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Would tin (II) sulfide be considered a covalent network solid?
Considering that tin has a Pauling electronegativity of 1.96 and sulphur 2.58, and that a bond is considered to be ionic with a Pauling EN difference of approx. 1.7 at the least, would tin (II) sulfide be considered covalent?
This would be astonishing, since you would expect a metal—non-metal bond to be ionic. Indeed, many (non-academic) websites (such as this one) state that SnS is ionic. Evidence for the ionic character of tin (II) sulfide is the fact that it precipitates from solutions containing Sn(II) ions when adding hydrogen sulphide (which forms S(-II) ions) – thus, the two kinds of ions form a ionic compound (right?). Furthermore, SnS dissolves in hydrochloric acid, while typical covalently bonded networks such as silicon dioxide do not.
Finally, this paper states on the first page (which is freely accessible) that evidence for tin(II) salts with typical symmetrical ionic symmetries is completely lacking.
Thanks in advance, and happy holidays!

"expect a metal—non-metal bond to be ionic" - no one should make such assumptions, it looks is very common myth fuelled by calling such compounds salts.

http://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently

This would be astonishing, since you would expect a metal—non-metal bond to be ionic. Indeed, many (non-academic) websites (such as this one) state that SnS is ionic.
Considering that tin has a Pauling electronegativity of 1.96 and sulphur 2.58, and that a bond is considered to be ionic with a Pauling EN difference of approx. 1.7 at the least, would tin (II) sulfide be considered covalent?

Yes tin (ii) sulfide is covalent and that website is rittled with non-ionic bonds. Percent ionic character can be estimated using equation:
$$\mathrm{\% I.C. = 1-exp[ -0.25(X_A-X_B)^2]}$$
$X_A-X_B > 1.7$ suggests (there are deviations) that percent ionic character is greater than $50\%$ and substance can be called ionic. Otherwise bonds are probably more metallic or covalent. For the electronegativity values of tin (ii) sulfide you get an percent ionic character of $9.16\%$.
$$\mathrm{\% I.C. = 1-exp[ -0.25(1.96-2.58)^2] = 9.16\% \ \ I.C.}$$
This means that tin (ii) sulfide bonds are about 90% covalent - compound is predominately covalently bonded.

Evidence for the ionic character of tin (II) sulfide is the fact that it precipitates from solutions containing Sn(II) ions when adding hydrogen sulphide (which forms S(-II) ions) – thus, the two kinds of ions form a ionic compound (right?).

Precipitating from soluble salts has nothing to do with the ionic nature and if anything would suggest the opposite as ionic substances typically dissolve well in water.

Furthermore, SnS dissolves in hydrochloric acid, while typical covalently bonded networks such as silicon dioxide do not.

Tin (ii) sulfide is not actually soluble in hydrochloric acid, but dissolves in it much like many metals, forming tin (ii) chloride and hydrogen sulfide which are both soluble. Silicon dioxide is oxophilic and much prefers bonding to the oxygen over the chlorine, and thus will not dissolve.
One thing to note is that tin is a metal by a narrow margin and in fact the $\beta$-tin is non-metallic and stable below $13.6^\circ C$.

There is no sharp border between ionic and covalent compounds, much like there is no sharp border between a boy and a man. Some compounds between a metal and a non-metal, like $\ce{LiF}$ or $\ce{NaCl}$, are predominantly ionic, some (think of $\ce{Mn2O7}$ or $\ce{TiCl4}$) have markedly covalent character, and the rest are somewhere in between. As you see yourself, $\ce{SnS}$ is ionic enough to have certain kinds of reactions typical of ionic compounds, but not so ionic in terms of electronegativity or crystal structure.
Now, people in some areas of chemistry would put forth some definitions to tell covalent from ionic. Your "difference of electronegativities of at least 1.7" is one of those. Another is based on $\ce{MgO}$ and would use it as a border. There are a lot of definitions, and it's not like they are all wrong or useless; in fact, they are necessary to keep our stuff in order. One has to remember, though, that they are all slightly different, and all inherently arbitrary (much like the legal age limitations for driving, voting, drinking, and other activities, if we'd talk once more in terms of the "boy or man" analogy).
This must have been said here like a dozen times before, but whatever. Repetition is the mother of learning, they say.


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Citric acid chelates calcium carbonate more than ethanoic acid, why does this result in a faster rate of reaction?
I performed an experiment using egg shells (calcium carbonate) to represent teeth and reacted the calcium carbonate with acids. When measuring the rate of reaction citric acid had a higher rate of reaction and from my research I saw that citric acid forms a more stable compound with calcium hence it chelates the calcium more than ethanoic acid.
In the mouth this would remove calcium ions from
$$\ce{Ca10(PO4)6(OH)2(s) + 2H+ (aq) <=> 3Ca3(PO4)2(s) + Ca^2+(aq) +2H2O(l)}$$
that equation hence the position of equilibrium shifts to the right so more tooth decay.
But with egg shells:
$$\ce{CaCO3(s) + 2H+ (aq) -> Ca^2+(aq) +CO2(g) +H2O(l)}$$
there is no equilibrium so if more calcium ions are chelated why does this increase the rate of reaction?

Well, it's also more acidic.

thanks but in terms of chelating does the chelation not have any effect?

Acetic acid has a $\mathrm{p}K_\mathrm{a}$ value of $4.76$ according to Wikipedia.
Citric acid has a $\mathrm{p}K_\mathrm{a,1}$ value of $3.13$ and a $\mathrm{p}K_\mathrm{a,2}$ value of $4.76$, also according to Wikipedia. It is about ten times as acidic as acetic acid is.
The greater acidity translates to a lower $\mathrm{pH}$ of the resulting solution. The lower $\mathrm{pH}$ translates to bases being protonated more and quicker. Considering that the decomposition reaction of carbonate is actually:
$$\ce{HCO3- + H+ <=> H2CO3 <=>> H2O + CO2 ^ }$$
The $\mathrm{pH}$ effect alone should be enough to explain a faster reactivity.
However, you can also invoke the following equilibrium which will be right-side shifted with citric acid (but obviously not so much with acetic acid):
$$\ce{CaCO3 (s) + cit^3- <=> Ca^2+ + CO3^2- + cit^3- <=>> [Ca(cit)]- + CO3^2-}$$
Where $\ce{cit}$ is citrate or the anion of citric acid. Complexation of calcium will remove free calcium ions from solution, and the solubility product only considers free ions in solution. Thus by complexing the free ions, more calcium can be dissolved.
I would still perform control experiments to make sure it is actually the complexation effect of citrate and not its acidity. For example, try reaction $\ce{CaCO3}$ with acetic acid in the presence of $\ce{Na3cit}$ or $\ce{Na2Hcit}$. These should be equally good complexing agents.

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Partial and total concentration in solutions

The total pressure of gas mixture equals the sum of partial pressures of its components.

Can this rule be true for liquid concentration?
I.e. can we say: the total concentration of mixture is equal to the sum of partial concentrations of its components?

It's true and false.
For example:
1 Solution with 1mol/L of NaCl and 1mol/L of KBr.
The total concentration of salts is 2mol/L.
But the ions concentration are:
Na+ 1mol/L
K+ 1mol/L
Cl- 1mol/L
Br- 1mol/L
Is it clear?

Thanks for your answer ,but can we say the total concentration of the reactants equals the total concentration of products ,

Can you give me an example eg a reaction?

@OmarAlBahra You can always say that the total concentration of reactant atoms equals the total concentration of product atoms. You cannot always say that the total concentration of reactant species equals the total concentration of product species: sometimes this is true, but often it is not.

Thanks @Brian ,like this reaction : ch3cooc2h5 +h2o give ch3cooh +c2h5oh

@OmarAlBahra Yes, that is a reaction where, technically, the total concentrations of reactants and products are equal, because the number of moles of reactants equals the number of moles of products.  However, water as a reactant/product is a unique case, since its concentration is so high the change due to reaction is negligible. So, for practical experimental purposes, the total concentration of products would be considered greater than that of reactants.

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Methods of dissolving a gas into a liquid
I know that carbon dioxide is dissolved into water in a soda fountain using a sort of a tubular manifold that mixes the water with pressurized carbon dioxide as it goes through it.
Is that the best way to dissolve a gas like carbon dioxide into cold (refrigerated) water, or are there better methods? Is there an official name for this kind of device, "aspirator," maybe? How is this done in the lab?

Have you heard of Condensation?


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In silanization, how to clean kerosene from glass surface completely?
I'm doing some silanization experiments using kerosene to dilute the silan solution. For instance I use a mixture of 10% silan in kerosene. However, after several tests I realized that I cannot completely clean the surface from the remaining kerosene after silanization, and in fact the contact angle that I read is more related to kerosene rather than silan. After silanization, I wash the glass surface with a stream of water and then put the surface in a bottle of acetone and then shake the bottle for 20 sec to clean the kerosene, but apparently it is not enough.
What is the most effective way to get rid of the remaining kerosene on the glass surface?

The best way would be to use some other anhydrous solvent rather than kerosene which is an organic goo from a chemist's point of view.

How about a low-boiling solvent such as pure  hexane, which could be baked off?

@DrMoishePippik I don't have hexane in the lab, but you think Acetone is not enough to clean the kerosene?

@MaxW From this problem, that kerosene cannot easily be removed from glass surface, I'm wondering whether it's possible for me to use a kerosene coating for my glass particles (which gives me a contact angle~45 instead of silanization)? I want to shoot each glass particle by a particle launcher. So, my particles are not too much in contact with water so that the kerosene could be removed. Do you think it works?

I'm baffled as to why you'd want some sort of atomic layer coating on a glass projectile.

@MaxW I'm doing an investigation on collision of a particle and a droplet and now I need to consider the effect of particle wettability on collision outcome. My particles are very small (d=2mm) and that's why I've chosen glass because I think that I can create different wettabilities on its surface. I appreciate if let me know in case you have any better idea.

What is the velocity of the collision?

@MaxW Please see this question


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What is the difference between a reaction of Lewis acids with Lewis bases and a redox reaction?
For example I came across this equation:
$\ce{2Al_{(s)} + 6HCl_{(aq)}} \longrightarrow \ce{2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_2_{(g)}}$
And thought this was a redox reaction but my book says it shows basic nature of aluminium?

Ohh I apologise, i meant the basic character of aluminium. I ll edit it right now

It's indeed redox - Lewis theory doesn't account for redox, but on grounds of Usanovich's theory Al as reductor would indeed be basic.

It's indeed redox - Lewis theory doesn't account for redox, but on grounds of Usanovich's theory ...

