Why does the unstabilised Wittig reaction selectively form cis alkenes? This question is meant for a simple unstabilised ylide.

The mechanism of the Wittig reaction, as given on ChemTube3d, involves a concerted formation of the oxaphosphetane (this is generally favoured over the traditional stepwise mechanism with betaine formation):

Here the puckered transition state has been clearly shown but involves the direct formation of the oxaphosphetane (without any betaine intermediate). In this transition state, the two possibilities are for the methyl groups to be either cis or trans to each other. Surely the transition state with the methyl groups trans to each other will be more stable. Doesn't this support formation of the trans product?

To my knowledge the mechanism of the Wittig reaction isn't fully resolved yet. But maybe I can give you some ideas about why the Wittig reaction with unstabilized ylides is (Z)-selective (well, with the exception of the Schlosser modification) instead of (E)-selective.

In the excellent book by Clayden, Warren, and Greeves, there is a section beginning p. 690 that describes the oxaphosphetane formation as a concerted, antarafacial [2+2]-cycloaddition reaction (similar to your "puckered transition state").

Here, the phosphonium ylide and the carbonyl compound approach each other at right angles. The substituents are arranged in order to minimise steric repulsions in the transition state. In particular, the phenyl group (Ph) points away from the PPh3 group, and then the R group on the ylide points away from the Ph.

The formation of the oxaphosphetane is irreversible and kinetically controlled. Hence, even though the trans oxaphosphetane is more stable, it is not formed.

(By the way, the book also has some justification why the Wittig reaction of stabilized ylides is E-selective.)

Edit:

Since the OP asked for it I will try to give some justification on why the ylide and the carbonyl group approach one another perpendicularly. For a proper description one would need the frontier orbitals of both compounds and then reason about the orbital interactions that bring about the [2+2] cycloaddition. Since I don't have those I will try to give some oversimplified description that should convey the general principle at work here. In the approximate orbital pictures I'm going to use I'll leave out the substituents for clarity.

To begin with, if one imagines the orbital interactions between the ylide and the carbonyl group, it is clear that the most important (primary) interaction for the reaction is the one between the HOMO of the ylide (nucleophile) and the LUMO of the carbonyl group (electrophile). What happens when both groups approach one another head-on or at right angles is shown in the following picture (for the perpendicular approach a top-view is used).

In both cases there are as many bonding interactions (black) as there are antibonding interactions (red). These situations are overall non-bonding and cannot lead to the desired product. This is the reason why [2+2] cycloadditions are usually thermally forbidden (by symmetry).

However, some compounds have additional orbitals that: (1) are similar in energy to the frontier orbitals drawn in the picture, and (2) possess the right symmetry to interact with the other orbitals. These "secondary orbital interactions" are weaker than the HOMO-LUMO interaction, but might provide enough bonding to tip the balance from an overall non-bonding to a bonding situation.

So, how does this translate to the situation of the Wittig reaction? Since O is more electronegative than C, the HOMO of the carbonyl group will have a larger orbital coefficient at O, while the LUMO will have a larger orbital coefficient at C. The situation is similar with the ylide. Since C is more electronegative than P, the HOMO of the ylide group will have a larger orbital coefficient at C. The ylide group is the nucleophile so the primary interaction will be between its HOMO and the LUMO of the electrophilic carbonyl group. Since the orbitals interact best where the coefficients are large, this primary interaction will be biased towards C–C bonding.

The question is now, is there a suitable secondary interaction? One can argue that P has a lot of orbitals (valence shell extension) and that the LUMO of the ylide might be some empty phosphorus orbital (or at least an orbital with a big coefficient on $\ce{P}$) - this could be a p-orbital perpendicular to the ylide's C=P π-orbitals, or a d-orbital, or something like that. This phosphorus orbital can then interact with the carbonyl group's HOMO, which has its biggest coefficient on O. So this bonding interaction will be biased towards P–O bonding. The situation is shown in the following picture:

The biases in the primary and secondary orbital interactions will lead to some distortions from the pure perpendicular approach which will become more pronounced the nearer the two species get to each other. For example, the ylide might rotate a bit so that its C and P atoms are closer to the carbonyl's C and O atoms respectively, in order to maximize the bonding overlap. And of course the perpendicular approach of ylide and carbonyl group will proceed in such a way that their respective biggest substituents are as far away from each other as possible (see diagram above).

After this quite lengthy answer I want to make clear that this description of mine is only a guess. The real reaction path is still a matter of debate and there are some compounds that rather react via a radical or ionic pathway. But I think it gives a good explanation for most of the observed behaviour and hope it helped the understanding. A more exact frontier orbital description of another thermal [2+2] cycloaddition, namely that of ketene with ethene, can be found in the textbook by Brueckner on p 653.


How to balance the reaction equation between copper (I) sulfide and nitric acid? What is the sum of stoichiometric coefficients of redox reaction: $\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$?

This exercise is from a multiple choice exam. I have balanced the equation:
$$\ce{1 Cu2S + 4 HNO3 -> 1 S + 2 NO2 + 1 Cu(NO3)2 + 2 H2O.}$$
Therefore I have calculated the sum of coefficients as $\sum a = 11$. This has not been offered as an answer and the correct answer is 20.

$$\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$$
Let's balance this reaction using the half-reaction method.
The oxidation half reaction (O.H.R) would be:
$$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -> 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$
The reduction half reaction (R.H.R) here is:
$$\ce{H\overset{+5}{N}O3 +e- -> \overset{+4}{N}O2} \tag{R.H.R}\label{rhr}$$
The final reaction would be $4\times\eqref{rhr} + 1\times\eqref{ohr}$.
Therefore the reaction would be partially balanced as:
$$\ce{Cu2S + 4HNO3 -> 2 Cu^2+ + S + 4NO2}$$
Now, we have $\ce{Cu(NO3)2}$ and not $\ce{Cu^2+}$, so we need to add the 4 more nitrate ions which implies 4 more molecules of $\ce{HNO3}$
So now, we have the equation as:
$$\ce{Cu2S + 8HNO3 -> 2 Cu(NO3)2 + S + 4NO2}$$
Now, we balance the hydrogens and oxygens, the left has 8 hydrogens and 24 oxygens, whereas the right has no hydrogens and 20 oxygens. Therefore, we need to add 4 water molecules to the right. Therefore, the final reaction becomes:
$$\ce{Cu2S + 8 HNO3 -> S + 4 NO2 + 2Cu(NO3)2 + 4 H2O}$$
Now, $\sum a = 1 + 8 + 1 + 4 + 2 + 4 = 20$ and not 23.


Do non-redox reactions exist? Redox reaction is a type of chemical reaction, and is the result of electron transfer between chemical species. But, all chemical reactions somehow involve electron transfer!

So, are there chemical reactions without electron transfer?

There are many, here are some examples

Ion interchange reactions of various types, such as precipitation

$\ce{NaCl + AgNO3 -> AgCl \downarrow + NaNO3 }$
Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions.

$\ce{PCl5 + 4 H2O ->  H3PO4 + 5 HCl}$
Reorganization of bonds between atoms of same type, such as catalytic benzene synthesis from acetylene

$\ce{3 C2H2 -> C6H6}$
Ligand exchange

$\ce{ [Fe(H2O)6]^{3+} + 6CN- -> [Fe(CN)6]^{3-} + 6 H2O}$
polar oligomerization

$\ce{ H2O + n CH2O -> HO-(-CH2-O-)_{n}-H}$


Which is more soluble in water, HF or SO2? I was wondering which is more soluble in water, hydrogen fluoride or sulfur dioxide? Since $\ce{HF}$ will hydrogen bond to other HF particles (as opposed to $\ce{SO2}$ dipole-dipole attraction).

Wouldn't $\ce{SO2}$ dissolve better as it would be easier for the $\ce{H2O}$ particles to dissociate?
Or would $\ce{HF}$ dissolve more readily as it will form hydrogen bonds with the $\ce{H2O}$ particles?

There are a couple different forces at play here. One big difference between the compounds is that $\ce{SO2}$ is much more volatile, with a boiling point of $\pu{-10 ^\circ C}$; in other words it's a gas. It does strongly hydrogen bond with water however, and is pretty soluble at $\pu{94 g/L}$.

$\ce{HF}$ is also pretty volatile, but much less so than $\ce{SO2}$ with a boiling point of $\pu{20 ^\circ C}$. It also is a more polar molecule than $\ce{SO2}$ and hydrogen bonds even stronger with water. In fact, $\ce{HF}$ is fully miscible with water, so clearly that would be the more water soluble compound.

To summarize, the water solubility of $\ce{SO2}$ is $\pu{94 g/L}$, whereas $\ce{HF}$ is fully miscible, thus $\ce{HF}$ is the more soluble compound in water.

Reference for all solubility and melting point data:
 Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.


Why is Ni[(PPh₃)₂Cl₂] tetrahedral? Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

  Planar-tetrahedral equilibria. Compounds
  such as $\ce{[NiBr2(PEtPh2)2]}$ mentioned above as
  well as a number of sec-alkylsalicylaldiminato
  derivatives (i.e. Me in Fig. 27.6b replaced by
  a sec-alkyl group) dissolve in non-coordinating
  solvents such as chloroform or toluene to give
  solutions whose spectra and magnetic properties
  are temperature-dependent and indicate the presence
  of an equilibrium mixture of diamagnetic
  planar and paramagnetic tetrahedral molecules.


How does a copper piece added in nitric acid change after nitrogen dioxide starts appearing? There's an experiment I have no way of performing but want to know its results. The instructions (in short) are:

Dip a piece of copper in a glass of concentrated nitric acid, $\ce{HNO3}$, using a pincette. When a brown gas starts forming, take out the piece of copper and rinse it with water. Right after that dip it into $\ce{HgNO3}$ solution. Leave it in there for some time and then take it out of the solution. Rinse it with water and wipe it with a napkin.
How does the piece of copper look?

What I've worked out is that the first part can be written as:
$$\ce{Cu (s) + 4HNO3 (aq) -> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)}$$
and that $\ce{NO2}$ is the brown gas.
What I'm  not sure about is why exactly I'm dipping the copper piece into a solution of $\ce{HgNO3}$ following the above reaction. And why am I rinsing it with water before dipping it into the $\ce{HgNO3}$ solution?

Metals often have corrosion on their surface, such as oxides.  By dipping in nitric acid and then rinsing with water you would achieve a true copper metal surface free of oxide or other corrosion products.


