how to define proper ppm concentrations for water solvent vs alcohol solvents? 1 ppm is 1 mg / kg
Water density: 999.97 kg/m³
methanol density: 791.80 kg/m³

solution 1 with water as solvent: at 1 $\ce{\mu}$g / mL is considered 1 ppm

solution 2 with methanol as solvent: at 1 $\ce{\mu}$g / mL, should it be considered 1 ppm?

If this is adjusted by the density of methanol, it wouldn't be 1 ppm. I heard it is a common practice that at the ppm / ppb levels, the density of the solvent is often omitted by only focus on the volume. Is this true?

I think this is why it's generally not advised to use relative units such as $\%$, $\mathrm{ppm}$ and $\mathrm{ppb}$. If they're going to be used, then it's best to indicate what's the ratio being considered, which is frequently done using $\mathrm{(v/v)}$, $\mathrm{(m/m)}$ or $\mathrm{(m/v)}$, $\mathrm{m}$ being short for mass and $\mathrm{v}$ for volume. For example:

$$\frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{1\ mL_{\ methanol}}} = \mathrm{1\ ppm\ (m/v)} = \frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{0.7918\ mg_{\ methanol}}} = \mathrm{1.263\ ppm\ (m/m)}$$

What's common practice may be heavily dependent on whom we're talking about. Since we usually prepare solutions in the lab by completing the volume up to a certain amount, one could say it's best to ignore the solvent density and use $\mathrm{(/v)}$ as a rule of thumb. The solute unity will depend mostly whether it's a solid being weighed or a liquid being measured.

As pointed out by @PLD, however, there might be cases where it's important to weigh the solvent, probably when very accurate measurements are necessary and specially if the solvent's volume changes considerably given changes in temperature.

One problem of using the $\mathrm{(m/v)}$ notation is that it doesn't use SI units, and might also be confusing. A more verbose way of indicating this is saying mass to volume. Doing some research for this I even found the "units" $\mathrm{ppm_v}$ and $\mathrm{ppm_m}$, which indicates the solvent's unity, but not the solute's, but definetly aren't as common and are also unclear.

The best solution here is simply to use the full unity, so instead of using $\mathrm{ppm}$ (which actually isn't accepted in many cases, for example most scientific journals), use $\mathrm{mg/mL}$ (or $\mathrm{mg\ mL^{-1}}$) or whatever was really measured. This doesn't leave room for double interpretations and indicates how the solution was prepared.


The difference between hydroxide and oxyhydroxide I cant discern these two terminologies. Would someone be very kind to help me on it?

I think hydroxide mean M(OH)x where M is a metal ion, while oxyhydroxide means something like MO(OH). Is that right? Oxyhydroxide equals to oxide-hydroxide.

Yes, you have correct idea.

Oxyhydroxide is any mixed oxide and hydroxide. For example Iron(III) oxide-hydroxide

While in hydroxide hydrogen and oxygen atom are joined with covent cond.


List in order of increasing radius List in order of increasing radius:

a) Rb, K, C5, Kr

For this one I got C, Kr, K, Rb

b) Ar, Cs, Si, Al

This I got: Ar, Si, Al, Cs

Does anyone know if this is correct? I don't have the solution for these in my book :S

There is a trend that follows an arrow from the lower left of the periodic table to the upper right.  As the atomic number increases along a period, the radius for each element decreases.  Here is a useful image of this:

Notice that, with the exception of the noble gases, you can draw an arrow from the lower left to the upper right.  The justification for what you see here is that as the principal quantum number increases, the probability that the electrons are more distance from the center of the atom also increases.  The higher the principal quantum number, the higher the energy of the electron and therefore its probable distance from the center of the atom.  Therefore, as you move down a group, the radius increases.

However, another factor is that as the number of protons increase in an atom, so does the positive charge which happens to pull the electrons toward the center of the atom.  So, there is the impact of the protons causing a decrease in radius (probability of the electron being closer to the center) and the probability that the electrons are in an orbital that is farther from the center.

The transition metals, however, have d orbital electrons.  These electrons enter into a shell that is closer to the nucleus and the change is less noticeable.  That is part of the reason why they do not seem to change as much as the other atoms, according to my understanding.


Why can you blow bubbles in milk but not hot cocoa, which is milk based? My daughter is asking this question and I’m assuming it’s something related to the cocoa added to milk but don’t know what change is happening molecularly and why it would affect the bubbles. Thanks!

Let me know if there is a better place to ask this question!

Milk contains lecitihin, which acts like soap to keep fat globules in milk suspended (although it does slowly separate into cream and skimmed milk, which you could demonstrate with non-homogenized milk).

Lecithin and soap work by having long molecules, one end of which is attracted to water (hydrophilic) and the other is attracted to oil or air more than to water (hydrophobic). This helps the water form a double layer, with air inside and outside and the lecithin holding things together. C. V. Boys Soap-Bubble and the Forces Which Mould Them is worthwhile reading and lists numerous experiments with detergent and water that you might try with your daughter's assistance.

Ah... but you ask why cocoa interferes with bubble blowing! Here are two tentative explanations:

The cocoa particles don't dissolve, so they "poke holes" in the soap/water/soap membrane, bursting the bubbles.
The cocoa adheres more tightly to the hydrophobic end of the milk lecithin than does air, preventing formation of the bubble layers.

You and she could experiment to see if either are correct. Try putting something other than cocoa in the milk (you might substitute dish-washing detergent in water for milk, because it makes longer lasting, cleaner and cheaper bubble stuff). Finely shredded cellulose is hydrophilic, and powdered candle wax is hydrophobic, for example. You could try finer powders, too. If you do experiment, please post your results here - I would be interested in your findings!


Electrode specification What factors go into selecting an appropriate material for an inert electrode for electrolysis?

It looks to me like metal oxides or platinum/platinum-coated titanium are the preferred options. However, I'm not really clear why platinum should be a good option since it doesn't have a particularly high standard electrode potential (1.188V). Lead oxide, gold, and silver all have higher standard electrode potential and are all cheaper, so why are they not preferred?

The exact choice of electrode really depends on the goal of the experiment and what will be in your cell:

The most basic requirement is that the solvent or electrode material not decompose at the potential you want to work at.
For each solvent/electrode combination, there is a potential window in which you can safely work.
In the most common situation—working in aqueous solutions, the choice of working electrode mainly depends on whether you want to observe oxidation or reduction.
Electrodes like platinum or gold are commonly used for oxidation reductions because they have high overpotential for oxygen evolution in aqueous solutions (platinum in particular, performs poorly for reduction as it is rather good at catalysing the hydrogen evolution process).
Conversely, for reduction reactions, mercury and bismuth have potential windows that extend to quite negative potentials, but will themselves oxidize as you move towards positive potentials.
There are also some materials that work pretty well for both positive and negative potentials, such as glassy carbon and boron-doped diamond.

For analytical purposes, you can get away with (and often prefer) small electrodes, so cost is less of a concern, but if the application is mass electrolysis, large electrodes are required to have a decent reaction throughput and more expensive materials may be impractical.
Cheap materials like stainless steel and graphite can be used as electrodes in some cases, though they are more limited in the systems in which they may be used.

As far as the reason to choose platinum over lead oxide, gold, or silver is concerned, the standard electrode potential isn't a primary consideration. (Silver's is not higher than platinum's, incidentally)
The standard electrode potential reflects the thermodynamic situation of a particular half-reaction, but is not terribly informative of how the material will behave in a given cell as it does not account for the kinetics of the reaction or any other factors.
Silver is usually not a good choice for a working electrode as it oxidizes quite easily in aqueous solutions, gold is actually a fairly common choice as it has a wider potential window than platinum in aqueous solutions, but it is easily fouled by organic/biological contaminants.
Platinum doesn't have the widest potential window, but is generally resistant to fouling and being reacted itself (giving it a long lifetime).

In a nutshell, electrode choice is dictated by many competing factors: cost, lifetime, potential range, mechanical considerations, etc.
In an analytical lab, screen-printed gold electrodes are quite common because they are relatively inexpensive, suitable for many different reactions, and can be discarded and replaced when fouled.
In an aluminum refinery, you'll find carbon and coke electrodes used in the Hall-Héroult process, as they need to be very large and withstand the high temperatures of molten cryolite.
In the cloralkali process, the anode is usually made from titanium as it must resist the highly oxidizing conditions that produce chlorine, but the cathode can be made from much cheaper nickel as that half of the cell is only used to reduce water to hydroxide and hydrogen gas.


Ammonia to tri substituted nitrogen I am told that an intramolecular proton abstraction can occur. For some reason this seems at least sterically unlikely. Is it possible sterically? Or is the proton abstraction best described as a equilibrium process between the nitrogen's hydrogens and water?

Until there's water present (I'm assuming that it's not the solvent), the proton can't come from water. The proposed intramolecular proton transfer would go through a 5-membered transition state, which is reasonable. If you don't like that mechanism, then two of the intermediates could transfer protons to each other. Depending on the concentration, one or the other or both could be occuring.

The exact nature of these proton transfers are difficult to prove and typically inconsequential, so you will often see "shortcut" mechanisms drawn which imply intramolecular proton transfer even when it is more unlikely than the one you show here.


Thermodynamics behind organic chemical reactions? I took organic chemistry a few years ago at my university, and although I went through the motions to pass the exams, I didn't understand the thermodynamic reasoning behind why the reactions actually occurred. For instance, I memorized that Claisen Condensation reactions occur between two esters in the presence of a strong base, but I don't fully grasp WHY the reaction happens in the first place.  Why are the products more stable than the reactants in cases like these? Does this reaction always occur whenever these components are in a mixture, or is there a sort of catalyst to trigger it? I feel similarly confused for many other organic reactions (e.g., Aldol condensations, Fischer esterification, Sn2, etc). I suppose I'm wondering if there is a general thermodynamic reasoning I'm missing in organic chemistry.

Whether or not the product is "more stable" would really depend on the sense in which you consider its reactivity. The $\mathrm{p}K_\mathrm{a}$ of a $\beta$-keto ester is about 11, so in that sense, it is much more reactive than either of the reactants. Once the $\beta$-keto ester is formed it is quickly deprotonated. It remains predominantly in this form because the anion is so stable, and the reaction is incapable of reversing. Thermodynamics does not always determine the result of an organic reaction. Many times, kinetics is the deciding factor.

