How to prevent Ag2O from decomposing to Ag on Aluminum I have silver oxide in an ink formulation. When I apply it (in small quantities) over aluminum and heat the resulting (to evaporate the solvent), the monovalent silver oxide seems to decompose to Ag. The temperature I am evaporating the solvent is around 30-40˚C. The decomposition of Ag2O to Ag occurs at a temperature over 200˚C. However, I am aware that aluminum is a good conductor of heat, so even though the temperature was only 30-40˚C, Ag2O could have deposed because the temperature on the aluminum surface could have been around (or higher than) 200˚C.

My question is, is there a way to solve this problem. How can I still use aluminum and not worry about decomposing Ag2O to Ag when evaporating the solvent? Do I have to further reduce the temperature and is this the only solution? I tried to research this concept but couldn't find anything.

Thank you so much for your help.

First,  as you mention, $\ce{Ag2O}$ starts to decomposes from ≥200 °C. Measure the temperature of the aluminum substrate with an IR thermometer or attached thermocouple to be sure you stay well below that. You can substantially speed drying of the ink by lowering the air pressure, e.g. using an aspirator pump, perhaps eliminating need to heat the substrate.

You may also need to anodize the aluminum prior to printing  to prevent chemical reaction with the $\ce{Ag2O}$ in the ink. Aluminum is much higher in the electromotive series and would rapidly replace silver in the oxide.


Claisen condensation Aren't two last steps superfluous? I just do not see how the fourth and last forms differ.

You are correct that there is no difference between the last fourth and last molecules shown.

However, the thermodynamic driving force behind the Claisen condensation is the formation of the highly stable enolate of the $\beta$ ketoester, or in your case $\beta$ diketone.

In order to form this stable enolate the original ester must have at least two $\alpha$ hydrogens so that there is at least one left at the end to deprotonate. Therefore the Claisen condensation will not work with esters having only one $\alpha$ hydrogen (and obviously not with esters with no $\alpha$ hydrogens).

Once you have completed the reaction to yield the enolate product, it can be acidified as shown in your reaction scheme to yield the neutral molecule.


How to determine which of the given redox reactions will occur? Which one of these reactions will occur?

  $$\ce{KMnO4 -> K2O + MnO2 + O2}$$
  $$\ce{KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O}$$
  $$\ce{KMnO4 + CO2 \longrightarrow K2CO3 + Mn2O7}$$
  $$\ce{KMnO4 + CO2 -> K2CO3 + MnO2}$$
  $$\ce{KMnO4 + SO2 -> K2SO4 + Mn2O3 + SO3}$$

And how am I supposed to know which one it is? It's not just this question, I struggle with these types of questions every time. There seem to be a lot of things to take into consideration.

I'm thinking whether it's under 3 or under 4. The oxidation numbers for the reaction 3 don't change at all. But in reaction 4, only the oxidation number of Mn changes (from +7 to +4) and the other oxidation numbers stay the same (if I'm right). So, I think reactions 1, 2 and 5 definitely happen, but I don't know about 3 and 4.

Both 3 and 4 are impossible. $\ce{CO2}$ does not react with the $\ce{KMnO4}$, because $\ce{KMnO4}$ is strongly oxidizing. $\ce{C}$ is in the highest oxidation state and can not be oxidized, hence, it is impossible to react.


Shifting of groups in carbocation rearrangement Phenyl has greater migratory power. However, why did methyl migrate in step 2 of the reaction?

In general, migratory aptitude of different groups follow following order:
$\ce{-Ph} > \ce{-C(CH3)3} > \ce{-CH(CH3)2} > \ce{-CH2CH3} > \ce{-CH3} > \ce{-H}$

For example, 2,3-diphenylbutane-2,3-diol (1) was allowed to react under homogeneous (25 mol% of p-toluenesulfonic acid) or heterogenious (20 mol% of graphene oxide) catalytic conditions using toluene as a solvent at $\pu{100 ^\circ C}$. Under both conditions, only pinacol rearrangement with phenyl migration was observed affording 3,3-diphenylbutan-2-one (2) in a 90 and 95% conversion. Methyl migration has not been observed [Reference].

Reference:

Pinacol Rearrangement and Direct Nucleophilic Substitution of Allylic Alcohols Promoted by Graphene Oxide and Graphene Oxide $\ce{CO2H}$: M. Gomez-Martinez, A. Baeza, D. A. Alonso, ChemCatChem, 2017, 9, 1032–1039 (https://doi.org/10.1002/cctc.201601362)


How to rationalise that when we dilute a weak acid, the pH increases but also the dissociation increases? When water is added to a weak acid like ethanoic acid, the number of ethanoic acid molecules that dissociate increases, however, pH increases (less acidic), too. Why is this so?

It's simply due to dilution.

If it's true that adding water increases dissociation (Gerard has done a nice effort in Does the number of H+ ions in solution go up on dilution of a weak acid?), concentration of $\ce{H3O+}$ inevitably decreases. As a consequence, $\mathrm{pH}$, defined as $-\log[\ce{H3O+}]$, increases.

Start with $\pu{1 L}$ acetic acid concentration $\pu{1 M}$,
$$\ce{[H3O+]} = \sqrt{K_c} = \pu{4.22E-3M},$$
that corresponds to $\pu{4.22E-3mol}$ of $\ce{H3O+}$.
If you double the volume ($\pu{2 L}$),
$$\ce{[H3O+]}=\sqrt{K_c/2}=\pu{2.98E-3M},$$
that corresponds to $\pu{5.96E-3mol}$ of $\ce{H3O+}$.

The amount of substance is higher, but concentration is less: dilution does play a role here.


Concentration and volumes In the reaction:

$$\ce{H2SO4 +CaCO3 \to CO2 + H2O + CaSO4}$$

Does altering the concentration of the sulfuric acid change the volume of the carbon dioxide produced [the volume of the sulfuric acid is constant]

The concentration is defined as the amount of particles per unit volume. If you have a greater concentration of sulfuric acid, there are more sulfuric acid 'particles' ready to collide and react with the calcium carbonate particles. This is a simple result from the collision theory of chemical reactions.

So, increasing the concentration of the sulfuric acid should increase the amount of carbon dioxide (and other products). However, this is not always the case.

Consider the case where calcium carbonate is the limiting reagent/reactant. This basically means that all of it gets used up before the other reactant gets used up. In a case like this, increasing the sulfuric acid concentration will not affect the amount of products, simply because there's not enough calcium carbonate.

It's like asking if having more flour is going to help you bake more cake. Often, it does. But if you don't have enough of the other ingredients, it doesn't matter how much flour you have! Obviously, this is an incomplete analogy because you could probably still make a cake, albeit a nasty one.


The conversion of energy during electroplating I've made a test and i've faced a problem with a part of a question which says: "What is the conversion of energy occuring in electroplating? " I answered that it is conversion from electric energy to chemical energy, since in electroplating, the anode is reduced to X^n+ and n+, where n+ ions are transferred through DC power supply into the cathode and then combines with X^n+ to give again X and plates the cathode, so the electric energy(transfer of electrons through the DC power supply) is converted into chemical one(change  X^n+ to solid). I want to know if my answer is correct or not and why? Helping is appreciated

That's a good start, but there is also ohmic loss, as much of the energy is also converted to heat in the electrolyte, electrodes or even wiring.

BTW, during WWII, silver was used to replace copper bus-bars in stationary machinery such as electrolysis apparatus, freeing the copper for use in war materiel and slightly improving the efficiency of the apparatus due to its better conductivity then commercial copper.


Volume of one equivalent of a gas How do we define an equivalent when we are not talking about any particular reaction? Like, we say that equivalent volume of Oxygen is 5.6L, so we have taken the n-factor to be 4, and in CO2 we take n-factor = 2.
What is the defining factor for an equivalent here?

The meaning of the word "equivalent" is relative to something. Fortunately, as a chemical term, that is still true. An equivalent is more completely called "a reaction equivalent" and is relative to the reaction. There are a few standard reactions (which are typically industry or discipline based) and it might be used usefully in limited (well constrained) situations. The problem is, especially when dealing with reactive gasses, that the words "uncontrolled" and "detonation" might also occur when explaining that you chose the wrong standard - the correct standard reaction is NOT always obvious or determinable in context. So, my best advice is that when someone uses "equivalent" ask what reaction, specifically, they are referring to. And when using the word yourself, always note the context. Examples: NCO equivalents, H(0) equivalents, H(+) equivalents may all refer to the same compound but would have different values.


How is "the surrounding" and its temperature defined to calculate entropy change during a reaction? Lets say supercooled liquid water at $263\ \mathrm K$ isobarically changes to solid ice at the same temperature. I wish to calculate the change in entropy of the surroundings and I happen to know the $\Delta H$ for the reaction.

This should be simple enough. Since the process is isobaric, $q_\text{rev} = \Delta H$ and $q_\text{surr}=-q_\text{reac}$

$\Delta S_\text{surr} = \frac{-\Delta H}{T} $

But what should T be here? Is it $263\ \mathrm K$? Or the temperature of the surroundings? Or are they both the same, ie;

Do we consider the temperature at which the reaction is taking place to be the temperature of the surroundings as well?

If so, why?

Thermodynamics deals with changes between systems at equilibrium, which is to say that the initial and final states of a process are regarded as equilibrium states. As explained in the comments, the surroundings are usually defined as an ideal reservoir of infinite size and thus infinite heat capacity and constant temperature, clearly an approximation, but the key point remains that stated in the previous sentence.

It does not make sense to define an equilibrium diathermal state in which the surroundings and the system have a different temperature. Since  in the stated problem the system undergoes exchange of heat with its surroundings, and the system is at an initial and final T of 263 K after the transition, then the surroundings have to be at 263 K throughout.

One potential point of confusion regarding this type of problem concerns the metastable nature of the supercooled state. This is not a true thermodynamic equilibrium but is regarded as sufficiently stable to be regarded as such. An additional point of confusion may be that heat is transferred to the surroundings even though system and surroundings are at the same T. This is possible because the surroundings have infinite size and heat capacity. One way to envision such surroundings is as a reservoir of a substance with a melting point of 263 K.


What are the steps to creating synthetic glucose? I need to know how to go about creating synthetic glucose with generally available materials. As I am aware of both photosynthesis and glucose in nature, I will not need lessons in how plants produce it. So, how may glucose be produced by artificial means?

A relatively simple process is described by Kobayashi et. al where they use cellulose, a structural component of plants, which is ball milled, mixed with activated carbons and HCl and hydrolysed to produce glucose in high yield (88%).

It requires only easily available ingredients and the ability to heat (and preferably stir) to 453K. Reference is given below.

High-Yielding One-Pot Synthesis of Glucose from Cellulose Using Simple Activated Carbons and Trace Hydrochloric Acid
Hirokazu Kobayashi, Mizuho Yabushita, Tasuku Komanoya, Kenji Hara, Ichiro Fujita, and Atsushi Fukuoka
ACS Catalysis 2013 3 (4), 581-587
DOI: 10.1021/cs300845f


What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in acidic solution? What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in an acidic solution?