A Lewis acid is an electron pair acceptor; a Lewis base is an electron pair donor. The simplest Lewis acid is the hydrogen proton. H$^{+}$ has no electrons by itself. It accepts electron pairs from bases in Lewis acid-base reactions. Here's an example of a Lewis acid-base reaction; note how a pair of electrons from the Lewis base attacks the Lewis acid.

Redox - a portmanteau of reduction and oxidation - involves the transfer of electrons. Hey, there are electrons being "transferred" above, right?
At first glance, it seems that all Lewis acid-base reactions would be redox reactions. However, there is an artificial line in the sand that one would do well to heed.
Redox reactions, as taught at the general chemistry level, involves changes in formal oxidation state. What's an oxidation state? It's the hypothetical charge that an atom would have if all bonding were ionic (as opposed to covalent, or electron-sharing). This means that the more electronegative atom in a bond would have complete possession of the bonding electrons.
If we go back to the reaction pictured above, then we see that electrons are not being "transferred" in the formal sense. Hydrogen is of much lower electronegativity than oxygen; hence, in a formal sense, it has gained no electrons - remember, when assigning formal oxidation states, we take all bonding as ionic.
Obviously, in reality, no bond is 100% ionic, but this is exactly the artificial line in the sand I was talking about.

For example I came across this equation:
  $\ce{2Al_{(s)} + 6HCl_{(aq)}} \longrightarrow \ce{2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_2_{(g)}}$
  And thought this was a redox reaction but my book says it shows basic nature of aluminium?

This indeed is a redox reaction, as aluminum is oxidized and hydrogen is reduced as we go from left to right.
It also shows that aluminum can react with hydrochloric acid to make hydrogen gas; this removes hydrogen protons from the solution and thereby makes the system more basic.
This is not an acid-base reaction in the Lewis sense, as electron pairs aren't being transferred. It's not an acid-base reaction in the Bronsted sense either, as hydrogen protons are not being moved around.

On grounds of Usanovich's theory Al as reductor would indeed be basic.

Usanovich defines an acid that accepts negative charge; a base as something that accepts positive charge. Therefore, Al is a base here; it becomes positive from left to right. This theory is an even more generalized view of acids and bases than Lewis acid-base theory.


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What are the reaction products of sulfur and potassium nitrate?
YouTube shows videos of dropping a piece of elemental sulfur into molten potassium nitrate and applying additional heat to accelerate the process. A white precipitate falls out to the bottom. There is never any mention as to what this material is. An internet search yields not a clue. Sulfur dioxide is obviously evolved, but KNO doesn't seem to exist if one relies on an internet search.

There are at least two videos (first and second) showing the reaction between sulfur and potassium nitrate. For the chemicals:
\begin{array}{cc}
\rm Species&\rm Melting~point\\\hline
\rm Sulfur &  \rm 115~^\circ C     \\
\rm Potassium~nitrate & \rm 334~^\circ C
\end{array}
It isn't clear if nitrogen, some nitrogen oxides, or both would be produced. My guess is that it would depend on particle size, mixing and so forth, thus that a mixture of nitrogen oxides would be very likely. Also the test tube starts with a tiny amount of oxygen gas from the air. One possible balanced reaction would be:
$$\ce{4KNO3 + 5S -> 2K2O + 5SO2_{(g)} + 2N2_{(g)}}$$
Looking at the products of black powder burning $\ce{K2SO4}$ and $\ce{K2S}$ can also be formed.

I suspect $\ce{K2SO4}$ is also possible.

Excuse me, I have been away from chemistry a sinfully long time, but how, in an environment "hot" enough to allow reactions, could $\ce{K_2O}$ and $\ce{SO_2}$ coexist whithout reacting?

@GyroGearloose - the notion is that the gases produced blow bits of sulfur or partially reaction products onto the walls of the closed tube so that there is an incomplete reaction. This isn't suggesting that most of the products are from incomplete combustion, just some.

When potassium nitrate is heated to its melting point, oxygen gas is evolved and potassium nitrite is formed. Any sulphur present would be rapidly oxidised to sulphur dioxide and possibly some sulphur trioxide. The residual nitrite could further react with excess sulphur to produce potassium derivatives  such as potassium sulphide and further sulphur dioxide.


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Predicting color of precipitate
Is there any way of predicting whether a precipitate will be colored? I know that for solutions, transition metal ions with unfilled d-orbitals will have color because of d-orbital splitting in the complex ions. However, this isn't applicable to precipitates because those are just ionic solids, not complex ions.
For example, $\ce{PbI2}$ is a yellow solid, and $\ce{Al(OH)3}$ is white. Is there a way to predict that $\ce{PbI2}$ will be colored while $\ce{Al(OH)3}$ will not be, or is it something we have to memorize to know?

For the most part you just have to know. You also have to realize that trace impurities can change the color. It also isn't simple since the perceived color also depends on particle size. There is the streak test for minerals which typically produces a different color than the bulk mineral specimen. Wikipedia link to "Streak test" -> https://en.wikipedia.org/wiki/Streak_%28mineralogy%29

Many examples need to be memorized.  However many substances have similar colours to their solutions because they are hydrated crystals, e.g. copper(II) sulphate is a blue precipitate but anhydrous copper(II) sulphate is colourless.  Since Al(3+) ions have no partially filled d-orbitals you'd reasonably expect it to have no colour, as most (all?) inorganic colour still involves d-orbital transitions.

There are a few wrong assumptions in your question that I would like to clear up.

Yes, d-orbital splitting is part of the reason for colour. However, $\ce{[FeF6]^3-}$ displays a beautiful d-orbital splitting which could display colour from the energy difference — but doesn’t. There is more to colour than d-orbital splitting; the most important terms are spin-forbidden and Laporte rules.
You somehow got the impression that ionic solids have nothing to do with coordination complexes. Please remove this idea from your brain; it is wrong. Ionic solids are equally coordination complexes as present in solution only that the environment and sometimes the composition of the complex is different. E.g, $\ce{CuSO4(aq)}$ is essentially a solvated sulphate ion plus $\ce{[Cu(H2O)6]^2+}$. When precipitated or crystallised as ‘$\ce{CuSO4 . 5H2O}$’, what you actually got was $\ce{[Cu(H2O)4(\mu{-}SO4)2].H2O}$ — clearly a coordination compound.

To answer your question, it depends on the nature of the compound you are precipitating — the colour can derive from cation, anion or their combination. For example, aluminium salts will be colourless unless the anion dictates otherwise. Likewise for zinc salts. For a very large number of salts where the cation is responsible for the colour, these will be the same as in solution. But sometimes ($\ce{PbI2, Fe4[Fe(CN)6]3}$ and others) you just have to know because only this specific combination will give this colour.


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Activating effects of amino and hydroxyl groups in different pH

How do the activating effects of substituents on an aromatic ring depend on the medium? Explain using the two compounds shown below.

I know that different groups on benzene ring can either deactivate (e.g. $\ce{-NO2}$) or activate (e.g. $\ce{-OH}$) the benzene ring towards attack by electrophiles. How does this activating or deactivating effect change in acidic and basic media?

@pcforgeek I'd think about the two compounds and whether the structures would change in acidic or basic media (e.g., do one or both functional groups gain or lose protons?)

Both the amino and hydroxyl groups activate benzene rings, as the lone pairs on N or O are in conjugation with the benzene π system.
However, in acidic media, the amino group is protonated to form an ammonium cation. Now, the lone pair on nitrogen is no longer available: it has been used to make a new N–H σ bond.

Consequently, the $\ce{-NH3+}$ group no longer acts as an electron donor via resonance. Instead, it is an electron-withdrawing group via the inductive effect (neutral amines are, too, but the resonance donation far outweighs the inductive withdrawal). For further details, you may wish to read this question: Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?

Phenols, on the other hand, can be deprotonated in basic media to form phenolates. Perhaps unsurprisingly, this has the opposite effect from before: the negatively charged substituent $\ce{-O^-}$ is even more electron-donating than the neutral hydroxyl group.

It's tempting to think that this is because "there are more lone pairs on oxygen", but that's not the truth. In order to have maximal overlap between a lone pair and a π system, they must be parallel to each other: on this oxygen, there can only be one lone pair which is parallel to the benzene π system. So, whether a substituent has two or three lone pairs doesn't make all that much of a difference.
The explanation is, instead, to do with the availability of the lone pair. We could think of the neutral phenol as being the same as the phenolate, but with an extra proton which is pulling one lone pair on oxygen towards it. This stabilises the lone pairs, and therefore makes them "less willing" - in a sense - to be delocalised into the benzene ring. (Technically speaking, the lower energy of the lone pair makes for a poorer energy match with the benzene ring LUMO.) In the deprotonated phenolate, though, there are three lone pairs around oxygen which repel each other. This means that they are higher in energy, and are more delocalised into the benzene ring.
As a result, phenolate reacts with very poor electrophiles such as $\ce{CO2}$ (Kolbe–Schmitt reaction) quite well.


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How many electrons will be shared?
Suppose we are to make a bond drawing of a covalent bond between two nitrogen atoms. How would we know how many electrons are shared in each (3)? Do we just logically look at it and see how many electrons are in each and figure out what we need to do to make it stable? I have two Lewis dots of each … But now what? How do I know which ones are shared?

With 2 electrons shared each, that leaves 1 pair and 1 single electron each, while each atom has only 7 electrons in the valence shell.  Unpaired electrons are very unfavourable, and 8 valence electrons is a stability bonus, so

Perhaps this short YouTube video can be of help to you: https://www.youtube.com/watch?v=FImG-s_-xts

By looking at the periodic table, you can see that Nitrogen is group 5 and therefore must have 5 valence electrons. So for 2 nitrogen atoms there would be 10 valence electrons altogether.
By the octet rule, there must be 8 electrons for each atom.

So right now in the image above, each nitrogen atom is bonded to 2 valence electrons(bond is equal to 2 electrons being paired )

Now I have added the 6 valence electrons to the nitrogen atom on the right so that it complies with the octet rule of having 8 valence electrons. If you were to count the number of valence electrons overall it would be 8. But we should have 10 valence electrons. So, I have added this onto the nitrogen atom on the left in the image below.
Although the nitrogen atom on the right has 8 valence electrons and complies with the octet rule, the nitrogen atom on the left does not as it only contains 4 valence electrons. To solve this problem, you would want to share more valence electrons with the left nitrogen atom. This would mean converting 2 valence electrons into a bond. THis has been illustrate in the next two images.

If you were to now count the number of valence electrons in each nitrogen atom, they would both equal 8. If you were to do some research on formal charge for the most ideal shape, this would be 0 and so is the ideal structure.