The Claisen condensation is catalyzed by the base, as it is recovered in the end, and though I am sure it occurs spontaneously without it, I doubt it happens very often. I don't think ethyl acetate would be as popular a solvent as it is if it quickly and spontaneously converted to ethyl acetoacetate.


Why do real gases deviate from ideal gas behavior? While studying Gaseous state in books, there was a common question which I had difficulty answering -

Why do real gases deviate from ideal behavior (3 points)?

Real gases deviate from ideal behaviour because their particles (atoms for inert gases or molecules) occupy some finite space and do exert interactive forces among them. Completely ideal behaviour is hypothetical because of the reasons above. At low pressure and high temperature, real gases behave approximately as ideal gases. In ideal behaviour, gas particles don't occupy space and do not have any interaction, as assumed in the kinetic theory of gases. But in reality this is not the case: we get errors by applying the ideal gas law.
That's why van der Waals corrected it by introducing suitable constants.


Adding lemon to lentils changes color from black to pink, why? I noticed that when I cook black lentils and add lemon juice to the water that's left, the water turns from black to pink!
What can be the reason for that?

In the black lentils used in Indian cooking (Vigna mungo), the natural water-soluble dye that causes the color change has been identified as quercitin-3-O-pentoside:

which belongs to a class of compounds called flavonols. When you pop hydrogen ions on and off some of the -OH groups, you can get a pure benzenoid form (colorless), a quinonoid form (pink), and an extensive quinonoid form (yellow). The color changes
are pH dependent, and you get pale pink at pH < 7, yellow at pH > 7, and
colorless at pH 7. When you add your lemon juice to the lentil water, you're getting the quinonoid form.

Other species of black lentils probably do something similar.

The pH-dependent color change occurs because when you have $-\mathrm{OH}$ groups on the rings instead of $-\mathrm{O}^-$ groups, the way electrons are confined changes; that changes the spacing between the molecule's energy levels, and changes the colors of light the molecule can absorb. (See my article here for more about how this works).

Many plants and household materials contain molecules that can change color or otherwise indicate changes in pH (See my article here for a list of some of them). Almost any flower, fruit, or plant part that is red, blue, or purple contains a class of chemical compounds called anthocyanins that change color with pH. The color of a flower or fruit depends on which anthocyanins are present, the pH of the pigment-bearing tissues, and the presence of other pigments, like yellow flavones.


Clarification needed regarding description of molecules without hybridization using pure p orbitals I need some clarification regarding this answer given by the user @Jan.

  @Jan says:

  Therefore, the a priori predicted most stable configurations of
  certain coordination numbers are:

  mono-, di- and tri-coordination: use 1, 2 or 3 p orbitals to form bonds; keep one lone pair in an s type orbital. (Predicted bond angle:
  $90^\circ$)

I could not understand the meaning of the first bullet point clearly.

I cannot agree that in mono, di and tri coordination pure $p$ orbitals are always used. Pure $p$ orbitals have bond angles of $90^{0}$. In that case, how can we explain the structure of compounds like $\ce{BF3}$ or ethene where $\mathrm{sp^2}$ hybridization is said to exist and bond angles are $120^{0}$ ?

Or did @Jan imply that we need to use pure $p$ orbitals in mono, di and tri coordination in case no hybridization takes place ? If so, how can we understand when hybridization will take place and when it will not? Please give some examples.

Yes, what I wrote on the topic in the previous question was very, very brief and while writing it I was not thinking of boron — an element that usually calls for exceptions due to it only having three valence electrons.

However nonetheless, the first thing you should do for any compound is to consider all atoms in ground state configuration and see whether you can form the required bonds from occupied orbitals in that state. Let’s take the example compounds $\ce{BF3, NF3, AsH3, C2H4}$ (tricoordinate) $\ce{CO2, H2O, H2Se, BeCl2, CH2}$ (dicoordinate) and $\ce{HCl, N2, CO}$ (monocoordinate) for the discussion. Any structural data referenced herein is taken from the corresponding Wikipedia articles.

Ideal cases

I consider the examples $\ce{AsH3, H2Se}$ and $\ce{HCl}$ to be ideal cases. Taking each atom in its ground state ($\ce{As}:{ [\ce{Ar}]}\,\mathrm{4s^2\,4p^3}; \ce{Se}{: [\ce{Ar}]}\,\mathrm{4s^2\,4p^4}; \ce{Cl}{: [\ce{Ne}]}\,\mathrm{3s^2\,3p^5}; \ce{H}{: \mathrm{1s^1}}$), arsenic has three half-filled p orbitas, selenium has two and chlorine has one. Each of these elements (except hydrogen) has a single lone pair in an s orbital and if there are further lone pairs, they are in p orbitals (arsenic has no more). Thus, we can construct $\sigma(\ce{s-p})$ bonds until all valences are satisfied.

Moving on to bond angles: well, $\ce{HCl}$ has no problems as there are only two atoms. But $\ce{H2Se}$ and $\ce{AsH3}$ also have hardly any problems since the central atoms are so large that the comparatively smaller ligands can be attached at almost perfect $90^\circ$ angles without significant steric repulsion. The bond angles repored ($\ce{AsH3}{: 91.8^\circ}; \ce{H2Se}{: 91^\circ}$) are in adequate agreement.

Ideal cases with steric distortion

The next set of examples along the route you should be going are $\ce{NF3}$ and $\ce{H2O}$. These compounds fulfil the general criteria outlined in the previous section. However, their central atoms are decidedly smaller. Thus, perfect $90^\circ$ angles are unobtainable as there would be a steric clash of the ligand atoms. The ligand atoms move apart to prevent steric repulsion; this is achieved by mixing in s-contribution into the bonding orbitals. $25~\%$ s-contribution would correspond to a tetrahedral bond angle of $109.5^\circ$ and $\mathrm{sp^3}$ hybrid orbials. The table below compares the bond angle and length data of the lighter homologues discussed here and the heavier ones discussed above (consider the fluorines in $\ce{NF3}$ to be pseudohydrogens).

\begin{array}{lcccc}\hline
\text{element} & \ce{N} & \ce{As} & \ce{O} & \ce{Se} \\ \hline
d(\ce{X-H})/\mathrm{pm} & 137 & 151.9 & 95.84 & 146 \\
\angle(\ce{H-X-H})/^\circ & 102.5 & 91.8 & 104.5 & 91 \\ \hline\end{array}

$\ce{CH2}$, carbene, is a dangerous case. As far as I know, no definite structures are reported (although Wikipedia mentions bond angles determined by EPR) since it is too reactive and since it is susceptible to dimerisation to form ethene. It also comes in three distinct forms: a singlet form, a linear triplet and a bent triplet. The most ideal structure of these three is the singlet carbene, which also fits nicely into this subsection.

For a singlet carbene, you would also start off from carbon’s ground state $[\ce{He}]\,\mathrm{2s^2\,2p^2}$. The two p orbitals each form a $\sigma(\ce{s-p})$ bond to the hydrogen atoms. The two hydrogens, however, need to move away from each other to ease steric repulsion; thus a little bit of s-contribution is mixed in for a final bond angle of $102^\circ$. A third p orbial remains unoccupied — note the similariy to $\ce{H2O}$ which has everything carbene has plus two electrons in a p-type lone pair.

Practically ideal cases which naturally lead to multiple bond

Herein, we shall discuss $\ce{N2}$ and $\ce{CO}$. For both, we can apply the method outlined above; I’ll start with the simpler $\ce{N2}$ case. We initially have one s lone pair and three, singly occupied p orbitals on each nitrogen. One of these forms a standard $\sigma(\ce{p-p})$ bond with the other nitrogen. Now, thankfully, the other p orbitals are nicely aligned and fit perfectly for generating two $\pi(\ce{p-p})$ bonds. Together, that gives us three bonds or a triple bond, composed of one σ and two π bonds.

The carbon monoxide case is essentially the same except that carbon would have one empty p orbital and oxygen would have one lone pair in a p orbital; only one π bond can be formed in the way outlined above. However, the unoccupied carbon orbital and the fully populated oxygen orbital can align likewise and form what could formally be described as a second, dative π bond. That renders nitrogen and carbon monoxide isoelectronic. Note, however, that the simplification introduced herein is not able to explain the big differences between $\ce{CO}$ and $\ce{N2}$ that arise from their actually rather different molecular orbital schemes.

Cases, in which an s type lone pair is impossible

I’ve cleared away all the simple cases. If you go through the list, the only ones that remain are $\ce{BF3, C2H4}$ and $\ce{BeCl2}$ — although we might as well include linear and bent triplet $\ce{CH2}$ in here. These all have in common that assuming an s type lone pair would not allow for sufficient electrons available for bonding.

If that is the case, the simple answer always suggests hybridisation (athough the MO description does not require it and nonlocalised molecular orbitals would not support it). At least one p orbital must be mixed in to allow for the generation of enough σ bonds; yet you should always try and make the p contribution of the hybrid orbitals as small as possible (in simplified terms) as that means the s lone pair ‘loses’ less energy en route. (Beware, the terms and images I used are inaccurate!) Any orbitals, populated or not, that are not required for σ bonds remain as p type ones. For $\ce{CO2}$ and linear triplet $\ce{CH2}$, that gives us an $\mathrm{sp}$ configuration of the central carbon with two unused p orbitals; in $\ce{C2H4}$, bent triplet $\ce{CH2}$ and $\ce{BF3}$, the central atoms are $\mathrm{sp^2}$ configured with one additional p orbital.

If possible — i.e. in $\ce{CO2}$ and $\ce{C2H4}$ — π bonds should then be formed. Each of the oxygens in carbon dioxide has one p-type semipopulated p orbital available for a π bond as does the second carbon of $\ce{C2H4}$. The carbenes cannot form π bonds due to the lack of partners, and thus they remain as diradicals. The linear one has two degenerate p-type radical orbitals, the bent one almost has an $\mathrm{sp^2}$ type radical orbital and one p type. (In fact, bent triplet carbene has a larger bond angle than the $\mathrm{sp^2}$ predicted $120^\circ$. This is in line with removing part of the s contribution of the radicals and supplying that to the bonding orbitals. Nobody expects you to correctly analyse the structures of carbenes, though.)