My suggestion is:

$\ce{5NO2-   +   2MnO4-   +   6H+   ->   5NO3-   +  2Mn^{+2}  +  3H2O}$

Is that right?

Your suggestion is certainly part of the picture, but as is usually the case with nitric acids, a number of side reactions can and often do take place depending on concentration and temperature. In acid solutions, for example, nitrite spontaneously decompose giving nitric oxide together with dinitrogen tetroxide. My feeling is that under optimum conditions your suggested reaction would predominate


What are possible side reactions for the generation of hydrogen via electrolysis? I recently completed making a simple water electrolysis setup to generate hydrogen which uses two graphite rods from a standard AA battery. The power source is a $24~\mathrm{V}$ $6~\mathrm{A}$ adapter. I am using salt water as electrolyte (The salt is iodized). Strangely, a minute or so after running, a black precipitate starts to form on the negative terminal.
Also, besides the hydrogen, another gas is generated which smells sour, not very sour, but somewhat like lemonade. I have no idea what is causing this.
I thoroughly cleaned the graphite rods before using them. Has anybody an idea about what reaction might be happening here?

First of all, check again your electrodes to see if this 'black precipitate' is formed at the negative electrode as you state; I predict it's actually happening at the positive electrode for the following reason:

Since you're using two graphite electrodes, it isn't a black precipitate that's forming, but is indeed a side reaction as you state. I suggest what you're actually making is oxygen gas at the positive terminal (as one would predict) which then reacts with the graphite electrode surface to generate $\ce{CO2}$. To verify this,take a look at your electrode—it should have started to break up, now having a non-uniform surface. Perhaps even a bit crumbly.

As LDC3 points out in his comment, you're forming $\ce{Cl2}$ as well as $\ce{O2}$ which is the characteristic smell you're noticing. To understand this we need to look at the standard electrode potential for each of these reactions. Forming $\ce{O2}$ from water at constant $\mathrm{pH}$ (compared to the standard hydrogen electrode) requires a potential difference of $-1.23~\mathrm{V}$ (this value is negative since this is the reverse of the tabulated version). Chlorine evolution on the other hand requires a p.d. of $-1.36~\mathrm{V}$. Remember that the larger (the less negative in this case) the reduction potential, the higher the tendency for reduction. So since $\ce{Cl-}$ reduction potential is more negative, it's less likely to undergo reduction and hence more likely to be oxidised (lose electrons) than water, which agrees with our chemical intuition.

So what we say here is that $\ce{Cl-}$ is preferentially oxidised over water, and as a consequence $\ce{Cl2}$ is formed preferentially over $\ce{O2}$. The reason why you have any $\ce{O2}$ forming at all is due to the electrolyte being the limiting reagent. Your supporting solvent (water) is in such abundance, essentially completely surrounding the electrode, that the system simply has to oxidise some water molecules some of the time.

Of course, experiment should rule the day, so revise your experiment to provide evidence for/refute my theoretical framework. Replace your positive terminal with a high purity metal electrode (copper should work and be readily available at high purity). As for the choice of electrolyte, test a couple of salts whose reduction potentials are less negative than water. Hint: perhaps look at sulphate salts and try and find a further reason why these are very useful in electrochemistry.


Is a liquid vitamin E supplement possible without additives? Looking at E vitamin products sold commercially, all “pure vitamin E oil” products seem to be for skincare rather than consumption as a supplement.

I have read that d-alpha tocopherol is the form of vitamin E with greatest bioavailability, but even liquid supplement products often contain sunflower or other oils in addition, which I understand to be a source of but of course are not entirely vitamin E.

Is a pure liquid supplement possible or practical for consumption as a supplement, and if not, why?

Vitamin E is not soluble in water. It is only soluble in oils. So it is no use swallowing pure Vitamin E, or this vitamin with water. It will poorly be absorbed by the intestine. You must absorb it in an oily solution.


Is it possible to design molecules of desired qualities? I was thinking if it would be possible to take into account every property of every atom known to man so that they can compute any type of compound they want. Is it possible to accomplish something like that with the present computation technologies or are the variables that are accounted in making of any kind of molecule and determining its properties too high?

Your title and the question itself arise two different issues, and I'll try to address both.

Yes it is possible to design molecules of desired qualities, only it's not all that simple. It requires a great deal of calculations, practice, empirical laws, heuristics, experimentation, guesswork, and magic. Eventually this body of knowledge grew so wide that people started to count it a science in its own right. It became known as chemistry.
Yes we can take into account every property of every atom and compute the properties of any compound they constitute. The problem, though, quickly becomes prohibitively expensive in terms of computation time. The reverse problem (to come up with a molecule of desired properties via calculations alone) is even more costly.

So it goes.


Why is 1,3-pentadiene favored in the reaction of pent-3-en-2-ol with sulfuric acid? Let's say we have the following reaction, a simple dehydration reaction.

Why do we get 1,3-pentadiene and not 2,3-pentadiene?

Hydrogenation of 1,trans-3-pentadiene (your top molecule) and the allene, penta-2,3-diene (your bottom molecule), both produce pentane plus heat.  Heat is given off because both of the dienes plus hydrogen is less stable than pentane.  With 1,trans-3-pentadiene, 53.1 kcal/m of heat is given off; with penta-2,3-diene 66.8 kcal/m of heat is given off (data from Frances Maron Fraser and Edward J. Prosen, Journal of Research of the National Bureau of Standards Vol. 54, No. 3, March 1955 Research Paper 2575, p. 143-148). Clearly the allene is of higher energy than the 1,3-diene by about 15.7 kcal/m.

The reaction conditions (strong acid and heat) suggest that an equilibrium mixture of possible products will result.  Around room temperature, each 1.4 kcal/m difference in energy between two isomers will shift the equilibrium between them by a power of 10 ($\ce{\Delta G=-RTlnK}$).  A 15 kcal/m energy difference between your two dienes suggests that the equilibrium will be ~ $\mathrm{10^{10}:1}$ favoring the 1,3-pentadiene. If any of the allene forms, it will rapidly convert to the 1,3-pentadiene.

Side Note: Why are cumulenes generally less stable than 1,3-dienes?

Conjugation which results in resonance stabilization is present in the 1,3-diene but not in the 2,3-diene.
Allenes (in fact, all cumulenes) have considerable strain built into them. If you view a double bond as a two-membered ring, it is easy to see this.  The bond angle in the two-membered ring is severely pinched down.  In turn, the bond angle external to the double bond tries to open up and relieve strain.  This strain relief, can't happen with the central carbon in an allene because it has a double bond on both sides.


How to identify a blow-out pipette? Is my pipette a blow-out pipette?
I suspect that it is a blow-out pipette, though the double lines aren't rings. My guess is it is a TC blow-out pipette. It's a big 50 mL pipette, longer than my arm.
I wonder, how am I supposed to know? I also have another one with a single dark band on the top.
EDIT: This video explains how blowing out a pipette is actually done.

Good to see classical analytical chemistry queries. These graduated pipettes or Mohr pipettes they are called, confuse even the relatively advanced users. First you have to check where the graduations end. If the pipette was meant to deliver 10 mL or 50 mL you should not see 10 or 50 mL mark near the tip of the pipette. That is one hint. Add a picture of the tip as well. Secondly, a double ring indicates that all liquid has to be expelled.
Notice that when you are doing very high precision work, one has to know the full drainage time as well after the liquid has been delivered. For example, after delivering 10 mL you need to wait for 5 s before blowing out the last drop. Very high quality glassware use to mention this time on the glassware.
In your blurred picture one can se Ex 5s, which means wait for 5 s before blowing out the last drop.
Volumetric pipettes should never ever be blown out, but they also have a drainage time, you have to wait for a certain time before dislodging the last drop purely from surface tension effects.


Why are pressure or volume not able to change the equilibrium constant? It is stated in my textbook that temperature is the only factor that can change the equilibrium constant. According to the Ideal Gas Law, temperature is dependent on the pressure and volume. Thus, shouldn't pressure and volume change the equilibrium constant too?

Just like any sweeping statements made by the General Chemistry textbooks, this statement is also not completely true. When you work at pressure extremes, as in modern day chromatography, such as 1000 or higher bar, large molecules can change their shape, in that case, one can see a change in retention factors which sort of indicates a change in equilibrium constant with pressure under constant volume.

Just for fun, some solvents can become solids at high pressures-but this happens at another high-pressure level.

Coming back to routine reactions, pressure will not affect equilibrium constants because whenever you try to adjust the pressure in a gaseous reaction, the equilibrium concentrations will change in such a way that their ratio remains constant-hence the equilibrium constant does not change.

One way to think about it as that the equilibrium constant is a ratio of forward and backward rate constant. Changes in pressure or volume will not change the kinetic energy of the molecules but only temperature can change the kinetic energy and hence affect the rate constants.

  According to the Ideal Gas Law, temperature is dependent on the
  pressure and volume.

Mathematically, this is incorrect. You would rather say that pressure and volume of a gas are functions of temperature i.e., the converse of what you wrote.


How do I determine the absolute configuration experimentally? Let's say I synthesized or isolated a chiral molecule and want to know the absolute configuration (R or S) of that molecule. I could obviously solve the structure by X-Ray crystallography, but that's a lot of work and I might not want to do that or might not have access to the necessary equipment.

What other ways exist to determine the absolute configuration of a molecule experimentally?

If your compound has been synthesized before, you can measure its optical activity and compare it with the data in the literature.

If your compound hasn't been synthesized before, it can be possible to differentiate between two enantiomers with NMR by measuring the spectra of your molecule in a chiral environment. Here's a review article on how to resolve enantiomers with NMR.

Otherwise, diastereomers can be differentiated easily. By looking at your molecule, you can (sort of) predict which spectrum it should give rise to by using the concepts of enantiotopic, homotopic, heterotopic and diastereotopic protons. We had to analyse and solve 2D-NMR spectra (1H and 13C) of diastereomers last year, and to a beginner, I have to say it was a challenge.

As always, it boils down to forming diastereomers.

EDIT

Take a look at these diasteromers:

(I've got finals coming up), it's easy to see that the hydroxyl will shift the ethyl's CH2 peaks towards lower field (higher ppm) in the second isomer. The interpretation of chiral media NMR spectra would have to be similar. My impression is that the acquisition of chiral media spectra is more involved, but the interpretation isn't. I might be totally off track, though.

Someone might want to share a more thorough answer.


What advantage is there in reporting the Lennard-Jones well depth in wavenumber units? The Lennard-Jones well depth $\epsilon$ is typically given in energy units ($\mathrm{kJ}$ or $\mathrm{kcal}$; sometimes per mole). Jasper and Miller, however, reported $\epsilon$ values in reciprocal centimeters ($\mathrm{cm^{-1}}$).
Why use $\mathrm{cm^{-1}}$?

It seems unnatural using $hc$ as a conversion factor for a classical
description.
Absorption peaks (e.g. in wavenumbers) are not related
to the LJ potential well, which is for non-bonded interactions.