There are no wavefunctions in DFT I have seen this phrase several times across DFT textbooks. However, I am not sure if it still holds. Was there a change or a theorem that proved it otherwise? Several programs display wavefunctions for DFT calculations. For example, Quantum Espresso has the option to collect wave functions. Some of the people that I have to talked to mention the Kohn Sham orbitals as if they were a real physical thing. Shouldn't it not be possible to obtain the wavefunction for a certain state? For example, I thought it was not possible to tell that a Kohn Sham orbital belongs to a wavefunction near the surface of a material.

I know that the Kohn Sham orbitals lack a physical meaning. They are simply wavefunctions that are used to obtain the correct density. They do not have an analogue of Koopmans' Theorem. In addition, "the exact wave function of the target system is not available in density functional theory" (Koch, 2001).

So why do people need/use wavefunctions (after all most DFT programs can group such wavefunctions)? I thought that they were Kohn Sham orbitals. Therefore, there should be seldom any use for them.

There are no wavefunctions in DFT.

I don't like that "wave function" is used here in the plural form and I feel like it reflects OP's misconception. The right way to say this is

  There is no wave function in DFT.

See, there exist only one wave function, the function that describes the state of a system in question and can be used to calculate its properties. When it comes to a many-electron system, the wave function is a function that describes the state of all the electrons present in the system interacting with each other. This is the wave function people have in their minds when they say the it is absent in DFT. And that is true: the whole point of DFT is to get rid of this function as a description of state since it is too complex to work with and difficult to visualise.

But that is not the whole story. In practise the Kohn-Sham DFT is used which introduces a fictious system of non-interacting electrons. Each electron in the fictious system is described by a Kohn-Sham orbital, a single-particle wave function, while the whole fictious system is described by the Kohn–Sham wave function, a Slater determinant constructed from a set of Kohn-Sham orbitals. Still there is no wave function in KS-DFT in the above mentioned sense:

KS orbitals don't qualify as the wave function just because they do not describe the state of a whole system;
the Kohn–Sham wave function doesn't qualify because it describes a different system (fictious non-interacting rather than real interacting).

So, the KS wave function is not the wave function, since it doesn't describe the (real) system. You can't use it to calculate the properties of the (real) system, you have to use the electron density to do so. In this sence, the statement that there is no wave function in DFT holds true for KS-DFT as well.

Can the KS wave functions (both single- and many-particle one) be somehow useful in any other way is a different story. See, for instance, the discussion of the usefulness of KS-orbitals here.


Why does the colour of gold sol change with particle size? Regarding the colour of gold sol, my book says

Finest gold sol is red in colour; as the size of particles increases, it appears purple, then blue and finally golden.

This is the image I found on Wikipedia which portrays the same phenomenon.

The same page also discusses the effect of size etc. on the colour, but I don't understand exactly what's happening.
Why shouldn't the colour be the same for all sizes? Isn't it characteristic of the molecules which make the sol as the light absorbed and reflected depends on them?
Does this mean that when we break down any other thing to very small particles, it's sol will change in colour?

The color of conductive nanoparticles depends on plasmonics. In brief, quantum behavior predominates in the few electrons of a nanoparticle, rather than as a continuous conduction band. Surface plasmons strongly influence the color and polarization of light in sols and colloids. For example, for hundreds of years, stained glass has been made with gold or selenium to get pinks and reds.


2-Bromo-1-chloro-4-nitrobenzene reacts with sodium ethoxide What are the major products?

I assumed that their is only one that is possible, which would be replacing the chloro group because the resonance structure supports that one. The bromo group doesn't seem to be able to be replaced. Would adding the ethoxide group, allow for the bromo group to replaced?

The reaction over here is a $\text{S}_{\text{N}}(\text{Ar})$ reaction. Normally, nucleophilic substitutions do not occur, but in presence of strong nucleophiles and electron withdrawing groups, it is possible.

Now, coming to your reaction, let's see what really happens:

Consider the resonation structures of 4-chloro-3-bromonitrobenzene:

You can see that the resonance structures produce a carbocation at the chlorine attached carbon, but not at the bromine attached one. Nucleophiles always tend to attack electrophilic centers, so it is the chlorine that will get replaced.

Continuing with the reaction, we will get:

So, the product would be 3-bromo-4-methoxynitrobenzene.


How do I know how many molecules a solute will split into when it dissolves in a solvent? I'm in high school and I'm learning about freezing point depression and boiling point elevation. I've been given this chemistry problem:

Acetic acid has a boiling point elevation constant of 2.93 and a normal boiling point of 118.1 Celsius. What would be the boiling point of a solution made by dissolving 100.0 grams of calcium carbonate (CaCO3) in 400.0 grams of acetic acid?

This requires me to use the equation for boiling point elevation, that is:
ΔTb = Kb · bsolute · i, or change in temperature = the ebullioscopic constant * the molality of the solution * the number of ions the solute splits into.

My question is how does one determine i for a given solute in a case like this? Thanks for any help you can give!

You need to know the type of reaction that is taking place and what types of bonds are being made and/or broken to determine how many ions the given molecule will split into.


Why alkanes containing even number of carbons can be packed more efficiently than alkanes with odd number of carbon atoms? Variation of boiling and melting points of alkanes with increase in the number of carbons

  The odd-numbered alkanes have a lower trend in melting points than
  even numbered alkanes. This is because even numbered alkanes pack well
  in the solid phase, forming a well-organized structure, which requires
  more energy to break apart. The odd-numbered alkanes pack less well
  and so the "looser" organized solid packing structure requires less
  energy to break apart.

  Source: https://en.wikipedia.org/wiki/Alkane#Melting_points

Why is that the alkanes containing an even number of carbons can be packed more easily compared to the alkanes containing an odd number of carbons?

Both types of alkanes can be laid next to each other; hence, why should the number of carbons even matter?

I started drawing and noticed something.

The diagram makes it pretty obvious that the alkanes with odd numbers cannot be packed efficiently.

I am not sure if this is the correct answer.

A recent article in PubMed (Angew Chem Int Ed Engl. 2016 Nov 2;55(45):14090-14095.) suggests that this odd-even effect also affects translation diffusion in the liquid state.

In the introduction of the PubMed article, it also addresses your particular question and confirms your intuition that it relates to the symmetry of the molecule. Interestingly, this was only experimentally confirmed in the last decade. To quote the article:

  Although this striking odd–even effect was known hundreds of years ago, its molecular origin was only revealed in the last decade. With the help of single-crystal diffraction and controlled crystal growth, the crystalline lattice of n-alkanes was determined.[5] It was confirmed that the intermolecular distances between the end groups, namely, the CH3 groups, are responsible for the alternation in the densities and melting points. In short, n-alkanes with an even number of carbon atoms pack more efficiently into ordered periodic crystalline structures and thus exhibit higher densities and melt at higher temperatures, whereas n-alkanes with an odd number of carbon atoms do not pack as effectively, and thus their densities are lower, and lower temperatures are needed for melting or freezing. A “packing effect” has been established to explain the odd–even effect as well as the related physical properties in the solid state.

Edit: For completeness, this is the article they link to that originally proved the effect experimentally: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-3773(19990401)38:7%3C988::AID-ANIE988%3E3.0.CO%3b2-0/full


How to make a CP2K input file with Avogadro? I am trying to learn to work with CP2K and now need to make its input file. CP2K has kind of adaptor for Avogadro2 but not Avogadro ( I don't know what is the relation between them !).  I also tried that adaptor plugin with Avogadro2 but it does not work for me ( tried on Mac OS X Yosemite and Windows 7).

I searched a lot but seems there is no way to make CP2K inputs with Avogadro . Is there any ?

If you look at the CP2K website, you'll see that the coordinates are only a small fraction of the input file. Most of it is text describing how a calculation should be run.

The key parts from a coordinates perspective look like this:

&CELL
  ABC 12.4138 12.4138 12.4138
&END CELL
&COORD
  O      12.235322       1.376642      10.869880
  H      12.415139       2.233125      11.257611
  H      11.922476       1.573799       9.986994
&END COORD

As it looks, these are the Cartesian coordinates of the atoms (in regular space) and the unit cell vectors. You can either save the file in .XYZ format or use an input generator like GAMESS or Gaussian to get these coordinates.

If you want a periodic boundary calculation, you just need to estimate the bounding box. You can do this manually, or you can go to Crystallography | Add Unit Cell and then tweak the volume. In Avogadro v1.1.x, you'll see the ABC lattice vectors needed for that CP2K block.

As it might seem, Avogadro2 will be v2.0 of Avogadro. It's still in development.


Carbaldehyde or formyl? Which is correct 5-methylbenzyl-3-carbaldehyde-1-oic acid or 3-formyl-5-methylbenzoic acid, MarvinSketch recognizes them both as correct?

The name ‘5-methylbenzyl-3-carbaldehyde-1-oic acid’ is not correct for several reasons. In the following, I focus on the question in the title concerning the choice of ‘carbaldehyde’ or ‘formyl’.

Both, ‘carbaldehyde’ and ‘formyl’, denote the substituent group $\ce{-CHO}$ and can be used, among other ways, to systematically name some aldehydes.

The part ‘carbaldehyde’ is a suffix. It can be used when the $\ce{-CHO}$ group is attached to a ring, for example cyclohexanecarbaldehyde. However, the preferred IUPAC name (PIN) for the analogous benzene derivative is ‘benzaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.

The part ‘formyl’ is a prefix. It can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.

The compound given in the question contains two characteristic groups, $\ce{-CHO}$ and $\ce{-COOH}$. The principal characteristic group must be expressed at the end of a name by means of a suffix or a class name, or implied by a trivial name. Since $\ce{-COOH}$ is the senior characteristic group in this case, the compound is named ‘3-formyl-5-methylbenzoic acid’ (not ‘3-carboxy-5-methylbenzaldehyde’). Note that the $\ce{-CHO}$ group cannot be expressed at the end of a name; it has to be expressed as a suffix (i.e. ‘formyl’).


Calculating Redox Potential; Coupled Reactions For the reaction of silver and hydrochloric acid to form silver chloride: would the following be the correct way to couple the below two half reactions?