Beryllium

You may have noticed that that element had gone missing at the end of the previous section. This is because it is the most special of all cases. All other alkaline earth metals behave as metals (form dications and practically ionic structures) while beryllium still features notable non-metal properties.

I think that the best description of $\ce{BeCl2}$ is via a four-electron-three-centre bond akin to $\ce{XeF2}$. That again requires an unhybridised central atom. The lowest of the three relevant MOs of the 4e3c bond would be formed of the fully bonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ combination; the second bonding MO would then be $\mathrm{p}(\ce{Cl}) + \mathrm{p}(\ce{Cl})$ where the nodal plane traverses $\ce{Be}$ — mixing in its 2p orbital probably increases the bonding character — and a final antibonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ orbital wherein the phase of beryllium’s orbital is of opposite sign with respect to the chlorine orbitals’. If that is the case, it should possibly be discussed with the ‘hypervalent’ cases of the other question.

I could be terribly wrong, though. Thankfully, nobody will request you to predict the MO scheme of $\ce{BeCl2}$ until at least the final bachelor’s examination.


Does the para position on benzene ring-polymers make the structure stronger? Aramid fibers are strong synthetic fibers. Kevlar is called a para-aramid, and it's outstandingly strong. Why does the para-position give Kevlar its strength?

When the Kevlar monomers are in the para-position, will this make the chains closer (shorten the length of the bond), and make strong interactions between the chains?

para isomers fits into the crystal lattice more perfectly than ortho and meta due to steric reasons and hence generally they have more melting point.
Also in certain aromatic compound ortho isomers form  intra molecular hydrogen bonding which reduces its boiling point (generally) compared to ortho and para where in these isomers would be forming inter molecular hydrogen bonding.
In the case of para aramid you can see that the two peptide linkage formed are with two other para amid molecules (always).Had it been ortho aramid then the two bonds will be more likely to get bonded to a single other ortho amid molecule.And the polymeristaion ends there (at the most polymer of para aramid will be having less length and has unsymmetrical structure).But in para amid it can polymerise to any lengh.
These are the two possible polymers for ortho.

there is no difference in the bond strength formed between two para amid molecule compared to two ortho amid molecule after all both are peptide linkages only (i don't know if the ortho aramid exists or not but what will happen is this for sure).(sorry the figure looks completely distorted.I am not really good with the paint application)
finally its the capability to polymerize to a great extent that makes it strong.
Its also worth mentioning that the functional (or substituent) groups at C1 and C4 (C1& C4-carbon numbering in benzene) can't bond themselves,as the bonds in any polyatomic group,can't bend and elongate as they want (after all the distance between C1 and C4 even if we consider diagonally,is itself quite long.And secondly thinking like this insane,say a bond bending in way they want.It will never happen with para compounds.The least may be meta compounds can.)


Why is the distance between the atoms shorter in the dicarbon anion compared to the cation? In general, the anions have bigger ionic radii than their corresponding neutral atoms, and the cations have smaller ionic radii then their corresponding neutral atoms. How come that does not apply to $\ce{C2-}$ and $\ce{C2+}$?

Looking up in a current compilation of bond lengths in the International Tables for Crystallography (IT), apparently there is indeed not so much deviation in the distance between two adjacent carbon atoms in a $\ce{C(sp)#C(sp)}$ pattern:

(IT, 2006 edition)

Some hints to render this excerpt easier to access.  The IT indeed discern different bond types (13 for all combinations of $\ce{C-C}$,  and 625 for all organic / organometallic compounds), and consequently, the source offers separate listings for, for example, $\ce{C(sp^3)-C(sp^2)}$ or $\ce{C(sp^2)-C_{ar}}$, too (first column). The second column is about the structure around the pattern in question highlighted in bold. For the $IT$, $\ce{C_{ar}}$ explicitly represents an aryl C in an six-membered ring; $\ce{C^{\#}}$ any $\ce{C(sp^3)}$ - linearly substituted, or not.

The third, fourth and fifth column are the unweighted sample mean $d$ in [Å], sample median $m$, and sample standard deviation $\sigma$, followed by the lower and upper quartile, $q_l$ and $q_u$, as well as the number of observations $n$.  Overall, the compilation is based on a subset of 10 324 X-ray and neutron diffraction data found in the Cambridge Crystal Database by September 1985, judged to be "good enough" (the source offers additional insight about the criteria applied).


What is produced when citric acid (powder form) is burned, is it safe? What is produced when citric acid (powder form) is burned?

Are the fumes harmful or safe?

As @paracetamol mentioned, if you burn it completely and properly, you should get nothing other than $\ce{CO2}$ and $\ce{H2O}$.

But what if you don't? Then it's a bit complicated. You can't say exactly what will form. There are chances of carbon monoxide formation, which is quite toxic. By binding to the heme complex iron in a hemoglobin, it renders it incapable of carrying oxygen to the body, causing dizziness, nausea, and likely death. Consider that.

Second thing is, there may be unburnt carbon particles that are emitted if you don't burn it well. Particulate matter such as these is highly dangerous to the lungs.

Thirdly, considering that atmospheric combustion is a highly complicated process, you can't really say what you could have produced from that oxidation. It may be toxic and may have crazy properties that even the safety equipment you have cannot protect you. That's something you should consider as well.


interconverting Van Der Waal's equation for real gases from number of molecules to density I want to ask a question about Van Der Waal's equation for real gases.
I was shown the following formula for real gases (VDW) which accounts for the liquification of water vapour which is not possible with an ideal gas model.
$$P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2} = \frac{\rho RT}{1-\rho b} - a \rho^2$$
which can be expressed using densities by making the denominators equal first:
$$P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2} = \frac{NkT}{V-Nb}\left(\frac{V}{V}\right) - \frac{aN^2}{V^2} = \frac{\frac{NkT}{V}}{\frac{V-Nb}{V}} - \frac{aN^2}{V^2} $$
but I appear to yield the result
$$P = \frac{\rho kT}{1-\rho b} - a \rho^2$$
where Boltzmann's constant $k$ has been switched for the gas constant, $R$. I can't find any sources that explain this derivation apart from "Molecular Driving Forces" by Dill, and he doesn't do a thorough conversion either.
How do I convert the numerator with Boltzmann's constant $k$ to the gas constant, $\rho$ which expresses VDW's equation involving densities?

Your analysis is correct in terms of number density.  But let's see how it plays out in terms of molar density.
Let n be the number of moles and A be Avagadro's number.  Then N=nA.  If we substitute this into your first equation, then I get $$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where $a'=aA^2$ $b'=bA$.  In terms of molar density $\rho=n/A$, this becomes:$$P=\frac{\rho RT}{1-\rho b'}-a'\rho^2$$


Why is the maximum adiabatic temperature found for an equivalence ratio a bit higher than 1? Why does the peak adiabatic temperature usually happen for an equivalence ratio a bit higher than 1? I think it had something to do with the probability of each fuel molecule finding a corresponding oxidizer, and the fact that the molecular weight of the fuel is generally higher than that of the oxidizer.

I actually found the answer in this 2011 paper, here is a quote from the abstract:

  The characteristic rich shifting of the maximum adiabatic flame temperature from the stoichiometric value for
  hydrocarbon/air mixtures is demonstrated to be caused by product dissociation and hence reduced amount of heat release.
  Since the extent of dissociation is greater on the lean side as a result of the stoichiometry of dissociated products, the peaking
  occurs on the rich side.


Nuclear equations for the decay of radon-222 to lead-206 I have to answer this on my worksheet. I understand nuclear chemistry equations a bit, as well as the process of alpha decay and somewhat about what happens when an atom changes its number of protons and neutrons, I understand the gamma symbol and the m symbol meaning the element is high-energy state but this has me stumped. It's an 8 step question. It reads:

  The inhalation of radon-222 and its decay to form other isotopes poses a health hazard.
  Write balanced nuclear equations for the decay of radon-222 to lead-206 in eight steps.
  Show step 1 below, and show steps 2-8 are on the next page.

  a. Step 1: radon-222 decays by alpha emission.

  b. Step 2: the daughter product in part a decays by alpha emission and is in a high energy state.

  c. Step 3: the high energy daughter product in part b decays by beta and gamma emissions and is in a high energy state.

  d. Step 4: the daughter product in part c decays by beta and gamma emissions and is in a high energy state.

  e. Step 5: the high energy daughter product in part d decays by beta emission.

  f. Step 6: the daughter product in part e decays by alpha emission and is in a high energy state.

  g. Step 7: the high energy daughter product in part f decays by beta and gamma emissions and is in a high energy state.

  h. Step 8: the high energy daughter product in part g decays by alpha and gamma emissions.

In any nuclear reactions the sum of the mass numbers $A_i$ (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges $Z_i$ (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state:

$$
\begin{align}
\ce{^{222}_{86}Rn &→ ^{218}_{84}Po + ^{4}_{2}He}\tag{a} \\
\ce{^{218}_{84}Po &→ ^{214\mathrm{m}}_{82}Pb + ^{4}_{2}He}\tag{b} \\
\ce{^{214\mathrm{m}}_{82}Pb &→ ^{214\mathrm{m}}_{83}Bi + ^{0}_{-1}e + γ}\tag{c} \\
\ce{^{214\mathrm{m}}_{83}Bi &→ ^{214\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{d} \\
\ce{^{214\mathrm{m}}_{84}Po &→ ^{214}_{85}At + ^{0}_{-1}e}\tag{e} \\
\ce{^{214}_{85}At &→ ^{210\mathrm{m}}_{83}Bi + ^{4}_{2}He}\tag{f} \\
\ce{^{210\mathrm{m}}_{83}Bi &→ ^{210\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{g} \\
\ce{^{210\mathrm{m}}_{84}Po &→ ^{206}_{82}Pb + ^{4}_{2}He + γ}\tag{h} \\
\end{align}
$$


Friedel-Craft reaction with amide substituent In this question, I know that aniline will not undergo FC reaction as it forms a complex with $\ce {AlCl3}$ and precipitates out. The answer given is (c) and I agree with it. However, the benzene ring with the amide substituent in (d) will be deactivated as the amide substituent will put a positive charge into resonance. So, will it give FC reaction? Should the answer be both c and d?

-NHCOCH3 is a slightly activating group due to the lone pair of N which shows +M effect. Hence it is Ortho para directing.