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Calcium reactivity vs electronic configuration
Calcium is an alkaline earth metal, so it is reactive. But, it has two valence electrons. Don't those two electrons fill up the first energy level? Isn't an atom with a full energy level considered stable? Why is it reactive then?

It also has 18 core electrons.

It's not a full level. It's but an s-sublevel.

The concept of stability is significant where exchange energy is involved which also forms the basis of Hund's rule,but for $s$-orbital exchange energy is not considered since there are no different spacial distribution unlike $p,d,f$-orbitals.Moreover Calcium is more electro positive atom due to less ionization energy so it would be more stable in compound form when it reacts with most of the electronegative elements.

The electronic configuration of Ca is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$
It has a total 4 Energy levels. The outermost energy level has 2 electrons called valance electrons. As, Calcium is metal and it can easy donate two electrons and becomes cation (Ca+2). Now, this cation is stable.
Calcium is reactive because it can easy donate electrons by providing energy to it called ionisation energy.

Calcium does not have four energy levels. It has 10 occupied electronic states at 6 different energetic levels. (the 2p and 3p states are triple degenerate). Furthermore, the reactivity of Ca is not explained because Ca can provide an energy termed the ionisation energy. Ionisation costs energy, it does not provide energy. The fact that Ca is reactive is because those two valence electrons are relatively unstable and can go to a lower energetic state by reacting. If under this reaction the electrons effectively migrate to another atom, Ca becomes ionised.


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What is the difference between crystal structures of proteins, organic and inorganic materials?
From the definition of crystal and the main differences between crystalline and amorphous material, it is known that crystal formation requires ordered bonding between atoms or molecules.
How can the protein with its irregular structure (when compared to organic and inorganic materials) make ordered bonding, since I often read about the crystal structure of the a protein and its ligand or inhibitor obtained by X-ray?
And is there any difference in crystal structure of an organic and inorganic material?

Molecular crystal isn't anything unusual...

Imagine an infantry unit of soldiers marching in file. Each of them may be quite irregular, but together they form a repeating pattern. And that's exactly what happens with protein molecules in a crystal.
When we say that a protein is irregular, we mean it on a different level. Indeed, one molecule of our protein is irregular if we are talking about the amino acid residues in it, of which there are probably a hundred: here goes Ala, then Gly, then all of a sudden Leu - what will come next? You can never tell. But wait, there is another molecule of the same protein sitting nearby, and it looks exactly the same as this one: Ala, Gly, Leu and so on. And there is yet another next to it, also similar, and many more of them, and together they form an ordered crystal structure which makes it possible for us to diffract X-rays on it and measure the intensities and then do some hardcore math and voila, here is your structure.

Ivan basically gave a nice and clear example of what is going in. I’m going to offer a more in-depth explanation.
If you were to consider possible crystal structures of just one type of atom, you can boil the possible structures down to a set of similar structures, the Bravais lattices. Only 14 of these exist with different constraints: i.e. the C-centred orthorhombic lattice requires $a \ne b \ne c \ne a$ and $\alpha = \beta = \gamma = 90^\circ$, with atoms placed at the corner of each unit cell and at the centre of the $a,b$ plane. If $a = b$, we could no longer describe this as C-centred orthorhombic; instead it becomes tetragonal ($a = b \ne c$; $\alpha = \beta = \gamma = 90^\circ$) with a smaller unit cell. Because each position is filled by a spherical atom, all symmetry operations allowed by the lattice itself are present in the final crystal.
Now assume, similar to what Ivan mentioned, that you replace the spherical atoms with an organic or inorganic molecule; say ammonia $\ce{NH3}$. Take another unit cell, and replace the spherical atoms with methylamine ($\ce{CH3NH2}$). And take a third and replace the with ethylmethylamine ($\ce{CH3-NH-CH2CH3}$). In the first case we still have a highly symmetric molecule which is nonetheless much less symmetric than a sphere is. Not all of the symmetry elements that are possible in a certain Bravais lattice actually remain when we do the replacement. We will end up with a certain space group — something less symmetric overall. There are 230 different possible space groups.
Note that ammonia has three intrinsic planes of symmetry in its molecular structure while methylamine has one and ethylmethylamine has zero. Therefore, the latter must crystallise in a less symmetric space group than the former two, simply because there is less symmetry present in the molecule.
If we move on to chiral molecules, say butan-2-ol, the symmetry has to be reduced even more. If we have an enantiopure substance, then in no way is it possible that a plane of symmetry transformes one molecule into another; the number of possible space groups reduces itself to the 65 Sohncke space groups — all those that do not contain centres or planes of symmetry.
The transition from a chiral small molecule to a huge chiral protein is then trivial. In the most simple of cases, just put the proteins in a nice orderly fashion next to one another to create a primitive lattice — voilà.

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'In MO theory, electrons are treated as spreading throughout the entire molecule'- isn't it exhibited in VB theory also?

First statement:

In the molecular orbital theory, the valency electrons are considered to be associated with all the nuclei in the molecule. - Concise Inorganic Chemistry by J.D.Lee.

Second statement:

In molecular orbital theory, electrons are treated as spreading throughout the entire molecule: every electron contributes to the strength of every bond. - Elements of Physical Chemistry by Peter Atkins.

Two different sources are actually saying the same thing in regard to MO theory: electron density is spread over the entire molecule, okay that is quite right & agreeable.
But the way the fact is presented as if it were the sole property of MO theory. Doesn't the electron density span over the entire molecule in VB theory also?
Check this pic taken from Peter Atkin's above mentioned book:

This is the electron density in $\ce H_2$ as predicted by VB theory; notice how the resultant wavefunction after bonding is spread over the entire molecule & this implies the electron density is spread over the entire molecule. So, even in VB theory, the electron density is spread over the nuclei of the molecule. So, why those special writings above?
Why is it that the spreading of electron density over the nuclei of the molecules specially in MO theory? Is not this exhibited even in VB theory also? Or am I missing something?
Can anyone please help me whether the spreading of the electron density is the sole property of Mo theory as is put up by the two authors? Is not it exhibited in VB theory also?

Aren't your other questions asking the same thing? http://chemistry.stackexchange.com/q/33879/4945 http://chemistry.stackexchange.com/q/33951/4945

@Martin-マーチン: Same thing? How can they be?? Of course, they are related to each other, but how can it be the same thing?

@Martin-マーチン: At one, the difference between the two theories is highlighted while the later deals why resonance isn't required for MO theory. Here, I'm asking whether the property of spreading the electron density over the entire molecule is solely exhibited in MO theory or as I am saying, also exhibited in VB theory also. Of course, I would appreciate if you elaborate how the previous queries have been reflected again here.

To be clear: the electron density does spread over the whole molecule in both models.  As ifilot's answer describes, the difference is in whether all AO's potentially contribute to a given MO, or whether just AO's on neighboring atoms are used to construct that MO.

In MO theory, you construct a molecular orbital by a linear combination of atomic orbitals. Because those atomic orbitals are part of different atoms, in this way, the valence electrons are associated with all the atoms in the molecule. So even if you consider a molecular orbital that is bonding between, let's say, atom 1 and atom 2 in the molecule, that molecular orbital has also contributions from all the other atomic orbitals (besides those that arise from atom 1 or atom 2).
In contrast, in VB theory, a bond between atom 1 and atom 2 can only have atomic orbitals of atom 1 and atom 2. The bond between two atoms is then considered to arise from two weakly coupled atomic orbitals. I have to point out though that modern valence bond theory now complements molecular orbital theory and no longer adheres to the valence bond idea that electron pairs are localized between two specific atoms in a molecule but that they are distributed in sets of molecular orbitals which can extend over the entire molecule.

Given the discussion below, I would like to add to my answer a simple example. Let us consider the $\ce{H2O}$ molecule. Within the framework of Hartree-Fock and using a STO-3G basis set, the $2\sigma$ orbital of water can be expanded in terms of the atomic orbitals as follows:
$$
\psi_{2 \sigma} = (-0.4330)\; \phi_{\textrm{O},1s} + (-0.4686)\; \phi_{\textrm{O},2s} + (0.2369)\; \phi_{\textrm{O},2p_y} + (0.2685)\; \phi_{\textrm{H1},1s} + (0.2685)\; \phi_{\textrm{H2},1s}.
$$
From this expansion you can see that multiple atomic orbitals contribute to the same molecular orbital. This is a molecular orbital that forms, in a way, a multi-atom bond between O and the two H atoms. You could also say that this molecular orbital contributes to both of the O-H bonds (in fact, more molecular orbitals than just the $2\sigma$ do).

which do not arise from atom 1 or atom 2- what? Could you please clarify that?

I was not very clear here. I have edited my answer. I meant to say: "besides those that arise from atom 1 and atom 2"

You meant to say there are contributions of atomic orbitals which are not of 1 & 2?

Exactly! :-) A good book that treats electronic structure calculations within the framework of MO is Modern Quantum Chemistry from Szabo and Ostlund (http://store.doverpublications.com/0486691861.html). If you read through chapter 2, I think my point becomes clear.

I've not enough money to buy this book now. Could you please elaborate it a bit here? I never thought molecular orbital is formed by atomic orbitals not from the participating atoms!

Any molecular orbital within MO theory is a linear combination of atomic orbitals. In mathematical terms: $\psi_{i}=\sum_{j} c_{j} \phi_{j}$ where $\phi_{j}$ are the atomic orbitals, $c_{j}$ the corresponding coefficients and $\psi_{i}$ the molecular orbital. $j$ runs over all the atomic orbitals and not only the orbitals of the atoms to which the bond adheres. So for the description of a molecular orbital, you use the complete basis set that consists of all atomic orbitals in that molecule. Hence, all these atomic orbitals contribute.

Okay. But so far, I have read the examples presented by the above books I mentioned, only the atomic orbitals that are present in the atoms participate only in the formation of  molecular orbitals. Why is it so? Examples include: $\ce H_2, C_2, O_2, O_2^{-1},F_2$ etc.

@user36790 Because these examples are diatomic molecules--the only atoms the AO's can come from are necessarily participating in a bond. For something larger,  say, ethane, a given $\sigma$ bond MO may have contributions from AOs centered on atoms all over the molecule.

@ifilot Probably worth up-editing this discussion into your answer.

Indeed, I have expanded my answer. I  hope it becomes clearer this way.

Isn’t LCAO theory a subtheory of MO theory?