$\ce{2Ag(s) + 2Cl^- ->2AgCl(s) + 2e^-} ~~E^0 =-0.222 ~V$

$\ce{2e^- + 2H_3^+O -> H_2 + H_2O}{ ~~E^0 =0.000~ V}$

Add the above:

$\ce{2H_3^+O +2Ag(s) + 2Cl^- -> 2AgCl(s) + H_2 + H_2O}{ ~~E^0 =-0.222~ V}$

Implying that the reaction is non-spontaneous as we have a "negative potential"?

$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$

As two half-reactions, it is seen that the silver is oxidized:

$2~Ag \rightarrow 2~Ag^{+} +2~e^- ~~ E^{\circ}= 0.80\,V$

And the hydrogen is reduced:

$2~H^+ + 2~e^- \rightarrow H_2 ~~ E^{\circ}= 0.0\,V$

To calculate the reduction potential you have to solve the following equation:

$\Delta E = E_{acceptor} - E_{donor} = E_{2H^{+}/H_2}- E_{Ag/Ag^{+}}$

$\Delta E =  0\,V - (0.80\,V) = -0.80\,V$

The reactions doesn't run voluntarily if $\Delta E < 0$


Rotational Constant for CO2 from an IR Vibrational-Rotational spectra I have been trying to calculate the rotational constants ($B$) for $\ce{CO}$ and $\ce{CO2}$ from IR Vibrational-Rotational spectra.

I know that for $\ce{CO}$ the peak spacing is approximately $2B$ (ignoring effects due to centrifugal distortion) however for $\ce{CO2}$ it seems the peak spacing represents $4B$ instead when I compare my results to literature?

Does anybody have any explanation as to why this is, or could provide reading on this subject to explain why the peak spacing for $\ce{CO2}$ is $4B$?

I'm assuming you are measuring the rotationally well resolved band at $\approx 2349 \pu{cm^{-1}}$ and you are correct.

The rotational band structure for the antisymmetric vibration centred at $\approx 2349 \pu{cm^{-1}}$ (also called parallel $v_3$ band) has alternate lines missing, which are those having odd J. $\ce{CO2}$ is a linear symmetric molecule ($D_{\infty h}$ point group) but nuclear spins on $^{12}\ce{C}$ and $^{16}\ce{O}$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $\Sigma^{+}_g $ resulting in the odd levels being absent.

The effect is due to the symmetry properties of the wavefunction, and  what happens on exchange of coordinates. For fermions ( multiples of spin 1/2 , eg. electrons, protons,) the wavefunction must be antisymmetric but symmetric for Bosons (integer spin 0,1,2 etc) which is the case for $\ce{CO2}$. Look for 'statistical weight', 'linear molecules', and 'nuclear spin' in a good spectroscopy book, Herzberg 'Infra Red and Raman Spectra' for example.


Potential energy surface when spin changes I'm trying to compute potential energy surface of S4 molecule (in MOLPRO 2015) along S2-S2 bond (it has a trapezoidal shape) from equilibrium to the dissociation threshold, all the other coordinates are fixed. S4 is a singlet (0 unpaired electrons) in the ground state but a quintuplet once it's dissociated (2 independent S2 molecules in the ground state, each has 2 unpaired electrons). I tried to do CCSD(t) calculations with spin 0 and 2. With spin=0 I get the right energy at the equilibrium but too high energy at the dissociation limit and it's vice versa for spin=2. I need a single surface to solve the Schrodinger equation in so I was thinking about just taking minimum between the two surfaces but I'm not sure if it's a correct way to do it.

This problem doesn't seem to be that rare so I think there should be a well known way to tackle it. I think MRCI can use a mix of the two states, adjusting coefficients in the process to make the transition smooth, but I'm not sure how to properly do it.

Dissoziation is a typical multi-reference problem, where single-reference methods like Coupled-Cluster usually fail. Therefore CASSCF+MRCI would be the better approach here anyway.

If you have a crossing Singlet and Quintuplet state, then you should calculate both. Different spin states do not mix, therefore there is no avoided crossing to be expected. The dissociation of the molecule will remain on its respective PES. To go from the at dissociation limit excited Singlet state to the Quintuplet you need some coupling. Something that switches the spin of the electrons. (I don't know what an actual process would be in this case, maybe some stimulated emission.)
So you should NOT just take the minimum of those two curves!

You should include both spin states (at least, maybe you need more) in the CASSCF state average, to get balanced orbitals. Make sure CASSCF results in smooth PESs for both states before attempting MRCI.
For MRCI you probably need to do independent calculations for both states, as Molpro does not allow different symmetries at the same time here. (But thats fine, as they should not mix anyway.)

Alternatively, you could try other multi-references approaches like MRCC or CASPT2, but I don't have any experience with those.


Activators for non-sulphide compounds in froth floatation I was recently studying about Metallurgy.

Concise Inorganic Chemistry by J. D. Lee claims the process of froth floatation can also be applied to non-sulfide ores such as Malachite ($\ce{CuCO3.Cu(OH)2}$) and Anglesite ($\ce{PbSO4}$) by using $\ce{Na2S}$ as an activator.

The coatings formed are $\ce{CuS}$ and $\ce{PbS}$ respectively. However, in a previous example, they used $\ce{CuSO4}$ as the activator over a $\ce{ZnS}$ surface and a coating of $\ce{CuS}$ was formed.

Therefore, going by the same co-relation in the case of Malachite and Anglesite, shouldn't the coating be a sodium compound instead?

A previous comment noted that sodium compounds are generally soluble and would not be strongly adsorbed on malachite and anglesite.

However, the other compound formed in the reaction between Na2S and a copper or lead salt would be CuS or PbS, both of which are very insoluble. Three cases can be visualized: 1) a sodium ion approaches the Pb or Cu salt and this aquated sodium ion forces off a lattice-bound dication, or 2) a sulfide ion approaches the Pb or Cu salt and either deposits on the crystal surface, forcing a nearby OH-, CO3-- or SO4-- to leave in order to maintain electrostatic neutrality, or 3) the low (but not zero) solubility of the Cu and Pb salts allows precipitation of PbS or CuS as tiny crystals which more frequently than not become bound to the original salt by epitaxial considerations.

In the case of CuSO4 applied to ZnS, ZnS is about 20 times more soluble than CuS, so an exchange of cations will occur, with the copper going into the insoluble phase.

Case 1 seems far-fetched. Cases 2 and 3 are similar, and may occur simultaneously. The key is that the insoluble product is what sticks to the material that you want to float. And the surfactant that forms the bubbles sticks to the activator (CuS or PbS).

It's interesting that ZnS is not easily flotable. ZnS in water may acquire an aquated layer, that is, it is activated by water, which is not selected by the surfactant, so it sinks. Both the copper carbonate and lead sulfate would seem to have surfaces more compatible with water than their sulfides, too, so converting the surface is required.


Reducing sugar among the given compounds The question is to determine which of the following sugars is not reducing?

I have the idea that in case of disaccharide sugars  in which the monosachharide units are bonded via the carbon containing aldehydic group are non reducing.However the present question deals with sugars having monosaccharide units.I have no idea on how to deal with such sugars.Any insights?Thanks.

The English wikipedia page on reducing sugars starts with

  A reducing sugar is any sugar that is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group. All monosaccharides are reducing sugars [...]

Unfortunately, the monosaccharides tend to hide the aldehyde and keto groups by undergoing intramolecular reactions between a carbonyl group and a hydroxy group. This reaction leads to a cyclic hemiacetal.

In order to figure out which of the structures in your question is a monosaccharide in disguise, we need to spot the hemiacetal moiety. With other words:

Which of the structures can open again to show a carbonyl group?

We can definitely rule out (C)! This is a hydroxymethyl-substituted cyclohexane with some hydroxy and a methoxy group.

What about (D)? Close, but no cigar. This is an acetal, which won't open that easy.

I'll leave the last two examples to you ;-)


Second order Jahn-Teller effect for d0 complexes I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?

My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.

EDIT:

I was reading an article on the second order Jahn Teller effect yesterday and found the following line:

I am not sure if I got all those rules correctly but it seems like I have to consider the product of both states and the product has to be of the same symmetry. If I think about the next low lying occupied orbital in an octahedral environment it would be the filled t1g Orbital. So This would mean I have to consider a t1g and t2g mixing here. I have no clue about those irreducible representations and point groups but from the tables you can find on the net I found that a multiplication of T1 and T2 in Oh point-group would end up in an A2 + E + T1 + T2. As this contains both T1 and T2 I guess it will somehow obey those rules.

Perhaps someone with more experience in symmetry and character tables might have suggestion about this.

After quite some research I finally found the original work on the problem. This comes from:
S. K. Kang, H. Tang, T. A. Albright, J. Am. Chem. Soc. 1993 115 1971 - 1981.

I won't discuss the symmetry parts, simply because I have no clue about symmetry and point groups and transformations. I copied the original description for the necessary relationships for a second order JT-effect to happen in the question, so if you know more about this you can probably confirm that.

So to my understanding what is happening for d0 compleces is the following. You really get a mixing of the lower and filled $\mathrm{t_{1u}}$ Orbital with the higher unoccupied $\mathrm{t_{2g}}$ orbital. Both of them split in a set of one orbital that is unchanged in energy, $\mathrm{a_{2}''}$ (for the original $\mathrm{t_{1u}}$) and an $\mathrm{a_{1}'}$ (for the original $\mathrm{t_{2g}}$), and two degenerate orbitals with $\mathrm{e'}$ symmetry that is stabilized and therefore lowered in energy for the original $\mathrm{t_{1u}}$ and elevated for the original $\mathrm{t_{2g}}$. According to the paper this is done by a distortion of the octahedron by a normal mode of $\mathrm{t_{2u}}$ or $\mathrm{t_{1u}}$ symmetry. The first is said to equal a rotation along a threefold axis in the octahedron to get a trigonal prism and change the $\mathrm{O_{h}}$ symmetry to $\mathrm{D_{3h}}$.

You can also see this in the appendix of the book:
T. A. Albright, J. K. Burdett, M.-H. Whangbo, Orbital interactions in chemistry, second edition, Wiley, 2013.

The next question is, why does this stop at the trigonal prism and not continue in mixing the other states as well? And the answer is, for some hypothetical hydrides this seems to happen. While in a d2 complex the new $\mathrm{a_{1}'}$ is filled and therefore an elevation in energy quite unvavorable a d0 complex should further distort. Then $\mathrm{a_{1}'}$ and $\mathrm{a_{2}''}$ mix and distort along a normal mode of $\mathrm{a_{2}''}$ symmetry that changes the trigonal prism $\mathrm{D_{3h}}$ to something they call pyramidalization with $\mathrm{C_{3v}}$ symmetry. This step can also be achieved directly from the octahedron by mixing all orbitals at once and twisting a triangular face of the octahedron.