Ahren W. Jasper, James A. Miller, Combust. and Flame 2014, 161 (1), 101–110. (preprint pdf)

I don't think there's any inherent advantage as one can always convert to another unit. It seems natural to me, however, because chemists generally have a good idea of the energy of vibrational and rotatial states in $cm^{-1}$, so being able to compare the depth of the well with some familiar chemical process is quite nice. For instance, if the well-depth were quite small, one could immediately identify that the system may not spend a lot of time there if it is comparable to the zero-point energy or something similar.

It's similar to how physicists report everything in electron-volts ($eV$). They do it because it's familiar and they have many physical processes they can use as a reference. For instance physicists know that the band gap of semiconductors is between $2-5\ eV$, so it's quite natural to report other energies in the same unit.


How to find a symmetry group of a system if all the symmetry transformations do not obey closure and don't form a group? For instance, consider a system with $p_x$ and $p_z$ orbitals at the vertices of a square (on xy-plane). A square by itself would have $D_4$ symmetry. However, because of the $p_x$ orbital; the $90^\circ$ rotation ($C_4$) and $270^\circ$ rotation ($C_4^{-1}$) are no longer symmetry operations. Now the rest of the $D_4$ members do preserve the symmetry, but do not form a group since the subset is not closed.
How does one go about formulating a symmetry group for cases like these?

Hmm, so I'm not entirely sure what notation you're using ($D_4$ sounded like Schönflies to me), but I'm sure you can figure out the equivalent.
In general I agree with Andrew's point that molecular symmetry is determined by atomic positions and not orbitals. But at least to me as of now, this seems to be a thought exercise more than anything (after all, no real atom has only two p-orbitals), so I'll play along. In Schönflies notation your system would be labelled $D_\mathrm{2h}$. The three $C_2$ axes and three mirror planes are illustrated below. I left out the identity operation $E$ and the centre of inversion $i$.


How many different possible structures (isomers) are there for the complex, [Mn(OH2)3(NH3)3] 3+? How many different possible structures (isomers) are there for the complex,
$\ce{[Mn(OH2)3(NH3)3]^{3+}}$?

I really don't know how to begin with this. I know the answer is 2, but i don't have the faintest idea how this is the case. Since $\ce{Mn}$ has a charge of $3+$, I know that its valence electron configuration is $3d^4$. I also know that $\ce{NH3}$ and $\ce{H2O}$ are monodentate, which means they contribute 1 lone pair.

Metal complexes with three each of two types of ligands $(\ce{MX3Y3})$ have two isomers: mer- and fac-isomers. For example, for $\ce{Co(NH3)3Cl3}$ there are two isomers:

  Fac for facial, in which all three ligands of each type are clustered on a single face of the octahedron and the three other three ligands are on the opposite face.

  Mer for meridional, in which the three ligands of each type are arranged meridianally in a row around the metal and the other three ligands are arranged likewise on the opposite side in a perpendicular meridian.


Why is the F-F bond in fluorine a p-p sigma bond? Won't the lone pairs hybridize to $\mathrm{sp^3}$ making it an $\mathrm{sp^3}$-$\mathrm{sp^3}$ $\sigma$ bond? If it still is $\mathrm{p^\sigma}$-$\mathrm{p^\sigma}$  bond what is the spatial location of the lone pairs?

Hybridisation is a mathematical concept chemists use to better understand spatial distribution of bonded atoms. It should only be used if we cannot explain the structure will without resorting to hybridisation.

The energy of a 2s orbital is lower than that of a 2p orbital in atoms with a fully populated 1s orbital. Therefore, there is a desire to keep its energy low. A bond between two p orbitals (of different atoms) will also lower the energy of one of these p orbitals (and create an antibonding σ* unpopulated one, too).

Furthermore, p orbitals extend into space in a directed manner: Along their coordinate axis, while s-orbitals are spherical in shape. The p orbital is already in a nice direction for bonding; any mixing with s orbitals will make it extend into space slightly less — potentially unfavourable for orbital overlap.

Finally, the lone pairs on each fluorine atom will be a set of two p type orbitals, each carrying a lone pair perpendicular to the bond axis, and an s type orbital. Experimentally, one should be able to detect two different energy levels for the two different lone pairs. (However, I didn’t go looking for papers and I can’t remember any lecturer mentioning any off the top of my head.)

So the $\ce{F-F}$ bond is indeed a $\mathrm{p^\unicode[Times]{x3c3}\ce{-}p^\unicode[Times]{x3c3}}$ bond because it makes the overall system energetically favourable.


Finding equilibrium constant from solubility product constant Given
  $$\ce{CuI(s) <=> Cu+(aq) + I-(aq)}\, ;\qquad K_\mathrm{sp} = 1.2 \cdot 10^{-12}$$
  write a reasonable chemical reaction that describes the decomposition of $\ce{CuI2}$ in aqueous solution, and show that this is a spontaneous reaction under standard conditions.

The answer found the Gibbs free energy change of the precipitation reaction (backward reaction of the one shown above) using the formula $\Delta G = -RT\ln{K_\mathrm{eq}}$ and substituted the value $1/K_\mathrm{sp}$ for $K_\mathrm{eq}$.

I do not understand why the inverse of the solubility constant can be used as the equilibrium constant of the backwards reaction.

Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?

Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?

Yes. The general rule actually is: For a given chemical reaction with equilibrium constant $K$:

$$\ce{aA\+bB->cC\+dD}$$

If we multiply the chemical reaction by $n$, we get $K_1$ for the following equilibrium:

$$\ce{\mathit{na} A\+\mathit{nb} B->\mathit{nc} C\+\mathit{nd} D}$$

as $K_1=K^n$.

Reversing the reaction is equivalent to taking $n=-1$, in which case we get: $K_2=\frac{1}{K}$

You must be wondering that:

  Why does the new equilibrium constant $K'=K^n$?

The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction:  $\ce{A +B —> C + D}$, $K_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$

Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.

So, if we multiply our reaction with $n$, it actually can be rewritten as:

$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K_\mathrm{c}'$ will be
$$K_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K_\mathrm{c}^n$$

Hence, proved.

PS: The previous answer proved the same result, but incorrectly. That proof has now been corrected thanks to the comment by @DavePhD below.


Finding out the composition of seawater/brine? As part of a project, I am working on precipitating salts from seawater and brine. I have a sample of seawater and I need to measure the concentrations of $\ce{Mg^2+, Ca^2+ and Li+}$ present in the sample. As well as being able to measure $\ce{MgCl2, CaCl2 and LiCl}$.

What analytical instrument would be best for this experiment? We have a decent budget but dont want to invest in equipment too large (i.e ICP-AES or ICP-MS) as the laboratory isnt very big.

I have looked into ion-chromatography and can get one for about 20k USD, anyone else got any better ideas or reasons not to go for ion chromatograph?

Your budget should allow you to measure  $\ce{Mg^{2+}, Ca^{2+}, Li+ and Cl-}$ in-house using ion specific electrode analysis. There will certainly be some training and maintenance involved, but anyone with a bit of a chemistry background will be able to learn this technique. Adding to the simplicity, and thus the ease of training, is the fact that your samples are all of a fairly similar composition.

Ion chromatography, which would basically require one set up for anions and one for cations, would break your bank in a hurry unless you have an experienced IC analyst with some time to spare and access to good used equipment.  Frankly, this just doesn't seem like a realistic approach.

One thing to keep in mind is that your budget will buy a lot of analysis by people with expertise in whatever analysis you choose to perform. They will even help you choose the best analytical technique for your needs.

My advice is to look at outsourcing first, particularly if you are going to be analyzing fewer than 10-20 samples per day.  If you will be analyzing a large number of samples, or if you need very quick turnaround times, then I suggest looking into ion specific electrode analysis.


Hydroboration oxidation reaction on trans,trans,trans-1,5,9-cyclododecatriene with carbon monoxide The given solution:

I got this question in an entrance exam coaching test paper in India. Product P is easily covered by the mechanism of simple hydroboration-oxidation reaction, and I've understood it well. However, product Q is what intrigues me.

Looking at the product and the reactants, I can guess that the carbon in $\ce{CO}$ functions as the source for the bridging carbon in the product Q. However, I could not figure out the mechanism for the same. It is not given in my textbook, and wasn't found via Google searches either.

I'v well understood how the intermediate product (bridged by boron) had formed. However, how is the carbon monoxide supposed to displace boron from there? I can guess that once a carbonyl group bridge is formed in the middle, the alkaline hydrogen peroxide will reduce the ketone to an alcohol (to get product Q), but I am not too sure.

You guessed it right: a ketone will be formed.
This reaction involves migration of alkyl groups from boron to the carbon atom of CO. There will be three migrations after which you will get your alcohol:

Image taken from National Programme on Technology Enhanced Learning (NPTEL). Lecture 26 : Boron Containing Compounds. Image link: Scheme 9

See also Handouts for Organic Chemistry Lectures given at Imperial College London, Chemistry, by Henry Rzepa.


Regioselective enolate formation Our professor only told us that A is favored, and B is not getting formed. I thought about the resonance structures and saw a contradiction to that what the prof told us. To be honest I think my argument with the resonance structure is not strong enough, because it does not explain the prefered product. Why is structure B disfavored?

@jerepierre: I checked again. You're right! I'm sorry for the wrong re-drawing. Although the above discussed reaction with deprotonation was also dictactical important.

The correct starting material is in fact a alpha,beta-unsaturated ketone which is reduced by Li/NH3!

This is the correct reaction:

Unsymmetrical ketones can produce two different enolates, as shown in your reaction scheme.  The enolate pictured in the top row of your drawing is referred to as the thermodynamic enolate, while the enolate shown in the bottom row of your drawing is the kinetic enolate.

More highly substituted double bonds are thermodynamically more stable than less substituted double bonds. Therefore, the thermodynamically favored enolate is the one that gives rise to the most highly substituted double bond in the enolate intermediate. The thermodynamic enolate will be formed when reaction conditions are chosen (see below) to allow equilibration of all enolate intermediates; product formation is then thermodynamically controlled.

The kinetically favored enolate is the fastest formed enolate, the one formed by abstracting the most acidic proton (proton acidity follows primary > secondary > tertiary).  In your reaction, removal of the proton on the secondary carbon would be kinetically favored (more acidic) over removal of the proton on the tertiary carbon atom (less acidic).

Reaction conditions can be selected to favor kinetic (aprotic solvents, strong bases, low temperature) or thermodynamic (protic solvents, weaker bases, higher temperature) enolate formation, giving the chemist control over selecting which product will be formed.

If your reaction were run under conditions favoring kinetic enolate formation, then we would expect to observe product formed from following the bottom reaction line.  If the reaction were run under conditions favoring thermodynamic enolate formation, then we would expect to obtain products from following the top line of your reaction scheme.

So if your professor said that pathway "A", the top line in your drawing, is favored, then the reaction is being run under conditions favoring formation of the thermodynamic enolate.