I am not quite sure. I think it is a matter of semantics. LCAO is an approximation used in MO theory, whereas MO theory by itself is a method to calculate electronic structure.

authors: Макс Харитонов


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Crossed Cannizzaro reaction

What is the major product of the following reaction:

I think methanol will be formed. Although formaldehyde is more reactive to nucleophilic addition, the rate determining step of the Cannizzaro reaction is the hydride transfer. So instead of a nucleophilic attack of $\ce{OH-}$ on formaldehyde, nucleophilic attack takes place on the aromatic aldehyde.
Is that correct?

Yes apparently, http://molcalc.org/calculation/f795351741aefc756661f83735c228f7#/solvation , http://molcalc.org/calculation/1358b63dfd8439726a48e1c402dd4cbd#/solvation (you can get the charges by clicking on the atoms)

No, the formaldehyde is used as a "sacrificial" reagent.  The formaldehyde is oxidized to formic acid and the other aldehyde is reduced to the corresponding alcohol. Since the formaldehyde carbonyl is very reactive, more of it will form the tetrahedral intermediate then the other carbonyl compound and although this step is an equilibrium, it stacks the deck in favor of formaldehyde being oxidized.

@Sujith Sizon bro.... our teacher was wrong. He made us learn the wrong products.

Background
It seems that you understand the mechanism of the Canizzarro and Crossed Cannizzaro reaction, but let me supply a link to an earlier answer on the Cannizzaro reaction itself for others who might want to review the reaction. Also, here is the mechanism for the Cannizzaro reaction from the same link (but note, as we'll discuss below, the first step is really an equilibrium).

In the Cannizzaro reaction an aldehyde disproportionates into the corresponding  alcohol and acid. The maximum yield of the alcohol or acid is therefore only 50%.  If you've worked in the lab for 2 weeks synthesizing 50 mg of the aldehyde, and if your next step is the Cannizzaro reaction to prepare the alcohol from it, then a 50% yield is fairly distressing!
Here is where the Crossed Cannizzaro reaction can come in handy.  If it's the alcohol that your after, then using formaldehyde as a sacrificial reagent allows you to achieve a much higher yield of the desired alcohol.  In the Crossed Cannizzaro the formaldehyde is always oxidized to the acid and the other aldehyde is always reduced to the alcohol - you could get a 100% yield of alcohol!
Your Question - Why is that?
The first step in either the Cannizzaro or Crossed Cannizzaro reaction is an equilibrium between the aldehyde(s) and hydroxide ion to form a tetrahedral intermediate as pictured below. In the case of Crossed Cannizzaro with formaldehyde, the formaldehyde carbonyl is extremely reactive and the equilibrium will lie strongly on the side of the formaldehyde tetrahedral intermediate, and consequently most of the other aldehyde will remain present as the aldehyde.

You are correct to note that the rate determining step is the second step where hydride is transferred from the tetrahedral intermediate to any remaining carbonyl compound. But look at the 2 possible transition states for the Crossed Cannizzaro with formaldehyde (pictured above) - they are virtually the same, just some minor differences.  Hence the 2 possible rate determining steps should have similar activation energies.  This means that the energetics of the rate determining step will not control the reaction outcome (e.g product distribution), rather the initial equilibrium will be controlling.
If most of the formaldehyde is converted to the tetrahedral intermediate, then most of the other aldehyde remains as the aldehyde; and if the 2 possible rate determining steps from the tetrahedral intermediates have similar rates (they do as explained above), then the most likely reaction will involve the formaldehyde tetrahedral intermediate (major tetrahedral intermediate present at equilibrium) and the other aldehyde (major remaining carbonyl compound present at equilibrium).
Hence, in a Crossed Cannizzaro the formaldehyde is always oxidized to formic acid and the other aldehyde is always reduced to the corresponding alcohol.

And one more thing. What if the reaction was between p-nitro benzaldehyde and p-methoxy benzaldehyde?

Most know that the formaldehyde carbonyl is much more reactive than most other carbonyls.  The rds transition states are pictured above so you can decide for your self if you think the energy difference between them is large enough to control product formation.  Here is a [link] (https://en.wikipedia.org/wiki/Cannizzaro_reaction) that notes that the formaldehyde is always oxidized.

If the reactants were p-nitro benzaldehyde and p-methoxy benzaldehyde, then p-nitro benzaldehyde is more reactive. But if nucleophilic attack takes place on p-nitro benzaldehyde , the hydride transfer step would be difficult.

It would be a mess, you'd get 2 alcohols and 2 acids.  But as you say, the nitro benzaldehyde tetrahedral intermediate would predominate at equilibrium.  Since the 2 rds transition states should again be comparable in energy, I'd expect a slight preponderance of the nitrobenzoic acid and methoxy-benzyl alcohol.

@ron: does this apply to benzoin condensation? Does the nucleophilic attack determine the yeild or the hydride transfer?

@AdityaDev There is no hydride transfer in the benzoin rxn, instead there is a proton removal step.  In the benzoin rxn all steps are reversible making it more complex to analyze.  You would need to consider how the aromatic substituent might affect each intermediate in the reaction.

@ron Is aldehyde more reactive because is less stereochemical hindered?\

@adosar If by "aldehyde" you mean formaldehyde and your talking about the Crossed Cannizzaro reaction, then the answer is yes. Hope that helps.

Sir, on everywhere it is saying that rate equation is third or fourth order depending upon involvement of hydroxide. But sir, as you showed in mechanism, the hydride step do not require any hydroxide and hence the rate equation should be always second order. Please tell why OH comes in rate law.

@An_Elephant  The hydride transfer step does require hydroxide.  Look at the transition state, it contains an aldehyde molecule, a formaldehyde molecule and a hydroxide molecule.  So the reaction follows third-order kinetics.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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How does VB theory explain the Si-O-Si bond angles in SiO2?
Silicon dioxide has a huge variety of structures. Most of them are built up from connected $\ce{SiO4}$ units — the $\ce{O–Si–O}$ angle is $109.5°$, accordingly. The VB/hybridization approach to this tetrahedron would consequently assign $4\text{ }\mathrm{sp^3}$  orbitals to the silicon atom.
Now, the interesting part about silicon dioxide is the flexibility of its $\ce{Si–O–Si}$ angles, i.e. the bond angle of the bridging oxygen atoms; they vary from $100$ to $180$ degrees (the latter occurs in hexagonal tridymite, see e.g. here), depending on the polymorph. In vitreous samples, these angles show a distribution centered around $145$ degrees or so (source). In common forms of crystalline $\ce{SiO2}$, the angle also is close to $145$ degrees (e.g. this source). How is this possible? Since the oxygen bonds to two silicon atoms, one would expect a $\mathrm{sp^3}$ hybridization, with two orbitals filled with lone pairs — hence, a $104°$ angle like in water, or a slightly bigger one (since silicon is less electronegative than hydrogen — thus, the $\ce{Si–O}$ bond is more polar than the H–O bond and the angle bigger). But never an angle as large as $170°$: that would almost point at an $\mathrm{sp}$ hybridization of the oxygen atoms. How is such an angle energetically more convenient than the typical bent one?

Most of the silicon dioxide polymorphs show deviation from the ideal (predicted) angles, but they're metastable and will eventually revert to $\alpha$-quartz. See The Quartz Page for a more thorough discussion. Also note that the angles are for Si-O-Si and not O-Si-O.

That the Si–O–Si bond angles arise from the arrangement of the SiO4 tetrahedrons is obvious, as well as the fact that the O–Si–O angles are always 109.5° in stable polymorphs (since we're talking about tetrahedrons). The question is: how does VB theory represent the bonding between oxygen and silicon with these extreme variety of angles? How does it justify the existence of non-109° bonds around an oxygen molecule? Why do these bond angles differ so drastically from other X–O–X bonds such as those in water, according to VB theory?

There is also the possibility that valence bond theory fails for a silicon dioxide. There are many compounds for which VB can't explain the structure.

After adding the bounty, I realized this question had a 150 point bounty two years ago with no takers. I included a link to a paper giving the bond angles in common polymorphs of crystalline $\ce{SiO2}$. I also added an citation for an article on bond angle distribution in glass to the question - maybe that helps.

In this paper, they modeled liquid $\ce{SiO2}$ and cristobalite using an ab initio approach treating it as an ionic system: https://www.researchgate.net/publication/220037831_An_ab_initio_parametrized_interatomic_force_field_for_silica

Tl;dr version: the O-Si bonds utilize sp hybrid orbitals so that “lone pairs” are in nearly pure p orbitals and, rather than being true lone pairs, engage in some sort of $\pi$ bonding, the exact nature of which is unknown.  Other hypotheses are that the bond is highly ionic or that it is favored due to oxygen-oxygen repulsion.
Long version: First, let me point out a slight inaccuracy in the posted question:

Since the oxygen bonds to two silicon atoms, one would expect a sp3 hybridization, with two orbitals filled with lone pairs —

