periodic Table of the Elements,” J. Chem. Educ. 1984, 61(2), 136-136 (https://doi.org/10.1021/ed061p136).