How are arsenic compounds removed from drinking water? I used to live near a volcano where the groundwater contains naturally occurring arsenic compound. Since $\ce{As2O3}$, $\ce{As2S3}$, and $\ce{As4S4}$ are predominantly occurring in drinking water, what are the processes for removing them?

The US Environmental Protection Agency (EPA) has a page dedicated to arsenic in drinking water.  It contains a section that suggests ways to remove it from water.

Disclaimer: I have no idea if the arsenic removal methods on the linked page actually work.  You should consult someone trained in the science of water purification for definitive information.


Protonation of Guanidine Why is the $\ce{sp^2}$ hybridized nitrogen atom in guanidine the most basic?

I know that as a general guideline, $\ce{sp^3}$ hybridized atoms are more basic than $\ce{sp^2}$. Bases are electron pair donors; having the electrons further away from the nucleus makes donation easier.

On the other hand, protonating the double bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule. Protonating the other nitrogens does not allow for the same degree of resonance stabilization.

And finally, in one textbook, I found mention of a +M effect - what is the +M effect? And while we're on that, what's +/- I? I understand that I stands for induction, but what's +I? What's positive induction?

One more note, since the double bonded nitrogen is $\ce{sp^2}$ hybridized, does that mean it has a greater negative charge density than the other nitrogens by nature of having greater s-character? Could this + charge density be giving the double bonded nitrogen more nucleophilic character?

Your misconception is directly at the beginning of your question and turns up multiple times. There are no $\mathbf{sp^3}$ nitrogens in guanidine.

Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced.

This is also why the $\ce{-NH2}$ nitrogens are not basic. Their lone pair is not sitting in an orbital all alone waiting for a proton. It is actively resonating with the π system of the other two. The only available lone pair guanidine has is that of the $\ce{=NH}$ nitrogen sitting in an $\mathrm{sp^2}$ hybrid orbital. Thus, a second protonation requires somewhat superacidic media.


Formal Charge in Coordinate covalent bonds I am in an honors chemistry course right now and I would like to know the formal charges of the atoms in CO, my confusion is that because the bond is partially coordinate covalent, are the bonded electrons still counted as #bonds/2 or should I count the number of electrons donated by each atom (making the formal charge 0 for both) Please help me understand and explain WHYnot just how. Thanks a lot.

The coordinate covalent bonds are still counted as 1 electron, since the pair is shared between both atoms. That means the Carbon will have a formal charge of -1 and Oxygen a formal charge of +1 on Carbon Monoxide.

Carbon FC: $\ce{(4 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = -1}$

Oxygen FC: $\ce{(6 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = +1}$


Lemon Battery works on water? My son and I were using a mini science experiment kit for creating batteries with copper, zinc, and lemonade, similar to the famous lemon battery project.  My son's hypothesis was that the liquid had to taste like lemonade to make a battery, and so he hypothesized (I thought correctly) that lime juice would work, but water and milk would not.

However, when we conducted our tests, the battery successfully ran a digital clock with all three liquids.  I washed the cups and copper and zinc sheets thoroughly between each test, but the non-acidic liquids continued to function as batteries.

Did I do something wrong, or is the liquid not required to be acidic to make a battery?

Sadly, the lemon battery is a bit of a cheat: it's not the lemon that's providing the energy at all. The very first battery ever was very similar to the lemon cell, but used damp cloth, or cardboard soaked in salty water (might be worth giving it a go, I've never tried).

The energy really comes from the metal of the two electrodes which have had lots of energy piled into them (heat etc) when they were refined. When you wonder where the energy comes from, think of the guys in the refinery and their furnaces, some of that is "in" the metal, and you're getting it back.

The purpose of the solution (or briney felt discs) is to give one part of this reaction (the electrons) a good run for their money, going through the electrical circuit, as they can't travel through the electrolyte, while the other part, the ions, can cross the other way, through the solution. They meet up at the other end, and the reaction is complete and the zinc slowly corrodes away.

So what you're testing is the quality of electrolytes. Salts and acids in solution make good electrolytes: salty water, lemons, potatoes, coke. Milk acts as an electrolyte because it contains some ions from salts (which nutritionists helpfully/confusingly call electrolytes).

Pure water is pretty bad as an electrolyte. There's one of two options. First, it's not absolutely useless as one, and maybe the clock requires so little power that water is good enough. The other option is that tap water is hard, or otherwise has salts (minerals etc) in it which increase the ion concentration greatly. Probably a combination of those two factors. It might be worth trying distilled water (if you have some for the car, for example) which is probably much poorer.

More about the "first battery" is here: it's lots of these metal-electrolyte-other-metal cells end-to-end. The zinc is slowly being corroded by the copper the necessary combination of a. ions (which are big, heavy old things that amble through the electrolyte) and b. poor, zippy, lightweight electrons which can't, and have to go "the long way round" through the electrical circuit. https://en.wikipedia.org/wiki/Voltaic_pile


Temperature in the Maxwell-Boltzmann Distribution I am attempting to generate the probability for the specific speed of a molecule using the Maxwell-Boltzmann distribution, but I cannot decide on which temperature I should use in the equation.

For example, I have a very, very cold liquid (let's say 70 K) at room temperature (we'll call that 296.15 K), but would I use the temperature of the liquid in the equation, or should I use the temperature of the air around the liquid?

If you are assuming a pseudo-steady state condition (where the bulk of the liquid is in thermal equilibrium with itself for the timescale of interest), and you want to study the bulk molecules of the liquid, then you want to use the temperature of the liquid.

Now, if you are going to be studying the surface of the liquid, using a Maxwell-Boltzmann distribution is much more questionable - it definitely isn't strictly correct, at least on time scales of any substantial length, but you might be able to get away with it for rough approximations.

If you gave some more details as to the specific application, I could probably give you a better answer.


What is the IUPAC name of the given compound? IUPAC name of the given compound is

(1) 3-Hydroxycyclohex-4-en-1-one
(2) 5-Hydroxycyclohex-3-enone
(3) 5-Ketocyclohex-2-enol
(4) 3-Ketocyclohex-5-enol

By what I understand by assigning the priority the parent is the ketone.
If I number carbonyl carbon as 1, then I should move towards the alcohol group because it has higher priority than the alkene and both are at number 3, so the alcohol should be assigned the number 3 and alkene as 4?
So, according to me, the answer should be (1), but in the key it is given as (2).

As for your other question, the simplified criteria for the numbering are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, you have correctly identified the principal characteristic group as ketone. It is expressed as suffix “one”. Thus, a low locant is assigned first to the ketone according to Rule (c), which yields “…1-one”.
Next, a low locant is given to the double bond according to Rule (e), which gives “3-en-1-one”.
Finally, a low locant is assigned to the remaining $\ce{-OH}$ substituent according to Rule (f).
Thus, the complete name is 5-hydroxycyclohex-3-en-1-one.
Rule P-14.4 applies to chains as well as rings.


Why is nitrosobenzene electrophilic on O in organocatalytic oxygenations? In 2003 Zhong1 and MacMillan2 reported the use of nitrosobenzene, PhNO, in the organocatalytic asymmetric α-oxygenation of carbonyl compounds:

which is all nice, except for the slightly counterintuitive fact that PhNO is acting as an electrophile on oxygen rather than on nitrogen.

Logically PhNO should be electrophilic on nitrogen (less electronegative than oxygen), and indeed, a 1994 review on the reactivity of PhNO lists numerous examples where this is the case.3 On top of that, if the enamine were to ad to nitrogen instead of oxygen, the phenyl group could be placed equatorial in the transition state:

Is there an explanation for the regioselectivity here?

References

Zhong, G. A Facile and Rapid Route to Highly Enantiopure 1,2-Diols by Novel Catalytic Asymmetric α-Aminoxylation of Aldehydes. Angew. Chem. Int. Ed. 2003, 42 (35), 4247–4250. DOI: 10.1002/anie.200352097.
Brown, S. P.; Brochu, M. P.; Sinz, C. J.; MacMillan, D. W. C.The Direct and Enantioselective Organocatalytic α-Oxidation of Aldehydes. J. Am. Chem. Soc. 2003, 125 (36), 10808–10809. DOI: 10.1021/ja037096s.
Zuman, P.; Shah, B. Addition, Reduction, and Oxidation Reactions of Nitrosobenzene. Chem. Rev. 1994, 94 (6), 1621–1641. DOI: 10.1021/cr00030a007.

Using proline as the catalyst provides enantioselectivity to the process, but also allows N vs O regioselectivity to be controlled (though not always perfectly).
Houk (and subsequently others) applied computational methods to the reaction in an attempt to understand why the nitrosobenzene was acting as an electrophile at oxygen rather than nitrogen, as would traditionally be expected (and as is often the case when no proline is used). The computational studies showed a clear preference for the transition state in which the basic nitrogen of nitrosobenzene was able to interact with the acidic proton of the proline:

Proline catalysis occurs via a proline enamine attack on the oxygen of the nitrosobenzene monomer with a simultaneous proton transfer from the carboxylic acid. Computational evidence shows that the observed heteroatom selectivity materializes through the difference in basicities of the oxygen and the nitrogen atoms.
Source: Cheong, P. H.-Y.; Houk, K. N. J. Am. Chem. Soc. 2004, 126 (43), 13912–13913. DOI: 10.1021/ja0464746.

This study further indicated that the transition state for attack at N (which you drew) is disfavoured by 2.5 kcal/mol:

Ref: Enantioselective Organocatalysis, 2007, Dalco (Wiley)


Vapour Pressure of A Solution I study in class 12, and have a doubt in vapour pressure that my school teacher  and my friends are not able to clarify.
My doubt is the reason why the partial vapour pressure of each individual component of a solution is not equal to their respective standard vapour pressures($P°$)?
My reasoning is that consider you have a pure liquid A in a closed container, the vapour pressure of it shall be equal to $P°_A$. Now when you add another volatile liquid B, a solution should be formed; however, the equilibrium of the reaction
$$\ce{A_{liq} <=>  A_{vap}}$$ should not be affected since the concentration of reactants and products remains the same since B is different from A. Eventually, the reaction of B must also reach equilibrium, independently, and thus the partial pressure of B  must be equal to $P°_B$ .
However, this does not seem to be the case. Could someone please help in clarifying my doubt through the viewpoint of equilibrium(Mathematically if possible) and not the standard explanation that the surface of the solvent get's blocked by the particles of the solute, hence the vapour pressure reduces.
Regards,
Arun

This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.
Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $p_i=p^*_i\chi_i$, where $p^*_i$ is the partial pressure of the pure component and $\chi_i$ is the mole fraction in solution.
The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $\chi_i^g$ is the mole fraction in the gas phase):
$$ \begin{align} p &= \sum_i p_i = \sum_i \chi_i^g p \\ &= \sum_i \chi_i p^*_i \end{align} $$


Free software to draw Newman projections I need to draw Newman projection of a molecule. Particularly 1,2-dicloroethane rotamers, but have no idea if there is any program to do that kind of things.