Now the final question was, why do we observe so many trigonal prisms then? And their final remark on this (unless I skipped something) that it's about the ligands. What ligands favor a second order Jahn-Teller effect for a d0 metal center? Basically only strong σ-donors. They mustn't have any π-donating effects. Since the HOMO LUMO gap has to be quite narrow the $\mathrm{t_{1u}}$ orbital has to be high. So no electronegative ligands. But also the $\mathrm{t_{2g}}$ orbitals have to be low, so a strong ligand splitting is favored but in a way that the $\mathrm{t_{2g}}$ is lowered. And most probably a strong nephelauxetic effect is advisable as well. This leads to alkyl and hydride as possible candidates. For the hydride many of the desired complexes are just unknown but the calculations show that is could distort to $\mathrm{C_{3v}}$. For the alkyl like many prismatic organometallic alkyl complexes we know the methyl seems to be to bulky to stabilize the pyramidal structure.

In the same sentence they also mention that for example alkoxides are not forming prisms. So this leaves me with the question how Mo-S or W-O structures in solid state still show the trigonal prism. But that has probably different reasons so I'll exclude them here.


Conformations of cyclohexane and flipping Conformations of Cyclohexane doesn't answers my question

This is what Organic Chemistry by David R. Klein states:

  Notice that the chlorine atom is on a wedge, which means that it is coming out of the page: it is
  UP. The methyl group is on a dash, which means that it is below the ring, or DOWN. The two
  chair conformations for this compound are as follows:

  Notice that the chlorine atom is above the ring (UP) in both chair conformations, and the
  methyl group is below the ring (DOWN) in both chair conformations. The confguration (i.e.,
  UP or DOWN) does not change during a ring ﬂip. It is true that the chlorine is axial in one
  conformation and equatorial in the other conformation, but a ring ﬂip does not change confguration. The chlorine atom must be UP in both chair conformations. Similarly, the methyl
  group must be DOWN in both chair conformations.

I am having a confusion in the last paragraph. Why should the Chlorine be up even after the ring flip? Moreover, why are we even discussing if it is above the ring or not, as it has already come in the equatorial position?

Essentially, it is stating that a ring flip retains the absolute stereochemical configuration. Ring flipping is a physical phenomenon of 6-membered rings, not a chemical reaction that interconverts stereoisomers. No bonds are broken, though many do get strained in the process.

I'm not sure what you mean by your question about the importance of being in the axial/equatorial positions. Both the methyl and chloro groups are vying to reduce 1,3-diaxial groups by placing themselves in the equitorial position, but doing so for one places the other in the axial position. I think a natural question that arises "is which one wins?"


How can I get xyz coordinates of atoms of a unit cell structure from CIF format files? I'm trying to run DFT or other 1st principle simulation of Pyrope ($\ce{Mg3Al2Si3O12)}$ and Grossular $\ce{(Ca3Al2[SiO4]3)}$ unit cell structure. So, I'm looking for unit cell xyz periodic box structure of Pyrope and Grossular.

I found some data from several places (here, here and here)

CIF format files from those databases do not give me proper atomic coordinates of unit cell. I downloaded CIF files for Pyrope from those databases, and converted with Open Babel, and even visualized with VEGA ZZ.

However, it displays only 4 atoms in the periodic box, while Pyrope or Grossular should have 20 atoms in the periodic box based on their formula (Pyrope ($\ce{Mg3Al2Si3O12)}$ and Grossular $\ce{(Ca3Al2[SiO4]3)}$).

This means either CIF of database is wrong, or I am. Or the software is unable to read CIF properly. I think I'm doing something wrong, but I'm not sure what is wrong with my CIF conversion.

If anyone knows how to retrieve the unitcell structure of crystals from CIF from those databases, please help me.

CIF contains coordinates for the atoms in asymmetric unit only (what Ivan called "the independent part" in the comments), plus the set of symmetry operation to fill the remaining space with those atoms.
What you want to do is to open CIF in any viewer supporting that format, make sure the unit cell is packed, e.g. the symmetry operations are applied along all three crystallographic axes $a, b, c$ from $0$ to $1$ and export the data to *.xyz format.
There are several programs that are capable of doing this.
For consistency I suggest using crystallographic data by Meagher [1] for both minerals at 25 °C from AMCSD:

Pyrope CIF
Grossular CIF

CSD Mercury

As Martin suggested, Mercury (Free, available on Windows, Linux, MacOS) is one of the most user-friendly and reliable crystallographic viewers.

Open the program and load CIF (I loaded pyrope.cif).
An asymmetric unit pops up:

Go to Calculate → Packing/Slicing… and under Packing area tick Pack option, and under Include Atoms area select … that Fit.

Now you have the content of the entire unit cell:

Save the data as XYZ by clicking File → Save As… → Save as type: XMol files (*.xyz).
This produces a file with the coordinates of all atoms of the unit cell.

Unfortunately, I'm not aware of a good method to eliminate atoms located on the periodic boundary with Mercury as packing coordinates can only be changed by a 0.1 Å step, which is rather crude.
However, there is a workaround I'm aware of with another program, VESTA.

VESTA

VESTA (free, available on Windows, Linux, MacOS) is yet another functional crystallographic viewer.

Open the program and load CIF (again, I loaded pyrope.cif).
By default VESTA not only recognizes atomic connectivity, but also populates the unit cell and the first coordination spheres of all atoms on top of that.

Since the goal is to cut off the content of neighboring unit cells and the boundary atoms, the interatomic bonds should be removed.
Go to Edit → Bonds… and delete all [three] bonds from the list.

Now we have the same atomic set as in Mercury after packing with all the boundary atoms present.

To eliminate the boundary atoms, you can decrease the boundary box by a tiny fraction.
On Style tab click Boundary… button and adjust upper and lower limits for  all three axes ($x_\mathrm{min}$, $x_\mathrm{max}$, $y_\mathrm{min}$, $y_\mathrm{max}$, $z_\mathrm{min}$, $z_\mathrm{max}$) making the box smaller by 0.001 Å on each margin.

After applying new boundaries you get the unit cell content without boundary atoms.

Save the data as XYZ by clicking File → Save As… → Save as type: XYZ file (*.xyz).

References

Meagher, E. P. The Crystal Structures of Pyrope and Grossularite at Elevated Temperatures. American Mineralogist 1975, 60 (3–4), 218–228.


Exocyclic vs. endocyclic double bonds in E1 elimination What will be the major product of this elimination reaction?

The exocyclic product has more substituent groups on the alkene which should make it more stable but I have heard that exocyclic alkenes are less stable than their endocyclic counterparts. Why is this the case and is it a strong enough effect to outweigh the stability provided by a fourth alkyl substituent?

Additionally, what influence will kinetics have on this reaction? I think that the endocyclic product should be kinetically favoured because there is less steric hindrance for deprotonation of the intermediate cation at the ring position than at the isopropyl position.

What will be the major product of this elimination reaction?

In the acid-catalyzed dehydration of 1-methylcyclohexanol, both 1-methylcyclohexene and methylenecyclohexane are formed in ca. an 85:15 ratio.  In your example we have two more substituents on the exocyclic double bond which will serve to further stabilize the exocyclic double bond and increase the amount of the exocyclic isomer. I don't know for sure how much it will push the equilibrium towards the exocyclic isomer, but I might guess that the product mixture would be at least 50:50, if not more of the exocyclic isomer - but I would expect a mixture.

  what influence will kinetics have on this reaction?

Well you won't see the kinetics unless you go out of your way and look for it.  Factors that influence which isomer is kinetically favored include:

statistical factors - there are 4 ring hydrogens that can be eliminated vs. only 1 on the isopropyl group; this favors kinetic formation of the endocyclic isomer
alignment of the hydrogen with the p-orbital on the carbocation carbon - the hydrogen on the isopropyl group can better align with the p-orbital than those hydrogens in the ring, however the cyclohexane ring can easily distort; so while this isn't a major factor it favors the exocyclic isomer
steric factors - I agree with your analysis, but would guess that the difference is small; still sterics would slightly favor the endocyclic isomer

It seems pretty close to call, but if I had to guess I'd say that the statistical factor wins out (and the steric argument further helps) and the endocyclic isomer would likely be the the kinetic product.

But again, in acid this is a fast equilibrium; the products will quickly equilibrate and the thermodynamically controlled product distribution will be observed.  Maybe if you run the reaction at low temperature with a strong dehydrating agent ($\ce{POCl3}$) you could capture some meaningful kinetic information.


Using significant figures when measuring with an instrument with marks other than powers of ten If an instrument does have markings that are powers of ten, then finding the precision of the instrument is easy. For example, if a graduated cylinder had markings for every $\pu{0.1 mL}$, or every $\pu{mL}$, or every $\pu{L}$, then it is easy to tell that those measurements will be to the hundredth mL, to the tenth of a $\pu{mL}$, and to the tenth of a $\pu{L}$, respectively. These would be valid measurements with the correct number of significant figures.
See this image, for example:

These markings are to the tenths and ones. This is simple.
However, what if that graduated cylinder went up by something like $\pu{5 mL}$? Would its precision be to the tens or to the ones (in other words, will the measurements be to the $\pu{0.1 mL}$ or to the $\pu{1 mL}$)? What would be the valid measurement?

I think the rule of thumb for the "reading error" being $\pm\,0.1$ of the smallest division applies here, so for your case, it'd be $5\;\mathrm{mL}\pm\, 0.5\,\mathrm{mL}$. See this page, entitled "Math Skills - Scientific Notation", from Texas A&M. Specifically, it is noted that

  A rule of thumb: read the volume to 1/10 or 0.1 of the smallest division. (This rule applies to any measurement.) This means that the error in reading (called the reading error) is 1/10 or 0.1 of the smallest division on the glassware. If you are less sure of yourself, you can read to 1/5 or 0.2 of the smallest division.


Why would this reaction occur? Won't it be unfavourable? I was just reading about how Nitrogen and Oxygen combine to form Nitric oxide.

My question is, why would this reaction even occur? Isn't the valency of Nitrogen 3- and Oxygen 2-, and therefore this reaction seems to form a product that still hasn't achieved the octet electronic configuration.
Could I please have some help?
Thank you in advance.