This is an incorrect interpretation of VB theory. Although we might approximate a molecule such as $\ce{H2O}$ as having an sp3-hybridized O atom, quantitative approaches to VB theory account for the fact that the contribution of s and p atomic orbitals to localized orbitals differs if the group differs. That is, lone pairs and bond pairs can have (and usually do have) different contributions from the s and p orbitals, ie different hybridization. A commonly cited example of this is $\ce{SH2}$, where the S-H bond orbitals are nearly entirely comprised of p orbitals, so the lone pairs are necessarily essentially sp hybrids.
Furthermore, one quantitative approach calculates the p orbital contribution based on the bond angle (Coulson's Theorem). In silicate, the Si-O-Si bond angle varies between ~140$^\circ$ and 180$^\circ$, depending on polymorph. We can calculate that the hybridization of these bond orbitals thus varies from sp (180 degree angle) to sp1.3 (140 degree angle), which is actually a relatively small variation in the s orbital fraction (43-50%). The lone pairs, therefore, are nearly pure p (93-100%) orbitals.
Although it is known that the $\alpha$-quartz polymorph (bond angle 144 degrees) is the most stable at low temperature, the observation of many polymorphs suggests that there is not much of an energetic penalty for the variation in bond angle within this range.
For simplicity, let’s consider the extreme case of the 180 degree bond angle, so that the O-Si bonds are formed from sp orbitals on O, and the lone pairs are in pure p orbitals. Both VB theory and MO theory indicate that this arrangement is only energetically favorable relative to a bent conformation if either the p orbitals are empty (as in $\ce{BeH2}$ for example) or if the electrons in the p orbitals are involved in $\pi$ bonding, for example the carbon atom of $\ce{CO2}$ or $\ce{HCN}$.
COnsistent with this idea, researchers have argued since at least as far back as the 1960's that the “lone pairs” on oxygen in silicate are in fact not lone pairs, but are involved in $\pi$ bonding to the silicon. This is supported by the short Si-O bond lengths and high bond strength which suggest a bond order greater than 1.
Historically, this interaction has been proposed to involve empty d orbitals on silicon, but more recently there has been a shift away from d orbital involvement, for similar reasons that d orbital involvement in phosphorus bonding has generally been disregarded. Interaction with empty $\sigma^*$ antibonding orbitals on Si, however, also does not appear satisfactory, as that would weaken the Si-O bonds (assuming that all Si-O bonds are equivalent, so all would have increased bond order from $\pi$ bonding, but decreased from addition of density to $\sigma^*$). Essentially, there would be a significant contribution of a resonance structure in which each Si has two Si-O bonds and one Si=O bond, similar to a carbonate.
Nonetheless, the idea that a large bond angle is favored due to some sort of $\pi$ bonding seems to be the prevalent view. Two other hypotheses, however, are still supported by some.
Other hypotheses
Ionic interactions
One alternative explanation is that there is very little difference in bond orbitals and lone pairs, which could result from a strongly ionic structure of $\ce{O^2-}$ anions and $\ce{Si^4+}$ cations[1]. On the oxygen, all of the electrons then are lone pairs, and the “bond angle” (which isn’t between any true bonds) is simply optimized to minimize repulsion between negatively charged oxygen atoms and between the positively charged silicon atoms. This would result in a tetrahedral arrangement of the negative charges around Si (as is observed) and a linear arrangement of the two Si atoms around each oxygen (the occasionally observed 180 degree angle). Since there aren’t any bonds, VB theory doesn’t really apply, and we consider the oxyanion as a lone atom with completely filled valence shell.
The problem with this argument is that free $\ce{Si^4+}$ is not observed under any circumstances, and most computational approaches suggest a high degree of covalency in the Si-O bond.
Oxygen-oxygen repulsion
A third hypothesis is that VB theory does not apply here because the minimization of energy of the Si-O-Si bond angle is not the only factor. Specifically, oxygen-oxygen repulsion favors a larger angle so as to increase the distance between silicon atoms and, therefore, the oxygens attached to adjacent silicon atoms.
Even at the small scale of a single molecule of $\ce{(HO)3Si-O-Si(OH)3}$, it has been argued that oxygen-oxygen repulsion is sufficient to cause a large Si-O-Si bond angle without any need for $\pi$ bonding[2]. The observed angle is a compromise between repulsion, which favors a 180 degree angle, and bond orbital energy, which favors a bent structure with bond angle closer to 110 degrees.
A problem with this hypothesis is that the Si-O-Si bond angle in disiloxane ($\ce{H3Si-O-SiH3}$) is also observed to be 144 degrees, even though there should be no oxygen atom repulsion effects and general steric effects should be much less than in silicates.
[1]Gillespie, RJ and Johnson, SA (1997) Inorg Chem 36:3031-3039. doi:10.1021/ic961381d
[2]Noritake F (2019) J Comput Chem Jpn 5:1. doi:10.2477/jccjie.2018-0016

"the occasionally observed 180 degree angle". Skimming the papers, it was not clear to me whether this was an artifact (disorder of say 175 degrees and 185 degrees, so to speak, averaged to 180 degrees) or real.

I appreciate that the answer discusses constraints on the angle other than the local bonding (it is a solid, after all).

@KarstenTheis - with respect to the 180 degree, I'm pretty sure I saw a paper with a table of polymorphs and one had 180 degrees, but until I can find that again, take a look at https://scripts.iucr.org/cgi-bin/paper?a19115. And w.r.t. other constraints, I was actually surprised that there wasn't more discussion in the lit about the long range order of a solid, perhaps because these wide bond angles are observed even in isolated small molecules like $\ce{(HO)3Si-O-Si(OH)3}$, and the computational work often focuses on them.


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Why is the antibonding pi bond higher in energy than the normal pi bond?
In my book, Organic Stereochemistry, the following picture is shown:

In the lowest diagram, in Figure 1.3, which shows a π* bond, the positive lobe of the p orbital of atom A is nearest to the negative lobe of the p orbital of atom B. By contrast, in the diagram above it, which shows a π bond, the positive lobe of the p orbital of atom A is nearest to the positive lobe of the p orbital of atom B. It is said that the π* bond is higher in energy, less stable, than the π bond. I would have thought it to be the other way around, due to electrostatic repulsion when two lobes of p orbitals which have the same charge overlap, and the electrostatic attraction when two lobes of p orbitals which have opposite charges overlap.
Why am I wrong? Why is it not this way around?

Firstly, the + and – signs you see do not indicate charge. They mark the sign of the electron’s wave function. In an atom, the electrons always have a negative charge, and the nucleus always has a positive one.
When a bonding orbital is formed, the electron density is concentrated in the area between the nuclei. The electrostatic attraction between the electrons and two (instead of one) nuclei, lowers the energy of the molecule. Furthermore, the electron density between the nuclei shields the electrostatic repulsion between the two positive nuclei to a certain degree.
In an antibonding orbital, the electron density is moved away from the internuclear area: the electron-nucleus attractions are minimized and the internuclear repulsion maximized. Thus, the molecule is destabilized.

When you say the electron density is concentrated in the area between the nuclei, isn't that in a σ bond where the electron density is in the AB axis. I thought in a π bond the electron density is zero along the AB axis and is actually above and below the AB axis.

Yes, I don't mean the axis, I mean the nuclei's neighbourhood: in a π MO, the electrons can easily interact with both nuclei. In a π* MO, this possibility is reduced drastically.

There's just one thing I don't understand, what is the significance of the sign of an electron's wave function? What does it determine? Why does one lobe of the p orbital have a positive wave function and the other lobe have a negative wave function? Also, in the second example from the top of the page, showing the overlap of a s orbital and a p orbital, why does the s orbital have a positive wave function?

@SaulMcShane http://chemistry.stackexchange.com/q/35212/16683

The wave function itself is no real physical quantity: what has a significance is its square modulus, which indicates the probability of finding the electron in a specific position (electron density). The sign of the wave function is important when combining orbitals in molecules – if you add two same-signed orbitals, a bonding MO will form; otherwhise, an antibonding MO will form. Regarding your doubt about why the orbitals have their respective signs… well, that's a long story, but you can find all the orbitals on the orbitron website.


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Bromination of Phenylacetone
Looking at the following reaction, I understand how the product forms, however, can somebody help me understand why two products are formed? Would the identical mechanism produce the two products?

Edit: This is how I understand the mechanism.

Can you draw a mechanism to show what you think happens to form the product. The mechanism contains an enol as one of the key intermediates, and drawing this intermediate should give you an indication of why you get both 'products'. The 'products' are infact two enantiomers of the same product (i.e. the reaction produces a racemic mixture).

@NotNicolaou I've added it.

Based on what you've drawn, you've answered your own question! xD     If you look at the enol, its planar with respect to C=C, attack can happen from above that plane, or below that plane, leading to both enantiomers of product formed via an identical mechanism.

@NotNicolaou Oh... That makes sense then. I expected there to be two different mechanisms. Thank you for the help.

Glad it makes sense. If you understand now, you can write and accept your own answer so anyone searching for the answer in the future can find it.

The mechanism that I have submitted in the edit of my post is correct. The reason both products are formed, is due to the the enol intermediate being planar with respect to the oxygen, allowing it to be attacked from either plane. This allows for the formation of a the so-called racemic mixture.

The first step involves the formation of enol by protonation of the carbonyl oxygen. Then the electrophile adds to the enolate, and it may approach in either direction, from the top or the bottom of the plane of the enol.And so, a racemic mixture is obtained.

I’m not sure why this answer is downvoted. From what I can tell, it’s correct.


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How to ionize a liquid and apply a force using electrohydrodynamics
I would like to create a contactless force on the surface of a liquid by ionizing it and placing the container between a cathode and anode, however I'm not sure how I would go about actually ionizing it.
The liquid is a photopolymer resin, although the specifics of it can be tailored to whatever we need so it makes the question rather broad. So I'm wondering if someone could give me some advice on how to create a surface force on a liquid using electricity.

Though it is possible to ionize the liquid, it is doubtful you want to do that: the effect of ionization would likely be equivalent to exposing the photopolymer resin to bright light.
A System for Cooling Electronic Elements with an EHD Coolant Flow from iopscience.iop.org describes setting up and testing a pump using isopropanol as the working fluid.
If you do decide to try electrohydrodynamic pumping, either of an atomized spray or EHD pumping of dielectric liquids, take precautions against ignition of the liquid and test whether ionization ruins the photoresist.


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What type of bonding is present in lithium tetrahydridoaluminate?
I can tell that the negative ion has one bond which is a co-ordinate/dative covalent bond.
However, overall, would you say that lithium tetrahydridoaluminate (lithium aluminium hydride) is ionic? If so, would you expect a lattice forming?

Yes, this is an ionic salt, there is really nothing special about it. Like very many ionic salts (sulfates, carbonates, nitrates) it also contains covalent bonds. Like nearly all ionic salts, it is solid and crystalline.

For the rcord this compund is called lithium aluminum hydride.

@A.K. The IUPAC name is lithium tetrahydridoaluminate but yes everyone calls it lithium aluminium hydride or just LAH.

@A.K. Can you explain the difference between hydrido and hydride? Is it simply due to IUPAC and non-IUPAC uses?

@A.K. The postdoc in our group once said that LAH be the better name for the solid form since it never is pure $\ce{Li[AlH4]}$ — but that if you are dealing with exactly that, call it lithium tetrahydridoaluminate.

https://en.wikipedia.org/wiki/Lithium_aluminium_hydride - not sure if bonding of hydride bridges to Li is more ionic or covalent.

I think we should provide answers on higher level of abstraction then simply ionic or covalent.


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Neighbouring group participation with and without retention of configuration in nucleophilic substitution of 2,6-dichloro-9-thiabicyclo[3.3.1]nonane

Treatment of the dichloride A under the conditions indicated gives B with retention of configuration, while treatment of dichloride C gives D with inversion of configuration.

I am having difficulty drawing out the inversion ring product to find out if the s lone pair can attack from anti to the $\ce{C-Cl}$ σ* bond on the vertical ring. If it does, then, there is retention of product in the first reaction after the $\ce{N3-}$ attacks.
Can someone please help using diagrams?

See also Converso, A.; Burow, K.; Marzinzik, A.; Sharpless, K. B.; Finn, M. G. 2,6-Dichloro-9-thiabicyclo[3.3.1]nonane:  A Privileged, Bivalent Scaffold for the Display of Nucleophilic Components. J. Org. Chem. 2001, 66 (12), 4386–4392. (http://dx.doi.org/10.1021/jo015632y) and Díaz, D. D.; Converso, A.; Sharpless, K. B.; Finn, M. G. 2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold. Molecules 2006, 11, 212–218. (http://dx.doi.org/10.3390/11040212); pdf

Is there a typo in compound C? Should those oxygens be there?