Do you know about some program available for Linux?

I refer to this kind of sketch:

You can use GChemPaint, which is available on the repositories of many popular distributions. Among its many components for drawing structures, it has a specific tool for Newman's projections.
You can add H, O and many other elements by simply right-clicking on the molecule, and it even allows you to search your molecule in databases like PubChem and ChemSpider.
A useful video tutorial can be found in spokentutorial.org

3.11 To Add a Newman Projection
To add a Newman projection, use the following tool:

This tool allows to add a Newman projection. The changes become permanent only when the mouse button is released.
To add a Newman projection, click on the background. Contrarily to other tools existing atoms are never used by this tool. It creates new atoms of the current element.
When the Newman projection tool is active, the toolbox looks like:

Options for the main bond

Bond length
The default length of the new projected bond. This has no influence ion the drawing, but might be important when exporting to a 3d model.

Bond order
The bond order.

Options for each atom

Bonds number
Number of other bonds for the atom.

First bond angle
The direction of the first bond. Angles are measured counter clockwise from horizontal right.

Angle between bonds
The angle between two consecutive bonds.

Warning
This version of GChemPaint does not correctly support Newman projections with eclipsed bonds.

Source: GChemPaint manual, pages 41-43.


Why does increasing the concentration of oxidizing agent decrease the E° in a galvanic cell? In a galvanic cell, when we add ions of the oxidation agent (in this example, Zn2+) to the anode solution, it decreases the cell potential.

However, if the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential? I know the Le-Chatelier and Nerst equation arguments, but I wanted a conceptual explanation.

I hypothesized that the higher Zn2+ concentration decreased the anode's affinity for solubilization, therefore decreasing the anode's desire to release electrons, thereby decreasing the cell potential, as less electrons will flow per unit of time.

If the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential?

The overall reaction is
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$

If this reaction were happening in a common reaction mixture (maybe with Zn and Cu powder on the bottom of a solution) we would write it as

$$\ce{Zn(s) + Cu^2+(aq) <=> Zn^2+(aq) + Cu(s)},$$

and would realize that the rate of the reverse reaction depends on the zinc ion concentration, thus affecting the equilibrium. However, the reaction at hand relies on electrons going through a single wire, and they would not go in both directions at the same time.

MaxW's comment is the key to the conundrum. There are reversible half reactions in either cell, and they determine the cell potential.

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

$$\ce{Cu <=> Cu^2+ + 2e-}$$

At each electrode, the half reactions are reversible, delivering electrons to the electrode and picking them up again. This back and forth establishes the potential difference that we can measure between the electrodes.

  I wanted a conceptual explanation

For some, a kinetics-based answer (like the one above) counts as conceptual (or particular as opposed to macroscopic?). Here is a second way of thinking about it, using a different but related system. Let's consider a concentration cell with zinc electrodes and zinc ions in both cells. With higher concentration of zinc ions in the right cell than the left cell, the reaction would be:

$$\ce{Zn(left) + Zn^2+(right) -> Zn^2+(left) + Zn(right)}$$

Using your logic, the potential would not depend on the zinc ion concentration in the left cell because that is one of the products. Now we add some zinc ions (in the form of zinc chloride or such) to the left cell, making its concentration higher than that of the right cell. The reaction now would be:

$$\ce{Zn(left) + Zn^2+(right) <- Zn^2+(left) + Zn(right)}$$

Again using your logic, now the potential would not depend on the zinc ion concentration in the right cell because that is now the product.

The question is how the half reaction on the right would "know" to reverse direction, and to stop influencing the potential via its zinc ion concentration. I'm hoping this reductio ad absurdum (I had to look it up) is a convincing argument that zinc ion concentrations in both cells are relevant.


Which has the greater N–O bond length, NO₂⁻ or NO₃⁻? Which has the greater N–O bond length, $\ce{NO_2^{-}}$ or $\ce{NO^_3^{-}}$?

I don't understand why the N-O bonds should be of different lengths in these two molecules (don't the same bonds have the same length)? I didn't quite know how to determine this and so I've tried googling the answer, but none of the answers already provided helped clear things up for me... Why does $\ce{NO_3^{-}}$ have a greater N-O bond length?

There are two ways to think of it.  First: draw out the (major) contributing Lewis structures for both ions.  Each iin has one pi bond, but it's shared between two linkages in $\ce{NO2^-}$ ion versus three linkages in the $\ce{NO3^-}$ ion.  So the $\ce{NO3^-}$ ion has less pi bonding in each linkage making those bonds weaker overall.

Second: In all of those Lewis structuresthere are nonbinding pairs of electrons on the oxygen atoms, plus a pair on the nitrogen in $\ce{NO2^-}$.  There are more such pairs in $\ce{NO3^-}$ than in $\ce{NO2^-}$.  These pairs tend to repel each other and the bonding pairs, and $\ce{NO3^-}$ requires more room, thus longer bonds, to accommodate the greater number of closely packed nonbonding electron pairs.

These two factors tend to occur together when the molecules have nonnetalluc atoms that fill their votes, as here.


Why is the magnesium(II) ion preferred over other ions in chlorophyll? Chlorophyll has a $\ce{Mg^2+}$ ion. Why is it preferred over other ions? For example, what happens if there is $\ce{Zn^2+}$ or $\ce{Ca^2+}$ or any other (divalent) cation instead of $\ce{Mg^2+}$?

Chlorophyll in plants has two main functions. First, it facilitates energy transfer of an absorbed photon's energy via the many other chlorophylls that constitute the antenna pigment-protein complex to the reaction centre. This means that with $> 95~\%$ efficiency the energy of a  single absorbed photon reaches the reaction centre which consists of a dimer of chlorophyll molecules (the special pair), two pheophytins ($\ce{2H}$ replace $\ce{Mg}$) and two other chlorophylls.  The second role is in electron transfer. Electron transfer begins from the excited special pair and the electron is passed to other nearby chlorophyll and pheophytin molecules and then to quinones/iron sulfur complexes depending on the particular type of reaction centre. The exciton interaction between the two chlorophyll in the special pair means that it can be treated as a single entity. This interaction lowers its excited state energy and so traps the energy arriving from the antenna.

It is possible to make chlorophylls with other metals instead of magnesium, but if these contain heavy or paramagnetic ones they will competitively quench the chlorophyll's excited state (via a spin-orbit coupling or 'heavy atom effect') and reform the ground state. Consequently very little if any energy transfer will occur. This means that insufficient energy will reach the reaction centre and so the efficiency of photosynthesis will be dramatically reduced.

Magnesium chlorophyll has a unique property which is that the absorption and emission spectra overlap exceptionally well meaning that energy transfer to nearby molecules by the Forster (resonance) mechanism is very favourable. As shown in the figure below the energy may move among $\approx 100\ \ce{Chl}$ molecules and reaches the reaction centre in a few ($ < 10\ \mathrm{ps}$) picoseconds. This means that light harvesting antenna have very poor fluorescence yield since most of the absorbed energy reaches the special pair, where is quenched, before a molecule has had a chance to fluoresce because the Chl fluorescence lifetime is $\approx 5\ \mathrm{ns}$. Changing the central metal will change the absorption and emission spectra, most likely in a way to be less effective at energy transfer simply because magnesium chlorophyll has almost the best possible overlap of absorption and emission spectra of any molecules.

The energetics of electron transfer are very delicately balanced. The electron has to move from the excited special pair to the final acceptor (say a quinone) through the steps $\ce{(Chl)_2^* -> Chl -> Phe -> Quinine }$ with as high an efficiency as possible (the whole process takes only $\approx \pu{250ps}$ in bacteria, and ten times faster in plants), but the rate of each back reaction on each step has to be as low as possible. Other metals in Chl will change the redox and even a change of $0.1~\mathrm{V}$ will be important, and so make this process less efficient. The reason for this is that in normal photosynthesis electron transfer is almost at the peak of the Marcus rate vs. free energy curve, thus increasing or decreasing the exothermicity will make the electron transfer less efficient.

All of this illustrates that photosynthesis is, via natural selection, highly adapted, and so small changes can have dramatic effects. Under different conditions, it may be possible for alternative schemes to do the same job.

The figure shows the antenna (light harvesting) chlorophylls in the protein complex of plant photosystem 1 and the reaction centre. (The protein residues have been removed for clarity). The reaction centre is at the centre, shown within the dotted line with the special pair shown edge on. The final electron acceptors are $\ce{FeS}$ complexes but they are not shown and would lie directly below the special pair. The molecules to either side of the special pair are the accessory chlorophylls through which the electron from the special pair travels.  The structure is modified from PDB entry 1JBO.


pH 10 to pH9 NaOH dilution I have a pH 10 NaOH solution. I need to bring down the pH to 9. So I used the $C_1V_1 = C_2V_2$ approach.

$C_1 = \mathrm{0.0001~mol/L}$

$V1$ is the necessary amount to produce 10 L of pH9

$C_2 = \mathrm{0.00001~mol/L}$

$V_2 = \mathrm{10~L}$

\begin{align}
C_1V_1 &= C_2V_2\\
0.0001~\mathrm{mol/L}\cdot V_1 &= 0.00001~\mathrm{mol/L}\cdot 10\mathrm{L}\\
V_1 &= \frac{(0.00001)(10)}{(0.0001)}\\
V_1 &= 1\mathrm{L}\\
\end{align}

I Applied this calculation in reality and used a pH probe and got a pH of 6.5. I think the drop is too sudden in this type of titration. What should I do?

Generally, you don't make pH of $7\pm2$ by dilution alone. The required dilution is so great that it would leave very little of your compound, and then any tiny impurity would affect the result greatly. Think of a droplet on the pH meter probe left from the previous measurement. Distilled water does not help much. All the same, it contains $\ce{CO2}$ from the air, so its pH is usually about 5.5, not 7 as one might expect.