Edit: Or even $\ce{N2O}$ or $\ce{NO2}$ for that matter?

You're right: $\ce{NO}$ is a free radical and the reaction between $\ce{N2}$ and $\ce{O2}$ can be achieved - supposed that enough energy is provided.

This may happen through an electrical arc discharge in the lab or, in the mesosphere (50-85km above earth) through ionizing cosmic rays. Note however that nitric oxide react with oxygen to eventually furnish nitrogen dioxide $\ce{2 NO + O2 <=> 2 NO2}$.


Why was lime used by the Nazi's? I am reading a book (Into That Darkness) about the Nazis, where there is the following passage:

  causing unimaginable suffering to the deportees who were either left waiting, naked and without food or water, in the open, sometimes for days, or else were crammed into railway cars the floors of which had been covered with lime and were left to suffocate on sidings

What is the meaning of “the floors of which had been covered with lime”?

Is “lime” short for something? Was the floor covered with this intentionally, or did this “lime” accumulate naturally? What would be the consequences of this to the people in the railway cars?

Is “lime” short for something?

In this case it is not a fruit Lime is a mixture of calcium hydroxide and calcium oxide produced from calcining limestone.

  What is the meaning of “the floors of which had been covered with lime”?...Was the floor covered with this intentionally, or did this “lime” accumulate naturally?

Not really the place to ask, but the lime likely was from previously treating bodies. As people were left to die, lime would be put on their corpse to keep the smell of decaying flesh down. There is also some belief that lime can speed up decay and reduce bacteria growth, but this is disputed. Any way not all of the lime would land on the body or it would fall off when moved. This process had happened so many times prior that the floor was covered in it. lime tends to build into layers as it is chemically similar to cement and will harden over time the same.

  What would be the consequences of this to the people in the railway cars?

Lime is very caustic and causes great respiratory distress when inhaled. It is used to capture the putrid scents of decaying flesh.

Today lime is still used at mass grave sites to capture the scent of decay and keep soil pH high. Low pH soil is an indicator for a mass grave as the decomposition products are acidic and lower the soil pH. Adding lime reduces this acidity masking (trying to mask) the presence of a mass grave.


Why are two three-membered rings preferred to a single five-membered ring in the reaction of a dihalide with alcoholic KOH? The given answer is 1,1-dicyclopropylmethanone:

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?

I think the product may be formed by the following pathway.
First, the carbanion will form when the strong base abstracts the proton from the α-position. The lone pair generated would obviously have two points of attack, either at the nearer carbon to form the cyclopropyl ring or at the farther carbon to form a five membered rkng where the whole controversy lies.
While apparently, the position five membered ring appears to be more stable than a small highly stained three membered ring, a very interesting phenomenon occurs if such a three membered, strained ring forms.
This is similar the basic resonace observed in cyclopropane.
The carbonyl functional group also triggers the reaction foreward towards formation of the product.
Let us begin by observing the resonance structures of the carbonyl functional group.
Clearly, out of the two contributing structures, positive charge may be seen residing on the carbonyl carbon while negative charge may be seen on the carbonyl oxygen.

A combination of these two resonance hybrids leads to generation of partial positive charge on carbon and partial negative charge on the net resonance hybrid.
Next, we have to observe the structure of the 1-methylcycloropyl carbocation.
Due to ring strain the bonds are bent and a special kind of resonance involving the σ C-C single bond is observed.

In fact the bonds are actually sp5 hybridized i.e. having large p character to ease the ring strain.
The enormous stablity can be alluded by the following resonce diagram.
Moreover, the immense stablity of tris(cyclopropyl)methyl carbocation, is actually due to this σ-resonace. It is actually the one of the most stable carbocations known.

As the number of cyclopropyl groups increase, the stablity increases.

Now to observe a combination of both.
In the final product the M- effect of the carbonyl group will favour the extended delocalisation of electrons of the σ-bond over the oxygen.
Thus, the enourmous stablity concept of cyclopropane may be attributed to the compound under question by the following diagram:

To draw the resonance structures in a simplified way we first show complete charge seperation between the carbon and oxygen of the carbonyl group. Then the structure becomes similar to a cyclopropylmethyl carbocation showing the required stablising resoance of σ-bond.
Due to this unique resonance effect, also known sometimes as dancing resonance, the product becomes exceptionally stable, more stable than the 5 membered ring obtained in the other case.
Hope this helps.
References:
What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?


Calculating bond dissociation enthalpy of F-F bond $$\ce{N2 + 3 F2 -> 2 NF3} \quad ΔH = \pu{-263 kJ mol-1}$$

  The bond dissociation enthalpies of the nitrogen triple bond and nitrogen–flourine bond are $\pu{946 kJ mol-1}$ and $\pu{272 kJ mol-1}.$
  Find the bond dissociation enthalpy of the $\ce{F-F}$ bond.

  A. $\pu{-423 kJ mol-1}$
  B. $\pu{-393 kJ mol-1}$
  C. $\pu{-141 kJ mol-1}$
  D. $\pu{+141 kJ mol-1}$
  E. $\pu{+423 kJ mol-1}$

I know that since bond dissociation enthalpy is endothermic it has to be one of the last two.

I tried substracting bond dissociation enthalpies of products from reactants and equating that to the enthalpy change but my answer is nowhere near the given ones.
Is my method wrong?

If you can't wrap your head around getting the answer directly, just calculate the enthalpy of reaction for the two bond dissociation energies you are considering.

Assume bond dissociation energy is 141 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{141 kJ/mol} - 6 \cdot \pu{272 kJ/mol} =  \text{negative value}$$

Assume bond dissociation energy is 423 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{423 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{positive value}$$

So it can't be 423 kJ/mol because the reaction is exothermic. If you complete the math for a bond energy of 141 kJ/mol, the numerical answer is -263 kJ/mol for the reaction enthalpy.

What if you actually wanted to estimate a bond energy?

In this case, you could call the bond energy $x$, and solve for it:

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot x - 6 \cdot \pu{272 kJ/mol} = \pu{-263 kJ/mol}$$

Solving for $x$ would give you the bond energy.


Comparing acidic strengths between benzylammonium ion and phenol Question

Compare the acidic stengths between benzylammonium ion and phenol.

I first tried to remove the H+ ion from both of them and tried to compare the relative stabilities of resulting conjugate bases, but I am not sure, which one would be more stable.
The first one would be a neutral compound and the second one would have resonating structures on which the negative charge gets mostly delocalised to carbon.
I have seen somewhere on the internet that benzylammonium ion has a $\mathrm{pK_a}$ value of 9.33 while phenol has a $\mathrm{pK_a}$ value of 10. If this data is correct, then how do we explain the acidity order?

The acid strength of each compound can be explained, but the acidity order is much more difficult to compare, because the two compounds are only remotely connected. It is misleading to conclude that the mere presence of a phenyl group somehow connects these molecules. The similarity of the pK$_a$s is likely a coincidence.
The question needs a clear definition of pK$_a$. We can ask what the pK$_a$ of phenol is, or the pK$_b$ of phenoxide. When we ask for the pK$_a$ of benzyl amine, we want the pK$_a$ of benzylammonium ion, not the pK$_a$ of benzyl amine going to C$_6$H$_5$CH$_2$NH$^-$. There is a bit of flexibility (or sloppiness) in the terminology here, and also a conflict in the values presented: the question quotes 9.33 for benzylammonium ion (I found 9.34 - no big deal), but the comment by the OP gives 8.82 for benzyl amine. The values should be identical because we know what the OP means. It appears that 8.82 is incorrect.
The difference between these molecules is far more important than any similarities. For example, one is a neutral molecule that you can put into a bottle with 6 x 10$^{23}$ more; the other is a charged piece of a molecule that you can hardly bring next to another one.
The phenyl ring has an enormous effect on the acidity of phenol, due to the resonance effect. It stabilizes the oxygen anion by more than 5 orders of magnitude, compared to, e.g., methanol (pK$_a$ = 15.5) or benzyl alcohol (15.4) or ethanol (15.9).
(Interesting that phenylethanol (pK$_a$ = 14.81) is more acidic than benzyl alcohol. This suggests that the methyl group in ethanol is more electron-donating than the benzyl group in benzyl alcohol; this agrees with the discussion in #3 below, for phenethylammonium ion.)
The phenyl ring has a much smaller effect on the amine molecule-ion; no resonance, just inductive. The pK$_a$ of benzylammonium ion (9.34) can be compared with, e.g.:

methylammonium ion (pK$_a$ = 10.66) Replacing the phenyl of benzylammonium ion with H reduces the K$_a$ by a factor of 21.

This is equivalent to saying that benzyl amine is less basic (pK$_b$ = 4.66) than methylamine (pK$_b$ = 3.34); therefore the protonated benzyl amine will give up an added proton more readily than methylamine. We can conclude that methyl is more electron-donating than benzyl.

ethylammonium ion (pK$_a$ = 10.71) Replacing the phenyl of benzylammonium ion with a methyl group reduces the K$_a$ by a factor of 23. Thus, ethylamine, like methylamine, is more basic than benzyl amine because methyl is more electron-donating than phenyl. (But we knew that.)

Phenethylammonium ion has a pK$_a$ = 9.73. Putting another CH$_2$ group between phenyl and nitrogen reduces K$_a$ by a factor of 2.5. This indicates that a benzyl group is less electron-donating than a methyl or ethyl group, in agreement with the comparison of the inductive effect between benzylammonium vs methyl- or ethylammonium ions and benzyl alcohol vs phenylethanol.

Benzyl amine and phenol are the neutral actors in this drama, and might react nicely in a one-to-one ratio. What would be the pH in solution? The graph shows the ratio of ion concentration to neutral molecule concentration for phenol and benzyl amine, separately, over the pH range of 5 to 9. They show contrasting slopes which cross at pH 7.33. At this pH, the ionic fraction of each compound is 0.21% of the whole - it doesn’t matter what concentration you started with if everything is soluble (benzyl amine is very soluble, but phenol, mp 43 C, is soluble in water slightly less than ~0.1 M in water at room temperature). Just add the same number of moles of each neutral compound to get 0.21% reaction and pH 7.33. What a coincidence! Practically the same pH as pure water.


Why are gas laws only applicable to gases? Why are the gas laws (such as Boyle's Law, Charles's Law, and Avogadro's Law) only applicable to gas and not liquids or solids?