@jerepierre i thought that too... But i dont think its a typo.

C looks OK, but shouldn't the oxygens also be on D?

A reacts at low T due to NGP.  The sulfur lone pair displaces the Cl, that's the first SN2 and you have inversion of configuration at carbon.  Then the azide displaces the sulfur, a second SN2 again with inversion.  The net result of 2 inversions is retention of configuration at the carbon initially bound to Cl.  In C there are no lone pairs on sulfur, no NGP, higher activation energy, higher temperature needed to get the azide to react via an SN2 with inversion of configuration.

Do you have a reference for the reaction C -->D or is this step hypothetical? Loong’s references do not address the chemistry of sulfone C. In earlier work in 1966, Corey and Block (https://pubs.acs.org/doi/pdf/10.1021/jo01344a001) prepared C and removed the chlorines with Ni/H2. In an adjoining paper, Weil, et al., (https://pubs.acs.org/doi/pdf/10.1021/jo01344a002) replaced the chlorines of C with CH3S- without addressing stereochemical issues.

The only difference between substrates A and C is disulfide vs sulfone nature at 9-position. Actually, that is good enough to display this major stereochemical outcome. As explained in Ref.1, the disulfide nature at 9-position of A has an advantage of having two lone pair on sulfur, one of which is proximate enough to attack backwardly on $\ce{C-Cl}$ bond ($\mathrm{S_N2}$ in nature) to give an episulfide ring (see following diagram and the extended comment of ron). The reaction center must have an inversion of configuration since the mechanism is $\mathrm{S_N2}$.

This episulfide intermediate undergoes ring opening backward attack by the intended nucleophile (e.g.,$\ce{N3-}$) to give the final product under 2nd $\mathrm{S_N2}$ mechanism. This stepwise replacement of chlorine by a nucleophile, the first substitution being considerably faster than the second, was first observed by E. J. Corey and E. Block (Ref.2). The nucleophile of their reaction was methoxy group of methanol (solvolysis). However, although they are the first group to suggest the 1,2-epi-sulfonium ion intermediates in these reactions, it was suggested for the initial dichloro compound A formation, not for the solvolysis by methanol. These types of neighboring group participation (NGP; or anchimeric assistance) have been observed previously for bridging carbon, nitrogen, and boron, all for the bicyclooctane series. In mechanistic point of view, this double inversion process preserves stereochemistry in substitution reactions on 2,6-Dihalosubstituted-9-thiabicyclo[3.3.1]nonanes (Ref.3).
Finally, C cannot follow the fast double inversion process due to the lack of a lone pair on $\ce{S}$ bridge, yet undergoes usual $\mathrm{S_N2}$ transformation (slow because the leaving group is $\ce{Cl}$) under the given conditions.
References:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).
E. J. Corey, E. Block, “New Synthetic Approaches to Symmetrical Sulfur-Bridged Carbocycles,” J. Org. Chem. 1966, 31(6), 1663–1668 (DOI: 10.1021/jo01344a001).
A. Converso, K. Burow, A. Marzinzik, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: A Privileged, Bivalent Scaffold for the Display of Nucleophilic Components,” J. Org. Chem. 2001, 66(12), 4386–4392 (DOI: 10.1021/jo015632y).


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What is the best material for my particles?
I'm investigating collision of a particle and a water droplet ($v_\mathrm{drop}=1~\mathrm{m/s}$, $v_\mathrm{particle} \approx 8\ \mathrm{m/s}$, $D_\mathrm{d}=2.8\ \mathrm{mm}$, $D_\mathrm{p}=2.0\ \mathrm{mm}$). Due to a head-on collision, particle penetrates into droplet and changes the droplet shape (collision outcome). I need to see that what effect particle wettability has on collision outcome. So, all other particle parameters, like density, should remain constant. I can use particles made from any material as long as I can create different contact angles on particle surface. Do you think which particles are the best for this purpose?
P.S: Currently, I've purchased glass particles and I'm trying to create different contact angles through silanization. But, any other suggestion is appreciated.

The experiment seems unlikely to be fruitful. For relative motion I'd think drag was the phenomena to investigate.

@MaxW This is one of our results.

In terms of the drop, I'd expected surface tension to be more important.


text:
Why do some diagrams of some molecules show the bonds of only some of their submolecules?
In the diagram of ATP below this paragraph there are no lines between the atoms in the $\ce{NH2}$ molecule or in the $\ce{OH}$ molecules; however, there are lines between all the other atoms. Does the omission of those lines represent some facts about  $\ce{NH2}$ and $\ce{OH}$ in the ATP molecule, or did the illustrator omit them in order to emphasize the rest of the molecule?

This is a common convention in order to make things clearer. Also, note the fact that not even all atoms are shown.

What atoms are missing? It's hard for me (a first year science student) to imagine an atom being so inconsequential that an illustrator could just as well omit it from the diagram.

Look at the ball-and-stick structure (second image from the top) and find that out for yourself. Now, it's not that these atoms are irrelevant and could just as well be absent; no, they must be there, but we simply draw the whole thing this way as a shorthand, much like we say "won't" instead of "will not", or like you just said "ATP" instead of "Adenosine triphosphate".

@IvanNeretin It seems to me that the point of a diagram is to show what the molecule is. When we say won't, we know that the listener knows that the meaning of won't is 'will not'. If we have to show someone what a molecule is, then they don't know what it is. A diagram might also be used (instead of "ATP") to aid communication to an audience familiar with the molecule, but in those cases, the depicted portions would only be the parts relevant to the topic. So I should have asked why do diagrams of ATP typically depict the atoms depicted in the diagram above instead of other atoms?

What you have shown is a variant of a structural formula.  I use the word variant as there can be more than one way of showing a structural formula. What you seem to allude to in your question about showing all the bonds relates to a displayed formula.

@Hal Chemists, like writers of english, know a bunch of standard abbreviations that simplify chemical structure drawings. When we omit hydrogens attached to carbons, we know that typical chemists will be able to add them by counting the number of bonds to carbons. And, in structural pictures, most carbon labels are omitted because we know that unlabelled vertices in the picture will be carbon.

Your example is one of a variety of shorthands that exist because every chemist after the third year of studies knows what is meant so they are too lazy to draw.
In a nutshell, the $\ce{NH2}$ group will always be a nitrogen with a lone pair, two single bonds (one to each hydrogen) and one bond to the molecular skeleton. Much like $\ce{-OH}$ will always be $\ce{-O-H}$. It is also common practice to omit any insignificant hydrogen atoms in organic molecules; in the image above, a total of eight hydrogens has been omitted because every chemist will know where they are (and that they are there).
In principle, the phosphate chain could also be shortened to $\ce{HO3P-PO2H-PO2H-{}}$ rather than writing out all of it. Any chemist will mentally ‘expand’ the molecule in the same (correct) way.
As you continue to study, you will notice when which groups are generally shortened, which rules to observe while doing so and which hydrogens are so superfluous that they are not drawn — and, most importantly maybe, how atoms are labelled that actually are missing hydrogens. it is beyond the scope of a single answer to note them all.

I see. So then, just as we omit the hydrogens attached to carbons because carbons give us reason to count on hydrogens being there, there are other constituents, x, that justify us in believing that if y is there. Thank you.

Third year chemist? Really I thought this was like second-third year high school stuff?

@DRF Take that as at the very, very, very latest, even if they stay in phys chem or comp chem.

Essentially scientists have better things to do than draw (an look at) every single atom.  This is a skeletal diagram of the molecule.  It is meant as a quick and efficient way to convey the structure of the molecule.  In fact if you look closely you'll notice that hydrogens bonded to carbons are not even shown.  This is done because chemists know they are there, but showing them would take time, clutter the molecule and obscure the functional groups that are more important to the properties of the molecule.  The $\ce{OH}$ and $\ce{NH2}$ groups lack lines because they are functional groups and again scientists know but have better things to do. Occasionally functional groups get even more short-handed.  the phenyl group which is a bonded benzene ring is often abbreviated $\ce{Ph}$ to conserve space on a paper and save time.  Hydrogen atoms will often be shown if they are bonded to a chiral center, but even they may be omitted and assumes to be opposite a show stereo group.

@A.K. I don't know if I just missed this or never asked, but I never got this explained to me in my High School chemistry class. Having it explained this way and looking at one of these images finally made it click for me, thank you!
(I was going down a rabbit hole after watching a chemistry tier list video lol)

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review


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Why ionic radii of Cu2+ is less than Zn2+?
Fully filled orbital has more effective nuclear charge than incompletely filled orbital. So atomic or ionic radii of elements or ions having fully filled orbitals should be less than that of elements having  incompletely filled orbitals (e.g. $\ce{F}$ and $\ce{Ne}$).
Now $\ce{Cu^2+}$ has a $\ce{d^9}$ configuration and $\ce{Zn^2+}$ has  a $\ce{d^10}$ configuration. So, $\ce{Zn^2+}$ has full filled d orbital and is thus stable. According to my assumption above, $\ce{Zn^2+}$ should have less ionic radii than $\ce{Cu^2+}$. But it is found that $\ce{Zn^2+}$ has more ionic radii than $\ce{Cu^2+}$.
Ionic radii of $\ce{Cu^2+}$ is $\ce{73}$ pm and that of $\ce{Zn^2+}$  is $\ce{74}$ pm. Why is this so?
(source)

are you sure that Zn2+ will have $d^{10}$ configuration coz then there will be no electron in the 4s sub-shell

@manshu Of course it will. It is the cation’s most stable electronic configuration.

@Jan Then why does nickel have $4s^2 3d^8$ configuration. even if both have same number of electrons

$d^{10}$ shell is very rigid while $d^9$ shell is polarisable and, since copper is typically square planar, it moves slightly from the ligands, allowing tighter association.

@permeakra Copper(II) is not typically square planer but typically a strongly Jahn-Teller distorted octahedron.

@manshu http://chemistry.stackexchange.com/a/4500/7475 Just because the electron count is the same does not mean the species are the same — zinc(II) has two more protons in its nucleus.

@Jan ionic radius is counted using closer atoms, so for the matter of ionic radius definition it is square planar. Besides, the distortion is rather large =).