When people want to have certain pH level more or less reliably, they use buffer solutions. Come to think of it, that's what nature itself does, because our blood is a buffer solution as well.


Is xenon electropositive in compounds of xenon? Do you agree with the below passage in my text?

Electronegativity is defined as the ability for something in a covalent framework to attract electrons also in covalent bonds.

Electropositivity is the inverse of electronegativity or the ability of something to donate/release electrons in a covalent framework.

How can a highly oxidized element be very electropositive? A highly oxidized element by definition has a minimal amount of electrons to be donated.

How can a highly oxidized element be electropositive?

An element can be electropositive, the highly oxidized element less so.

There are a number of things that don't sound quite right in the passage you quote.  For example, the Pauling electronegativity for xenon has been assigned (2.6, if you'd like to see the methodology used to caculate the electronegativity of a noble gas, see this earlier answer). Also, depending on the compound, hypervalent (or hypercoordinate) bonding, rather than use of d orbitals, might be used to explain the bonding in some noble gas compounds


Can paper burn without oxygen or air? If you put a paper inside an oven vacuum packed and go beyond $250~\mathrm{^\circ C}$, will it burn? There's no oxidizer. If not, what will happen?

No, the paper will not burn without oxygen being present.

Paper is made primarily of cellulose which is a polymer of glucose.

If you heat paper in a vacuum the cellulose simply decomposes to $\ce{H2O}$, $\ce{CO2}$, $\ce{CO}$ and carbon.  As the paper decomposes it will "char" or turn brown to black as the cellulose polymer degrades.  Here is a link to an abstract describing the process in more detail.


Changing Zinc acetate dihydrate to Zinc acetate For some Chemistry work in the laboratory, I needed Zinc acetate but laboratory only has Zinc acetate dihydrate. Is there any way to convert hydrate form to anhydrous form? Maybe by heating or any other method?
If by heating, how much time and temperature I should give?

Wikipedia is your friend...Zinc Acetate

    Decomposes at 237 °C (459 °F; 510 K)
    (dihydrate loses water at 100 °C)

So heating to 120-130 °C for several hours would remove the water, but not be hot enough to decompose the salt. The dried salt will be hygroscopic of course so store it in a desiccator.


The most acidic hydrogen among ethane, ethene, ethyne and allene Among
$$\ce{CH3-CH3}\qquad\ce{CH2=CH2}\qquad\ce{HC≡CH}\qquad\ce{CH2=C=CH2}$$
I need to find out the one having highest acidic hydrogen.
As per my observations, there are two equivalent resonating structures possible in case of conjugate base of $\ce{CH2=C=CH2}.$ However, the correct answer is given as $\ce{HC≡CH}$ due to inductive effect of sp hybridization of carbon.
But the priority order in determining stability of a conjugate base is:

Aromaticity
Number of equivalent resonating structures
Resonance
Hyperconjugation
Inductive effect

As such, the correct answer should have been the allene. Where did I go wrong?

The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic site.
I think you over resonance, its an easy mistake, but my notes from Orgo I show that the difference in pKa from an sp2 to an sp hybridization are roughly a factor of 2.
TL:DR, Induction beats resonance.


What is the difference between tempering and hardening metals and their end products? I think I've heard hardening is when you dunk red-hot metal into cold water, and tempering is when you take that hardened metal, heat it slightly, and then let it cool slowly. However what is the difference between their end products?

As the names imply, hardening makes the metal more rigid but more brittle, and tempering (from "temperate", moderate), forgoes some hardness for increased toughness. Iron alloys are hardened by rapid quenching and toughened by annealing, but some copper alloys are made more ductile by quenching and are hardened by working.

In brief, annealing allows atoms to migrate to a less-strained position, whereas quenching lock atoms in place; this is used to make tempered glass, as well as to harden metals.

Extremely rapid quenching can produce a metallic glass, where precipitation of constituents is prevented.

Edged-tools may need the body to be ductile, so that it does not shatter, but require for the edge to be hardened so that it retains sharpness. See http://www.smt.sandvik.com/en/products/strip-steel/strip-products/knife-steel/hardening-guide/purpose-of-hardening-and-tempering/ or http://www.stormthecastle.com/blacksmithing/blacksmithing-a-knife/stock-removal-method-of-knifemaking-part-4.htm.


Non-wetting fluids for precise volume measurements I need to precisely measure the internal volume of an open steel fixture.  I was planning to do this via displacement, but I'm wondering what the best fluid would be.  Ideally it should have no surface tension so that it readily penetrates every facet of the fixture.  It would also be convenient if it was non-wetting which (if I understand that term correctly) means that it would not adhere to the steel when removed.

One suggestion I heard was to use water-soluble cutting fluid.  Are there any other readily-available fluids with these characteristics, or well suited to this application?

For a very precise volume measurement, you might want to consider using mercury because of its high density. (You would actually measure the weight of the fluid.)


Bonding and Phase Changes? When water boils into water vapor, its temperature (average kinetic energy) does not change because the inputed energy is used to break its bonds. But how does that work?

Also, in a liquid, aren’t intermolecular bonds broken and reformed constantly anyway? Why does a phase change require a time interval for breaking bonds?

Temperature involves kinetic energy; however, bonds and intermolecular forces (IMF's) involve potential energy.  The potential energy of the system increases when water vaporizes, but kinetic energy does not change.

In a liquid, for each IMF broken, another is made, resulting in a net change of zero.

A phase change requires a time interval because it takes time to add energy to the system.


Geometry optimization: dependence on initial geometry One would think that initial configuration should not significantly impact the final optimized geometry. Starting with two moieties, the final "optimized" geometry is not unique and depends highly on how they are placed initially. Is there a way to avoid this? Otherwise one could never be sure that they have ended up with the global minimum.

Starting with two moieties, the final "optimized" geometry is not
  unique and depends highly on how they are placed initially. Is there a
  way to avoid this?

Yes and no. No, because, as OP anticipated, starting from an input geometry the optimisation algorithm will indeed find the closest (from the algorihtm point of view) local minimum and not necessarily the global one. And yes, because by performing an extensive conformational analysis, one in principle can find structures that correspond to all possible local minima and then identify the global one.


Which substance has the highest temperature range between melting and boiling point Which substances exist that are normally liquid and that will not freeze nor boil even at relatively low/high temperatures as compared to the freezing and boiling points of water? Or how can I identify such substances online?

If it is relatively non-reactive, non-toxic, cheap and has high thermal conductivity that is also a bonus!

Gallium melts at 30 °C but doesn't boil until 2200 °C.  If 30 °C is a bit too warm to count as "room temperature" or "normally" for you, I found an old paper that recommends tetralkyl silanes such as tetradodecyl silane as lubricants that are liquid over very wide temperatures.

Addendum:

Dowtherm A is a eutectic mixture of biphenyl and diphenyl ether.  According to its manufacturer:

  These compounds have practically
  the same vapor pressures, so the
  mixture can be handled as if it were
  a single compound. DOWTHERM A
  fluid may be used in systems employing
  either liquid phase or vapor phase
  heating. Its normal application range
  is 60°F to 750°F (15°C to 400°C),
  and its pressure range is from
  atmospheric to 152.5 psig (10.6 bar).

[...]

  The viscosity of DOWTHERM A
  fluid is low and changes only
  slightly between the melting point
  of the product and its top operating
  temperature.


How does current start to flow when galvanic cell is first connected? Using the Daniell Cell as an example, where in the steady state we expect zinc to be oxidised and copper to be reduced spontaneously due to the overall reduction in Gibbs free energy, how does the system "know" to start moving electrons from the anode to the cathode? What is the underlying physical mechanism?
When the two half cells are first connected via the external circuit, it seems that they would not yet "know" anything about each other; each half cell would not already "know" it is anode or cathode (e.g. zinc could be the cathode in a different galvanic cell) and so how would the system immediately start driving the electrons in the right direction?
In other answers on this site (e.g. https://chemistry.stackexchange.com/a/28638/98580) I've seen statements like "the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell".
However, as a physicist this is somewhat unsatisfying, as the phrase "induce reduction" seems to bundle up a whole lot of physical processes in the half cells and the external circuit into a very high level concept. Fundamentally, it must come down to electrical forces and thermodynamics to explain the overall behaviour of the system.
My understanding is that prior to connecting the external circuit, each electrode is in equilibrium with its electrolyte, undergoing both oxidation and reduction at the same time; so each electrode could both push and pull electrons to/from a connected wire. The moment the external circuit is connected, do electrons start flowing from the zinc electrode to the copper electrode because zinc has a greater "tendency" (than copper) to push electrons onto a connected wire, and copper has a greater "tendency" (than zinc) to pull electrons from a connected wire? That would then sound like the galvanic cell gets started via a statistical process and is sustained by electrical charging of the electrodes thereafter.

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. Furthermore, electrons are swept out of the conductor by the electromagnetic force, which propagates on the surface of the conductor at a large fraction of the speed of light in vacuuo, C. The fraction is the velocity factor, perhaps 0.5 to 0.75 C.
So the answer to your question is not as clear-cut as it might seem. You could say the reaction propagates at the speed of the EMF, C times the velocity factor in the connecting wire. However, one then must take into account the buildup of slow-moving ions, creating a space charge slowing things down.


Do bulky bases participate in substitution reactions? I see how bulky bases influence the regioselectivity of E2 and E1 reactions. I do not see very much content on the involvement of bulky bases in substitution reactions.

I know that bulky bases like t-BuOK are generally not involved in substitution reactions because the t-BuO− ion would prefer to take a β-hydrogen instead. Are any  substitution reactions possible, and if so, what mechanism would be seen in substitution reactions involving bulky bases?

It is very tough, but cases of t-butoxide acting as a nucleophile are known, e.g. J. Org. Chem. 2011, 76 (24), 10129–10139:

  n-Butyl-tert-butylether. To a stirred solution of 10 g of potassium tert-butoxide (89 mmol) in 100 mL of anhydrous dimethylsulfoxide was added 12.1 g of n-butylbromide (89 mmol) dropwise at 10 °C. The reaction mixture was allowed to warm and stand under stirring for 8 h at room temperature. The mixture was distilled under vacuum at −20 °C. The crude was redistilled at normal pressure to collect 6 g (52%) of the fraction distilled at 75−78 °C as a colorless liquid.

Mechanistically, I don't see any other real possibility by which this occurs apart from an SN2 reaction. Of course, there will be significant amounts of competing elimination, and that is probably why the yield is not amazing.


Acid-catalysed isomerisation of phenyl epoxide to aldehyde How does $\ce{H3O+}$ cause the formation of an aldehyde from the following epoxide?

I thought that the epoxide would undergo opening to form a 1,2-diol, as illustrated below. Am I wrong?

You are correct if this epoxide is a alkyl peroxide. However, this is a 2-phenyloxirane (styrene oxide or phenylethylene oxide),  which can stabilize a protonated intermediate by resonance, specifically with 2-hydroxyphenyl group. Thus, it can isomerize to phenylacetaldehyde through 1,2-hydride shift (similar to pinacol-pinacolone rearrangement). The following diagram illustrate how styrene oxide converted to phenylacetaldehyde (Ref.1):

Note that styrene oxide is not stabilized by extra hydroxy group here. That's probably the reason the authors have used slica-supported phosphotungstic heteropoly acid in cyclohexane (no other nucleophile there).

The identical isomerization has been used in the synthesis of 2-phenylnaphthalenes from styrene oxides (Ref.2).

References:

Vinícius V. Costa, Kelly A. da Silva Rocha, Ivan V. Kozhevnikov, Elena V. Gusevskaya, "Isomerization of styrene oxide to phenylacetaldehyde over supported phosphotungstic heteropoly acid," Applied Catalysis A: General 2010, 383(1–2), 217-220 (https://doi.org/10.1016/j.apcata.2010.06.005).
Kishor V. Wagh, Bhalchandra M. Bhanage, "Synthesis of 2-phenylnaphthalenes from styrene oxides using a recyclable Brønsted acidic $\ce{[HNMP]+HSO4−}$ ionic liquid," Green Chemistry 2015, 17(8), 4446-4451 (DOI: 10.1039/C5GC01170A).


Reaction between copper (II)chloride and aluminium foil When I reacted copper chloride and aluminum foil instead of getting the brown color I got green color. Can anyone share their procedure where the experiment actually succeeded?

This is an easy experiment to do at home, so I did it about an hour ago. The first figure shows the starting chemicals and "equipment":

I did not have any copper chloride, so I used copper sulfate pentahydrate, $\ce{ CuSO4.5H2O}$, and $\ce{NaCl}$. (The sodium and sulfate ions are just spectator ions here.)

I added the two chemicals to the $\pu{25 mL}$ of water in the little beaker, stirred until they were fully dissolved, then poured the solution onto the aluminum foil. The second figure shows what happens about every 2 minutes:

The third figure shows what happens around the 10 minute mark: the solution has eaten through the aluminum and the brown copper is evident:

Finally, Fig. 4 shows the result after the solution has leaked away:

So, the redox reaction works as expected and as shown in various videos on the web. Note that copper sulfate pentahydrate crystals are blue, as are aqueous solution of it. The copper chloro complex ion, $\ce{CuCl4^2-}$, is green.

EDIT: Per the comment from @AJKOER, I ran the experiment again, with powdered iron (Fisher, 100 mesh, electrolytic) and copper sulfate solution. The next figure shows the starting materials. I put a strong magnet under the little beaker to prevent iron, which was deliberately in excess, from going over in the decantation that would follow.

Pouring the copper sulfate solution into the beaker resulted in a vigorous reaction and quite a bit of heat. I stirred the reaction mixture and let it go to completion. The magnet still stuck strongly to the bottom of the beaker, indicating that there was still substantial powdered iron remaining.

The mixture was decanted to another beaker, with some mostly useless filtering (no good filter paper at hand), and allowed to settle for about an hour. The copper particles produced in the reaction are very small and settle out very slowly. Some of the supernatant solution was transferred to a sample cell, illuminated from the left via an LED flashlight and photographed. This is shown in the next figure:

Despite the light scattering, the green color of the ferrous sulfate solution is evident. The photo will be updated after more particulate has settled out.


How is it possible to safely use carbon monoxide as a fuel? Is there anything that will allow carbon monoxide to be used as a fuel? It is quite toxic, but has attractive properties otherwise (only produces CO2 with no water, also no smoke or soot). Aside from a permanent spark, or a pilot light, I can't see any way of doing this.

Is there some catalyst that will help the reaction? Apparently, platinum can be used as catalyst between hydrogen and air.

During WWII many cars were powered by a gasogene engine, which transformed wood or charcoal into a synthetic gas composed by $\ce{CO}$, $\ce{N2}$, $\ce{H2}$ and $\ce{CO2}$. Thus the use of $\ce{CO}$ (albeit not as a pure gas) as a fuel is not something new (the former use of piped coal gas for house heating and cooking has already been mentioned above). I am not sure however about how safe those gasogene engines were.


What happens in the NH4Cr2O7 with HgSCN reaction? Apparently this video is a reaction of $\ce{NH4Cr2O7}$ with $\ce{HgSCN}$.

The title of the video suggests the reaction opens a portal to hell, but I am pretty sure that is not the case.

What exactly goes on in the reaction?

I can't comment on the metaphysical aspects ;)

This experiment is known as Pharaoh's snakes, and it is indeed impressive.

Upon heating, mercury thiocyanate decomposes to yield mercury vapour, carbon sulfide, and dicyan (ethanedinitrile):

$$\ce{2Hg(SCN)2 ->[\Delta] 2 Hg + 2CS2 + (CN)2}$$

As a result, the solid starting turns to a foam-like mass.

The effect is amplified due to the presence of ammonium dichromate, which oxidizes the carbon disulfide to sulfur dioxide and carbon dioxide.

The experimenter is apperently aware that quite a lot of rather harmful components are released and therefore has decided to run the show in the fumehood and closes the window immediately after igniting the mixture.


How can off-diagonal Hamiltonian matrix elements be nonzero when the overlap matrix is diagonal? While studying about Hückel theory, I got accustomed to the approximation of making the overlap matrix an identity matrix; that is making the off-diagonal elements zero as $S_{AB}= S_{BA}= 0\;;$ this implies the use of orthogonal base states of AOs. The off-diagonal elements of the Hamiltonian matrix are still taken as constants that may be non-zero: $H_{AB}= H_{BA}= \beta_{AB}$, where $\beta_{AB}$ is a negative quantity.

Then I wondered why it isn't the case that $H_{AB}= H_{BA}= 0$ strictly, as it represents the expectation value - the average energy contribution of the overlapping region of AOs $A$ and $B$.  But, it would seem that overlapping is not actually possible, as is evident from the fact that $S_{AB}$ is zero.

In this question, when I asked about this, ifilot replied:

  Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij}=δ_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle ϕ_i|\hat H|ϕ_j\rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a non-zero outcome.

I know he is right in this point; but I'm still having trouble seeing how this is possible.

As Peter Atkins in his book Elements of Physical Chemistry wrote:

  [...] The integral $H_{AB}$ depends on both $\psi_A$ and $\psi_B$, and we can interpret it as the contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

Evidently, this phrase makes clear that $H_{AB}$ is non-zero iff there is overlap between the AOs.

So, how can $H_{AB}\ne 0$ and $S_{AB}= 0$ both be true at the same time? And, what are the physical implications? While the former means there is overlap, the latter means the opposite; it seems really contradictory.

I believe the confusion lies in some ambiguity in the use of the term overlap.

For the $S_{AB}=0$ approximation of Hückel theory, overlap is used in the specific mathematical sense of the "overlap integral":

$$
S_{AB} = \left<\varphi_A|\varphi_B\right> =
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\varphi_B\!\left(\vec r\right)\mathrm{d}\vec r}
$$

However, this does not mean that at each point in $\mathbb{R}^3$ only one of $\varphi_A$ or $\varphi_B$ is allowed to bear a non-zero value.  In fact, at most points in $\mathbb R^3$, $\varphi_A$ and $\varphi_B$ will both bear non-zero values. Since $\varphi_A$ and $\varphi_B$ will in general have "co-located regions of non-zero values," the two functions can be described qualitatively as overlapping, in the same sense as in your second quote. It is the overlap of the functions in this more general, qualitative sense, that allows $H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>\neq0$.

(To note,  in order for Hückel's approximation to be exact, the orbitals would have to be selected such that the specific integral $S_{AB}$ over all $\mathbb R^3$ equals zero$^\ddagger$. I believe Hückel did not actually do this, but instead just coerced to zero any nonzero $S_{AB}$.)

To help explain more explicitly why this is, I'll first write out the integral implied by the Dirac notation for $H_{AB}$:

$$
H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>=
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\hat H\!\left[\varphi_B\!\left(\vec r\right)\right]\mathrm{d}\vec r}
$$

Then, I'll introduce two (real-valued) one-dimensional functions defined on the closed domain $x=[0,1]$ to use for demonstration purposes:

$$
\varphi_A\!\left(x\right) = \sin\!\left(2\pi x\right) \\
\varphi_B\!\left(x\right) = \sin\!\left(3\pi x\right)
$$

$\hspace{10mm}$

It is straightforward to show that $S_{AB}=0$ for these two functions.  One way is to plug the overlap integral into Wolfram Alpha. Another option is to do it by hand.  In the below figure, the green curve is a plot of the product $\varphi_A\!\left(x\right)\cdot\varphi_B\!\left(x\right)$, and the orange curve is a plot of the cumulative integral of this product:

$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\varphi_B\!\left(x'\right)\mathrm{d}x'}
$$

$\hspace{10mm}$

As can be seen, the integral over the entire domain is zero, satisfying $S_{AB}=0$.

Now, let's introduce a simple "Hamiltonian"$^\dagger$:

$$
\hat H = x
$$

The next figure is a replicate of the first one above, with $\hat H\!\left[\varphi_B\right]$ also plotted as the dashed red curve:

$\hspace{10mm}$

As in the second figure above, one last plot shows the local value of $\varphi_A\!\left(x\right)\hat H\!\left[\varphi_B\!\left(x\right)\right]$ along with the cumulative integral:

$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\left[x'\cdot\varphi_B\!\left(x'\right)\right]\mathrm{d}x'}
$$

$\hspace{10mm}$

The value of $H_{AB}$ here is clearly not equal to zero.  (Wolfram Alpha concurs.) Thus, it is absolutely possible to have $H_{AB} \neq S_{AB} = 0$.

To note, the principle exhibited here on a closed domain in $\mathbb R^1$ applies equally well to orbitals on an infinite domain in $\mathbb R^3$. Note also, if I were actually performing a quantum mechanical calculation here, I would have been a lot more careful about normalizing my wavefunctions (viz., ensuring $S_{ii}=1$).

$^\ddagger$Numerous methods are known for generating orthogonalized orbitals with $S_{AB}=0$ strictly, from a general set of orbitals where $S_{AB}\neq0$ occurs for some non-null subset of the $\left(\!A,\!B\right)$. To my knowledge, the most commonly used orthogonalization for atomic orbitals is Löwdin's method (see this PDF). The Gram-Schmidt orthogonalization (usually in its "modified" form) is used in a variety of other areas of quantum chemistry. Wikipedia also has a pretty solid article on orthogonalization methods in general.

$^\dagger$You'd never use this as a Hamiltonian operator. Further, I'm not sure if it technically even is a Hamiltonian operator. (I think it'd work as a dipole moment operator?) But, whatever, leave me alone, it's for purposes of illustration.


Interaction of the oscillating magnetic field of electromagnetic radiation with a molecule or atom Whenever we read rotational, vibrational and electronic spectroscopy, we find details on the interaction of the electric dipole moment of the molecule with the oscillating electric field of electromagnetic radiation. The selection rules are also derived on the basis of transition dipole moment operator on wavefunctions. I was talking to spectroscopists today and asked that why do spectroscopy textbooks (and authors) do not mention anything about oscillating magnetic field of light waves. The magnetic field also carries the same energy as the oscillating electric field. Doesn't that interact with electrons as well?* They said that interaction of the magnetic field with a molecule's electric dipole is very weak and hence nobody mentions about it. Does anyone have a good reference or an equation which shows that the oscillating magnetic field interacts very weakly with the molecule's electric dipole moment? They were mentioning Fermi's golden rule and at this stage I lost them.

__
* Electron spin resonance is an example.

The magnetic dipole transition intensity is typically about $10^5$ times smaller than the allowed electric dipole but as this can vary by ten orders of magnitude, second order transitions (with terms $x^2,y^2$ etc ) as well as magnetic dipole transitions can occasionally be important. The electric dipole matrix element  is $\langle a|\hat r|b\rangle$, where $\hat \mu =e\hat r$, the  second order dipole $\langle a|\hat r \hat r|b\rangle$ and magnetic dipole $\langle a|\hat J|b\rangle$ where $\hat J$ is the angular momentum operator
$\displaystyle \sim \langle a|z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} |b\rangle $.
The transition intensity is the square of these terms and the selection rules differ for each type of transition. A derivation is given over several pages by Steinfeld 'Molecules and Radiation' chapter 1, 2nd ed (publ MIT Press).


What is the structure of the Nucleus? The structure of an atom is well known to me as something like this:

But what is the accurate structure of the Nucleus ? What is the arrangement of the Protons and the Neutrons or any of the other nucleons? I will be glad if anybody give an image of its structure.
Thanks!

The nucleus is the very dense region consisting of protons and neutrons at the center of an atom.
The diameter of the nucleus is in the range of 1.75 fm ($1.75×10^{−15}$ m) for hydrogen (the diameter of a single proton)2 to about 15 fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about 23,000 (uranium) to about 145,000 (hydrogen)

Source:Wikipedia


Software for generating a lattice from a unit cell for molecular crystals? I am looking for a software package which can take a unit cell (and lattice parameters, etc.) and from that generate a crystal lattice. Ideally it will have functionality for orthorhombic, hexagonal, and other types of unit cells which are slightly less trivial than a cubic lattice.

Finally, it should work for molecular crystals.

I have seen the ASE lattice class (https://wiki.fysik.dtu.dk/ase/ase/lattice.html), but it isn't so clear to me how to define a lattice for a system when the unit cell is quite large.

As an example, I have a unit cell for a clathrate hydrate which contains 68 water molecules and I want to generate a lattice from this unit cell. How would one usually go about doing this?

My guess is that one way around this is to use some kind of MD package and just give the unit cell along with the lattice parameters and the type of lattice. Then, after applying periodic boundary conditions, somehow get the package to print out the coordinates of atoms beyond just the original unit cell. This feels a bit hacky to me, but if this is normal and someone knows a relatively simple way to do this, then let me know. Answers involving either open-source or purchasable software are acceptable, although open-source is always preferred :)

As an aside, it seems to me that it may not be too hard to just write code to do this for my specific cases, so if anyone has a good reference for generating the lattice given the type of unit cell, lattice parameters, and a set of Cartesian coordinates, that would also be helpful.

In addition to the presentations about CCDC Mercury and Vesta by @andselisk here and later mine about Avogadro could be extended.  As described earlier, Avogadro's capability to read .cif files about crystallographic models allows you to extend these  via the slab builder to any extent you need (GUI entry Crystallography -> Build -> Slab) and to export them either as normal .xyz file, or (via the extension tab, template assisted) in one of multiple formats recognized by quantum chemical programs (e.g., Gaussian, GAMESS-UK, MOPAC).

It equally is possible to start from scratch with a cubic unit cell of 3 Å.  Subsequently, you would alter the lattice parameters ($a, b, c; \alpha, \beta, \gamma$) and enter atoms one, after the other, either in cartesian coordinates (either Angstrom, Bohr, nm or pm), or in fractional atomic coordinates as typically in .cif files. You may wrap the atoms into the cell, and of course impose one of the space group symmetries to fill your cell, too.  You may symmetrize your constructed unit cell and reset (within tolerances you determine) into a standard setting, too. Equally than just reading a .cif,.pdb, or .xyz, these constructed unit cells may be extended with the slab builder, and equally may be exported in one of the computer chemistry formats mentioned earlier, too.


What's the reaction type of the Hydrogen shift? Rearrangement of cycloheptatrienyl carbene yields heptafulvene via a hydrogen shift.

I see that the reaction is favorable because there is a conjugative effect with the $\pi$ system of the ring in the product.

Does this rearrangement happen by a radical TS or is the mechanism more of a sigmatropic rearrangement?

My arguments for

radical TS: The electron of the carbene (let's assume a singlet carbene here) lies close to the hydrogen, it could be a planar cyclic transition state.
[1,2]-Hydrogen shift: From my chemical intuition I think the $\ce{C-H}$ bond is more a polar bond and therefore I would hydrogen describe more as a proton (positive partial charge). I would say it's a [1,2] sigmatropic rearrangement, 2 electrons are involved, and therefore suprafacial and describing the carbene $\mathrm{sp^2}$ as HOMO and the hydrogen as LUMO.

What's the correct mechanism for this rearrangement?

This reaction can be viewed as a 1,2-sigmatropic hydrogen shift or as a 2+1 (if we are counting atoms, 2+0 if we are counting electrons) cycloaddition (adding the carbene's empty p-orbital across the $\ce{C-H}$ bond - analogous to its addition to an olefinic pi bond to produce a cyclopropane). However the reaction is most commonly viewed as and referred to as an "insertion" reaction because the carbenic carbon is inserting itself in between the carbon and hydrogen atoms that comprised the $\ce{C-H}$ bond.

(image source)

Here's what is known.

Thermal decomposition of most diazo compounds produce singlet carbenes.

Singlet carbenes add stereopecifically to olefins.  That is, addition of a singlet methylene carbene to trans-2-butene produces only the corresponding trans-dimethylcyclopropane

(image source)

Further, insertion of singlet carbenes into stereogenic centers results in retention of stereochemistry at the chiral center.  These facts suggest that both the cycloaddition and insertion reactions of simple singlet carbenes either proceed in a concerted fashion, or if any intermediate is formed it is extremely shorted-lived in order that stereochemical information be preserved.

Simple carbenes are typically mildly electrophilic.  The HOMO (the filled $\ce{sp^2}$ orbitals) and LUMO (the empty p-orbital) for a singlet carbene are shown on the left side of the following figure.

(image source)

Therefore, we would expect the interaction of the carbene LUMO (the empty p-orbital) and the $\ce{C-H}$ bond HOMO to be controlling, as the energy separation here is much less than the other possible HOMO-LUMO interaction between the carbene HOMO (the filled $\ce{sp^2}$ orbital) and the much higher in energy $\ce{C-H}$ antibonding orbital, the LUMO.

This controlling HOMO-LUMO interaction involving the empty carbenic p-orbital in the rate determing step is consistent with the observed electrophilic nature of the carbene.

All of this information taken together suggests that interaction of the empty carbenic p-orbital with the electrons in the $\ce{C-H}$ bond is the critical step.  A geometry like that shown below is likely preferred.  As the insertion proceeds, the trigonal carbenic carbon rotates.  The rotation starts to bring the carbenic $\ce{sp^2}$ orbital into alignment with the p-orbitals in the 7-membered ring.  Also during the rotation the $\ce{sp^2}$ orbital starts its rehybridization into a p-orbital. When completed we have created the final product, heptafulvene.  The reaction is likely concerted and certainly does not involve any long-lived intermediates. Note too that the heptafulvene is formed with increased electron density on the former carbenic carbon in accord with the dipolar nature (dipole moment ~0.5 D due to resonance stabilization) of the molecule.


reaction with Barium Chromate, Potassium nitrate I'm confused, in the reaction:$$\ce{K_2CrO_{4(aq)} + Ba(NO_3)_{2(aq)}\rightleftharpoons BaCrO_{4(s)} + 2KNO_{3(aq)}}$$
I got a white precipitate in an opaque yellow solution and I was told that the white precipitate was $\ce{BaCrO_4}$ but while searching on Internet I found that its color is yellow, and that the $\ce{KNO_{3(aq)}}$ is white, so my question is what is true? or what causes the color of the final solution?

If you used distilled water to prepare the solutions of potassium chromate and barium nitrate, it is more likely that you got a yellow precipitate of $\ce{BaCrO4}$ in a yellow solution of excess $\ce{K2CrO4}$. The yellow color originates from $\ce{CrO4^{2-}}$, and a small amount of finely divided precipitate may look white at first sight (namely when the yellow color of the solution is more intense), when it is actually pale-yellow (it can look like this).