Gases, liquids and solids have different interactions between their components

Ultimately they are applicable to gases because they only apply to gases. The reasons why they work are worth understanding, though, because they are critical to the development of chemistry which might not exist if not fro some of the convenient properties of gases.

The most important law is that a given number of atoms or molecules as a gas always occupies the same volume if temperature and pressure are fixed. This was what allowed the generation of data that formed the basis of modern chemistry: things combine in fixed ratios. This led to the experimental proofs of the existence of atoms which would be a lot harder to do if we had to rely on measuring the volumes of solids or liquids where the volume tells us nothing useful about how much stuff makes up the substance. It is easy to measure the volume of a gas.

But why is this true? The essence of the reason why gases make up a fixed volume whatever they are made of, is that, to a good approximation, in gases the component parts are small, far apart and have only elastic interactions. If you do the statistics of such a system a whole bunch of apparently relevant factors cancel out (like the mass and velocity of the particles) to give a system which obeys the key gas laws independently of the molecule or atom making up the gas.

But this only works in gases. Liquids have strong interactions between their molecules; not strong enough to prevent individual molecules wandering around randomly but strong enough to hold them to a fixed volume without needing a container (gases will expand to infinite volume without a container to prevent them diffusing into the atmosphere). Solids have even stronger interactions that hold their individual molecules or atoms in a well defined location. The specific properties of solids and liquids depend on the details of those forces and, since there are a wide variety of forces and force strengths, the collective properties are a lot more complex.

In gases, the ideal gas-law approximations are good because the molecules of the substance have enough energy to break the attractive forces that would otherwise make them solids or liquids. Once free of those attractive forces they tend to occupy a much larger volume and their behaviour becomes better explained by the (simpler) gas laws as we can start ignoring the complexity of the attractive forces.

But the gas laws are an approximation. If they were not, nothing would ever liquefy or solidify. But the hotter a gas gets the more it behaves ideally as the less the attractive forces are relevant to the behaviour of the gas.


How to create .CIF file from DRX data in a paper, for compound not found in crystallographic database I tried to recover a .CIF file with the crystal structure for the compound discussed in this 2006 paper, strontium hydroxide octahydrate - $\ce{Sr(OH)2·8H2O}$ - from Cambridge Crystallographic Data Centre (CCDC), but couldn't find it. The paper itself has tables with information recovered from x-ray diffraction data, as shown below.

Is there any free software or open source tool where I can just type the data from these tables to assemble a crystal structure like the ones I'm used to download from the database?

You can't find this entry in CCDC's CSD, because the scope of this database are small organic, or organometallic molecules' crystal structures.  (For large organic molecules, analogue studies' data may be found, e.g., on wwpdb.org.)  Because your compound is considered as inorganic*, you'll find it in the ICSD, the inorganic crystal structure database.
Note, the paper includes the statement:

«Further details of the crystal structure investigation may  be  obtained  from  Fachinformationszentrum Karlsruhe, 76344 Eggenstein-Leopoldshafen, Ger-many (fax: +49-7247-808-666; e-mail: crysdata@fiz-karlsruhe.de,   http://www.fiz-informationsdienste.de/en/DB/icsd/depotanforderung.html) on quoting the depository number CSD-417100.»

Because CCDC and ISCD established in 2018 a search interface in common, and entries' labels in both databases differ from each other, it is easy to find the record here.  It is not as comfortable as their own GUI and API interfaces (example ICSD), but it is possible to download the record's .cif.
Note that complementary to these two databases, there are others like e.g., the American Mineralogist Crystal Structure Database and the COD.  Entries 2100155, 2100156, and 2100157 of the later for example point to the .cif about $\ce{Sr(OH)2 ·8H2O}$ of a study by Ricci et al. at 20, 100 and 200 K, published 2005 in Acta Cryst. B.

*) A simplified definition of «organic compound» is that it is molecular and contains at least one carbon atom excluding carbon oxides ($\ce{CO, CO2}$, and others), their hydrates, and simple metal carbonates ($\ce{H2CO3, HCO^-_3, CO^{2-}_3}$).


Conditions for molecular dissymmetry In my textbook, under the topic "Asymmetric and Dissymmetric Compounds", the following statement is given:

  A molecule which does not possess plane of symmetry, centre of symmetry, and alternating axis of symmetry is called dissymmetric.

Why the definition didn't include the "axis of symmetry" as a condition for molecular dissymmetry?

From Wikipedia:

  A chiral compound can contain no improper axis of rotation ($S_{n}$), which includes planes of symmetry and inversion center. Chiral molecules are always dissymmetric (lacking $S_{n}$) but not always asymmetric (lacking all symmetry elements except the trivial identity). Asymmetric molecules are always chiral.

That means that a dyssymetric molecule lacks of $S_{n}$ axis(planes of symmetry and inversion centers) but can have $C_{n}$ axis of rotation. For example the following molecule is dissymetric: it does not have any plane of symmetry or inversion center but it has a $C_{2}$ axis of rotation(the "axis of rotation" is a synonyms of the "axis of symmetry" you mentioned in your question).


NMR: How the relaxation times T1 and T2 depend on the correlation time / amount of molecular tumbling. Why do nuclei that have a smaller / faster correlation time have a higher / slower T1 / T2?

From my understanding:

Fast Brownian motion creates a wide range of B.local (local magnetic field created by dipole of spin X near spin Y) fluctuation frequencies
Slow motion gives rise to fewer frequencies but more intense fluctuations
Fluctuations at the right frequency cause transitions between energy levels / relaxation / re-establishment of equilibrium
The larger a molecule the slower it tumbles (correlation time is slower)
The slower the tumbling the faster the relaxation

To me, logically, it'd make sense that if a molecule tumbles faster then the relaxation should be faster, not the other way round. Can anyone explain this?

Thanks

Relaxation is often very counterintuitive and isn't a particularly easy subject to understand, so it's not always useful to argue based on what seems to be logical. I'm not sure where you're studying this from, but it seems to be slightly lacking in the detail needed to properly understand what's going on. Here is a brief overview of what's going on, but I strongly recommend consulting a textbook which will explain it in more detail. Chapter 9 of James Keeler's Understanding NMR Spectroscopy (2nd ed.) is a very good reference. You can find a pdf of the corresponding chapter in the previous edition here.

The link between rate of tumbling (correlation time, $\tau_c$) and the "amount of fluctuations at the right frequency" is given by the spectral density

$$j(\omega) = \frac{2\tau_c}{1 + \omega^2\tau_c^2}$$

This function tells you how much the random fields are fluctuating at each frequency $\omega$. For relaxation, the important frequency to look at is the Larmor frequency $\omega_0$, i.e. the central quantity is

$$j(\omega_0) = \frac{2\tau_c}{1 + \omega_0^2\tau_c^2}$$

By differentiating this with respect to $\tau_c$ you can find that this is at a maximum when $\tau_c = 1/\omega_0$. The rate of spin-lattice (longitudinal) relaxation is proportional to $j(\omega_0)$, so it is fastest when $\tau_c = 1/\omega_0$. That means that $T_1$ will be smallest when $\tau_c = 1/\omega_0$. When $\tau_c$ decreases below this, or increases above this, spin-lattice relaxation will be slowed down and $T_1$ will increase.

On the other hand, spin-spin (transverse) relaxation depends on the sum $j(\omega_0) + j(0)$. $j(0)$ is simply equal to $2\tau_c$, and you can show (again with calculus) as $\tau_c$ increases, the total sum $j(\omega_0) + j(0)$ increases continually. So $T_2$ decreases when $\tau_c$ increases.


The Trend Between Boiling Point and Solubility in Organic Chemistry Is it true in assuming that a higher boiling/melting point means that an organic compound will be more soluble in water?

I'm trying to distinguish between the solubility of aldehydes vs. ketones based on boiling point. I received a question in a test today about it, comparing propanal with propanone, based solely on boiling points given, and my structural understanding of the two. I thought propanone had lower solubility of the two, as the ketone group is surrounded by carbons, while the propanal is exposed. Scouring through the internet, I'm not so sure my answer was correct!

As andselsik pointed out, definetley no in the general sense. Just think of some practical examples like PTFE (Teflon) - which melts at about 330 °C but is definetly not water soluble.

However there is one connection I'd like to point out that allows to relate the enthalpy of melting to the solubility:

Consider phase equilibrium between a soluble compound C and water. At equilibrium, we have equality of chemical potentials:

$$\mu^{solid}_c(p,T,x_s)=\mu^{liquid}_c(p,T,x_l)$$

Assume that $x_s$ at equilibrium is 1 (there would be a tiny fraction of water in the solid, but its fair to neglect that). Then the LHS is just the chemical potential of the pure solid:

$$\mu^{*,\:solid}_c(p,T)=\mu^{liquid}_c(p,T,x_l)$$

Now using the solvent convention for the chemical potential of the liquid mixture:

$$\mu^{*,\:solid}_c(p,T)=\mu^{*,\: liquid}_c(p,T)+RT\mathrm{ln}(x_c^l \: \gamma_c)$$

Rearranging this and recognising that a chemical potential difference of a pure substance across a phase boundary is the gibbs energy of that phase transition gives

$$\Delta g_{c;\: melt} = \Delta g_{c;\: S \rightarrow L}=\Delta \mu_{c;\: S \rightarrow L} = RT\mathrm{ln}(x_c^{l} \gamma_c)$$

and

$$\Delta g = \Delta h + T \Delta s$$

using the limiting behaviour of the activity we find that for $x_c , \gamma_c \rightarrow  1$, the RHS tends to zero; we can reexpress the entropy of the phase change by setting $\Delta g$ in the above equation zero, and substitute the result into the equation obtained from equality of chemical potential:

$$\Delta h_c^{*,\:melt} \Big(\frac{T}{T^{melt}}-1\Big) = RT\mathrm{ln}(x_c^l \gamma_c)$$
...
$$x_c^l\gamma_c=\mathrm{exp} \Bigg( \frac{\Delta h_c^{*,melt}}{R} \Big(\frac{1}{T^{*,melt}}-\frac{1}{T}\Big)\Bigg)$$

The activity coefficient itself is obviously also a function of the composition, so this does not give us a straightforward way to calculate the solubility of compound C (it does however provide us with a route to $\gamma$, which is normally harder to measure than a solubility) but it shows that phase change properties have all sorts of implications!