Nuclear chemistry home made A boy in Memphis built 1 year ago a nuclear fusion reactor at his home . How did he achieve so high temperatures in order for the fusion to take place?The fusion that took place was 2 atoms of hydrogen-2 were fused into helium.What I know is that hydrogen-2 can only be made by fusion of 2 hydrogen-1 atoms and then the unstable radioactive diproton which is produced is converted into hydrogen-2 and 1 positron.Where did he find so much energy? And if we pressure a gas very hard,can  we fuse 2 hydrogen atoms into 1 diproton?

Yes, this really happened, probably (it's not a peer-reviewed effort).

The reactor is a Farnsworth fusor, a very low-density, low-rate inertial-confinement reactor. They're surprisingly easy to build and get working; many people have done so, and verified their function by measuring neutron output. An accelerating voltage of 10 or 20 kV produces particle energies equivalent to over 100 million degrees C.

There's active debate about whether a fusor of this design could ever produce more power than it consumes, but there's been some interest in using them as compact neutron sources (that you can actually switch on and off).

You asked about fusing hydrogen-1 to hydrogen-2. These experimenters don't attempt that; they just buy deuterium, which is also readily available.


Instrumental methods of analysis I am taking the instrumental methods of analysis course. Our job is to find the unknown  mass of $\ce{Fe}$ which is added, by measuring the absorbance of different solutions with ortho-fenantroline.

Is there any way I can know if my calculated mass of $\ce{Fe}$ is correct or not?   Is there any way I can prove it, since I know the absorbances?

Am I right that you are planning to simply calculate the $\ce{Fe}$ concentration using the Lambert-Beer equation?

The obvious proof would be to prepare $\ce{Fe}$ solutions with known concentration(s) to compare the absorbances. Using these solutions, you can also make sure that the assumption of linearity over the concentration range is actually fulfilled.


What does it mean for meth or any other compound to be 'pure'? In Breaking Bad, Walter White can cook $99.1\%$ pure meth. In general what does it mean for a substance to be pure? In this case, what exactly does it mean for meth to be pure? Does higher purity mean less unhealthy? More addictive? Giving more euphoria for the user?

For comparison, is the purity of prescribed methamphetamine or methylphenidate to people with ADHD or narcolepsy close to $100\%$? My guess is that purity is how close Walt's product is with legal meth.

When a compound is "pure" it means the contents of that compound are exactly what we claim them to be.  If something is $99\%$ pure, that means $99\%$ of it is the expected material (in this case, meth), and 1% is other non-meth compounds.

Purity matters in chemistry and biology because the other compounds can be very nasty.  In theory, if you had a bag of 100% pure meth, and a bag of $50\%$ pure meth, you could simply double the dose with the second bag and end up with the same amount of methamphetamine in your system.  However, you would also have a double dose of $50\%$ "other" stuff.

Sometimes "other" stuff isn't a big deal.  It all depends what that "other" stuff is.  I don't know much about the methamphetamine trade, but the effect of purity is a big deal in the ethanol world.  Moonshiners know to discard the "head" and the "tail" of the distilation, the first and last parts of the distilation, because they contain fusel oils (which taste really bad), and methanol (which can make you go blind in large enough dosages).  They want very "pure" ethanol.

To you talk about the purity of prescription drugs, typically what matters more for the drug companies is that they fully understand what "other" things are in the pills or injections, and they are confident they are safe.  That's why, when you look at many pill vials they will list their "inactive ingredients." They want to make sure you are very confident that the only things you are putting in your body are the "right" things.  Going back to the alcohol industry, "pure" ethanol is actually $95.6\%$ ethanol by volume, and $4.4\%$ water.  That is the limit for how "pure" ethanol can get with fractional distillation.  To have ethanol in any higher percentages than that you have to go to exotic processes to create anhydrous alcohols.  However, that is deemed "pure" because we still know what's in it.  We know its $95.6\%$ ethanol and $4.4\%$ water.

Of course, $100\%$ purity isn't always the desire.  There are myriad alcoholic spirits out there.  All of them are sold at 80 proof ($40\%$ alcohol by volume), so they contain the same amount of ethanol.  Its the "other" stuff that makes each spirit unique.  You pay a great deal of money for the particular impurities that make a great well aged scotch.  You pay very little for the impurities in a cheap tequila that leave your head pounding the next morning.


What is the difference between an ionic and molecular activity model? (Example: Diopside-Hedenbergite) Original question:

When looking at a Diopside-Hedenbergite solid solution, what is the difference between an ionic and molecular activity model? Why does the moleclar model yield higher values?

Post-Edit question:

The two activities calculated below seem to me like I'm just comparing some random values. What is the true activity of the Fe- or Mg-mineral?

So far I got:

Dipside and Hedenbergite are Pyroxenes. Pyroxenes have to metal sites (M1 and M2):

\begin{align}
M^1M^2[Si_2O_6]
\end{align}

The difference between Hedenbergite and Diopside is:

\begin{align}
Hedenbergit (Fe^{[M1]}Ca^{[M2]}[Si_2O_6]) \longleftrightarrow Diopside (Mg^{[M1]}Ca^{[M1]}[Si_2O_6])
\end{align}

The activity of an element is assumed to be equal to the concentration:

\begin{align}
a=\color{Red}{\gamma}x \\
a=x
\end{align}

In the next step one has to calculate the concentrations on the M1 site because it's the site where the ion exchange happens:

\begin{align}
x[Fe]=\frac{[Fe]}{[Fe]+[Mg]}
\end{align}

\begin{align}
x[Mg]=\frac{[Mg]}{[Fe]+[Mg]}
\end{align}

The molecular mixing model apparently compares the activity of the Fe- and Mg-mineral on the M1 site:

\begin{align}
a[Mg]=x[Mg]
\end{align}

\begin{align}
a[Fe]=\frac{1-x[Mg]}{x[Fe]}
\end{align}

Whereas the ionic mixing model compares the activities of all the cations of the minerals:

\begin{align}
a[Mg]=x[Mg]^{[M1]}x[Ca]^{[M2]}
\end{align}

\begin{align}
a[Fe]=x[Fe]^{[M1]}x[Ca]^{[M2]}
\end{align}

Why are there two activities? Shouldn't there be one true activity? What does the one activity mean and what does the other mean?

The activity of a chemical species is a key concept in thermodynamics of solutions. While the thermodynamic potential is the “true thermodynamic variable”, i.e. the one that plays a primary role in statistical mechanics, chemists do tend to like activity (which is univocally related to chemical potential) because it is close to another quantity which has clear physical meaning: concentration.

Now, relating the activity (or chemical potential) of a species to its concentration in a mixture is impossible to do in a general way. Models have been developed to provide approximations, however, the best known being the ideal solution, where activity equals molarity (concentration). Your “molecular mixing model” appears to be a model of ideal solid solution. However, one cannot expect this approximation to hold in case of strongly non-ideal solutions, and ionic solutions are prominent in that category. The “ionic mixing model” you describe thus appears to be another such approximation, more suited to these ions in a solid solution. I have never encountered it before, but I found a series of lecture slides where it seems to be very neatly introduced (see around slide 31).

While I cannot fully address the question, I hope this helps!


IBrF2 - van der Waals repulsions - and number of unique bond angles I understand that the molecule $\ce{IBrF2}$ is T-shaped, with three ligand atoms and two non-bonding pairs.

However, how many unique bond angles can this molecule exhibit?

My professor argues for the existence of only 2 unique bond angles in this molecule, with the somewhat $< 180^\circ$ axial $\ce{F-I-F}$ bond angle and the two equivalent and somewhat $< 90^\circ$ $\ce{F-I-Br}$ bond angles.

However, can't we also arrange this molecule as to have this axial arrangement of atoms: $\ce{Br-I-F}$?

My professor argues that this arrangement is impossible due to van der Waals repulsions. I suppose this makes sense; the two lone-pairs $\ce{<->}$ large bromine atom repulsions would make this shape unfavorable.

Impossible? I'm not sure about that. Dammit, nothing in chemistry is impossible.

What do you think? I'm arguing for the existence of either two or three bond angles because we can either have a $\ce{F-I-F}$ axial arrangement or a $\ce{Br-I-F}$ axial arrangement. Plus, axial bonds are longer than equatorial bonds. So that would lower the energy of the $\ce{Br-I-F}$ axial conformer. Side note: why are axial bonds longer than equatorial bonds?

In sum:

I suspect the molecule $\ce{IBrF2}$ does not exist, and is purely conjectural (theoretical) at this point.  If it ever had been synthesized or observed, it would have made the cover of Science or Nature.  "First neutral ternary interhalogen compound found!" (A handful of anionic  ternary interhalogen species are known.)

But my hunch is that, if the molecule did exist, the conformation your professor describes would be favored because of symmetry. Having the same atom in (roughly) opposite directions maximizes the amount of coupling because the opposite orbitals have same energy and can combine better than if it were two different atoms.  Hence greater stabilization.

  Side note: why are axial bonds longer than equatorial bonds?

This is true in binary compounds like $\ce{PCl5}$, and apparently a sufficient explanation is that each axial chlorine wants to avoid 3 other chlorines at 90$^\circ$ angles, while the equatorial chlorines have 2 at 90$^\circ$ but the other 2 are further away anyway, at 120$^\circ$.  In other words, if you just want to maximize the average Cl-Cl distance, subject to the constraint that the P-Cl distance is reasonable, you're better off with a slightly elongated bipyramid instead of one inscribed in a sphere.

In the hypothetical $\ce{IBrF2}$ of course the axial I-F bond would be shorter than the "equatorial" I-Br bond, since Br is a bigger atom.


Why Is This The Worst Beer Cup My wife bought me a special cup for me to drink beer out of.  The problem is that, if I pour any beer in it, it goes crazy with foam.  I will get just a few ounces of beer in there and the rest of the 16 ounce cup will be full of foam.  The cup is hand spun from clay.  The texture is still pretty rough.  So, if there is any glazing, it is not really noticeable.  The cup also has little brown specks in it here and there.  Does anyone know the reaction that is taking place to produce all this foam?  Is there a way to fix it?  Let me know if I need to supply any more details.  You should assume you are talking to someone whose last chemistry class was high school.

The rough texture is a physical catalyst (provides nucleation sites) for the evolution of carbon dioxide from dissolved carbonic acid. Pottery glaze fixes the foaming issue.


Aromaticity of heterocyclic compound This is a question of JAM 2017. According to the answer key the answer is 4. But I found only the last 3 as aromatic . So which one is the 4th aromatic compound? Please explain

Actually not all of the last three are aromatic.  Look carefully at the third compound; the heteroatom is boron, not nitrogen, and so does not contribute the electron pair that would be needed to make six pi electrons in an aromatic ring.  That compound  is instead antiaromatic.

So JAM 2017 is apparently rendering both of the first two compounds as aromatic.  This, however, is dubious.  The first compound in the list is a pyrone, which not everyone would call aromatic.  To make an aromatic contributing structure in pyrone we would have to render the carbonyl group as its zwitterionic structure ($\ce{C^+-O^-}$) and then distribute the positive charge around the ring which has an electronegative oxygen atom.  We'd have to weigh the favorable effect of aromaticity against the unfavorable effect of putting a positive charge into an electronegative ring.