After quick lurking, I came with plausible explanation, based on polarisability of electronic shell. $d^9$ and less occupied electronic shells are much more polarisable than $d^{10}$, especially considering that ionic radii are typically measured for octahedral neighborhood, matching natural symmetry of electronic shells. I'm going to write an extended answer a bit later.

To answer your question, it is good to see the full picture. I used data from the Wikipedia data page on ionic radii to plot the dependence of the ionic radius of dications by the number of d-electrons. The plot includes 1st d-row data for high-spin and low-spin states (these are terms from crystal field theory and are properly introduced elsewhere). It also includes data for Pd-Cd, Pt and Hg.

As we can see, a rather complex dependence is observed. However, it is clear, that the increase in ionic radius from the d8 to the d10 state is not a random fluctuation, it is observed for all three rows. To understand the nature of this dependence it is critical to understand that the values are for an octahedral ligand field (the values would be difference for tetrahedral and other coordinations). In such a field d-orbitals become unequal — three of them are lower in energy while two are higher. In high-spin complexes the difference is not enough to overcome electron pairing energy, while in low-spin ones it is. As result, for low-spin complexes we see a clear minimum at d6, corresponding to the three lower d-orbitals being fully filled, while for high-spin complexes we see changes in the trend for the 4th (first on higher orbital) 6th (all orbitals half-filled) and 9th (first paired on higher orbital) electrons. Now, let's check out the orbitals mentioned (thanks to chemwiki)

$\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ are the orbitals with higher energy. As it can be seen, they are located between the metal atom's center and the ligands. As such, electrons on said orbitals have clear effect on metal-ligand distance, so for low-spin state the observed ionic radius monotonously descreases up to 6 electrons (fully occupied lower orbitals) and then monotonously increases. For the high-spin state, the increases are also located on steps when an electron is added to higher orbital, but in this case the electrons are added on d4-d5 and d8-d10 configurations, so the increase happens in two steps. For the 2nd and 3rd rows, only few last steps on the curve are observed as most elements in these rows tend to form metal-metal bonds in low oxidation states and thus their true ionic radii are unknown.
And now, why are these dependences of limited use. Consider the graph of $\ce{M-Cl}$ distances for chlorides of the first d-row below. The data was obtained from the open crystallography database. No correlation with the plot above can be established, despite most of the chlorides except for $\ce{ZnCl2}$  having their metal in sixfold coordination. This is because said chlorides are not isostructural. Layered $\ce{MoS2}$ and $\ce{CdI2}$ structures together with chains of squares and 3D-lattices occur. Furthermore, for iron two different structures (layered and rutile-like ones) were found with, obviously, different $\ce{M-Cl}$ distances (the layered one is plotted. As a fun fact, the longer distance is observed in the higher-pressure rutile modification). An obvious conclusion for the plot is that only for isosctructural fragments we can conclude something based on electron structure of the central metal atom.


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Ionization energy comparison between K+ and Cl-
So I have encountered a question which looks very suspicious to me.
If you have $\ce{Cl-}$ and $\ce{K+}$, Can you say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$?
We know for a fact that when you put more electrons it makes the the pulling force of the protons a bit scattered (because of the incremental of the electrons in $\ce{Cl-}$) so the ionization energy of chlorine should go down.
The opposite happens to $\ce{K+}$ when I remove electrons the pulling force of protons is stronger (Because it pulls less electrons) than it used to be so the ionization energy goes up!
So What I am proposing here that we can't just say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ just because that happened. $\ce{Cl}$ had a really high ionization energy compared to $\ce{K}$ and the decrement of the ionization wouldn't be that big to make it less than the ionization energy of $\ce{K+}$.
(The answer was that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ in the exam paper but I questioning that)
So if there is a chart or something for this that would be awesome. I hope I can get some help here.

Gravity ain't involved here. Are you looking for a chart of ionization energies? If so, they're all over the place - Google it. As it reads now, your question is a little unclear: if you can edit it for clarity to specify what the gist of your question is, that will help you and the community here get an answer (or more) to your query.

Oops didnt mean gravity sorry.
I know that there are a lot of charts out there but I cant seem to find any chart that has cl- in it.  The question is simply is: Who has a higher Ionization energy k+ or cl-? and explain why.
I put my argument up there. So that who will answer the question can check it out.

Look at it this way: what is the electronic configuration of $\ce{K^+}$? What is the electronic configuration of $\ce{Cl-}$?

The general reaction we are looking at when discussion ionisation enthalpy is:
$$\ce{X -> X+ + e-}$$
With the electron removed into space (i.e. totally disappearing). An electron is a negatively charged particle, so removing it from something negative will be easier than removing it from something positive. Thus, you have a very stable $\ce{K+}$ ion and are trying to rip away another electron from this (bad) — and you need to rip it out of a core orbital — all valence orbitals are already empty, the last one when you removed that previous electron to generate $\ce{K+}$ (very bad). A combination of bad and very bad gives us a very high ionisation energy for $\ce{K+}$.
Discussing chloride, first of all we have an anion, so the removal of an electron is generally good for removing excess charge. Second, we again have something that has a full octet — but this time the octet is a valence octet, not a core octet, so removing the electron will be a lot easier. We have a combination of good and relatively good, so the ionisation energy of $\ce{Cl-}$ will be rather low.

The simplest way to approach this is to look at the total number of electrons, which is 18 for both $\ce{Cl-}$ and $\ce{K+}$ (the electron configuration of Argon).
In one case you have 17 protons in the nucleus pulling all these electrons together, in the other 19. It is obvious that the $\ce{K+}$ nucleus will be pulling harder, and hence has the higher ionization energy. You can safely predict that the ionization energy of $\ce{Ar}$ will be in between both values.
Numerically, however, the approximate proportion between the ionization energies is not 17 : 18 : 19, but rather (17-17) : (18-17) : (19-17) due to screening by the remaining electrons. Indeed, ionization energy is mainly driven by $\frac{Z_\rm{eff}}{r}$, where $Z_\rm{eff}$ is the effective (screened) nuclear charge and $r$ the radius. The radius will be similar for $\ce{Cl-}$, $\ce{Ar}$, and $\ce{K+}$, but $Z_\rm{eff}$ is 0, 1, and 2, respectively. If we assume a radius of 0.9, the ionization energies will be roughly 0, 1.1, and 2.2 Rydberg ($\pu{0 eV}$, $\pu{15 eV}$, and $\pu{30 eV}$).
This can be understood by considering that the ionizing electron from $\ce{Cl-}$ does not feel any electrical force as it moves away from the neutral $\ce{Cl}$ atom, it is only bound by electron affinity, which is an order of magnitude lower ($\pu{3.5 eV}$). The ionizing electrons of $\ce{Ar}$ and $\ce{K+}$, on the other hand, are strongly attracted by the positive ion they leave behind.


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What do you call acids that can disassociate more than 1 H+
What do you call acids that can disassociate more than 1 H+? Is it like Class n acids for acids that can disassociate n H+?

diprotic acid = 2, triprotic =3, polyprotic acid.

http://chemistry.stackexchange.com/questions/16804/what-is-the-word-for-how-polyprotic-an-acid-is


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Textbook chemistry vs Real life chemistry and impurities
Long ago when I took chemistry courses in high school and lower, chemistry seemed so simple and elegant. All equations and exercises involved pure elements, pure compounds, with a very clean simplified diagrams.
Now that I look at real world, the world is full of impurities. Even water we drink is not pure $\ce{H2O}$, but rather full of "other stuff", and no two samples of water are "absolutely identical".
This makes me wonder. To what extent can you overlook these impurities when doing science, in predicting outcomes of experiments.
For example, you can make naive formulas for how much salt can dissolve in water until it stops dissolving salt. However, in real world,

Pure water does not exist as something easily available, and if it does, from what I read, it doesn't stay pure for very long.
Table salt, while mostly sodium chloride, often contains iodine, and "other stuff" that I am not sure about.

So surely these impurities impact the behavior of your experiments, and depending on the experiment, you can get reactions you never expected or wanted, or behavior can be drastically different.
My question is, how do scientists control experiments when dealing with unpure compounds? How do they know that their reactions to produce "table salt" don't create unwanted chemicals that will blow up in small quantities, or create some new kind of poison that they will only be noticed in a few days? Perhaps create some kind of acid that burns through the glass of their beaker or cause some other hazardous effects. Are they naive and ready to face the unexpected or are there absolute certainties in which they can control it (can you really control everything?)?

Well, you should use stuff pure enough for your purpose - there's whole spectrum of cases.

Shi* in *hit out...    In all seriousness, research chemists can buy chemicals in very pure forms from chemical suppliers to avoid any unwanted reaction. Sometimes this is necessary. Other times there's no need to worry.    People trying to do chemistry in their basements is a whole other story.

But how pure is pure enough. Surely we can't microscopically examine every single particle and bond of our substance to ensure it is absolutely made solely of what you know is in it. I mean this can be debatable, since specific  reactions can be used to test it and not require such examination, but even so, it seems hard to believe that the space of what could be there is that predictable.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I edited your post to include MathJax for $\ce{H2O}$ — you can learn more about it in the [help], this meta-post or this one.

When you buy a chemical from a chemical supplier it comes with documentation that lists various impurities and their concentration. This is done for each batch of the chemical they say.  Obviously they can't do it for EVERY possibly impurity. But they do it for common ones and ones that are likely to be there due to the way the chemical was produced

Surely not all chemicals come from chemical suppliers. And isn't it possible for smallest of unexpected impurities to cause significant differences in results? Starting to get carried away since I'm now curious how cooking things doesn't inevitably poison us if we do it often enough (since we can't account for all possible reactions that can happen), and we don't buy our food from chemical suppliers.

All chemicals used by chemists (university chemists and industrial/pharmaceutical chemists) come from chemical suppliers. In UK universities at least there are strict rules about where chemicals are allowed to be bought from and by whom.

Sidetracking even more and now curious about why chemicals are looked at as unhealthy, if they are more predictable and manageable than organic unpure substances.

Do you get that, for everything, there is amount so small that it doesn't matter?

For the most part yes, but in chemistry surely once in a while something small does matter, and matters a lot. That is what worries me, that rare case when something small makes a huge impact. Perhaps these small things are easier to predict though.

Yeah the stuff you eat is not likely to be pure. But it also probably isn't contaminated with cyanide or anything that will kill you, so it doesn't really matter. "Why chemicals are looked at as unhealthy" - 1) Nobody adds dangerous chemicals to food, or else they would have a huge lawsuit hanging over them. I mean, there are exceptions with some unscrupulous people, but the vast majority of additives are fine (in moderation). 2) I wouldn't trust the general public's opinions on "chemicals".

Well, scientists usually purchase the purest grade of chemical available if they want to use it for a reaction. (Leading to the absurd case that the less pure reagent grade is often dismissed for the purer analytical grade.) Usually, special care is taken for these chemicals to remove especially those impurities which could hurt most — you really don’t want any palladium in your sample of sodium bromide; that could do some really bad things.
That directly leads to a second point: No chemist would ever use ‘table salt’ or tap water unless in dire need. As you state, the impurities are basically incontrollable.
With that in hand, usually chemists have a good idea of what they just added and how that influenced the reactions. But often some things still go wrong. One of the most famous examples in organic chemistry is the Nozaki-Hiyama-Kishi coupling, which was initially reported as requiring only a $\ce{Cr^{II}}$ salt — usually $\ce{CrCl2}$. However, yields of other groups were much much worse than those originally reported by Nozaki and Hiyama. It wasn’t until eleven years after initial publication that — independently of each other — the original authors and Kishi both discovered that $2~\text{mol-}\%$ $\ce{Ni^{II}}$ are required as a co-catalyst for the success of the reaction. The original chromium salt batches of Nozaki and Hiyama were obviously contaminated by enough nickel(II).
An inorganic example I once heard of in a solid state chemistry course was a crystal structure published with experimentals in a solid state reaction. (I can’t remember the exact compound.) Other authors tried to reproduce the reaction but got much worse yields than the $100~\%$ that are expected in solid state reactions. The solution was that the square unit cell contained a hydride anion in its centre, almost invisible due to its low electron density. This hydride was introduced to the reaction from the silicon grease the authors used for the glass joints.

Thanks for the amazingly written answer. Still, the suppliers can only avoid what they already know to cause some really bad things, can they really avoid all of these miniscule impurities that slip by, or does it happens too rarely to matter?

@Dmitry It happens to rarely matter. Say if I had sodium chloride, the most common impurities would be other alkaline metals and other halides plus other typical anions such as carbonate. These don’t disturb the reaction. Of course, this isn’t always the case: Monochloroborane $\ce{BH2Cl}$ can contain detectable amounts of dichloroborane $\ce{BHCl2}$ which would usually disturb the reactions you are doing. But then again, it usually just ends up being reduced product yield, so ‘okay’.


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Product of intramolecular aldol condensation of 3-methyloctane-2,7-dione

What is the major product formed when 3-methyloctane-2,7-dione is treated with sodium hydroxide?

My main issue is where the base deprotonates. Two enolates are possible (I neglected the other two because upon cyclisation, they form seven-membered rings which are less stable than the five-membered rings otherwise formed). Enolate 2 should be more stable than enolate 1 due to the greater degree of substitution:

However, these two enolates lead to different products. So, which product should be formed? I know that if a strong base like LDA was used, a proton will be abstracted from one of the less hindered terminal carbons. But hydroxide ion is small and should be able to abstract more hindered protons.

For the enolates that can be formed, the others have already given comprehensive answers: All four enolates are possible and since we are under thermodynamic control (weak base), we expect o observe them all. Numbering the chain carbons according to IUPAC, and noting the non-carbonyl end of the enolate, I expect the enolate ratio to be $3>6>1\approx 8$.
But your question asked for the reaction’s major product. First of all, the two terminal enolates (carbons 1 and 8) are to be dismissed: The seven-membered ring is much less stable. Then, the reaction mechanism shows us which enolate leads to a thermodynamically much more stable product than the other:

The more stable enolate 2 unfortunately cyclises to give a β-hydroxyketone 5 which has a quarternary carbon between the ketone and the newly-formed hydroxy group. No elimination is possible here. The other enolate aldol product 3, however, can eliminate to the corresponding cyclopentene 4. This should likely be given as the reaction’s major product since all steps up to the elimination are reversible. (The elimination is, too, but it is still most stable.)
Indeed, Stork and Williard report that the cyclopentene 4 is obtained when 3-methyloctane-2,7-dione is refluxed with ethanolic KOH for 2 hours, although no yields are given as it is only mentioned in a footnote.[1]

Stork, G.; Williard, P. G. Five- and six-membered-ring formation from olefinic α,β-epoxy ketones and hydrazine. J. Am. Chem. Soc. 1977, 99 (21), 7067–7068. DOI: 10.1021/ja00463a053.

To quote directly from Wikipedia:

In the deprotonation of an unsymmetrical ketone, the kinetic product is the enolate resulting from removal of the most accessible α-H while the thermodynamic product has the more highly substituted enolate moiety.

Enolate 2 in your question is therefore the most stable as it is the most highly substituted. However enolate 1 should not be completely written off, there are many examples of this kind of substitution of enolate doing thermodynamic aldol reactions. The two terminal enolates are, under these conditions, highly negligible, both due to the lack of stabilisation and also since they would react to form the seven-membered ring, as you point out.
Unlike enolates formed via LDA where enolisation is complete, thermodynamic enolates are in equilibrium with the carbonyl, hence there is also the possibility of getting some of enolate 1 forming and going on to snap into a five membered ring.  However, as Jan points out, drawing both 5-rings there is a clear preference for formation of one product over the other.


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Why is the thioester bond weaker than a regular ester bond?
What are the contributing factors to the fact that thioesters are less stable than their oxygen counterparts?

Stable with respect to what? Stability can be kinetic or thermodynamic, and also depends on the reagent you're talking about. For sure there are many examples where you can keep a thioester intact whilst manipulating an ester.

For most applications, though, it is the inability of sulphur to participate well in the resonance across to the carbonyl. Thioesters thus look much more like isolated (more reactive) ketones.

+1. No arguments with that explanation. I'm just keen to hammer home the point that stability should always be qualified. Alkenes are seen as being fairly 'stable' in a chemical sense but we all know what methane + a spark is capable of.

Why is the thioester bond weaker than a regular ester bond?

Sulfur is a much bigger atom than oxygen or carbon. As a result, resonance donation from the sulfur into the carbonyl carbon is limited, and the C-S bond doesn't have as much multiple-bond character as the carbonyl carbon-oxygen bond in an ester.

What are the contributing factors to the fact that thioesters are less stable than their oxygen counterparts?

Kinetically or thermodynamically?
Stable with respect to what?
If we're talking about thermodynamics, and stability with regard to hydrolysis, then yes, thioesters are generally less stable than esters.
It is more favorable for water to attack a thioester's carbonyl carbon because it has a greater partial positive charge than a corresponding ester's. This is due to the resonance issue discussed above. A thioester also has a more stable leaving group, as the larger size of sulfur allows negative charge to be distributed better.

authors: Дарина Симеонова


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Is rotation of a mirror image allowed in order to determine chirality? Doesn't rotation alter the mirror image of the molecule?
So, chiral molecules are those which have non-superposable mirror images.
trans-1,2-Dimethylcyclopentane is chiral; 2-butanol is chiral; both don't superpose on their mirror image.
Then my book cited 2-propanol as achiral. How?
The mirror image doesn't seem to superpose with the molecule. How can it be achiral then?
Then the book provided a pic where the author showed how 2-propanol is achiral; he rotated the mirror image & then superposed! Done! 2-propanol is achiral!
But really how can rotation be allowed? After all the mirror image is a stereo-isomer; if it is rotated, wouldn't it be changed?
I'm not getting, why rotation is allowed before superposition. Not every-time, the mirror image is rotated before superposition.
Can anyone please explain how rotation is allowed before superposition?

superposition is pertinent only to the chiral centers.

Molecules in the nature do rotate all the time, come to think of it.

As a rule of thumb, in order for a molecule to be chiral, it must have a chiral center. There are some exceptions. A chiral center is a generalized extension of an asymmetric carbon atom, which is a carbon atom bonded to four different entities.
Carbon-2 in 2-propanol is not a chiral center because it is only bonded to 3 unique substituent groups: ($\ce{-H}$), ($\ce{-OH}$), and 2 ($\ce{-CH3}$).

he rotated the mirror image & then superposed! Done!! 2-propanol is achiral!!

You seem to misunderstand. Mirror images of chiral molecules are non-superposable, and mirror images of achiral molecules are superposable (because they are the same molecule).
Rotating the molecule only means orienting a molecule differently in 3-D space, not changing anything about the molecule itself. I'd recommend playing around with a molecule building set to solidify this in your mind.

so we can rotate the mirror image at any angle and it won't make a difference,right?

https://chemistry.stackexchange.com/a/64170/15410

When checking out chirality or achirality you are always allowed and supposed to rotate the molecule freely in all dimensions. Essentially, what you are trying to prove, is that the mirror image of a molecule can be achieved by rotation alone.
The thought behind this is simple. Chirality is macroscopically proven by optical activity. If you reduce the experiment to its bare basics, you have a polarised light beam that passes through a solution and hits a single molecule to be deflected either to the right (dextrorotatory) or to the left (levorotatory). A second molecule it hits which has the same chirality will move it in the same direction by the same amount. If, however, it hits a molecule which is exactly the mirror image of the first, the effect will be cancelled and the plane of polarisation is moved back into the vertical. In a solution containing only one enantiomer of a chiral substance, the beam will never hit a mirror image and thus always be deflected in a certain direction.
If we continue this thought experiment, we realise that molecular motion in solution is random. Thus, a molecule can — if we give it conciousness — move around freely and let it try to assume exactly the mirror image of the previous one. Therefore, we are allowed and encouraged to rotate (but not to use another mirror image or inversion!) our molecule until it ‘fits’.
By the way, most if not all achiral compounds can also be rotated before you apply the plane of symmetry. Oftentimes if not always, you can rotate them in a way that you see the plane of symmetry present in the molecule itself. If you managed to do that, you won, because just use a plane parallel to that one for mirror imaging and the image will automatically be identical to its template without further rotation.
The plane of symmetry in isopropanol cuts through the $\ce{CHOH}$ bit of isopropanol, dissecting each of those four atoms into two half-spheres. One methyl group will be transformed exactly onto the other by this plane.

I know that this post is like 2 years old now, but can we rotate about a sigma bond before superimposition? I don't want to ask a separate question as it might be marked a duplicate...

@Abhigyan Yes you can. Both before and after. If rotation around sigma bonds will determine that your molecule is achiral, it does not matter whether you rotate until you find a plane of symmetry or if you create the mirror image first and check by rotation if it fits later.

A chiral centre has 4 different atoms or groups attached to it. 2-propanol has two $\ce{CH3}$ groups attached to the second carbon, therefore it does not exhibit stereoisomerism.
Rotating a molecule about an axis so that it can be superimposed on its mirror image does not change its chemistry. If a molecule can be superimposed on its mirror image without having to make/break bonds to reform it, then it does not exhibit stereoisomerism.