What makes sulfur-containing compounds such good antioxidants? Many sulfur-containing compounds (like glutathione and cysteine groups) have high antioxidant activity. Why?

There are several factors that combine to make sulfur compounds such effective antioxidants.  First, sulfur compounds often contain S-H bonds (mercaptans) and are excellent radical scavengers because sulfur radicals are thermodynamically more stable than the analogous oxygen radical.  That is to say that the following equilibrium lies far to the right
$$\ce{ROS^{.} + RS-H <=> ROS-H + RS^{.}}$$
where ROS represents a "reactive oxygen species" such as hydroxyl radical ($\ce{HO^{.}}$), superoxide radical ($\ce{O_2^{-~.}}$, etc.  Sulfur compounds (sulfides and disulfides, for example) are also very easily oxidized to their corresponding sulfoxides.
$$\ce{R-S-R +ROS -> R-SO-R + deactivated ~ROS}$$
If multiple mercaptan groups exist within a molecule (for example multiple cysteine residues in an amino acid) they can be oxidized to the corresponding disulfide while reducing an oxidizer.
$$\ce{2~ RS-H + [oxidized~molecule] -> RS-SR + [H-reduced~molecule-H]}$$
This oxidative process provides sulfur compounds with a second pathway to trap or transform, and consequently inactivate, ROS.

In addition to being very effective ROS scavengers, sulfur compounds also bind strongly to many metal ions.  These same metal ions often play a role in the generation of ROS.  By binding to them, sulfur compounds effectively remove them from the ROS generating process thereby further lowering the concentration of ROS in the cell.


Aurora borealis - less favored energy transitions higher in atmosphere? This question describes how some energy transitions become more prominent as you higher in the atmosphere. Short summary: Some electron transitions are "forbidden" by quantum mechanical selection rules, and so have a larger time delay from time of excitation until time of photon emission. Thus, low in the atmosphere (high pressure), it's more likely that the species loses the excitation energy by collision than emission.

However, as you go higher in the atmosphere, you see red (due to energy transition in atomic oxygen). The characteristic green line (also due to atomic oxygen, but different transition) disappears.

Why don't you see a mix of the two colours? After all, from the collision rate argument you would expect fast transitions (like the green of atomic oxygen) to become even more prominent.

Why is it that some transitions are less likely as you increase in height?

(I have a feeling that the answer to this question will also explain why nitrogen does not play a more central role in the colour of the aurora.)

Why don't you see a mix of the two colours? After all, from the collision rate argument you would expect fast transitions (like the green of atomic oxygen) to become even more prominent.

There can be a mix of colors.  However, according to Auroral electron energy derived from ratio of spectroscopic emissions Journal of Geophysical Research Volume 79, Pages 5181–5186, the ratio of the atomic oxygen versus molecular nitrogen bands are a function of "a characteristic energy of the precipitating electron flux".

  Why is it that some transitions are less likely as you increase in height?

Well, consider Atmospheric Composition and Vertical Structure.  Figure 3 shows how dramatically pressure drops as altitude increases.  This means a change in altitude of 15 kilometers could reduce the concentration of the species of interest by a factor of 10.

In addition, dissociation of the $\ce{O2}$ and $\ce{N2}$ to atomic O and N increases with altitude.  See especially Fig. 6 which plots the points where monoatomic oxygen starts to exceed diatomic oxygen, and where monoatomic oxygen starts to exceed diatomic nitrogen in concentration.


Free Radical Reaction of Methane and Chlorine The problem statement:

  Lets say you have $1 L$ of $2 M$ methane and the same amount of chlorine. Lets also say that both are liquids since those are most likely to react. Now the only way they can both be liquids at STP is if the temperature is as cold as an antarctic winter so this is not aqueous. Gases more often bump the wrong way and solids don't react unless it is oxidization or dissolving.

  Now the initiation step is forming the first molecule of $\ce{HCl}$ and Methyl.

  Now the methyl and chlorine atom really want to react and form chloromethane

  Relevant equations

  $\ce{CH4 + Cl2 -> HCl + CH3Cl}$
  (this continues up to tetrachloromethane)

  $\ce{2 CH3Cl -> Cl2 + C2H6}$
  (this can continue for much longer than the previous one can)

The attempt at a solution

$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl}$

$\ce{2 Cl + 2 CH3 -> 2 CH3Cl}$

$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2}$

$\ce{2 HCl -> 2 H2 + 2 Cl2}$

This obviously can't happen because then we have more chlorine than we started out with. Why? Well that $1 M\; \ce{Cl2}$ from ethane plus $2 M\; \ce{Cl2}$ from $\ce{HCl}$ is $ 3 M\; $$\ce{Cl2}$ and we started with $2 M \ce{Cl2}$. Just like the number of each element the molarity has to be balanced. What I find confusing is figuring out the molarity of each compound at each step of the process not the compounds per se.

OK, let's focus on the stoichiometry of your last 4 equations:

$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl~~~~}$  this looks OK

$\ce{2 Cl + 2 CH3 -> 2 CH3Cl~~~~~~~~~~~}$ this looks OK

$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2~~~~}$ this should be $\ce{1 CH3Cl + 1 CH3Cl -> 1 C2H6 + 1 Cl2}$

$\ce{2 HCl -> 2 H2 + 2 Cl2~~~~~~~}$ this should be $\ce{2 HCl -> 1 H2 + 1 Cl2}$

I'm not sure I agree with all of your chemical reactions, but at least now you don't wind up with more chlorine then you started with.


Solubility increases as Temperature increases.. Ks increases with temperature meaning dissolving is endothermic... But what about NaOH? Question: I was self-studying and came across some confusion across a question. My confusion lies in this: Solubility increases as temperature (T)  increases, Ks increases with T, meaning dissolving is endothermic. But what about $\ce{NaOH}$?
The problem I am putting forward is more to do with the equilibrium Ks, especially using the Le Châtelier's principle. We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that. We also know that dissolving $\ce{NaOH}$ is exothermic, so we don't need to discuss that. However the two statements (observations) contradict the Le Chatelier principle, because if the dissolution of $\ce{NaOH}$ is exothermic, the solubility should decreases with increase in temperature! Why doesn't it?

Let me address this statement first:

  We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that.

While this is frequently the case, there are plenty exceptions, and only one exception is required to prove the statement false. Lets consider the solubility of $\ce{Ca(OH)2}$.

$\begin{array}{rr}
\pu{0^oC}: & \pu{1.89g/L}\\
\pu{20^oC}: & \pu{1.73g/L}\\
\pu{100^oC}: & \pu{0.66g/L}
\end{array}$

$\ce{Li2SO4}$ is another example that follows this trend of decreasing solubility with increasing temperature, and there are plenty others.

Although the heat of a chemical reaction often follows Le Chatelier's principle, there are exceptions.  Based on the facts as laid out above and in the OP, the dissolution of $\ce{NaOH}$ is a case in which Le Chatelier's principle does not apply to the heat of the reaction; the dissolution of $\ce{NaOH}$ is exothermic, yet it's solubility increases with temperature.


What attributes cause a substance to be nonconductive or conductive I understand what causes elements to be conductive or nonconductive. But what causes a substance say rubber to be nonconductive at a molecular level?

Conductivity requires mobile electrons. Metals tend to have loosely held valence electrons so metals are generally conductive. Metals have loosely held valence electrons for two reasons: metals tend not to be very electronegative (able to stabilize electron density) and metals tend to be somewhat big atoms by nature of many metals having a d-group of electrons. D electrons are further from the nucleus on a time average basis than s electrons or p electrons.

However, conductors do not have to be metals. Consider graphite. Graphite is $\ce{sp^2}$ hybridized. Graphite is made solely of carbon atoms - and carbon isn't a metal! The mobile pi electrons in graphite due to resonance allows graphite to be a good conductor.

I should also note that conductivity does not always imply mobile electrons; solutions can be conductive too and this is due to mobile electrolytes (ions). For example, you will find that deionized water is completely unconductive. However, regular tap water is generally conductive due to the presence of ions (e.g. chloride ions from chlorination). Gatorade is also very conductive because it is especially formulated to contain a lot of electrolytes.


Does a Na2SO3 solution oxidize to Na2SO4 in an oxygen rich atmosphere? Let's assume I have a pure solution of water and $\ce{NaOH}$. When this solution comes in contact with an atmoshpere containing inert gases and $\ce{SO2}$, then $\ce{NaHSO3}$ and $\ce{Na2SO3}$ will be formed into the solution.

The question is, if the atmosphere now contains also $\ce{O2}$, would these reactions happen?

$$\ce{2Na2SO3 + O2 <=> 2Na2SO4 + 2e-}\\
  \ce{2NaHSO3 + O2 <=> 2NaHSO4 + 2e-}$$

What would happen to the freed electrons?

I'm not sure what is the stable form at room temperature, since we have both sodium sulfite ($\ce{Na2SO3}$) and sodium sulfate ($\ce{Na2SO4}$) in our lab and they're both stable, in air. I do know that the sulfite will oxidise to sulfate with the addition of bleach.

That said, your redox reaction is not quite accurate. In sulfite, sulfur is in the $\ce{S^4+}$ oxidation state. In sulfate, it is $\ce{S^6+}$. What happens in the reaction is that the sulfur gives away two electrons, which then go to an oxygen that changes from $\ce{O^0}$ to $\ce{O^2-}$.

One way of writing it would be $\ce{S^4+ + 1/2O2^0 -> S^6+O2^2-}$. Total charge is preserved and there are no free electrons.


How to know about radical character of a diradical molecule? I understand that the radical character is quantified by the occupation number of the lowest unoccupied natural orbital obtained from UHF calculations or from CASSCF calculations.
My issue is that I already have an organic diradical whose ground state is high-spin state. In this case how can I quantify the radical character ?
Is there a way to know about the mixing percentage from higher spin states which in turn could give us information about the radical character of the molecule ?

Take a look at this is the article, which provides quantitative characterization of radical character:
M. Head-Gordon. Characterizing unpaired electrons from the one-particle density matrix. Chem. Phys. Lett., Vol. 372, Issues 3–4, 2003, p. 508-511

Based on natural orbitals, the analysis in this paper is applicable to any correlated method - it needs only one-particle state density matrix.