As for the second compound, we would need to "polarize" two carbonyl groups and concentrate two positive charges into the small ring, and that is not happening even with aromaticity.  There is a saving grace, however: if the $\ce{N-H}$ protons are tautomerized onto the oxygen atoms, converting the "keto" structure to its "enol" form, then the tautomerized molecule is 2,4-dihydroxy-1,3-diazine, which would be unambiguously aromatic.

What it boils down to is that aromaticity is often not the simple yes/no question often suggested in introductory chemistry texts.  Given this ambiguity, one could potentially defend an answer of 2, 3, or 4 (but not the last three for the two larger numbers) to the JAM 2017 question.


Mechanism of reaction between triphenylmethanol and acetyl chloride I am reading an old method for making triphenylchloromethane (Org. Synth. 1943, 23, 100):

$\hspace{30 mm}$

When I draw out the structures of the molecules, I see this:

But the $\mathrm{S_N1}$ mechanism doesn't make sense to me since $\ce{OH-}$ is a bad leaving group, and there is nothing to protonate it.

I find it unlikely that the substitution proceeds via the hydroxyl group directly leaving. If there is any evidence that this reaction proceeds via direct $\mathrm{S_N1}$ of $\ce{OH-}$ by $\ce{Cl-}$, I would be very interested to see it. When an alcohol and an acyl chloride are mixed, I think it is much more probable that they will initially form the ester via nucleophilic acyl substitution:

$\hspace{27 mm}$

Presumably, this ester is rather unstable because of steric repulsions between the three phenyl groups and the ester group. So now, the $\ce{-OAc}$ group ($\ce{Ac} = \ce{COCH3}$) can leave as a relatively stable carboxylate ion to give the resonance-stabilised carbocation:

$\hspace{35 mm}$

The carbocation is then intercepted by chloride ions that were formed earlier to give the product you want. So, the second part of the reaction is a $\mathrm{S_N1}$ substitution of $\ce{OAc-}$ by $\ce{Cl-}$.

$\ce{OAc-}$ ($\mathrm{p}K_\mathrm{aH} = 4.76$) is a much better leaving group than $\ce{OH-}$ ($\mathrm{p}K_\mathrm{aH} = 15.7$).

There is another pathway possible, since $\ce{HCl}$ is formed from the initial reaction of the alcohol with the acyl chloride. If $\ce{HCl}$ is present in appreciable quantities, it can protonate the original alcohol and you could perhaps have a direct $\mathrm{S_N1}$ of $\ce{H2O}$ by $\ce{Cl-}$.

$\hspace{10 mm}$


Kinetics of molecule reacting with another of itself For the following reaction:

$$ \ce{O3 + O3 -> 3O2} $$

The rule (which makes sense for combination of two different moluecules) states that the rate constant R should be:

$$ R \propto \ce{[O3][O3]} = \ce{[O_3]^2} $$

But in the case of a combination of the same molecule this seems counter-intuitive. If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

No. You seem to be really close to understanding why this would not be the case, though. To help bridge the gap, let's consider the kinetics of a general bimolecular reaction:

$$\ce{A + B -> C}$$

We know that: $$\ce{\frac{d[\ce{P}]}{dt}} \propto \ce{[A][B]}$$

and that doubling $\ce{[A]}$ doubles $\ce{\frac{d[\ce{P}]}{dt}}$ because when we substitute $[2\ce{A}]$ for $\ce{[A]}$, we get: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A][B]}=2\ce{[A][B]}$$

If $\ce{A}=\ce{B}$, however, then it follows that we must also substitute $[2\ce{A}]$ for $\ce{[B]}$, which then gives us: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A}][2\ce{A}]=4\ce{[A][A]}=4\ce{[A][B]}$$


Which ions are accounted for in total and net ionic equations? There are a number of things I don't understand about ionic equations.

First off, when you do net ionic equations, is it correct that you're only focusing on precipitates? The way I understood it, you put "no reaction" if there are no solid or gaseous products. To clarify, if all of the products of the reaction are aqueous, we look at it as if no reaction had taken place.

Then I went and looked in my book, and that made it seem like all products that weren't covalent were spectator ions, i.e., covalent molecules were the only ones you note.  Now I'm even more confused.

After reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little guide to ionic equations:

Total Ionic Equations

A total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction
$$\ce{2 Na3PO4 + 3 CaCl2 -> 6 NaCl + Ca3(PO4)2 v},$$
one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following:
$$\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -> 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$

Net Ionic Equations

Net ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation:
$$\ce{2 PO4^3- + 3 Ca^2+ -> Ca3(PO4)2 v}$$

The ions that were eliminated are termed spectator ions, because they do not participate in any notable chemical reaction.

If something is still unclear, maybe have a go at Kristy M. Bailey's
website (Oklahoma City Community College), which explains it pretty neatly.


Retrosynthesis of spirocyclic compounds? I have been trying to complete this retrosynthesis problem for a few days but can’t figure out how to get from Ester starting compound to a spirocyclic compound. I have tried removing the Ester but can’t figure out what best to replace it with.
I have access to any reagents and starting materials as long as the nominated substance is used.

Step 1 - Form the anion of your starting material ($\ce{NaOEt/EtOH}$ will do this nicely) and react with phenethyl bromide ($\ce{PhCH2CH2Br}$).

Step 2 - Add the product to hot polyphosphoric acid$\ce{^{[1]}}$ to cyclize it to the required product.

Reference

Cyclization of Aryl-Aliphatic Esters with Phosphorus Pentoxide in Phosphoric Acid by Richard C. Gilmore, Journal of the American Chemical Society, 1951, 73 (12), 5879-5880, DOI: 10.1021/ja01156a521


Difference between catenation and allotropy Can anyone tell me what the difference between catenation and allotropy is? I would like the basic explanation.

Catenation refers to the ability of an element to form long chains. The most notable example is carbon (catenation of carbon is what gives rise to the entire field of organic chemistry).

Allotropy is an entirely different concept; it refers to the natural existence of an element in different forms, i.e. bonded differently. For example, solid carbon exists in the forms of diamond and graphite (and more).

In a sense, the two are related because the ability of an element to catenate often leads to the formation of many different allotropes, since the chains can be linked in different ways. Apart from carbon, another example of an element that can catenate well and forms multiple allotropes would be phosphorus. On the other hand, nitrogen and oxygen prefer to form triple and double bonds respectively, instead of single bonds; this means that catenation is unlikely and that is why these elements have few allotropes (nitrogen exists as diatomic $\ce{N2}$, oxygen as diatomic $\ce{O2}$ and ozone).


Finding equation of line of the P-T graph for a gas dissociation reaction Gas A (1 mol) dissociates in a closed rigid container of volume 0.16 lit as per following reaction.

  $$\ce{2A (g) -> 3B (g) + 2C (g)}$$

  If degree of dissociation of $\ce{A}$ is 0.4 and remains constant in entire range of temperature, then correct P Vs T graph is

  Slope 0.75 and passing through origin
  Slope 0.8 and passing through origin
  slope 1.1 and passing through origin
  slope 0.4 and passing through origin

\begin{align}
\ce{2A &&-> 3B& + &2C}\\
1-2x&&3x&&2x\\
\end{align}

where $2x$ is dissociation constant. Substituting value of $x$ we get moles of A, B and C to be 0.4, 0.6 and 0.4 respectively.

I am confused about how to proceed after this. How to proceed?

The overall notion is that gases A, B and C are all behaving as an ideal gases which can be represented by the mythical ideal gas X. Then the moles of X are equal to moles of A+B+C. Then we can use the ideal gas equation:

$$pV=nRT$$

which simplifies to:

$$p = \dfrac{nR}{V}\times T$$

Given that the degree of dissociation of A is 0.4 and we started with 1 moles of A, then 0.6 moles of A remain and 0.4 moles got converted.

$$\ce{2A -> 3B + 2C}$$

or

$$\ce{A -> }\dfrac{3}{2}\ce{B + C}$$

let B and C equal mythical gas X and

$$\ce{A -> }\dfrac{5}{2}\ce{X}$$

but since A = 0.6 moles of X, the total moles are

$$n = 0.6 + 0.4\times\dfrac{5}{2} = 1.6\text{ moles}$$

We know that n = 1.6 moles and that V = 0.16 L so:

$$p =\left(\dfrac{n}{V}\right)RT = \dfrac{1.6\text{ mole}}{0.16\text{liter}}RT = 10RT\dfrac{\text{moles}}{\text{liter}}$$

Now the problem doesn't really specify what value of the ideal gas constant, $R$, to use nor what units the temperature, $T$, should be.

The temperature must be an absolute scale since the line passes through the origin. Thus of the main choices of Kelvin and Rankine, using Kelvin seems reasonable for temperature.

but there are still options for the pressure units of R

(a) $8.314\text{ L}\cdot\text{kPa}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(b) $8.314\times10^{−2} \text{ L}\cdot\text{bar}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(c) $62.36 \text{ L}\cdot\text{torr}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(d) $8.206\times10^{−2}\text{ L}\cdot\text{atm}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$

For $10R$ to be between 0.4 and 1.1 the only answers that fit are (b) and (d). Thus the pressure could be either in atm or bar. Presumably the OP's book would have stated a preference for R.


If SOCl2 reacts with alcohols via SNi, why doesn't POCl3? A bit of a follow-up to Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols?.

The reaction with $\ce{SOCl2}$ is also used instead of $\ce{PCl3}$ and $\ce{PCl5}$ when retention of stereochemistry is required. Phosphorus trichloride and phosphorus pentachloride both lead to inversion.

The mechanism proceeds via a $\mathrm{SN_i}$ pathway, or internal nucleophilic substitution. This step has been highlighed in red; an intimate ion pair$^{[\text{see below}]}$ is given in square brackets. Adding a nucleophilic solvent such as pyridine considerably increases inversion with $\ce{SOCl2}$ via an attack at the sulfur atom. Green marks blocking of the intramolecular substitution; blue is for inversion.

As far as I know, $\ce{POCl3}$ has not been seen following an $\mathrm{SN_i}$ mechanism (even without pyridine). Why?

If possible, please provide a qualitative mechanistic reason. Something that is readily explained by waving hands. Regardless of whether such an approach is achievable, quantitative (calculational) answers are also most welcome.

References discussing intimate ion pair formation

F. A. Carey, R. J. Sundberg. $(2007)$. Advanced Organic Chemistry
Part A: Structure and Mechanisms, $4$th edition, pp 269$-$276. ISBN: 0-306-46242-7
W. A. Hughes, E. D. Cowdrey, C. K. Ingold, S. Masterman, A. D. Scott. 'The Mechanism of Elimination Reactions. Part 1. Unimolecular Olefin Formation from Alkyl Halides in Bulphur Dioxide and Formic Acid'. Journal of the Chemical Society, $(1937)$, 1271$-$1277. DOI: 10.1039/JR9370001271
E. S. Lewis, C. E. Boozer. 'The Kinetics and Stereochemistry of the Decomposition of Secondary Alkyl Chlorosulfites'. Journal of the American Chemical Society, $(1952)$, 74, 308$-$311. DOI: 10.1021/ja01122a005
D. J. Cram. 'Studies in Stereochemistry. XVI. Ionic Intermediates in the Decomposition of Certain Alkyl Chlorosulfites'. Journal of the American Chemical Society, $(1953)$, 75, 332$-$338. DOI: 10.1021/ja01098a024
C. C. Lee, A. J. Finlayson. 'Rearrangement In The Reaction Between Thionyl Chloride And $3$-Methyl-$2$-Butanol. Canadian Journal of Chemistry, $(1961)$, 39(1): 260$-$261. DOI: 10.1139/v61-030
C. C. Lee, J. W. Clayton, D. G. Lee, A. J. Finlayson. 'Rearrangement Studies With $\ce{^14C-XIII}$: The Thermal Decomposition Of $1$-$\ce{^14C}$-$2$-Butyl Chlorosulfite'. Tetrahedron, $(1962)$, 18 1395$-$1402. DOI: 10.1021/ja01098a024
H. Patin, G. Mignani, C. Mahe, J-Y. Le Marouille, A. Benoit, D. Grandjean. 'Ferrocenyltrithiocarbonates: I. Direct access from α-ferrocenylcarbinols by a $\mathrm{SN_i}$ mechanism. Absolute x-ray structure determination of (R)-ferrocenylmethylmethane S-methyl-trithiocarbonate'. Journal of Organometallic Chemistry, $(1980)$, 193, 1, 93$-$103. DOI: 10.1016/S0022-328X(00)86079-9
J. L. Kice, G. C. Hanson. 'Mechanisms of SNi reactions. Effect of aralkyl group structure on ion-pair return in the decomposition of aralkyl thiocarbonates'. Journal of the American Chemical Society, $(1973)$, 38 (7), 1410$-$1415. DOI: 10.1021/jo00947a037
M. B. Smith. $(2013)$. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, $7$th edition, pp 311, 486$-$487, 490, 598, 1316. ISBN: 978-0-470-46259-1
 The $\mathrm{SN_i}$ (substitution nucleophilic internal) mechanism: retention of configuration. Powerpoint presentation, Università degli Studi di Napoli Federico II.
James. $\ce{SOCl2}$ and the $\mathrm{SN_i}$ Mechanism. Master Organic Chemistry, Alcohols. webpage

The step in the red box is wrong. You probably get dissociation to form an ion pair. The pair quickly collapses, leading to chloride attack on the cation before any dissociation or diffusion occurs so that you get the retention of stereochemistry.

For phosphorus based chlorination, this ion-pair mechanism is probably not favorable, though it's not clear to me why this is the way it is.


Synthesis of phenoxyacetone from phenol How can phenoxyacetone be synthesized from phenol?

My Attempt:

We would first need to deprotonate phenol by addition of NaH in THF and then add bromopropene.

I'm not quite sure where I would go from here. Any help would be much appreciated!

Your suggested method seems excessively complicated. Substitution of chloroacetone with phenol under basic conditions should give O-alkylation in reasonably good yield. The phenol will be deprotonated and the nucleophilic phenoxide ion will then undergo an $S_N2$ reaction with the primary alkyl chloride.


Why hasn't chemistry brought about a greater level of human enhancement(as of yet)? So, this question had me thinking, this as someone who comes from the field of machine learning. Why haven't chemicals enhanced our minds and bodies as much as was predicted in the early years?

From the 40s till about the 70s I would say, many scholars and general talking heads predicted that chemistry would not only revolutionize food, industry(which it has) but also completely redefine the limits of humans by extending life exponentially as well as give super-human like mental and physical attributes.

Seemingly, this has not happened yet. While chemistry has completely changed a lot of what humanity does, it has never quite lived up to expectations(much like mine). So what is the disconnect? Is it that we just need more time? I am suspicious of that as chemistry is an very old field by academic standards and from what I read has been seeing diminishing returns in research.

I ask this as someone with only a rudimentary understanding of organic chemistry. Thanks!

This is not a limitation related to chemistry, but related to the lack of understanding of the human body and mind.
There are a couple exceptions, such as the Cochlear implant (wich basically transforms a deaf person into a hearing cyborg), pacemakers, and some drugs that can enhance mental capabiliyties.


Cathode + Anode + Rechargeable battery I'm confused by the following from this webpage:

The cathode is a metal oxide and the anode consists of porous carbon. During discharge, the ions flow from the anode to the cathode through the electrolyte and separator; charge reverses the direction and the ions flow from the cathode to the anode.
On discharge, the anode undergoes oxidation, or loss of electrons, and the cathode sees a reduction, or a gain of electrons. Charge reverses the movement.

This says that the metal oxide electrode is always the cathode and the porous carbon electrode is always the anode. As far as I know, that designation should be correct on discharge but backwards on charge. The anode is always whichever electrode is doing the oxidation, and the cathode is the electrode doing the reduction.
My naive understanding would be this:

The cathode is the electrode with the reduction half reaction. The anode is the electrode with the oxidation half reaction. This is true for both charge/discharge.
The anode is the one that produces electrons and cathode receives electrons. This is true for both charge/discharge.
When switching between charge/discharge, the redox reactions reverse, and the cathode/anode designations also switch to preserve cathode==reduction and anode==oxidation.
During discharge, the battery functions as a galvanic cell, where the redox reaction produces electric energy, electrons go down their electrical gradient from the negative electrode to the positive electrode. The anode is the negative electrode, the cathode is the positive electrode.
During charge, the battery functions as an electrolytic cell, where electric energy drives a nonspontaneous redox reaction, electrons go up their electrical gradient from the positive electrode to the negative electrode. The anode is the positive electrode, the cathode is the negative electrode.
In a lithium ion battery, the positive electrode is the metal oxide and the negative electrode is the porous carbon. The anode/cathode designations switch depending on whether the battery is charging or discharging.

Please help me understand this.

I have asked experts -> I am right. The common convention is to label the physical electrodes as cathode/anode based on discharge even though that will be backwards when the battery is charging. That is a poor, confusing, and illogical naming convention, but it is what it is.

The more clear and logical naming convention would be to name the physical electrodes either positive or negative since those labels apply in both charge/discharge directions. Some people use this logical naming convention, but the illogical convention seems more prevalent.


Spontaneity and nature of attack of fluorine gas on aluminum What is the nature of the reaction of attack of fluorine gas on aluminum metal? Is it spontaneous in nature?
I have studied reactions of halogens on aluminum but it had no information about the attack of fluorine and I would like to know about this reaction.

Aluminium has a tenacious oxide layer so the reaction of gaseous fluorine with aluminium metal is firstly the reaction of fluorine with aluminium oxide.

So finally $\ce{F2}$ will react with $\ce{Al2O3}$? Also will $\ce{F2}$ react with fresh $\ce{Al}$ metal surface.

First and foremost! Do not do this reaction unless you are properly trained and have the appropriate safety equipment. Fluorine is one of the most dangerous substances out there, and one of the presumed products, $\ce{OF2}$, is also terrifying.
If you can calculate the free energy change for a reaction, you will know whether it is spontaneous. To do this, you need to have more information than you currently provide. We need to know the products of the reaction in order to write a balanced chemical equation. Then we can look up appropriate thermodynamic values for each substance in the reaction in order to calculate the free energy change. Below is what I presume to be the balanced equation for the reaction of fluorine with aluminum oxide:
$$\ce{Al2O3 + 6F2 -> 2AlF3 + 3OF2}$$
Using the NIST Chemistry Webbook, we can look up the enthalpies of formation and standard entropies of each substance so that we may calculate the enthalpy and entropy changes.
$$\begin{array}{l|c|c|c|c|c|c}
& \ce{Al2O3} & \ce{F2} & \ce{AlF3} & \ce{OF2}  \\ \hline
\Delta_f H^\circ\ \mathrm{(kJ/mol)} & -1675.7 & 0 & -1510.4 & 24.52 \\
S^\circ\ \mathrm{(J/K)} & 50.92 & 202.8 & 66.5 & 247.46 \\
\end{array}$$
We can use these data to calculate the free energy change of the reaction:
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_rS^\circ$$
If the free energy change is negative at that temperature, the reaction is spontaneous.


Does tetramine copper dissolve cellulose? I saw a Youtube video which showed that cotton (cellulose) dissolves in deep blue solution of tetramine copper made by adding 20% ammonia in copper carbonate salt. I tried to replicate the process using $\ce{CuSO4}$ and copper chloride. None worked. Where have I gone wrong?

As @iad22agp states, it is not sufficiently alkaline. Schweizer's reagent is tetraamminediaquacopper dihydroxide, $\ce{[Cu(NH3)4(H2O)2](OH)2}$. See Schweizer's reagent.

That is not Doktor Schweizer, I presume.


Cauchy-Schwarz integral screening different inequalites Eq. (9.12.23) in [1] states that:

$$\sum_{abcd}c_{ab}c_{cd}g_{abcd}>0\tag{MEST 9.12.23}\label{91223}$$

with $c_{ab}=c_ac_b$ for $c_a$ being some distribution coefficients and $g_{abcd}$ being the two-electron integrals in AO basis. Now for $c=a$ and $d=b$ giving:

$$c_{ab}c_{ab}>0$$

Which seem obvious as long as $c$ are real numbers.
It can thus be seen that to fulfill Eq. (\ref{91223}) that:

$$g_{abab}>0 \tag{MEST 9.12.24}$$

Which fulfill the requirements for the Cauchy-Schwarz inequality, thus giving:

$$\left| g_{abcd} \right| \leq \sqrt{g_{abab}}\sqrt{g_{cdcd}}\tag{MEST 9.12.25}$$

Now since $c_{ab}=c_{a}c_{b}$ this implies that:

$$c_{aa}c_{bb}>0\tag{1}$$

which in turns further implies that:

$$g_{aabb}>0\tag{2}$$
Which surely also fulfill the requirements for the Cauchy-Schwarz inequality, giving:

$$\left| g_{abcd} \right| \leq \sqrt{g_{aabb}}\sqrt{g_{ccdd}}\tag{3}$$

Now further doing some index juggleling will lead to the following six different inequilities:

$$\left| g_{abcd} \right| \leq \sqrt{g_{abab}}\sqrt{g_{cdcd}}\tag{Ineq. 1}\label{ineq1}$$

$$\left| g_{abcd} \right| \leq \sqrt{g_{acac}}\sqrt{g_{bdbd}}\tag{Ineq. 2}$$

$$\left| g_{abcd} \right| \leq \sqrt{g_{adad}}\sqrt{g_{cbcb}}\tag{Ineq. 3}$$

$$\left| g_{abcd} \right| \leq \sqrt{g_{aabb}}\sqrt{g_{ccdd}}\tag{Ineq. 4}$$

$$\left| g_{abcd} \right| \leq \sqrt{g_{aacc}}\sqrt{g_{bbdd}}\tag{Ineq. 5}$$

$$\left| g_{abcd} \right| \leq \sqrt{g_{aadd}}\sqrt{g_{ccbb}}\tag{Ineq. 6}$$

I think I have ever only seen Eq. (\ref{ineq1}) described before, surely this could be because the rest are obvious extensions, that I just haven't thought of as being obvious before.
Or that I have made some flaws in my reasoning.
In practise are all of the inequilites used?
If not anyone got clues about why not?
Especially Ineq. 4-6 is somewhat different from Ineq. 1-3.

[1] : Molecular Electronic-Structure Theory, Trygve Helgaker, Poul and Jørgensen Jeppe Olsen

As you have already stated in the question, (Ineq. 1) is the standard inequality used in Cauchy-Schwarz integral screening. (Ineq. 2) and (Ineq. 3) are not valid inequalities, which is clear if you select atomic orbitals such that a & b are spatially disjoint from c & d and the proposed upper bounds vanish. (Ineq. 4), (Ineq. 5), and (Ineq. 6) follow from a different use of the Cauchy-Schwarz inequality where the inner product is interpreted as being over the six spatial integrals and requires the Coulomb kernel to be pointwise non-negative so that you can take its element-wise square root,

$$ \begin{align} |g_{abcd}| &= \left| \int d\vec{r} d\vec{r}' \phi_a(\vec{r}) \phi_b(\vec{r}) V(\vec{r},\vec{r}') \phi_c(\vec{r}') \phi_d(\vec{r}') \right| \\
&= \left| \sqrt{V}\phi_a \phi_c \cdot \sqrt{V}\phi_b \phi_d \right| \\ & \le \| \sqrt{V}\phi_a \phi_c \| \| \sqrt{V}\phi_b \phi_d \| = \sqrt{g_{aacc}} \sqrt{g_{bbdd}} \end{align}. $$  (Ineq. 4) is not useful for integral screening because it is strictly weaker than (Ineq. 1) since $g_{abab} \le g_{aabb}$ (for real orbitals). Whereas $g_{abab}$ vanishes as orbitals a & b become spatially disjoint, $g_{aabb}$ only decays algebraically in the average distance between a & b. (Ineq. 5) and (Ineq. 6) are not strictly weaker than (Ineq. 1), but they have the same problem of being weakly decaying bounds that don't screen very much. For example, they are tight and thus tighter than (Ineq. 1) when a=b and c=d, but this is a class of integrals that will only be screened out for extremely large spatial separations between a & c.


Z-matrix of a series of 5 colinear atoms How can I write a Z-matrix for the following complex?

  O==C==O  H--O
               \
                H

I have managed to write:

O
C   1  a
O   2  a     1  180
H   3  d     2  180    1  ?
O   4  b     3  180    2  ?
H   5  b     4    t    3  ?

but I don't know how to specify the dihedral angles indicated with ?. Are they all zero? Or completely unspecified? If so, will I be able to do a coordinate optimization with these variables or will it be a problem?

You cannot have bond angles of 0 or 180 degrees in a Z-matrix. This is because the dihedral becomes degenerate. To solve this, you can add 'dummy atoms' which provide an auxiliary point of reference to remove the 180 degree angle. These atoms are usually denoted $\ce{X}$ or $\ce{Xx}$ in computational chemistry suites. Usually, dummy atoms are placed at 90 degrees to a pair of atoms and the third collinear atom is then defined with a bond angle of 90 degrees and a dihedral of 0 or 180 degrees with respect to the previous 3 atoms.

Example

To illustrate what I'm talking about, here's a schematic of your pair of molecules, with three dummy atoms:

This could have a Z-matrix like:

O   1
C   1    a
Xx  2  1.0  1  90
O   2    b  3  90     1  180
Xx  4  1.0  2  90     3  0
H   4    c  5  90     2  180
Xx  6  1.0  4  90     5  0
O   6    d  7  90     4  180
H   8    e  6  104.5  7  180

Notes

There are no hard and fast rules about how many you should use (obviously we could simply define subsequent dihedrals in terms of the leftmost dummy atom) however it's best to strive for locality so that small changes in some variable don't correspond to large displacements elsewhere, in which case rounding errors or large gradients can ruin your day. You could also have defined all of the dihedrals in your molecule with respect to that convenient non-collinear hydrogen, however this is the Z-matrix equivalent of spaghetti code.

You will note that I use different letters for each variable. This is because forcing the $\ce{H-O}$ and $\ce{C=O}$ bonds to be of equal length is probably an unrealistic constraint for a geometry optimisation of this system.

Finally, note that dummy atoms do not take part in the electronic structure of a molecule, so feel free to put them in anywhere and set the dummy atom 'bond' length to whatever you want. 1.0 Ångström is traditional as far as I know.

P.S.

Molden has a complete, if idiosyncratic graphical Z-matrix editor.


Why is Charcoal such an excellent adsorbent? Charcoal (also activated charcoal) is known to adsorb a huge variety of substances including a variety of paints, dyes and many different kinds of ions. Moreover, the amount of adsorption at normal pressure and temperature is high. I know physisorption requires high capacity of Van-der-waal bonding with the adsorbate which is favoured by surface atoms with large, easily-dispersed electron clouds and chemisorption requires actual chemical bond formation which requires highly reactive surface atoms. Charged species get adsorbed to form a double layer when the surface is charged, either due to ionization or due to polarity.

  But charcoal possesses none of the structural specialities required to
  be a great adsorbent. Neither it has charged centres (unlike colloidal
  proteins, or metal sols) nor it has a very high reactivity (Charcoal
  is almost the final product of combustion of wood, bone, sugar etc and
  is supposed to be immune to further oxidation even at considerable
  temperature). Why is it then an excellent adsorbent?

Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface area to mass ratio.


What is the highest possible expanded octet? Often called "hypervalent", chemicals like phosphorous pentachloride and sulfur hexafluoride are possible due to the fact that their central atoms form covalent bonds with more than four other atoms, giving rise to uncommon arrangements.

In the case of $\ce{PCl5}$, the phosphorous atom forms 5 bonds, giving rise to trigonal bipyramidal shape according to the VSEPR model. With $\ce{SF6}$, the Sulfur atom forms 6 bonds, giving rise to octahedral shape.

My question is, which central atom can achieve the highest number of bonds through the use of an expanded octet, and what is its shape?

The greatest I can think of is iodine heptafluoride, whose shape is pentagonal bipyramidal due to its fluorine atom of 7 bonds.

I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention hypervalency, what you seem interested in is hypercoordination, or even more generally, just compounds with high coordination numbers (hypercoordination is used specifically when the number of ligands in a compound is larger than "normal"). Hypercoordination and high coordination numbers are entirely independent of hypervalency or VSEPR theory altogether. Regardless of the precise electronic structure in a compound, coordination numbers can often be determined far more directly, especially when the compounds can create crystal structures for x-ray crystallography or neutron diffraction.

Usually there is little focus given to compounds with more than six ligands, as the vast majority of compounds will have atoms surrounded by six or less ligands. However, there are some representatives for higher coordination numbers. Some books will mention iodine heptafluoride, $\ce{IF7}$, as a seven-coordinate compound, with its pentagonal bipyramidal structure. It is still possible to add a fluoride and obtain an example of coordination number 8 in the octafluoroiodate(VII) anion, $\ce{IF8^{-}}$, which has an interesting square antiprismatic geometry (take a cube and twist a face by 45°).

For coordination number 9, representatives can be found in the transition metal hydrides, such as the nonahydridorhenate anion, $\ce{ReH_9^{2-}}$, and the lowest energy configuration in this case is a curious tricapped triagonal prism.

Is it possible to go higher than coordination number 9 in coordination compounds? While there are examples, at the present it seems that none of them contain solely monodentate ligands, that is, individual ligands which only bind to the centre once. This is because either there would have to be a lot of crowding over the central atom, creating repulsions between the ligands, or because to allow enough space the ligands would have to stay relatively far from the central atom, making their bonds weak.

However, if a single ligand is allowed to bond to the centre through more than atom simultaneously, then the coordination number can keep increasing without requiring the presence of too many ligands. Actinides have a very rich coordination chemistry and are capable of generating several impressive compounds, such as uranocene with coordination number 16, but it seems the current record is held by actinide elements surrounded by four cyclopentadienyl rings, each with five carbon atoms, reaching an amazing coordination number of 20 in tetrakis(cyclopentadienyl)thorium(IV) ($\ce{Th}\mathrm{(\eta ^5-}\ce{C5H5}\mathrm{)_4}$ or its uranium analogue.

To finalize, it's interesting to note that though there are no examples yet of compounds with coordination number 10 or above containing only monodentate ligands, their expected geometries can be calculated even for much higher coordination numbers (under certain assumptions). For very high coordination numbers, calculations and physical reality will likely diverge, but perhaps a few more coordination numbers with the expected geometry will be unlocked by the study of ultraheavy element chemistry. Of course, the calculations linked here don't help much to study the geometry of coordination compounds with polydentate ligands, as they rely significantly on the geometry of the ligand itself

Edit: This very relevant article raises some interesting points. For example, endohedral fullerene compounds could be thought of as a central atom surrounded by a single ligand in the shape of a cage, so one could possibly make a case for structures with coordination numbers of 60, 70, 80 or even more. The article also calculates the possible existence of a compound containing 15 monodentate ligands, the cation $\ce{PbHe_{15}^{2+}}$, though it would be very weakly bound (as you might expect from a helium compound) and possibly restricted to the gas phase.


Why the energies of orbitals in the same subshell decrease with increase of atomic number? Why the energies of orbitals in the same subshell decrease with increase of atomic number? For eg., energy of 2s orbital of hydrogen atom is greater than energy of 2s orbital of lithium atom. What is the reason?

Short answer: more protons in the nucleus means higher nuclear charge which means stronger attraction between the electron and the nucleus.

Long answer: The energy of the orbital is driven by the Hamiltonian operator $\hat{H} = \hat{T} + \hat{V}$ where $\hat{V}$ is the potential energy. For a simple 1-electron atom/ion, the potential energy is the electrostatic attraction between the positively charged nucleus $Q$ and the negative electron $q$: $V = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r}$. Lithium has 3 protons in the nucleus versus the 1 proton for hydrogen, so $Q$ is three times higher and the potential energy of the 1s orbital in $\ce{Li^{2+}}$ is roughly three times more negative than the 1s orbital in $\ce{H}$ (remember $q$ is negative).

If you have more than one electron, the potential energy becomes more complicated because you have to include the electrostatic repulsion from all the other electrons as well, but the general trend still holds, all else being equal.


Laser cutting polyurethane I have a laser cutting machine and have always read that PVC (or types of vinyl in general) are very dangerous to cut because it releases chlorine gas. Chlorine gas = toxic (plus the $\ce{HCl}$ thing), makes perfect sense. This is well-known information, and it appears all over informational sites about laser cutting materials.

But I now have someone telling me that cutting polyurethane foam is way more dangerous since it produces $\ce{HCN}$ (hydrogen cyanide). Now this information is in a few limited places and mentioned kinda as a footnote, but it is not nearly as widespread as the Chlorine gas risk that is talked about elsewhere. (I know that the chlorine gas presence during cutting of PVC effectively makes $\ce{HCl}$ gas which clearly does damage to the laser system itself which might explain the prevalence of the information about it)

I know that polyurethane is made with isocyanates, which I'm wondering if that is leading to some people seeing that word and getting confused and assuming that there is cyanide in the foam.

What is confusing about this is that polyurethane is generally considered a very "laser friendly" material to cut (stated all over the place on laser hobbyist and professional websites under acceptable materials). Producing the same gasses that were used in gas chambers seems pretty darn dangerous and like there should be red flags up everywhere.

Can anyone here speak to the chemistry of this, firstly how much if any HCN would be produced by using a 10000nm laser (meaning intense heat based, but nonflaming cutting). And if properly ventilated with an industrial air handling system designed to move the full volume of air from the cutter outside many times per minute what are the risks involved? Does HCN bond to surfaces and stick around? Or is it fully ventilated as a gas?

Clearly, I'm not a chemistry expert, so looking for someone who knows more to help me understand what is real and what is just scary sounding jargon.

From SDS:

Special hazards arising from the substance or mixture: carbon oxides ($\ce{CO}$, $\ce{CO2}$) nitrogen oxides ($\ce{NO}$, $\ce{NO2}$ etc.) hydrocarbons, isocyanate vapours and hydrogen cyanide can be released in case of fire.
The polyurethane polymer formulation greatly changes so the fumes that can form at high temperatures (above the melting temperature). A suction and exhaust system is required. To measure the exact composition of the fumes very specific analysis are required.


Coordination compounds with alkene ligands I came across this compound: $\ce{[Pd(η^2-C2H4)Cl2(H2O)]}$

What does that '$\eta^2$' sign mean? I am guessing that it means that the ligand is bidentate because if it wasn't, I don't think it is possible (well it would be rare) for a metal to be only co-ordinates by 3 ligands.

If this is the case, how can the carbon atom in ethene bond to Pd as it doesn't have a lone pair? Does the double break and hence the 2 carbons bond to the metal?

Ethylene is a monodentate ligand and there are 4 ligands total: ethylene,  two chlorides, and water.

Regarding hapticity, you can find a great answer here: What does the eta notation mean in the naming of a transition metal complex?

While a chapter can be dedicated to the nature of organometallic bonding, I think it is practical to say that it is the two electrons in the pi-bond that are involved in bonding to the metal center.


Is dihydrogen the only example of overlap between two s orbitals? I heard that $\ce{H2}$ is the only example of s–s orbital overlap. Can anyone give an example which contradicts this statement?

You don't need particularly exotic things to get a counterexample. Metal–metal bonding in mercury(I) compounds is quite prevalent, where the electronic configuration is $\ce{Hg+}$: $\ce{[Xe](4f^14)(5d^10)(6s^1)}$, and the bonding involves 6s–6s overlap. The prototypical example is the $\ce{Hg2^2+}$ cation, which features in mercury(I) chloride $\ce{(Hg2^2+)(Cl^-)2}$; this cation is valence isoelectronic to $\ce{H2}$.

Fundamentally, it is a relativistic contraction of the 6s orbital which makes such bonding common in Hg(I): see also Why does mercury form polycations?. The bonding in alkali metal dimers is qualitatively similar, but significantly weaker.

If we play the "every bond has orbital overlap" card (i.e. not just covalent bonds, which I think is perfectly fair), then the delocalised bonding in Group 1 and 2 metals arises from overlap of ns orbitals on each atom to form bands, as has already been alluded to in the comments. Even simpler would be lithium hydride, which likely has some degree of 1s–2s orbital overlap, even though the bonding may be primarily ionic.


How would you go about solving this pH problem? What is the $\mathrm{pH}$ of a $0.24\ \mathrm M$ solution of benzylammonium chloride ($\ce{C6H5CH2NH3Cl}$)?
  $K_\mathrm b$ for benzylamine is $2.2\times10^{-5}$
  Your answer must be within $\pm 0.4\ \%$

I was able to determine that $[\ce{H3O^+}]=-1.8\times10^{-5}\ \mathrm M$ using $K_\mathrm b$. Then I used the formula $\mathrm{pH}=-\log[\ce{H3O^+}]$ to determine that the $\mathrm{pH}$ is $0.0000414465$. However, this answer keeps being marked as incorrect.

What am I doing wrong?

The $K_\mathrm{b}$ value of benzylamine, the conjugated base of benzylmmoniumchloride, is given. The acid constant of benzylammoniumchloride can be inferred via the identity
$$ \mathrm{p}K_a + \mathrm{p}K_b = 14. $$
Thus, we have
$$ \mathrm{p}K_\text{b,benzylamin} =  4.65$$
$$ \mathrm{p}K_\text{a,benzylammoniumchloride} = 9.35 $$
$$ \Rightarrow K_\text{a,benzylammoniumchloride} = 10^{-9.35}$$
As furthermore
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus] \cdot [\mathrm{PhCH_2NH_2}]}{[\mathrm{PhCH_2NH_3^\oplus]}} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
and
we see that
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
This leaves us with an equation with only one unknown variable and is easy to solve as $[\mathrm{PhCH_2NH_3^\oplus}] = 0.24~\mathrm{\frac{mol}{L}}$ is given:
$$
[\mathrm{H^\oplus}] = \sqrt{[\mathrm{PhCH_2NH_3^\oplus]}\cdot K_\text{a,benzylammoniumchloride}} \approx 1.0354 \cdot 10^{-5} ~\mathrm{mol/L}
$$
And as the pH is the negative of the logarithm with base ten we have:
$p\mathrm{H} \approx 4.985$


Heat Dissipation of Chemical Reaction When a chemical reaction dissipate heat to the surrounding, the molecules that participate in the chemical reaction will have high kinetic energy and will transmit some of the kinetic energy to the surrounding and the remaining kinetic energy will increase their temperature.

When we say that heat is dissipated to the surrounding we are talking about "some of the kinetic energy" right? Also when we write that heat dissipated is equal to $\Delta H$, it is included the remaining kinetic energy?

Can somebody confirm whether my understanding is correct or clarify where I went wrong if it is not?

Sorry, but I am far from chemistry field.

Let establish some thermochemistry terminology first:

Work: the total energy transferred by a system to another. There are several types of work. In thermodynamics, the most common one is pressure volume work, given by: $w = -p_{ext}\Delta V$

Internal Energy: the total kinetic and potential energy of a system

Enthalpy ($H$): is the total energy of a system. It also can be thought as the amount of energy that is required to make all the bonds in the system plus all the energy that is required to make room for the system. Hence it should be clear that it is given by the formula: $\Delta H = \Delta U + \Delta {(pV)}$

Heat Transfer (q): the amount of heat energy (kinetic energy) that is transferred from a system given by the formula: $q = \Delta U -w = \Delta U + p_{ext}\Delta V ~~$ (when only pressure volume work is done)

In your question, you are actually referring to $q$ (heat transfer), not $\Delta H$. This is because $\Delta H$ also accounts for the change in pressure and volume.

However under certain conditions, $q$ and $\Delta H$ can be equal. If the external pressure is constant (which is a common case as most experiments are done in the open atmosphere which has a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p_{ext}\Delta V = q$$


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

This is what the CIF is meant to be rendered like (shown in PyMOL):

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

note how so many bonds are not shown properly. I suspect if I could convert the molecule to SDF or CML or some other 3D chemical format I'd be able to fix this issue and Avogadro will be able to render the file properly.

I have tried doing the file conversion with OpenBabel but the files generated look, when viewed in PyMOL or Jmol (which render the CIF fine), just like the CIF does in Avogadro with all the disconnected atoms.


Melting points of benzoic and stearic acids I wonder, why does benzoic acid have a higher melting point than stearic acid?
Here is the data:
Benzoic acid melting point = 122.3 °C
Stearic acid melting point = 69.3 °C
I do understand that organic acids can bond with both London dispersion forces (van der Waals) and hydrogen bonds (due to the -COOH group). I have also learned that the longer carbon chains, the higher melting point due to the increase of the London dispersion forces.
However, in the case with benzoic acid vs stearic acid, I can't figure out an explanation as to why benzoic acid has a higher melting point. I suspect it is because of the resonance structure in the benzene ring in benzoic acid?

Puzzles are more puzzling when the pieces almost fit together, but for some unexplained reason, just don't fit. In other words, a comparison should be made between two almost  (but not quite) similar things.
In this case, stearic and benzoic acids are quite different. They have some carbons (quite different, and not even the same number!), and one similar group of four atoms ($-COOH$) - not very much similarity.
How about we compare the melting points of octadecane and stearic acid - and their structures? There's a lot of similarity there, and a little dissimilarity. Octadecane melts at 28-30 C (Ref. 1) vs. stearic acid at 69.3 C (Ref. 2). Obviously the difference is caused by the added stickiness of the carboxyl groups for one another, because they form dimers in the crystal, a structural feature (Ref. 3) lacking in the paraffin.

That establishes that a carboxyl group raises the melting point of a paraffin chain. An x-ray structure for $C_{24}H_{50}$ (octadecane not available) looks just like the one for stearic acid, but without the carboxyls (Ref. 4).

Now let's compare benzoic acid with toluene. The numbers of carbons are the same, and we just convert one $CH_3$ group to a $COOH$, as in the previous comparison. Toluene stacks in the solid (Ref. 5) with energies about 3-4 kcal/mol, which isn't much, considering that toluene melts at -95 C. But benzoic acid stacks nicely (Ref. 6) and melts at 122 C.

Now there's a puzzle: why does a carboxyl group raise the mp of a paraffin by 40 degrees, but raises the mp of an aromatic ring by 217 degrees? Looking at the formulas, looking at the drawings of a molecule (or two) doesn't help. Looking at actual data (because those little molecules have a mind of their own, and we have a hard time thinking like they do) like x-ray crystal data, suggests that putting a little planar group ($COOH)$ onto a long chain might strengthen intermolecular bonding a little, but extending a ring's planarity and adding in some hydrogen bonding, which amplifies the extent of planarity could do some powerful intermolecular stickiness.
I guess the wiggly straight chains don't have such good contact with their neighbors as the benzoic acid dimers do. Who would have guessed, without the crystal structure data?
Will puzzles never cease? If octadecane melts at 28-30 C, and its mp is raised by turning a methyl into a carboxyl, you might wonder what a phenyl group would do. Apparently, not much. The mp of 1-phenyloctadecane is 29 C (Ref. 7).
Ref.1 https://en.wikipedia.org/wiki/Octadecane
Ref.2 https://en.wikipedia.org/wiki/Stearic_acid
Ref. 3 https://www.researchgate.net/figure/Crystal-packing-of-left-the-A-super-14-and-right-the-A-1-13-two_fig1_233992144
Ref. 4 https://www.researchgate.net/figure/Crystal-structure-of-C-24-H-50_fig1_270344720
Ref. 5 https://www.x-mol.com/paper/6024417
Ref. 6 https://www.researchgate.net/figure/Selected-intermolecular-contacts-in-the-benzoic-acid-crystal-viewing-the-dimers-from-the_fig2_51049984
Ref. 7 https://pubchem.ncbi.nlm.nih.gov/compound/1-Phenyloctadecane#section=Experimental-Properties