How can a waterproofing material be "water-based"? I've been trying to determine how a particular waterproofing material, which claims to be "water based", can still be hydrophobic. First of all, the product in question is EPRO ECOLINE-R, which also claims to be an asphalt emulsion. My first guess is that the curing process causes a change in some way to become hydrophobic once applied. But I just can't wrap my head around how a product, which is designed to resist water absorption, can also be water based? I can find no information on the exact chemical makeup, other than "petroleum hydrocarbons".

One might need to have more information about the product you've posted to say what occurs in the specific example. But, the general concept could be similar to Latex (water-based) paint. Where the water proofing chemical is dispersed by an emulsion, and once the water evaporates, the proofing material is restructured by its, now unhindered, intermolecular forces. These forces might be strong enough, once the proper structure exists, to resist water coming back and trying to interrupt the interactions.

Below is the Wikipedia explanation for my example of latex paint.

  The term "latex" in the context of paint in the USA simply means an aqueous dispersion; latex rubber from the rubber tree is not an ingredient. These dispersions are prepared by emulsion polymerization. Such paints cure by a process called coalescence where first the water, and then the trace, or coalescing, solvent, evaporate and draw together and soften the binder particles and fuse them together into irreversibly bound networked structures, so that the paint cannot redissolve in the solvent/water that originally carried it.


How to determine pressure and concentration of carbon dioxide in blood plasma? A man suffering from untreated diabetes mellitus is admitted to a hospital. Doctors fear that his blood $\mathrm{pH}$ may have dropped because of ketoacidosis. Analysis of his blood reveals that $[\ce{HCO3-}] = \pu{16 mM}$ and $p_\ce{CO2} = 30.$ If $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is $6.1,$ determine whether the patient runs a risk of acidotic coma. (Note: in plasma under physiologic conditions, concentration of $\ce{CO2}$ and $p_\ce{CO2}$ are related by the solubility constant for $\ce{CO2}$ in plasma which is $\pu{0.03 mM/mm Hg}.$

We use the Henderson-Hasselbalch equation, then

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log[\ce{HCO3−}][\ce{H2CO3}]$$

Moreover, here $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is given whereas the acid is $\ce{H2CO3},$ and also if we consider second dissociation that is $\mathrm{p}K_\mathrm{a2}$ then also it is wrong because $\ce{HCO3-}$ will dissociate into $\ce{H+}$ and $\ce{CO3^2-}$ so in this dissociation there is no relation of $\ce{CO2}.$  And also why solubility constant is given, how can we find out $\ce{CO2}$ from solubility constant?

Since $p_\ce{CO2}$ is $\pu{30 mmHg}$, and we are given the $0.03$ of $\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}$ ratio, we can calculate the concentration of the $\ce{CO2}$ in plasma.

$$\pu{30mmHg}~\ce{CO2}\times 0.03\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}= \pu{0.9 mM}~\ce{CO2}$$

$$\ce{CO2 + H2O \xrightarrow{\text{carbonic anhydrase}} H2CO3}$$

$$\text{Thus, } \pu{0.9 mM}~\ce{CO2}= \pu{0.9 mM}~\ce{H2CO3}$$

$$\ce{H2CO3 <=> H+ + HCO3-}$$

$$\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}} = K_{\mathrm{a}_\ce{H2CO3}}$$

$$\log{\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$-\left(\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}\right)=-\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$

$$\ce{[HCO3-]}=\pu{16 mM}$$

$$\ce{[H2CO3]}=\pu{0.9 mM}$$

$$ \mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}=6.1$$

(PS: It is actually $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, not that of $\ce{HCO3-}$. Note that $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is $10.3$.)

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$
$$\mathrm{pH}-\log{\frac{16}{0.9}}=6.1$$

$$\mathrm{pH}-1.25=6.1$$

$$\mathrm{pH}=7.35$$

  The $\mathrm{pH}$ of blood is maintained within extremely narrow limits of 7.36-7.4 for venous blood and 7.38-7.42 for arterial blood. (Source: Upadhyay - Biophysical Chemistry)

So, we can say the $\mathrm{pH}$ of the blood of the patient is slightly acidic than normal.

I hope I've answered your question.


Why are there no hydrogen atoms in the crystal structure of a biomolecule? I am taking a physics and chemistry at the nanoscale class at my university and for a project I must learn how to use Visual Molecular Dynamics (VMD). One of the models involved in this project is apo-core-streptavidin in complex with biotin at pH 4.5 (1SWE.pdb).  The crystal structure can be found protein data bank. There are no hydrogen atoms in this file/model and I was wondering why that is.  I can't find the answer to this anywhere online.

Typically crystal structures are determined by X-ray diffraction off of electrons within a bond and around the nuclei of each atom. Since H only has 1 electron, and its often involved in a polar type bond (hence doesn't spend much time by the H nucleus when bonded) it is notoriously hard to "see" with X-rays.

In order to have X-rays diffract accurately off H atoms the sample crystal needs to be large and very regular; this is not usually the case with proteins. As a workaround information about the H position can be calculated by the other atoms hybridization. This is often unreliable especially with all the hydrogen bonding in proteins which pulls them out of their idealized positions.


What is the balanced equation for H2O2 + KI + dish soap I am not sure how to balance $\ce{H_2O_2 + KI + dish soap}$.   Is the dish soap involved in the equation or just to show the produced gas?  Is it a double replacement or does the $\ce{KI}$ just act as a catalyst and $\ce{H_2O_2}$ is deconstructed to $\ce{2H_2O + O_2}$? It has been a while since I had chemistry.

Ah!! The elephant toothpaste experiment.
First thing first, this reaction is actually a decomposition experiment as the following equation.

$$\ce{2H2O2 -> 2H2O + O_2}$$

Second, $\ce{KI}$ acts as a catalyst in this experiment. You also can use $\ce{MnO4}$ too as the catalyst for this experiment.

Third, the dish soap makes the product of the experiment more soapy and bubbly. So, you can see it will create look like a toothpaste. If you didn't put the dish soap, it will only look like gas bubbling out if the solution.


High voltage AC water electrolysis not working I was trying to use a $500~\mathrm V$ AC power supply to split water, but nothing happened. I  tested everything with a $9~\mathrm V$ battery, and it all worked fine.

Why did it not work with the high voltage? (Do not worry, I was taking the proper precautions, such as insulation and three pairs of gloves, and I did not touch any live wires.)

From the comments, mcocdawc:

  Was it DC or AC?

OP:

  AC. I was trying to see if it was possible.

There's your problem. Alternating current spends half its time pushing electrons forward through the circuit, and half the time pushing them in reverse through the circuit.  Thus, on average the net current pushed through the cell is zero.

So, leaving aside any metal dissolution/deposition, you were forming tiny, tiny amounts of electrolysis gases at the surface of your electrodes while there was current moving through the cell in one direction, but immediately consuming them again when the current reversed. The total number of coulombs passed in each direction was presumably so small that no gas bubble nucleation occurred before the polarity reversed.

Regardless: STOP USING $\mathbf{500V}$ FOR ELECTROCHEMISTRY EXPERIMENTS. It's way more potential than you need, and terrifically unsafe. In most cases, anything above $40~\mathrm V$ is overkill, and for most experiments that $9~\mathrm V$ battery you mentioned will more than suffice.


How can I prepare a PBS (Phosphate Buffer Solution)? pH=7.4 Phosphate-buffered saline (abbreviated PBS) is a buffer solution commonly used in biological research. It is a water-based salt solution containing disodium hydrogen phosphate, sodium chloride and, in some formulations, potassium chloride and potassium dihydrogen phosphate.

Google is your friend!

The preparation of pH 7.4 PBS (with KCl and KH2PO4) is described on this site   here. Preparation of many other buffer systems are also described.

Here are the conditions (lifted from the linked site)

Prepare 800 mL of distilled water in a suitable container.
Add 8 g of NaCl to the solution.
Add 200 mg of KCl to the solution.
Add 1.44 g of Na2HPO4 to the solution.
Add 240 mg of KH2PO4 to the solution.
Adjust solution to desired pH (typically pH ≈ 7.4).
Add distilled water until volume is 1 L.


Identify products of Propiophenone using nmr I've been analyzing a compound "propiophenone", using NMR and IR, and
I can't tell which of the nmr spectrums relates to which isomer of C9H10O2 is which.

At first I thought the two isomers were ethyl benzoate and phenyl propanoate, which would make the 1.5 peak from the CH3 group and the 2.7 and 4.2 peaks from the CH2, and the CO stretch in ethyl benzoate is about 1730. But then I didn't know why the stretch in the other isomer would be 1770 and I thought that the ethyl benzoatate would be the major isomer, but I was told that the isomer D (with a stretch at 1730) is the minor isomer.

As this is clearly a homework type question, this answer will just give you some helpful pointers to head you in the right direction, and leave you to piece together the answer.

My usual approach for these types of problems is to have a look at the NMR data first, and ignore all the IR data. We might use that later for double checking.
From both spectra, it is possible to identify the following spin systems:

-CH2CH3
-C6H5 (monosubstituted benzene).

We also know we must have a -OC(O)- group in there also. That satisfies our molecular formula.

Looking at Isomer C first:

CH2 group at 2.7, indicative of it being adjacent to the carbonyl of an ester
Ortho aromatics at 7.0  and upfield of the meta, indicative of substitution with an electron donating (or aromatic activating) group, Ar-O-
I'll let you piece together the structure of C.

Looking at Isomer D:

CH2 group at 4.4, indicative of being adjacent to a heteroatom, so -O-CH2
Ortho protons at 8.0  and downfield of the meta , indicative of substitution with an electron withdrawing (or aromatic deactivating) group; Ar-C(O)-
I'll let you put this structure together also.

And what about the IR data? We know the following, which is consistent with what we propose from NMR:

Al-C(O)-O-Ar 1740-1770cm-1
Ar-C(O)-O-Al 1715-1730cm-1

The general rule with IR absorptions is that conjugation lowers the frequency - I'll leave you to rationalise this with some fancy arrow-pushing from the aromatic ring to the carbonyl group.


Bond cleavage of 3 membered cyclic ring on heating I was told to propose a mechanism for the following conversion:

One obvious step is the conversion of cyclic ketal to ketone because of acidic conditions. But I am unable to understand how the three membered ring opened to give an alkene? (However, I can understand  that the three membered ring is unstable)

I feel that somehow, a carbocation is being generated and E1 type elimination is taking place.

What is happening exactly? And does this type of thing take place with any three membered ring under given conditions?

Possible mechanism for cyclopropyl ring opening of given compound in acid medium: