Comparing bond angles in carbonyl dichloride and carbonyl dibromide Background
I was studying VSEPR and hybridization theory and one of the popular questions in many textbooks was comparison of bond angle of $\ce{NH3}$, $\ce{NF3}$ and $\ce{NCl3}$.
$\ce{NF3}$ had smaller bond angle as electrons were nearer to the more electronegative fluorine atom, thereby reducing bond pair - bond pair repulsion (of N-F bonds) and thus allowing contraction of bond angle.
Further, in $\ce{NCl3}$ bond angle was larger than both $\ce{NH3}$ and $\ce{NF3}$ due to steric repulsion between large Cl atoms. So the order was $\ce{NCl3}$>$\ce{NH3}$>$\ce{NF3}$.
Now, a question that I later saw in one of my textbooks was the analysis of bond angles of $\ce{COCl2}$ and $\ce{$\ce{COBr2}$}$, which have trigonal planar geometry. According to my textbook, $\ce{COCl2}$ has smaller $\ce{O-C-Cl}$ angle, less than $\mathrm{120^{\circ}}$, (and thereby greater $\ce{Cl-C-Cl}$ angle). The angles for $\ce{COBr2}$ were not given.
My Approach
Now if we apply the reasoning given for $\ce{NCl3}$ and $\ce{NH3}$, $\ce{NF3}$ in this question comparing $\ce{COCl2}$ and $\ce{COBr2}$, we would find that the $\ce{C-X}$ bond pair electrons would be more polarized towards X, if X is Cl, and thus analogously to $\ce{NF3}$, it should contract leading to a smaller bond angle than in $\ce{COBr2}$.
Also, due to steric repulsion, $\ce{C-X}$ bond angle in $\ce{COBr2}$ should be greater than $\ce{COCl2}$.
Solution given in the book
The author has stated that there are four electrons in the $\ce{C-O}$ bond
Thus, bond pair repulsion between electrons of $\ce{C-O}$ bond and $\ce{C-X}$ bond would be much greater than any repulsion between the electrons of the $\ce{C-X}$ bonds.
As electrons in the $\ce{C-X}$ bond would be closer to carbon atom (and thus $\ce{C-O}$ bond pair electrons) in $\ce{COBr2}$, therefore a greater amount of repulsion would be present between them. This repulsion would lead to contraction of Br-C-Br bond angle leading to an angle less than that of $\ce{COCl2}$.
However, no data was provided in support or against it. Also no reasoning was given why they would be greater or less than $\mathrm{120^{\circ}}$, only the above reasoning for comparing the bond angle was provided.
I have not been able to find any bond angle data. My teachers are also not sure. Also, the textbook I referred to is infamous for having lots of errors.
Which reasoning is correct, or is it something else entirely?
What is the actual data?

The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles of $\ce{COCl2}$ and $\ce{COBr2}$.
To circumvent these issues, a better method(less prone to failures) would be using Bent's rule. Now, there has been a lot of discussion on this website over what this rule actually is (see here) and why it is physically useful (see this). The Wikipedia article on the same has a wonderful intuitive discussion under the "Justification" tab for why this rule exists. However, I will not be delving into the origins of this rule, and just state outright that the basic premise of this rule is that not all hybrid orbitals are equivalent under a given hybridization scheme. The percentage s-character and p-character (that is, the amount of s-orbital or p-orbital present in a given hybrid orbital as a percentage of the whole) varies as per the electronegativities of the substituents.
A statement of this rule was provided by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

The idea is that the amount of p character will increase (alternatively, the amount of s character will decrease) in those bonds which have more electronegative substituents (that is, the atom other than carbon in the bond), while simultaneously the amount of p character will decrease(alternatively, the amount of s character will increase) in bonds towards less electronegative substituents in such a manner that the net sum of percentage s and p characters over all the bonds in the molecule comes out to be $100$%. This is essentially a redistribution of the orbital energies across bonds to minimize the energy of the system. Further details on this rule can be found in the links mentioned above.
Once you are aware of this rule, applying it to your case becomes a fairly simple task. I will try to compare the $\ce{Cl-C-Cl}$ and $\ce{Br-C-Br}$ angles here, the orders for the $\ce{O-C-Cl}$ and $\ce{O-C-Br}$ angles can be easily inferred from there. From a classical hybridization perspective, the carbon in both these molecules uses $\ce{sp^2}$ hybrid orbitals for bonding, and all the three hybrid orbitals thus generated should possess equivalent atomic orbital's character ($33.33$% s character and $66.66$% p character approximately). But as per Bent's rule, since $\ce{Cl}$ and $\ce{Br}$ are both more electronegative than carbon, they will increase p character towards their direction (that is, in the $\ce{C-Cl}$ and $\ce{C-Br}$ bonds respectively). Since $\ce{Cl}$ is even more electronegative than $\ce{Br}$, the p character increase in the $\ce{C-Cl}$ bonds will be more than that what will happen in the $\ce{C-Br}$ bonds.
Hence, we can say that more p character will be directed in the two $\ce{C-Cl}$ bonds as compared to the two $\ce{C-Br}$ bonds. This directly bears a relation to the bond angle: more is the p character in the constituent bonds, lesser will be the bond angle(you can see this from a fairly easy observation: $\mathrm{sp}$ hybrid orbitals are generally at an angle of $180°$ with $50$% p character, the $\mathrm{sp^2}$ hybrid orbitals are generally at an angle of $120°$ with $66.66$% p character and the $\mathrm{sp^3}$ hybrid orbitals are generally at an angle of $109.5°$ with $75$% p character).
Hence, Bent's rule predicts that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ should be lesser than the $\ce{Br-C-Br}$ of $\ce{COBr2}$. Experimental data says that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ is $111.83 ± 0.11°$, while the $\ce{Br-C-Br}$ of $\ce{COBr2}$ is $112.3 ± 0.4°$. Hence, the prediction made by Bent's rule is confirmed.
Notice that the difference between the given bond angles is quite small (around $1°$). This is also somewhat expected, as the electronegativity difference between chlorine and bromine is also quite small (approximately $0.2$).
The advantage of employing this method is that you don't have to consider conflicting factors like those you stated in your approach. The rule explains these angle variations based on Coulson's theorem, from which the following relation pops out:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
for $\theta \in (90^\circ,180^\circ)$

($\theta$ stands for the bond angle while ${P}$ and ${S}$ stand for p character and s character respectively)
As you can clearly see, when the p character goes up, the value of $\cos \theta$ also goes up if you closely observe the simplified form $\cos \theta = 1 - \frac{1}{P}$. This means that the value of $\theta$ will decrease for the given range of $\theta$ (as $\cos \theta$ is a decreasing function for $\theta \in (90^\circ,180^\circ)$)
Hence, you can generally predict things quite accurately with Bent's rule without getting confused with contrasting factors, as they are usually taken care of while determining the atomic orbital characters.


What should be the reaction between lead oxide and dilute sulfuric acid? I know for a fact that dilute $\ce{H2SO4}$ is a non-oxidizing acid which makes it a reducing agent. I also know for a fact that $\ce{PbO2}$ is a strong oxidizing agent. So, I thought Redox reaction should have occurred between them, but my teacher says that no reaction occurs between the said reactants because $\ce{PbO2}$ does not react with low concentration of acid. If it were so, why does $\ce{PbO2}$ react with dilute $\ce{HCl}$. Since, both dilute $\ce{HCl}$ and dilute H2SO4 are reducing acids of low concentrations, both should be eligible for the Redox reaction.

I have tried this on the internet but all I get are reaction balancing links .JD Lee's Concise Inorganic Chemistry has nothing about this also.

YOur logic is wrong.

  I know for a fact that dilute $\ce{H2SO4}$ is a non-oxidizing acid...

sort of, but so far so good.

  ...which makes it a reducing agent.

This is where you are wrong, just because it is non-oxidizing does not make it reducing. Concentrated $\ce{H2SO4}$ is an oxidizer, diluting it does not make it less of an oxidizer, but limits the strength/mechanism.

A similar example is mixing nitroglycerin and diatomaceous earth doesn't make nitroglycerin less explosive, it just makes it harder for it to explode (dynamite).


Why the boiling point decreases with increase in branching in alkyl groups? I found out the answer is that

with increase in branching, the molecules attain a spherical shape with less surface area. As a result, interparticle forces become weaker resulting in lower boiling point.

My problem is that if it attains a spherical shape then how is its surface area less because sphere has the largest surface area?
So please anyone can help me by answering in simple language?

It's not just how much surface area, but, crucially, the surface area in contact with other molecules. Bear in mind that we are looking especially at van der Waals' forces here (or London dispersion forces, or instantaneous dipole forces, or whatever other name you have for them).

If you have a lot of long chain molecule together, almost all of their surface area is very near the next molecule. Examine these sausages. See how almost their entire length on the left and right side of each sausage is in contact with the next sausage. Going back to molecules, this represents a large surface over which the intermolecular forces can act.

On the other hand, branched molecules do not fit as tightly together. See this image (no pun intended)

Extra branches on the molecules mean that they do not stack as well, so for a given molecule, there is less surface area of that molecule in close contact with another molecule (therefore weaker van der Waals).


Question about null space balancing method I'm trying to use the null space method to balance the following equation:
.
I obtained the following composition matrix:
$$\begin{bmatrix}
3 & 8 & 1 & 12 & 4 & 2 \\
1 & 0 & 0 & 1 & 0 & 0 \\
4 & 4 & 3 & 40 & 3 & 1 \\
0 & 2 & 1 & 3 & 1 & 0 \\
0 & 1 & 0 & 12 & 0 & 0
\end{bmatrix}$$
where rows are in order H P O N Mo. I take the rref of this matrix, and augment it with one row of zeroes except the last element is 1. Taking the inverse of that matrix, I get $$\begin{bmatrix}
\frac{1}{51} & \frac{4}{17} & \frac{35}{51} & -\frac{1}{51} & -\frac{56}{51} & 1\\
\end{bmatrix}.$$
After scaling the elements by 51 I would get the end result of coefficients: 1, 12, 35, 1, 56, 51 which does not at all balance the equation. A correct solution would be 1, 12, 21, 1, 21, 12.
I have used the exact same steps to balance other equations. What am I doing wrong?

Fortunately this issue was just caused by a small oversight. In the midst of so many coefficients, you accidentally skipped the second nitrogen atom in $\ce{NH4NO3}$, which changes the fifth element in the fourth row from 1 to 2. The correct composition matrix is then:
$$\begin{bmatrix}
3 & 8 & 1 & 12 & 4 & 2 \\
1 & 0 & 0 & 1 & 0 & 0 \\
4 & 4 & 3 & 40 & 3 & 1 \\
0 & 2 & 1 & 3 & \color{red}{\textbf{2}} & 0 \\
0 & 1 & 0 & 12 & 0 & 0
\end{bmatrix}$$
After complementing with the bottom row $\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and inverting the matrix, we obtain:
$$\frac{1}{96}\begin{bmatrix}
3 & 79 & 2 & -9 & -14 & -8 \\
36 & -204 & 24 & -108 & -72 & -96 \\
39 & -477 & 90 & -213 & -246 & -168 \\
-3 & 17 & -2 & 9 & 14 & 8 \\
-51 & 417 & -66 & 249 & 174 & 168 \\
0 & 0 & 0 & 0 & 0 & 96
\end{bmatrix}$$
Finally, transposing the rightmost column and dividing its coefficients by 8, the result is  $\begin{bmatrix} -1 & -12 & -21 & 1 & 21 & 12 \end{bmatrix}$, exactly as expected.
Thank you for the linked article! Hopefully others may find it useful. If you're into linear algebra applied to chemistry, I recommend you take a peek into chemical reaction network theory!


Specific reduction of certain functional groups Consider the compound $\ce{Alkyl-(4 oxocyclopent-2-ene)carboxylate (A)}$. For example consider-

  Now consider the operations:

  $\ce{1. (A) + (i)BH3/THF (ii) H+}$

  $\ce{2. (A) + H2/Pt}$

  $\ce{3. (A) + H2/Pd-C}$

  Products

  In $1.$ ester group reduced to an alcohol.

  In $2.$ double bond reduced to a single bond, and

  In $3.$ double bond reduced to a single bond, ester group reduced to an alcohol.

However, on the next page addition of $\ce{(i)BH3/THF (ii) H+}$ selectively reduced carboxylic acid when both ester and carboxylic acid were present.

My questions:

$(i)$ I know that $\ce{(i)BH3/THF (ii) H+}$ can reduce carboxylic acid, but I am unaware if it can reduce ester.

$(ii)$ I only knew that, the reduction of ester in presence of carbonyl group could be done by first protecting the carbonyl group (converting it to cyclic ketal by reacting with ethylene glycol) and then using $\ce{LiAlH4}$. Is reaction $3.$ correct?

$(iii)$ Why aren't product $2.$ and $3.$ same? I've always seen the two work same.

Note:

My chemistry teacher gave me few notes for Chemistry Olympiad preparation, so I can not cite source. I don't know how to make compounds, so will completely reside on IUPAC nomenclature.

1.The selective reduction of the carboxylic group is explained in this article. Link:https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114. It could reduce esters but is not favourable at all and difficult.So if u only have an ester grp and its treated with bh3, then you can reduce it. However if both carboxylic and ester grps are present, carboxylic acid gets reduced .

Regarding reduction with h2/pt and h2/pd-c not much can be said. The thing is at higher temperatures and pressures(extreme conditions)both could reduce esters. Generally esters are NOT reduced via cataytic hydrogenation (it is just not favourable).However reduction has been performed in mild conditions via other complexes like ruthenium.(https://pubs.acs.org/doi/pdfplus/10.1021/op4003278


Can ethyl acetate isolate cinnamaldehyde from cinnamon oil? I recently watched a video on extracting cinnamaldehyde from cinnamon using dichloromethane as a solvent.  I would like to attempt the process.  However, I would prefer not to use haloalkanes because of the health risks.  Would ethyl acetate be a suitable replacement solvent for dichloromethane in extracting cinnamaldehyde from cinnamon oil?

Yes, ethyl acetate can be used to isolate cinnamaldehyde from cinnamon oil, though there are better non-haloalkane options. In a study that aimed to optimize the extraction of cinnamaldehyde from cinnamon powder, it was found that from methanol, ethanol, ethyl acetate, and water, methanol was the most efficient. The exact method used in the study was:

Collection of plant material: Bark of cinnamon was collected from a local market at Visakhapatnam, AP. The bark was cleaned and dried under sunlight for $24\ \mathrm{h}$. The dried bark was powdered and used as a raw material and stored in an air tight container. Cinnamon powder was sieved by using different particle sizes ranging from 354 to 125 microns.
Preparation of the extract: Cinnamon powder ($2\ \mathrm{g}$) was added with ethanol ($25\%$) and methanol ($25\%$) in different flasks and the volume was made $50\ \mathrm{mL}$. The solution was soaked for $1\ \mathrm{d}$ and $3\ \mathrm{d}$ respectively. After the soaking time, the solution was filtered using Whatman No.1 filter paper and the filtrate solution was heated to $78^\circ \mathrm{C}$ and $65^\circ \mathrm{C}$ and made up to $50\ \mathrm{mL}$ with distilled water and hexane and incubated for $2\ \mathrm{h}$.


Is it a redox reaction if the oxidation state of the element has changed only in one of the products? I’m not sure on which elements to compare to ensure that this is a redox reaction:

$$\ce{2 H2SO4 (aq) + 2 NaBr (s) -> Br2 (l) + SO2 (g) + Na2SO4 (aq) + 2 H2O}$$

I’m trying to indicate whether this involve oxidation/reduction. I’m looking at the oxidation state of $\ce{S}$.

What do I compare $\ce{H2SO4}$ with: $\ce{SO2}$ or $\ce{Na2SO4}$?

I tried both. Oxidation state of $\ce{S}$ in $\ce{SO2}$ is $+4$ and in $\ce{Na2SO4}$ is $+6$. One of them tells me it’s reduced, and the other tells me that no redox reaction is happening.

Which do I compare with and why? I’m not sure.

One sulphuric acid is used to form sodium sulphate and the other one is used for oxidizing.
So you have a reduction:

$\ce{4H+ + SO4^2- +2e- -> SO2 + 2H2O}$

and an oxidation:

$\ce{2Br- -> Br2 +2e^-}$

The sodium sulphate is formed with:

$\ce{H2SO4 + 2NaBr -> Na2SO4 + 2HBr}$

This last one is not a redox reaction.


What is the mechanism behind the decomposition of enediol phosphate intermediate to methyl glyoxal? I can't seem to figure out the mechanism for the decomposition of enediol intermediate (in glycolysis) to methyl glyoxal? How does the phosphate group get replaced by a hydrogen?

This reaction can occur only when the $\ce{C-O}$ bond to the phosphate group
  lies in a plane that is nearly perpendicular to that of the enediol so
  as to permit the formation of a double bond in the intermediate enol
  product.

In solution, the enediol intermediate readily breaks down with the elimination of the phosphate at C3 to form the toxic compound methylglyoxal:

The spontaneous decomposition of the enediol intermediate in the triose phosphate isomerase (TIM) reaction to form methylglyoxal through the elimination of a phosphate group.

In order for this elimination to occur, the bond to the phosphate group must lie, as shown above (a), in the plane perpendicular to that of the enediol.

This is because, if the phosphate group were to be eliminated while this bond was in the plane of the enediol as shown in (b), the $\ce{CH2}$ group of the resulting enol product would be twisted 90° out of the plane of the rest of the molecule. Such a conformation is energetically prohibitive because it prevents the formation of the enol’s double bond by eliminating the overlap between its component $\mathrm{p}$ orbitals.

Reference

Voet and Voet Biochemistry


What's the difference between "extent of reaction" and "position of equilibrium"? My chemistry textbook defines the terms in question as follows:

  Extent of reaction: the relative amounts of products compared with reactants. The extent of reaction is indicated by the value of the equilibrium constant.

  Position of equilibrium: the relative amounts of products and reactants at equilibrium. The position of equilibrium varies depending on the extent of the reaction.

I don't understand the difference between these two concepts and would appreciate it if someone could explain both how they are different and how they are linked/influence each other.

Thanks.

Two differences. Firstly, the extent of reaction $\xi$ can be mathematically defined, whereas the "position of equilibrium" is for the most part qualitative, in the sense that we only say it "moves left" or "moves right", but not "the position of equilibrium is $x$".

More importantly, the extent of reaction does not actually tell you anything about the equilibrium. In this respect, it is more analogous to the reaction quotient $Q$ rather than the equilibrium constant $K$. If you measure the extent of reaction $\xi$ when the system is at equilibrium, then it does provide a measure of the equilibrium position. This is just like how if you measure $Q$ at equilibrium, it is equal to $K$. However, in principle you can measure $\xi$ (or $Q$) at any stage of the reaction, not just when it is at equilibrium.

Both $\xi$ and $Q$ are mathematical ways of describing how much products there are versus how much reactants, but they are not equivalent, they have different forms. See Wikipedia for more information.

So, I don't agree with the statement that "the extent of reaction is indicated by the value of the equilibrium constant" on two counts. Firstly, the extent of reaction is not directly "indicated" by $Q$ – they both measure the same thing, but not in the same way. Secondly, and crucially, $Q$ is not the equilibrium constant $K$: it is only equal to $K$ at equilibrium, which this sentence doesn't capture.


Conductometric titration curve when KCl is gradually added to AgCl solution The situation given is that we are performing conductometric titration where KCl is gradually added to AgCl solution and we have to plot the variation of conductivity with the volume of aq. KCl added. I am new to conductometric titration and I have covered the cases of titration of acid v/s bases but I have never come across a case like this.

My thought was that since AgCl is a weak electrolyte it's initial conductivity would be low. And now we add KCl to it which is a strong electrolyte so the conductivity should increase. I also think that common ion effect would come into play and the dissociation of AgCl would be pushed back further but I'm still unsure about how an end point would be obtained here. Can someone guide me? Any help would be appreciated.

I'm a bit puzzled by the phrase "AgCl solution," since there wouldn't be a solution but rather colloidal size particles of AgCl in the aqueous solution. I don't think you could have a bottle of colloidal size particles of AgCl since once dried the particles would stick together in a lump.

For AgCl, a tiny amount would dissolve hence there would be conductivity slightly better than pure water.

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$

However AgCl also form a complex with chloride:

$$\ce{AgCl(s) + Cl-(aq) <=> AgCl2-(aq)}$$

So as you add KCl the curve would show one slope as KCl builds up enough concentration to dissolve the AgCl dissolves, then a second slope as the AgCl dissolves, then a third slope after all the AgCl has dissolved.

For ecah change the curve won't look like two straight lines intersecting at a point but more like a parabola. So you'd need to extend linear portion of each and find the end point at the intersection of the linear extensions.


Can someone explain the article below? This is the article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5611670/
I'm not a Chemist, so I can't understand the article very well.

The conclusion states that the molecules tested should not be called "piezolytes". Is this a nomenclature issue only? My only question is: does the presence of such molecules in deep sea organisms help in keeping proteins in their naturally folded state, resistant to denaturation due to high pressure or not?

From the abstract of the article in question:

  The results suggest that the pressure stability of a protein in solution is not directly affected by the presence of these proposed piezolytes, and so they cannot be granted this distinction.

So the answer is "no", the supposed piezolytes do not appear to have the purported function. They do stabilize proteins against thermal denaturation, but not against pressure-induced destabilization.


Voltammetry on homemade supercapacitor I would like to get some voltammetry curves with my homemade supercapacitor in order to be sure that I have only capacitive effects (capacitive current) in my cell and no redox reactions (faradic current).
One of my electrode is my carbon material and the other one is stainless metal.

I only know the 3 electrodes system, do I really need it in this case ? How should I process for my experiment ?

Testing capacitors is almost always done with a two electrode setup.
There are some details here about interpreting the data, but the experimental setup is that you just connect the counter and reference electrode inputs together to one half of the capacitor and the working electrode input to the other.
A three electrode setup is used when we only care about the interfacial potential of a single electrode (the working), and thus use a measurement near it by the reference electrode to control the counter electrode accordingly.
In a capacitor, we are generally interested in controlling the voltage across the entire device and observing its behaviour as a whole, so we don't need a reference electrode. (though depending on your specific device, it may be useful to observe only one electrode for any number of reasons)


How can I remove a PEG shampoo from a natural horn-comb without damaging it? I washed by accident a horn-comb with a shampoo that contains PEGs, now how can I remove these types of chemicals from the horn-comb without damaging its surface?

Soap, water and a dash of elbow grease. PEGs are all surfactants which will wash away with water and if they don't then an alcohol-water mix (vodka) will certainly remove any PEG residue, probably along with some of the natural oils in the horn comb that has built up over time and usage.

This may leave the comb with whitened patches or streaks on it.  These are not a PEG residue but fine cracks in the horn, interstices in the layers of keratin that the horn is made of, that used to be filled with oil which the PEG shampoo has removed.  Any number of natural oils will improve the appearance of the horn, including, but not limited to, natural scalp oils (i.e. use the comb for its intended purpose), linseed oil (a favourite of cricketers for their bats), castor oil, beeswax, olive oil (a bit sticky) or a light application of any proprietary furniture polish.  Choose something unscented if you don't want Mom to find out. And start with the least viscous options you have, these will achieve greater penetration of the fine cracks in the horn and "seal" them with a wax if possible.

I've seen antique tortoiseshell respond very badly to cleaning with detergents, but the damage is really illusory and can be restored with the gentle application of oils and waxes.


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?

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.


Constitutional isomers for four carbon thioesters Question:

  How many isomers are there for thioesters with the formula $\ce{C4H8OS}$?

Answer given is four.

I can see that the following two are possible constitutional isomers:

But what are the other two?

One way to break things down in by the number of carbon atoms on either "side" of the thioester.

There are 4 carbon atoms to work with, of which one will be taken up by the thioester moiety.  Thus there are three left to work with.

On the carbonyl "side" of the thioester, between zero and three carbons are possible.  However, on the sulfur "side" of the thioester, at least one carbon atom is required; otherwise the compound is not a thioester but a thioacid.  Thus the possibilities for carbon atom distribution are:

one carbon on S side, two on CO side.
two carbons on S side, one on CO side.
three carbons on S side, zero on CO side.

You have drawn the only structures for the first two possibilities.

For the last possibility, there are two possible ways to attach a three-carbon alkyl chain to the thioester sulfur:

as a $n$-propyl group
as an isopropyl group

Thus, the four possible thioesters are:

methyl thiopropanoate
ethyl thioacetate
n-propyl thioformate
isopropyl thioformate


Does cooling a potato change the nature of its carbohydrates? A talk-show guest of Joe Rogan claimed that cooling a potato after cooking creates "resistant starch" that's better for human consumption

https://www.youtube.com/watch?v=niwqfwA2Lb8

The claim is also made here: https://www.healthline.com/nutrition/cooling-resistant-starch

This seems to contradict everything I know about chemistry, since cooling only slows down all chemical processes.

Is there any validity to the claim? What is this "resistant starch"? Why does cooling the potato after cooking create it?

Does cooling a potato change the nature of its carbohydrates?

Yes, retrogradation is a reaction that takes place when the amylose and amylopectin chains in cooked, gelatinized starch realign themselves as the cooked starch cools.

Resistant Starch:

  There is weak evidence that resistant starch can improve fasting glucose, fasting insulin, insulin resistance and sensitivity, especially in individuals who are diabetic, overweight or obese. In 2016, the U.S. FDA approved a qualified health claim stating that resistant starch might reduce the risk of type 2 diabetes, but with qualifying language for product labels that limited scientific evidence exists to support this claim. Because qualified health claims are issued when the science evidence is weak or not consistent, the FDA requires specific labeling language, such as the guideline concerning resistant starch: "High-amylose maize resistant starch may reduce the risk of Type 2 diabetes. FDA has concluded that there is limited scientific evidence for this claim."

U.S. National Library of Medicine - "Studies on effect of multiple heating/cooling cycles on the resistant starch formation in cereals, legumes and tubers" (non-paywall source), (July 2009), by Yadav BS, Sharma A, and Yadav RB:

  Abstract
  'Resistant starch' (RS) is defined as starch and starch degradation products that resist the action of amylolytic enzymes. The effect of multiple heating/cooling treatments on the RS content of legumes, cereals and tubers was studied. The mean RS contents of the freshly cooked legumes, cereals and tubers (4.18%, 1.86% and 1.51% dry matter basis, respectively) increased to 8.16%, 3.25% and 2.51%, respectively, after three heating/cooling cycles (P< or =0.05) with a maximum increase of 114.8% in pea and a minimum of 62.1% in sweet potato (P< or =0.05). Significant positive correlations were observed between the RS content and amylose (y=0.443x-5.993, r=0.829, P< or =0.05, n=9) as well as between the percentage increase in RS and insoluble dietary fiber content (y=2.149x-24.787, r=0.962, P< or =0.05, n=9). A differential scanning calorimeter study showed an increase in the T(0), T(p), T(c) and DeltaH values of the repeatedly autoclaved/cooled starches. The intact granular structure was also observed disappear, as studied using scanning electron microscopy.


Fluorescence quenching by palladium(II) vs. platinum(II) complexes Why do we observe much stronger (in most cases complete) fluorescence quenching by $\ce{Pd^2+}$ than $\ce{Pt^2+}?$ I work with complexes of both metals and there is no real 100% explanation I can find why this is the case. I am looking for a full quantum mechanical explanation (if possible). If you know a piece of literature, please point me towards that.

The theory of this quenching effect was developed about $70$ years ago by Förster, who showed that the quenching is not due to diffusion or collision. It is due to a resonance phenomena. The fluorescence yield is quenched by a solute whose mirror image of the adsorption spectrum overlaps the emission spectrum of fluorescence. It is is proportional to the overlap integral of these two spectra.
The quenching effect due to an agent X on the fluorescence is given by Stern-Volmer equation : ${I_0/I = 1 + \gamma [X]}$. Now T. Förster has shown that the quenching constant $\gamma$ is equal to $\gamma = \frac{2\pi ^2}{3} \frac{N}{1000} R_0^3$, where $N$ is Avogadro number, and $R_0$ is the critical Förster distance, where $R_0^6$ is proportional to the overlap integral $J(\nu )$ of the two spectra, and to the sum of all energy transfer rate constants (solvent to solutes and solute to solute).
This theory will not be developed here.
But, as you are looking for a piece of literature, I recommend the following references, where this theory is explained.
See : T. Förster, Discuss. Faraday Soc. $27, 1, (1959)$ and : T. Förster, Ann. Physik $2, 55, (1947)$.


Why are 3 equivalents of organolithium needed in synthesis of ketone from carboxylic acid? For the reaction of a carboxylic acid and organolithium reagents to synthesise a ketone, Clayden's, on p 219, mentions:

  Notice that three equivalents of organolithium are needed in this reaction: one to deprotonate the acid, one to deprotonate the hydroxyl group, and one to react with the lithium carboxylate.

It seems to me that "the acid" and "the hydroxyl group" refer to the same thing. A carboxylic acid molecule only has one acidic proton and it belongs to a hydroxyl group. Thus, shouldn't these two terms refer to the same thing and only two equivalences of the organolithium reagent be necessary?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

The text box you are quoting from (p 219) refers to this specific reaction. The compound in question possesses an α-hydroxyl group, which needs to be deprotonated too.

Notice that three equivalents of organolithium are needed in this reaction: one to deprotonate the acid, one to deprotonate the hydroxyl group, and one to react with the lithium carboxylate. These chemists added a further 0.5 for good measure.

However, the scheme directly above this incorrectly indicates that 3 equivalents of MeLi are required for a generic carboxylic acid:

In the absence of any other acidic groups, only 2 equivalents of organolithium are needed. In practice, however, more than 2 may be added.
Note that the 3 protons are still needed to get to the protonated species shown directly afterwards. In practice the proton source (e.g. saturated aqueous $\ce{NH4Cl}$, or just plain water) is usually added in excess.


Does Amoxicillin (as trihydrate) degrade in stomach acid? I crushed my antibiotics (amoxicillin + potassium Clavulanate). Does it deform in the stomach acid and therefore become useless? Or does the acid have no effect and the amoxicillin goes directly into the intestines and blood stream subsequently?

Because amoxicillin is available as oral tablets, it can be ascertained that it is not significantly degraded to a useless form in the stomach.  In fact, it has an oral bioavailability of 95% (meaning 95% of the orally ingested material goes into the blood stream rather than being destroyed in the stomach). The term for an oral medication being destroyed before it enters the blood stream is "first pass metabolism". So oral amoxicillin can be said to undergo minimal first pass metabolism. Oral potassium clavulanate also has a high oral bioavailability.


To what microscopic quantity does Gibbs energy correspond to? Is the Gibbs energy of a system equal to the potential energy of a system? The answers to this and this questions seem contradictory. Also, this does not seem to tell that Gibbs energy is equal to the potential energy. By potential energy, I mean the Electric potential energy between the atoms, electrons and nuclei.(I'm not considering other potentials due to nuclear forces or gravitational forces.)

If Gibbs energy is not potential energy, then what is the physical significance or physical interpretation of Gibbs energy at the molecular level? I know that Gibbs energy is equal to the maximum non-expansion work that can be extracted. I just want to know what does that mean at the molecular level.

As others have said, the Gibbs function (or Gibbs energy) makes sense at a macroscopic, thermodynamic level (instead of a microscopic level), but if you need to picture it, you can elaborate what it means from the deconstruction of its mathematical definition.

Internal energy

Let's start with internal energy, U. Internal energy is the mechanical energy (potential + kinetic) of all the particles in the system. Take into account that it isn't reducible to the energy of separate particles - these particles are interacting (for chemical systems, through intermolecular forces) and the potential and kinetic energy are intrinsically related (though quantum mechanics). The only energy that you aren't considering is the kinetic energy of the system as a whole - if your chemical system is moving through space or rotating around its centre of mass, that energy isn't relevant here (it's also trivially easy to calculate via Newtonian mechanics) - and potential energies that arise from external factors that affect the system uniformly - so if your whole system is subject to gravity, that potential energy isn't computed into the internal energy (also trivially easy to calculate via classical physics). In other words, the internal energy discounts anything that you could calculate if you picture your system as a point mass (as you do in high school mechanics); it deals with whatever is happening inside that point mass.

This mechanical energy is conserved, but the balance between its components will depend on the exact positions of all atoms; there would be constant transferences of energy from kinetic to potential, from inter- to intramolecular interactions, etc. It's easier if we start picturing the system in statistical terms - as a huge amount of molecules, moving in random directions at random speeds (following known statistical distributions), of which thermodynamic functions (such as Gibbs') capture an average.

So, in the internal energy you are taking into account: 1) the kinetic energy of molecules in your system, 2) the potential energy of intermolecular interactions, 3) the kinetic energy of intramolecular movements (rotations, vibrations), 4) intramolecular potentials (arising from the electronic structure of molecules).

Enthalpy

Since we are considering the movement of molecules relative to the centre of mass of the system, we know that the total movement of the system is zero - the centre of mass remains fixed. If you add all the momenta of the molecules, you get a zero.

However, there's a possible collective movement that, while averaging to zero, implies a macroscopic physical change: movements towards the system or away from it. It is possible, if you add all the momenta, to get a total momentum of zero and, at the same time, that more molecules are moving away from the centre of mass than towards it - if you picture it in macroscopic terms, that means that the system as a whole remains at the same position, but is expanding (think of a balloon). And, likewise, if more molecules move towards the centre than away from it, the system is immobile but contracting. Both these collective movements can, potentially, change the volume of the system - they are a microscopic interpretation of pressure.

Note that, ff the system actually changes volume, the intermolecular potential energy is affected - if the system expands, its molecules are further apart and its interactions are weaker; for a contracting system, the opposite is true. The energy involved in that process is purely a mechanical work - captured in the term $pV$. When we work with solids or liquids, changes in volume are usually negligible; when we work with gasses, we usually store them in rigid containers - so changes in volume are also usually negligible. Therefore, it's convenient for chemists to define a function that discounts this mechanical work from the internal energy - that's exactly what enthalpy, H, is.

So, in the enthalpy you are taking into account: 1) molecular movement (except expansion and contraction), 2) intermolecular potentials, 3) intramolecular movement, 4) intramolecular potentials.

Entropy

The molecular interpretation of entropy, S, is always contentious (disregard any explanation that includes the words "chaos" or "disorder"), but we can discuss it simplistically. Entropy is more complex to visualise because it's an intrinsically statistical magnitude, and it ultimately relates to how energy is transferred between the different forms (inter- and intramolecular, kinetic and potential). What entropy implies is that when more than one configuration produces equivalent states, that makes the state more likely - effectively stabilising it, lowering its energy.

If you took microscopic photos of your system at different points, you'd get random positions and states for its molecules - but some situations would be more likely, from a purely random viewpoint, than others. If you have a two-component system, it's very very unlikely that their molecules will be spontaneously segregated in different parts of the system - if you place molecules randomly, it's far more likely that they will be well mixed throughout the system, so, more entropy for the latter. Molecules with many internal movements (like large molecules with many vibrations) also produce more equivalent states - you could get many photos in which the only difference is what particular bond vibrates more strongly, while that's impossible for, say, diatomic compounds. An ideal crystal at T=0K, where atoms are fixed exactly in periodic positions, without defects, and without vibration because they don't have any kinetic energy, has only one way to be arranged - so it has minimum entropy.

The ability to populate these configurations is closely tied to temperature. Temperature provides the energy that produces these movements and allows the system to explore more configurations - through Boltzmann's distribution. Since entropy indicates how many configurations there are for a given state and temperature provides the energy to do so, the "entropic stabilisation" term depends on temperature, and, conversely, systems with higher entropy are more sensitive to entropic stabilisation. That's what the term $TS$ measures.

This is important because that "entropic stabilisation" means that a fraction of the energy of the system (precisely $TS$) is tied up in providing these alternative configurations, and cannot be extracted directly from the system. Unless chemical reactions or other structural changes take place, that energy is inaccessible.

Gibbs function

We can then describe a function that computes 1) the internal mechanical energy of the system, 2) discounts the energy involved in expansion/compression work, and 3) discounts the energy that is inaccessible due to entropy. This is what the Gibbs function, G, means at a microscopic level. Note that this doesn't make for a clear-cut distinction between what microscopic forms of energy are included in the Gibbs function - it isn't as simple as "kinetic out, potential in", because some things (mainly entropy) are statistically defined and only make sense for a collective of (interacting) particles.


Why does magnesium have an exceptionally low melting point? Why does magnesium have a lower melting point than both calcium and beryllium? It does not seem to fit into the group trend.

Following up on my comment, even though I'm not convinced that this is a 'real' answer to the question.  Consulting the SGTE database of elemental free energies [1] (with various updates over the years), one can obtain the melting temperatures for the hcp, fcc, and bcc phases of beryllium, magnesium, calcium, strontium, and barium (rough guesses looking at plots — consider good to approx. $\pm\pu{5 K}):$

$$
\begin{array}{lrrrcc}
\hline
\text{Element} & T_\mathrm{m}(\text{hcp})/\pu{K} & T_\mathrm{m}(\text{bcc})/\pu{K} & T_\mathrm{m}(\text{fcc})/\pu{K} & \text{Phase at RT} & \text{Melt phase} \\
\hline
\ce{Be} & 1544 & 1560 & 1018 & \text{hcp} & \text{bcc} \\
\ce{Mg} & 922 & 755 & 705 & \text{hcp} & \text{hcp} \\
\ce{Ca} & 940 & 1114 & 1063 & \text{fcc} & \text{bcc} \\
\ce{Sr} & 900 & 1050 & 1030 & \text{fcc} & \text{bcc} \\
\ce{Ba} & 635 & 1000 & 709 & \text{bcc} & \text{bcc}\\
\hline
\end{array}
$$

Several things to note, amongst them the variety of which crystal structure is most stable for the different elements at different temperatures. Also, the hcp phase's melting point does almost(!) monotonically decrease going down the table.  The biggest anomaly would actually seem to be the relative instability of the bcc and fcc phases of magnesium (it is the only one that remains hcp up to the melting point).

Overall, it would appear that the group 2 elements display a wide variety of preferences for crystal structure, and hence bonding energy configurations.  Is $\ce{Mg}$ unusual? Not really…

References

Dinsdale, A. T. SGTE Data for Pure Elements. Calphad 1991, 15 (4), 317–425. DOI: 10.1016/0364-5916(91)90030-N.


how to determine the relative differences in activity between metals I am doing a lab for school and I do not know what this question is asking. In this lab we are making electrochemical cells from different metals.

You have a standard pot of liquid electrolyte into which are dipped non-touching same-spaced paired strips of different clean metals.  The external circuit between the strips is a sensitive voltmeter (with extremely high electrical resistance - hard by zero current flow).  You measure the potential (noting direction!) between all combinations of paired strips, (n)(n-1)/2, and then draw certain conclusions.

Why are amalgams saturated with respect to excess solid metal interesting substrates for this experiment (e.g., aluminum)? How do you construct a standard hydrogen electrode?


What causes the old book smell? I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question):

One thing I notice is that there is a distinct smell that comes from the pages of older books. (the smell is not unpleasant) - it seems that the older the book, the more distinct the smell.

What is the chemistry behind the old pages in a book smell?

This is a very interesting question. Provided that the materials used in making papers aren't the same around the globe, this is a very broad case of study. However, a study has been conducted in which the main goal was to identify the compounds that are the cause of the smell; VOCs:[1]

  Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air.

So, quoting the article:

  Using supervised and unsuper-vised methods of multivariate data analysis, we were able to quantitatively correlate volatile degradation products with properties important for the preservation of historic paper: rosin, lignin and carbonyl group content, degree of polymerization of cellulose, and paper acidity.

  It is a result of the several hundred identified volatile and semi-volatile organic compounds (VOCs) off-gassing from paper and the object in general. The particular blend of compounds is a result of a network of degradation pathways and is dependent on the original composition of the object including paper substrate, applied media, and binding.

  The 15 most abundant VOCs present in all chromatograms were selected for further analyses: acetic acid, benzaldehyde, 2,3-butanedione, butanol, decanal, 2,3-dihydrofuran, 2-ethylhexanol, furfural, hexadecane, hexanal, methoxyphenyloxime, nonanal, octanal, pentanal, and undecane.

Strlič, M., Thomas, J., Trafela, T., Cséfalvayová, L., Kralj Cigić, I., Kolar, J., Cassar, M. (2009). Material degradomics: on the smell of old books. Analytical Chemistry 2009, 81 (20), 8617-8622.
The main link to the article might be behind a paywall, so for the interested, the relevant ResearchGate article is accessible.


Why does NMR deal with protons? I ask this b/c I know the NMR deal with magnetic fields (MF) and it can change energy of proton’s magnetic energy. But my question is that electrons also have a spin and M F. Just like how IR and Mass deal with electrons, why don’t NMR also talk about electrons? Why just focus in the nucleus.

NMR deals with the nucleus of an atom (not just protons, but all nuclear isotopes with spin>0). Electrons also have a magnetic moment, and the technique used to probe this phenomenon is EPR/ESR. While NMR uses frequency in the MHz range (MHz), ESR uses higher energies in GHz range.

Lots of common references available on the internet to learn more, but you might be interested in:

https://physics.stackexchange.com/questions/171256/whats-the-difference-between-nmr-and-epr
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance


How is the chirality of sidechains written in the IUPAC name? I circled what I believe to be the chiral centers and I added the locants for the longest carbon chain. One of the chiral centers is in a side chain, so that carbon doesn't have a locant. How should its absolute configuration be included into the IUPAC name?

You have correctly ideentified and numbered the parent structure (the longest chain), which is undec-3-yne.
Including the substituent groups, the complete name is 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne.
Note that names for simple substituent groups derived from acyclic hydrocarbons with the free valence not at the end of the chain are generically called ‘alkanyl’. A low locant is assigned to the free valence (i.e. the suffix ‘yl’) in accordance with the numbering of the chain. In this case, butan-2-yl is such a substituent group.
Also note that the locant ‘2’ in butan-2-yl refers to a different structural element compared to the other locants in 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne. Therefore, parentheses are used around the substituent prefix to separate the locants.
If you need to indicate the stereochemical configuration, the same numbering and separation of locants is used, for example: (5R,10R)-10-bromo-5-[(2S)-butan-2-yl]-2-methylundec-3-yne.


Extracting other mineral from salt in evaporated seawater I recently stumbled upon this news:

Seawater contains 47 minerals and metals. Starting with the most abundant and proceeding to the least abundant, these are chloride, with a concentration of $\pu{18 980 ppm}$ (parts per million) in seawater, sodium ($\pu{10 561 ppm}$), magnesium ($\pu{1 272 ppm}$), sulfur $(\pu{884 ppm})$, calcium ($\pu{400 ppm}$), potassium ($\pu{380 ppm}$), bromine ($\pu{65 ppm}$), inorganic carbon ($\pu{28 ppm}$) and strontium ($\pu{13 ppm}$). Then follow boron ($\pu{4.6 ppm}$), silicon $(\pu{4 ppm})$, organic carbon ($\pu{3 ppm}$), aluminum ($\pu{1.9 ppm}$), fluorine ($\pu{1.4 ppm}$), nitrogen in the form of nitrate ($\pu{0.7 ppm}$), organic nitrogen ($\pu{0.2 ppm}$), rubidium ($\pu{0.2 ppm}$), lithium ($\pu{0.1 ppm}$), phosphorous in the form of phosphate ($\pu{0.1 ppm}$), copper $(\pu{0.09 ppm})$, barium ($\pu{0.05 ppm}$), iodine (also $\pu{0.05 ppm}$), nitrogen in the form of nitrite (also $\pu{0.05 ppm}$) and nitrogen in the form of ammonia (once more $\pu{0.05 ppm}$). Thereafter, we have arsenic ($\pu{0.024 ppm}$), iron ($\pu{0.02 ppm}$), organic phosphorous $(\pu{0.016 ppm})$, zinc $(\pu{0.014 ppm})$, manganese ($\pu{0.01 ppm}$), lead ($\pu{0.005 ppm}$), selenium ($\pu{0.004 ppm}$), tin $(\pu{0.003 ppm})$, caesium ($\pu{0.002 ppm}$), molybdenum (also $\pu{0.002 ppm}$) and uranium ($\pu{0.0016 ppm}$). Then come gallium ($\pu{0.0005 ppm}$), nickel (also $\pu{0.0005 ppm}$), thorium (also $\pu{0.0005 ppm}$), cerium ($\pu{0.0004 ppm}$), vanadium $(\pu{0.0003 ppm})$, lanthanum (also $\pu{0.0003 ppm}$), yttrium (also $\pu{0.0003 ppm}$), mercury (once more $\pu{0.0003 ppm}$), silver (also $\pu{0.0003 ppm}$), bismuth ($\pu{0.0002 ppm}$), cobalt $(\pu{0.0001 ppm})$ and,finally, gold ($\pu{0.000008 ppm}$). Altogether, there are some 50 quadrillion tons (that is, $\pu{50 000 000 000 000 000 tons}$) of minerals and metals dissolved in all the world’s seas and oceans.

Which is quite surprising for me (thought those white crystal was only $\ce{NaCl}$)
Assuming I already have TONS of general salts (mixed) from simple seawater evaporation, is there any way to separate each mineral? Let's say by pouring the salts in magnet (for material that is having magnetic properties), etc

(lithium salt pond, which seen exactly same as $\ce{NaCl}$ pond)
Apologize for shallow understanding of the topic. Any suggestion is highly appreciated since I can't find the answer elsewhere.
P.S: This question have been posted earlier in Physics

One material extracted from seawater is magnesium.

In the United States, magnesium is obtained principally with the Dow process, by electrolysis of fused magnesium chloride from brine and sea water [emphasis added]. A saline solution containing $\ce{Mg^{2+}}$ ions is first treated with lime (calcium oxide)[sic; calcium hydroxide would seem more likely] and the precipitated magnesium hydroxide is collected:

$\ce{Mg^{2+}(aq) + CaO(s) + H2O → Ca^{2+}(aq) + Mg(OH)2(s)}$

The hydroxide is then converted to a partial hydrate of magnesium chloride by treating the hydroxide with hydrochloric acid and heating of the product:

$\ce{Mg(OH)2(s) + 2 HCl → MgCl2(aq) + 2H2O(l)}$

The salt is then electrolyzed in the molten state. At the cathode, the Mg2+
 ion is reduced by two electrons to magnesium metal:

$\ce{Mg^{2+} + 2 e^− → Mg}$


Is the reaction of calcium carbide with water a Brønsted-Lowry acid-base reaction or Lewis acid-base reaction? Consider the following reaction:

$$\ce{CaC2 + 2H2O -> Ca(OH)2 + 2C_{2}H_{2}}$$

If it can be considered as an acid-base reaction, is it a Brønsted-Lowry (B.L.) reaction or a Lewis reaction?

If it is B.L., of course it is also Lewis; but I cannot understand if it can be B.L. in this case, since 2 protons are actually passed to the double carbon atom, but there is also the $\ce{Ca}$.

We may regard the calcium hydroxide and calcium carbide  as essentially ionic except for the covalent bonding within the anions.  So there is no Lewis reaction involving the calcium.

That leaves the proton transfer, which can be described fully by the B-L theory.  You actually have two reactions, successively forming $\ce{C2H-}$ and then $\ce{C2H2}$.


How can I remove water vapor from an airstream without a cold trap? I'm trying to perform vacuum distillation at home with a rotary vane vacuum pump.

In order to not ruin my vacuum pump, solvent vapor (in my case, water) should be removed from the air before it reaches the pump inlet. The only method that I've encountered for accomplishing this, is using a cold trap to chill the air below $\mathrm{-50^\circ C}$ to deposite the water. To this end, the cold trap is filled with a cryogenic liquid such as liquid nitrogen.

Unfortunately I do not have access to liquid nitrogen, dry ice, etc. Is there any other affordable way to remove water vapor from the air before it reaches the pump?

Could I use a container filled with crushed silica beads or a liquid to absorb the water in the air? Or would using a very hydrophobic membrane work?

You won't easily be able to use some other liquid to pull the water vapor out of the air. I think the cheapest option is a hydrophobic membrane or filter that you feel comfortable using. A commercial desiccant system will be the highest performance next to cryogenic cooling but may not be cost effective (depends on the scale you are working at). You can also try salt in place of silica beads for absorption drying versus adsorption drying; I imagine the silica will be more efficient but salt will be cheaper. When using things like silica, salt, or commercial desiccants you may run into issues of too much pressure drop as the gas travels through the packed column. You will have to figure out how often the desiccant media needs to be regenerated/replaced to make sure your drying is adequate.


Painting corroded 9V battery contacts with vinegar I have an old radio with corroded 9V battery contacts, probably from a battery that corroded though I don't know.

I wanted to clean up the contacts with 5% vinegar since that works with pennies, but this is not something I can dunk in a small container since it is attached (with a very short wire) to the radio).

So I want to thicken vinegar to a gel, and not change the pH.  I would rather do this slowly, I can leave the solution on the contact for as long as necessary.

A high school chemistry teacher suggested vaseline as a thickener.  I just wanted to get a second opinion that vaseline will not somehow cause a problem?  Glycerin was also mentioned.  Could either react in any way with the metal electrode?

If I use either, presumably the acididy will decrease by dilution with the thickener.  Is this linear, or does the thickening of the solution have any other affect on acids?  I can try treating the surface for a few hours, or a day, then how could I get it off?  My guess is a cotton swap with water and detergent to clean off the gel?

I am not 100% sure what the surface of the contact is, it could be a nickel alloy, it certainly is a formerly shiny silvery metal, now somewhat dulled with green raised bumps on it from some former chemical reaction.

If you think I should just apply vinegar, or anything else for that matter, with a cotton swab I could do that, I just want to be able to apply something and clean up the surface with a minimum of effort.

Using vinegar, petroleum jelly, or glycerol to clean the contacts would leave some residue.  Vinegar might dissolve it somewhat more quickly than water, but you have to be sure you get all the vinegar off the contact, or it could cause more corrosion.  Petroleum jelly - I think someone was kidding you.  Vinegar & petroleum jelly don't mix.

Much easier: use a small piece of emery paper or steel wool to grind the deposit off.  Keep sanding until the metal is shiny.  Use a damp rag to remove the powder.  Don't let the ground-off powder fall into the circuitry.

Edit:  added steel wool.


Why the term 'unified' in "unified mass unit"? Why different unit symbols - amu, u and dalton are used for representing atomic mass? And what does the term 'unified' mean in "unified mass unit"?

Thank you.

Your question is historically important. Nobody has the energy to write an answer which would require several pages of an article. In short, it is a can of worms. For almost 5-6 decades chemists and physicists used a different scale for the atomic masses, so the atomic masses in the publications/books used by physicists did not exactly match what the chemists used for their own publications. Note that it is just not a matter of academic egos, the entire chemical industry is dependent on atomic weights (for calculations and trade). Money matters. This is why both the governments and the scientific community wanted to sort this out.

Both parties were not willing to budge because that would have affected their millions of pages of research. However a compromise was reached to use a common scale by two three sensible men i.e., they proposed to choose C-12 in the 1960s instead of oxygen. This process sort of "unified" the community of physicists and chemists and it did not affect the values of either party by much.

More history is given here, The AME2003 atomic mass evaluation (I). Evaluation of input data, adjustment procedures, Nuclear Physics A 729 (2003) 129–336

There are scores of other articles from the 1960-70s on the choice of C-12. Use Google Scholar. The original work is in French I guess, I did not check

T.P. Kohman, J.H.E. Mattauch and A.H. Wapstra, J. de Chimie Physique 55 (1958)
393.


Why 1 mole of H2 occupied the same volume occupied by 1 mole of O2? I'm asking a question about the volume occupied by gasses in standard temperature and pressure.
My textbook said that a mole of any gas occupies $\pu{22.4 L}$ at standard temperature and pressure.
But $\ce{O2}$ is greater than $\ce{H2}$ in size

I asked my teacher why the volume occupied by 1 mole of $\ce{H_2}$ is equal to the volume occupied by $\ce{O_2}$ at standard temperature and pressure, but he seemed to not know the answer, and I remain confused.
Please answer in simple words.

The volume occupied by a gas has no relationship to the size of the molecules of the gas
This is one of the principles of the gas laws in chemistry. It is based on observation: in the early days of chemistry it is what chemists observed to be true. And it is a very important chemical law.  But, of course, that doesn't explain why it is true.
There are several ways to gain some appreciation of this without too much theory. consider water as a liquid compared to water as a gas (I'm simplifying a bit but not by much). One mole of liquid water occupies about 18mL of space and this space is largely dominated by the size of the water molecule (again reality is a little more complex as the specific way the molecules interact makes a small difference to the volume occupied). As water vapour (the same substance but as a gas) 1 mole of water occupies over 22L of space (ignoring the fact that water is not a gas at STP, but bear with me as this is the space it would occupy if it were). That is well about 1200 times more space. This implies that very little of the space occupied has anything to do with how big the water molecule is.
The same broad idea applies to all gases.
So why does the molecular size not matter? A detailed understanding requires a fair few calculations about how gas molecules behave. But, the bottom line involves thinking about the molecular cause of pressure and the meaning of temperature. Pressure is caused by gas molecules bouncing off the walls of a vessel. Temperature is the average energy of those molecules which is a function of their speed and their mass. Doing the calculations shows the relationship between volume, pressure and temperature. This calculation explains the properties of gases but the volume of the molecules has no part.
Strictly speaking the ideal gas laws assume that the size of molecules don't matter. But this is a good approximation when the gases are far away from the point where they become liquids. The simplified theory matches actual observations well. And the fact that, in typical gases, the volume occupied by the molecules is less than 0.1% of the volume occupied by the gas shows why the molecular volume is usually irrelevant.


Why do polysaccharides consisting of alpha-glucose have helical structures? Why do the polysaccharides consisting of alpha-glucose such as starch and glycogen have helical structure, while the one such as cellulose consisting of beta glucose don't display such structures?
The textbook said that is because of the angle of 1-4 linkage, but what kind of angle is it that makes helical structure?

Note: This is an off the top of my head answer (I might be horribly wrong), but it has prompted me to do some research. I will add more details, and/or make corrections once I am done. In the meantime, I would love to hear what others have to say on this.

Pictured below is the 2D structure of amylose

This primarily composed of $\alpha$-(1-to-4)-D-glucose units. Although the $\alpha$-(1-to-4) links are capable of relatively free rotation about their torsional angles, hydrogen bonding between the $\ce{O}$3 and $\ce{O}$2 encourages helicogenesis.

Single helical amylose has hydrogen-bonding between $\ce{O}$2 and $\ce{O}$6 atoms on outside surface of the helix and the ring oxygen pointing inwards.

Contrat this with cellulose which is a linear polymer of $\beta$-(1-to-4)-D-glucopyranose units.

The fully equatorial conformation of $\beta$-linked glucopyranose residues stabilises the chair structure, minimising its flexibility.

Pictured below, the 2D structure of cellulose and displaying the intrastrand hydrogen bonding

Interactive 3D visualisation for amylose and cellulose.


Why is water a dipole? Water ($\ce{H2O}$) is a dipole. The reason why is simply because it is not symmetrical, there are more electrons on the oxygen side than on the hydrogen side, and the electronegativity of oxygen.

But why isn't $\ce{H2O}$ symmetrical like $\ce{CO2}$? Why isn't $\ce{H2O}$ non-polar like $\ce{CO2}$? Does it have anything to do with the orbitals?

Yep, it has to do with the orbitals.

$\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below).

On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an angle to each other. They add up to give a net dipole moment (shown with grey in the diagram).

The colors indicate electron density, red is more dense/blue is less dense. Dipole moment is from low density to high density.

OK, so why do these molecules have differing shapes? This is where orbitals come in. I'll try to explain as much as I can without going into orbitals.

Carbon has an outer shell electronic configuration as $2s^22p^2$. Out of these four electrons, two are used in $\pi$ bonds, and two in $\sigma$ bonds. If you don't know what those are, just look at it like this for now: A set of bonds between two atoms will have one and only one $\sigma$ bond, with the rest of them $\pi$ bonds. So, any single bond is made up of just a $\sigma$ bond, a double bond is made up of one $\sigma$ and one $\pi$, and a triple bond is made up of one $\sigma$ and two $\pi$ bonds. What these types of bonds actually are can be explained if you know what an orbital is.

Now, what VSEPR says is that the geometry of the molecule is only decided by the $\sigma$ bonds and lone pairs on the central atom. You count up the $\sigma$ bonds and lone pairs (lets say they add up to $x$), and decide the geometry based on that. The geometry is the most stable configuration of $x$ hybrid orbitals. In simple terms, if we took $x$ balloons and tied them together, the directions the balloons point in help us correspond to where the bonds and lone pairs lie:

In $\ce{CO2}$, we have two $\sigma$ bonds and two $\pi$ bonds (as each double bond has one of each type). Each bond takes up one electron from carbon, so we have no leftover electrons for forming any lone pair. Since we have two $\sigma$ bonds and 0 lone pairs, $x=2$, giving us the structure given by the first set of balloons, which is linear. And $\ce{CO2}$ is indeed linear:

Now let's take water. The central atom (Oxygen) has a valence configuration of $2s^22p^4$, that is, 6 electrons. In water, since we have two single bonds, we have one $\sigma$ bond each (and no $\pi$ bonds). So we have total two $\sigma$ bonds.

But this leaves us with $6-2=4$ unpaired valence electrons. These form two "lone pairs" (pairs of electrons which do not bond). With two lone pairs and two $\sigma$ bonds, $x=4$. This gives us a tetrahedral structure (third in the balloon diagram). Two of the four points in the tetrahedron are occupied by the lone pairs, and two by bonds:

(Note that the angle 104.5 is not the same as the angle in perfect tetrahedra, 109.25--this is due to the lone pairs repelling each other)

So finally, we have the following "bent" structure for water:

From the structure, as shown above, it is very easy to check if the molecule has a dipole moment.


Instability of heavy hydrides Question:

Why do hydrogen compounds from the third period down (that are electron deficient or have completed octets) readily decompose?

Thoughts:

Apparently, heavier group 14 elements form hydrides which decompose readily because they have valence orbitals with which $\ce{O_2}$ and $\ce{H_2O}$ can coordinate. But I've tried and tried and cannot think of why this makes $\ce{CH_4}$ more stable than $\ce{SiH_4}$. What is the difference between the two (except for the fact that silane contains a larger central atom)?

As for the remaining groups on the periodic table, I cannot fathom how this trend holds.

There are two factors at work here. One is that going down within a group of the periodic table, say from C to Si to Ge, gives you more electropositive (metallic) elements, so they behave more like metal hydrides and are easier to hydrolyze. A second factor is that with increasing atomic radius as you go down to heavier elements, the overlap between hydrogen's valence orbital(s) and those of the element will be poorer, and the resulting bonds weaker.


What's the strongest known organic acid? What's the strongest known organic acid? Is it maybe trifluoromethanesulfonic acid or tautomer of pentacyanocyclopentadiene?

EDIT: Since the question was reactivated I thought about formalising it. I can think of a few categories of compounds: neutral (actually created or only studied in silico) and cationic (also two options). Second Dave's answer and ANM's one do the trick for two groups, but two other options remain (still prefer answers without boro/silico/etc. stuff).

According to "The Strongest Isolable Acid" Angewandte Chemie International Edition 43 (40): 5352–5355, carborane acids are the strongest, with $\ce {H(CHB11Cl11)}$ being the strongest.

Depending upon ones preferred definition of "organic" this might not be considered organic, but the derivative $\ce{H(CHB11Me5Br6)}$ is strong enough to fully protonate benzene.

$\ce{H(CHB11Me11)}$ is stronger than $\ce{H(CHB11Me5Br6)}$ and only slightly weaker than $\ce {H(CHB11Cl11)}$, so if $\ce {H(CHB11Cl11)}$ is disregarded as no being organic, then $\ce{H(CHB11Me11)}$ is strongest.  If $\ce{H(CHB11Me11)}$ still isn't organic enough, then protonated benzene or a protonated alkane.

(A later article and the article in Nicolau's comment report $\ce {H(CHB11F11)}$ and $\ce {H(CEtB11F11)}$ which may be even stronger than the above)


Trying to understand SMILES(?) patterns In the example calculation from the conformer search package fafoom is a mysterious pattern that I'd like to understand.

The SMILES code of the given example structure is:

CC(=O)N[C@H](C(=O)NC)C

There is a first not so hard pattern for the smart_cistrans pattern identifier

C~[$(C=O)]-[$(NC)]~[C]

which tells the program that peptide bonds behave like cis-trans isomers. That is relatively easy to see but I don't really understand it to adapt it on further problems.

But there is another identifier for torsion angles that for me is even harder to understand:

[*]~[!$(*#*)&!D1]-&!@[!$(*#*)&!D1]~[*]

The [*]'s at the beginning and at the end look like defining any atom ... but the cursing signs in the middle don't tell me anything.

Dear reader, if you understand how those patterns are generated, I would ask you break them down piece for piece that I can use them afterwards on my own.

First off this is SMARTS not SMILES.

Think of SMARTS as like a regular expression language for molecules (e.g., SMILES).

So let's break down some of those characters:

~ is "any bond"
! is "not"
$ indicates a recursive SMARTS expression.
@ indicates "any ring bond"

The $(*#*) means "two atoms with a triple bond between them". The !D1 means and not an atom with one explicit bond.

Taken as a whole, [*]~[!$(*#*)&!D1] means any atom with a bond to an atom that doesn't have just one explicit bond or a triple bond.

Basically, the pattern is trying to find a single bond in the middle with some rotatable side-groups:
[*]~[...]-&[..]~[*]

There's a very nice tool called SMARTS Viewer that can help to debug and construct SMARTS.

Seems like there are also now more polished commercial versions called SMARTStools through BioSolveIT including an editor.


Aromaticity of annulenes Why are some annulenes such as [14] and [18]annulenes are aromatic even though they are not fully planar as cited by Wikipedia? What does it mean that they can achieve planar form?

It means very little, at least in terms of stabilization.  Perfectly planar or not, larger aromatic rings bring less stability than smaller ones, so they react more like mere conjugated polyenes than what we think of as "aromatic rings".  Wikipedia:

When the annulene is large enough, [18]annulene for example, there is enough room internally to accommodate hydrogen atoms without significant distortion of bond angles. [18]Annulene possesses several properties that qualify it as aromatic.[1] However, none of the larger annulenes are as stable as benzene, as their reactivity more closely resembles a conjugated polyene than an aromatic hydrocarbon.

Cited reference:
1.
Oth, Jean F. M.; Bünzli, Jean-Claude; De Julien De Zélicourt, Yves (1974-11-06). "The Stabilization Energy of [18] Annulene. A thermochemical determination". Helvetica Chimica Acta. 57 (7): 2276–2288.


While filling electrons, we follow Aufbau principle, but not while removing them. Why is this so? I recently came across a question Why is the vanadium(3+) ion paramagnetic?, where the asker is wondering how $\ce{V^{3+}}$ is paramagnetic (he used Aufbau in reverse to remove the electrons), while the correct answerer to that question remarked that removing electrons must follow the order from outer shell to inner shell. For example, in the case of $\ce{V^{3+}}$ the electronic configuration of $\ce{V}$ is $\ce{[Ar] 3d^3 4s^2}$. The asker used reverse-Aufbau, which is $\ce{[Ar] 3d^0 4s^2}$; while the actuality was $\ce{[Ar] 3d^2 4s^0}$.

My question: Why does filling electrons follow a certain rule, while removing them follows a different rule?

Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom.

When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with different interactions between nucleus and electrons, which affects the energy of the orbitals.

Also keep in mind, that the orbital picture is just a simplification. Strictly speaking (and from a purely theoretical point of view) orbitals do not exist, they are merely a mathematical crutch to solve the Schrödinger Equation. Although in practice this picture works surprisingly well, hence people use it a lot.

A proper quantum chemical calculation will give you the correct ground state, but you might have problems identifying it as something like $\ce{[Ar] 4s^2 3d^0}$. It will be a mixture (linear combination) of many electron configurations, although $\ce{[Ar] 4s^2 3d^0}$ might be the most important one for $\ce{V^{3+}}$.


Do cyclic oxygen molecules exist and their theoretical stability? Do oxygen molecules ever become cyclic shapes like carbon and other elements?

Why doesn't oxygen form cyclic compounds? How stable would they be if they do exist. Especially more than 3 oxygens. Why are there just 2 allotropes of oxygen known to mankind.

Unlike the likes of carbon, oxygen-oxygen single bonds are highly unstable. Because oxygen is such an electronegative compound, it can only become satisfied with bonds where it hoards a majority of the electrons. Because of this, compounds such as peroxides ($\ce{RO-OR}$) will readily decompose into two radicals. This is why hydrogen peroxide is used as a radical initiator for chemical reactions. Since a single $\ce{O-O}$ bond is so unstable, it would be extremely unlikely for a larger chain to exist. Peroxides by themselves are highly unstable. Not only do they decompose into radicals, but they also rapidly decompose into water. Peroxides have a single $\ce{O-O}$ bond, and are somewhat stabilized by donating groups on either side of the peroxide. Now, imagine a chain that looks like this: $\ce{O-O-O}$. The central oxygen is not only bonded one inductively withdrawing oxygen as in peroxides, but two of them. In addition, it is not stabilized by electron donation from any source. Despite this, it is predicted that small amounts of cyclic ozone might exist in the atmosphere. Cyclic ozone is so much less stable than regular ozone for the reasons I listed above, but it should exist. Any larger oxygen ring will most likely not form.

Carbon does not have this same problem. Since carbon has a mid range electro-negativity, it is stable forming bonds with itself. In addition, carbon can form four covalent bonds, normally two of which are with hydrogens (as in a cycloalkane). Carbon is perfectly stable with this electron density, leading to the stability of cyclic alkanes.


How do I find mass equivalence of a photon? The problem has a flame emitting a wavelength of $589\ \mathrm{nm}$, and asks for the mass of one photon of that wavelength. It tells me that $1\ \mathrm J = 1\ \frac{\mathrm{kg\cdot m^2}}{\mathrm s^2}$.

I converted to $\mathrm J$ using $E=\frac{h\cdot c}{\lambda}$. I'm not quite sure what to do past that. I have $3.36\cdot 10^{-19}\mathrm J$. I know the answer is supposed to be $3.75\cdot 10^{-34}$.

$$E = \frac{h\cdot c}{\lambda}$$

$$1\ \mathrm J = 6.24150934 \times 10^{18}\ \mathrm{eV}$$

then you have
$$E = mc^2 = \frac{h\cdot c}{\lambda}$$


What's the standard deviation in bond angle for water molecules at a particular temperature? Suppose I have a beaker of water at 20 °C. I want to know the standard deviation of the bond angles between the hydrogen atoms in the water molecules. How do I figure this out?

If I knew that the bond angle was the only possible degree of freedom of the molecules, then I'd be able to use the Boltzmann distribution to calculate this, if I knew the energy of the molecule as a function of bond angle. But I don't know the energy of the molecule as a function of bond angle, and also there are a lot of other degrees of freedom (e.g. the molecules can have overall kinetic energy, or they can be spinning rapidly, or the other vibrational modes of the water molecule can contain some of the energy.

Is there an easy way for me to look up the energy of the water molecule as a function of the positions of the hydrogen atoms? (I guess I could just calculate this myself using my favorite quantum chemistry software.)

I tried to answer by proposing a simple model. The model didn't work as well as I hoped, but the general procedure might serve as an answer.

A harmonic model

As a first approximation, you might consider a simple model for the energy of a single water molecule as a function of its angle. One possibility might be a harmonic function

$$E(\theta) = \frac{k \theta^2}{2}$$

where $k$ is an unknown parameter and $\theta$ is the angle. This is very simple and leaves out many complexities, of course.

If you had $E(\theta)$, the energy of a single water molecule at a particular angle, and if you assume an independent particle model (which may break down for a liquid), you could apply Boltzmann weighting

$$p(\theta) = \frac{1}{U}\exp\left(-\frac{E(\theta)}{R T}\right)$$

where $p(\theta)$ is the probability of finding a water molecule at a particular angle and $U = \int_0^\pi d\theta \exp\left(-\frac{E(\theta)}{R T}\right)$ is a normalization factor.

If you had a probability distribution $p(\theta)$ describing the spread of angles at a certain temperature, getting the mean and standard deviation is easy:

$$\langle\theta\rangle = \int_0^\pi d\theta\ \theta\ p(\theta)\\
\langle\theta^2\rangle = \int_0^\pi d\theta\ \theta^2\ p(\theta)\\
\sigma_\theta^2 = \int_0^\pi d\theta\ \left(\theta - \langle\theta\rangle\right)^2\ p(\theta) = \langle\theta^2\rangle - \langle\theta\rangle^2$$

So this would be how to get $\sigma_\theta^2$.
Below is my numerical attempt.
Beware: this model is too simple.

Numerical results

I did some calculations (mostly with SymPy) with the above and I came up with the following:

$$p(\theta) = \frac{1}{U} \exp\left(-\frac{k \theta^2}{2 R T}\right)\\
U = \sqrt{\frac{\pi R T}{2 k}} erf \left(\frac{\pi}{2} \sqrt{\frac{2 k}{R T}} \right)$$

$$\langle\theta\rangle = \frac{k}{R T U}\left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right) \exp\left( -\frac{\pi^2 k}{2 R T} \right)\\
\sigma_\theta^2 = \frac{R T}{k} - \left[ \frac{\pi}{U} \exp\left( -\frac{\pi^2 k}{2 R T} \right) +
\left(\frac{RT}{kU}\right)^2 \left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right)^2 \exp\left( -\frac{\pi^2 k}{R T} \right)
\right]$$

By substituting constants and temperature ($\pi = 3.14159265$, $R = 8.314459848$ J K$^{-1}$mol$^{-1}$ and $T = 293.15$ K) and setting the average above to as close as possible to 104.45° and solving numerically the complete least squares problem, I found an approximate solution for $k$: $384.05105591$ J/rad².

Sadly, this value is associated with an average angle of around 96°, which means the model is very poor.

Using this best $k$ gives us a $\sqrt{\sigma_\theta^2}$ of

$$\sqrt{0.508770928037803} = 0.713281801280393 \text{ rad}$$

which is around 41°. Again, a too simplistic model: the best prediction gives angles as 96°$\pm$41°!

A bad model

By expanding the average in series, I found the following

$$\langle\theta\rangle \approx 1.570796325 + 0.000530047067361739 k + O(k^2)$$

The second order term above was $- 7.86976333117302\times 10^{-7} k^2$, so I dropped it.
Observe the first term is pretty much 90° in radians. All other terms decay too quickly.

By doing the same expansion for the standard deviation the following comes out:

$$\sigma_\theta^2 \approx - \frac{4.54747350886464 \times 10^{-13}}{k} + 0.411233515772252 + 0.000388544793228128 k + O(k^2)$$

The next term in the series above was $- 4.34803406790969 \times 10^{-7} k^2$ so I dropped it. Again, a (wrong) zeroth order term dominates.


Why can't a primary cell be recharged? The explanation I have found is that primary cells have irreversible reactions, so passing current in the opposite direction cannot recharge the cell. But as far as I understood it, no reaction is really irreversible - what one calls irreversible simply has very high values of the equilibrium constant, so for all intents and purposes proceeds to completion. Nevertheless, some small - maybe undetectable practically - amount of reactant does remain and the reaction does reach equilibrium. So even in a primary cell, when it is used up, equilibrium has been achieved, so recharging should be possible - it might take long to build up the initial value of potential since the required concentration of the electrolytes will be very different from that at equilibrium, but it should be possible. Why then is it not?

In some cases, a primary cell can be recharged, e.g. Mn-alkaline cells can take a partial charge. However, there are a number of reasons primary cells cannot be fully recharged, and that secondary cells have to be specifically designed to allow recharging.

Gas evolution: once most of a cell is charged (though all material has not been converted to its original state), the electrolyte is electrolyzed (apologies for the pleonastic redundancy), usually evolving hydrogen and oxygen, causing the cell to rupture. Nickel-metal hydride cells, for example, get around that issue by incorporating a catalyst to recombine them, $\ce{2 H2 + O2 -> 2 H2O + heat}$.
Physical change in the reactants: the standard "dry cell" (Leclanché cell) is made with an outer shell of zinc. As the cell discharges, parts of the shell are dissolved, eventually leaving holes. Reversing the current flow will cause zinc to redeposit, but not necessarily filling in the weakened areas. Even lead-acid cells designed for storage have limited lifetime because of the buildup of large, insoluble crystals of lead sulfate (sulfation, aka sulphation in GB), which may break off from the lead plates, preventing reversal of the chemical reaction during charging.
Physical damage to the insulator: metal dendrite may grow during charge-discharge cycling, piercing the dielectric between cells, even in well-designed secondary cells.


Why is it that energy of electrons in higher orbits is greater than the ones in lower orbits? There is this picture I found online which indicates that the energy level is higher as we go up and more electrons are added to the atom.

What I'm having trouble understanding is why the energy level is higher and not lower. The electrons are further and further away from the nucleus, at infinity their energy would be zero. Is it correct to say that the electrons' energy is negative and as we go further it increases, going to zero at infinity?

That is the only way I can see all this make sense.

What I'm having trouble understanding is why the energy level is
  higher and not lower. The electrons are further and further away from
  the nucleus, at infinity their energy would be zero.

You are correct. The innermost electrons are the most tightly bound to the nucleus and the outermost ('valence') electrons have the least energy, explaining why they can be removed (in chemical reactions e.g.)

But these energies are always negative and use the principle that infinitely far away from the nucleus the electron's energy would be $0$.

This way defined, the energy of the hydrogen electron in the ground state ($1s$) e.g. is $-13.6\ \mathrm{eV}$. The energy of the electron in excited states ($2s$, $2p$, $3s$ etc) is less negative, until for infinite distance (a free electron) it is zero ($0$).

Let's take a numerical example to demonstrate this. The ground state ($n=1$) of hydrogen's electron is $E_1=-13.6\ \mathrm{eV}$ and the first excited state ($n=2$) is $E_2=-3.4\ \mathrm{eV}$. What are the energies needed to remove the electron from these states?

Since as the energy at infinite distance is $0$, in the case $n=1$, then:

$$\Delta E= 0-(-13.6)=13.6\ \mathrm{eV}$$

And in the case $n=2$, then:

$$\Delta E= 0-(-3.4)=3.4\ \mathrm{eV}$$

So it takes less energy to remove an electron in a higher orbital than a lower one. That fits our observations.

Equally, the energy needed to excite the electron from $n=1$ to $n=2$ is:

$$\Delta E_{1\to 2}=-3.4-(-13.6)=10.2\ \mathrm{eV}$$


Will gaseous ionic compounds be free moving ions? I knew while learning about electrolysis that if the ionic compound is molten it becomes free moving ions.

If that is the case, what will happen if I continued heating till it reaches the boiling point so that the ionic compound evaporates?

Will it still be free moving ions?

Also, shouldn't the result be more efficient at electrolysis than in the liquid state due to increased mobility of ions? If not, why?

Usually not. Boiling point rarely exceeds 4-5 thousand kelvin. A typical ionic bound energy is about 5 eV. 1 eV is roughly 11 thousand kelvin, so ions in low-temperature vapors exist as molecules. When the temperature becomes enough to break ionic molecules, it is enough to strip one or two electrons from atoms, so high-temperature vapors will be plasma with electrons and positive ions.


Why is Diels–Alder adduct exo when we leave it for a while? Recently we did an experiment of Diels–Alder reaction and we left the adduct overnight. I've read a paper of Woodward that says the adduct of Diels–Alder thought to be endo, whereas in our experiment it turned out to be exo.

As you probably know, the Diels–Alder (DA) reaction is reversible — the easiest example is the reaction of cyclopentadiene, which polymerizes at room temperature, which — if needed as monomer — has to be cracked just in advance.

In terms of the selectivity you observed, the endo-oriented product of the DA reaction often is the kinetic one, the exo-oriented product the thermodynamic preferred one.  Which is, if the rate of a DA reaction is to be increased while still obtaining endo-preference, an increase of the reaction temperature is skipped in favour a Lewis-catalysis, addition of a perchlorate, or high (external) pressure.


Solubility of NaCl in water I learned that solubility of substance increases with increase of temperature. But I came to know solubility of NaCl doesn't increase when temperature is increased . So what is the reason for this phenomena?

Solubility of solid in a liquid depends on the nature of the solid.

It depends on whether the dissolution process is exothermic or endothermic. Using Le Chatelier's principle, if the temperature is increased for an exothermic process then dissolution decreases and if the temperature is increased for an endothermic process then dissolution increases.

I am going to stick to the case of $NaCl$, if you want to know more about Le Chatelier's principle, click on the link below and look for the 2nd answer https://physics.stackexchange.com/questions/311409/why-do-most-gases-dissolve-easier-at-colder-temperatures/311429#311429

There are some solids for whom change in temperature doesn't affect the degree of dissolution.

There are some substance which on heating, change on a particular temperature from one polymorphic form to another (like $\alpha$ to $\beta$ as in the case of ammonium nitrate) or from one hydrated form to another ($CaCl_2.6H_2 O -> CaCl_2.4H_2 O$) or from hydrated to anhydrous form ($Na_2SO_4.10H_2O -> Na_2SO_4$).

Generally, dissolution process in a solvent is a disintegration process. This process needs energy. In such cases, energy is absorbed. But in some cases, besides the process of breaking or ionization, there is hydrate formation. During hydration heat is evolved. The net result is that heat is either evolved or absorbed.

There are also cases in which heat of separation of ions in just equal to the heat of hydration and there is very little heat effect as in the case of $NaCl$. The heat of solution of $NaCl$ is very small as the heat of ionization is nearly equal to the heat of hydration. Therefore, temperature doesn't affect the dissolution of $NaCl$ as the heat of hydration almost equalizes heat of dissolution.


How can -G/T and -A/T be thought of as 'disguised' entropies? I understand that entropy is defined as $\mathrm{d}S=\mathrm{d}q/T$ for a reversible change but I fail to see why $-G/T$ and $-A/T$ can be thought of as 'disguised' entropies. This is a question on my problem sheet at university and my tutor has hinted: "You need to think about the total entropy change for a process".

The fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$ may be divided by  (-T):

$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$

In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:

$$\Delta S_{total} ={\Delta S_{surroundings} + \Delta S}_{sys}$$

$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.

$\ce{\Delta G}$ is a disguised entropy change, because $\ce{\Delta H}$ is intrinsically an entropy change too, as explained above.


Can organomercury compounds synthesize accidentally from household cleaning products? I had broken a CFL partially on carpet last week. It was a spent bulb, and the outer glass shell broke but not the fluorescent tube - apparently the tube did have a small hole in it which found out after all my other steps when I saw it in the garbage.

When I cleaned it up first, I used a mild spray cleaner which contained methyl esters and various surfactants - I found this out now, not then, or I wouldn't hzve used it. Wiped it all up with a paper towel, then vacuumed in case I missed small pieces. I did ventilated at that point.

About two hours later, my cat made a mess on the rug. At that point, I sprayed down the carpet with resolve (which says it contains hydrogen peroxide). Wiped that all down with a towel, then revacuumed...and realized I hadn't cleaned out the dust cup. I think I then emptied that outside and cleaned out the HEPA filter.

My concern now is what I may have exposed myself to in that botched cleanup. Any ideas on what may have formed? Do I need to be concerned?

Unfortunately, you seem to have already broken some of the protocols of cleaning up a broken CFL according to the EPA's website.

According to this site:

  Vacuuming is not recommended unless broken glass remains after all other cleanup steps have been taken.  Vacuuming could spread mercury-containing powder or mercury vapor.

Also according to data from a study of the effect of the breakage of CFLs performed by the Maine Department of Environmental Protection:

  Mercury concentration in the study room air often exceeds the Maine Ambient Air Guideline (MAAG) of 300 nanograms per cubic meter (ng/m3) for some period of time, with short excursions over 25,000 ng/m3, sometimes over 50,000 ng/m3, and possibly over 100,000 ng/m3 from the breakage of a single compact fluorescent lamp.

and also:

  Vacuuming a carpet where a lamp has broken and been visibly cleaned up, even weeks after the cleanup, can elevate the mercury readings over the MAAG in an un-vented room.

At the end of the report, it states that although mercury vapor is extremely toxic:

  ...it is unclear what the exact health risks are from exposure to low levels of elemental mercury...

It seems as though the best course of action you could take now would be to ventilate the room and if possible your house extremely well for several hours, and to avoid vacuuming any more.


Is there a definition of rotatable bond? Some intramolecular bonds are rotatable in the sense that the torsion angle around this bond is flexible (for example, $\psi$ and $\phi$ angles in a protein backbone). Others are essentially fixed (for example, the $\omega$ angle in the peptide bond in a protein backbone).

Is there an operational definition, or a published study where a general definition is given, which would allow me to recognize in a generic molecule (or at least an organic molecule) whether a given bond is rotatable or not?

If there is no general theory for this, what are the most common indicators of whether a bond can or cannot be rotated? Are there some rules of thumb?

Is there an operational definition, or a published study where a
general definition is given, which would allow me to recognize in a
generic molecule (or at least an organic molecule) whether a given
bond is rotatable or not?

It depends on your purpose. Why do you want to identify some bonds and call the rotatable? For instance, PubChem, one of the biggest databases of molecules, defines rotatable bond as follows:

Rotatable bond is defined as any single non-ring bond, bounded to
nonterminal heavy (i.e., non-hydrogen) atom.

To some extent all this boils down to the following IUPAC definition of free rotation:

free rotation (hindered rotation, restricted rotation)
In a stereochemical context the rotation about a bond is called 'free'
when the rotational barrier is so low that different conformations are
not perceptible as different chemical species on the time scale of the
experiment. The inhibition of rotation of groups about a bond due to
the presence of a sufficiently large rotational barrier to make the
phenomenon observable on the time scale of the experiment is termed
hindered rotation or restricted rotation.

At usual lab conditions a rotation around a single non-ring bond, bounded to nonterminal heavy atom will be indeed almost free, thus, PubChem definition can be used as a very simple rule-of-thumb to identify such bonds. But in general it depends on conditions. And purposes.

With respect to protein backbones being rotatable or not: I'm not a speacialist in this area, but Wikipedia article on peptide bond claims that (emphasis mine)

Significant delocalisation of the lone pair of electrons on the
nitrogen atom gives the group a partial double bond character. The
partial double bond renders the amide group planar, occurring in
either the cis or trans isomers.

So, this partial double bond character might contribute in protein backbones being not quite rotatable. Besides, the Wikipedia article also claims that (emphasis again mine):

In the unfolded state of proteins, the peptide groups are free to
isomerize and adopt both isomers; however, in the folded state, only a
single isomer is adopted at each position (with rare exceptions).

So, at least in the unfolded state protein backbones are rotatable. The fact that only two configurations (cis and trans) are possible is irrelevant. The notion of rotatable bond does not imply having few specific possible configurations.


Does 2-chloro-2-methylpentane react with NaI faster in acetone or water? Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?

I thought that since it is a tertiary halide it will proceed via SN1 reaction. So it must be faster in water as it is protic solvent.

But the answer given is Acetone. Why?

It  is simple this is the famous finkelstein reaction.

As you can see in the image NaCl is formed which is insoluble in acetone hence it precipitates and the reaction cannot go reverse.

But if we take water the reaction will shift in backward direction as NaCl is soluble in water.

 That is the reason we take acetone.


Will adding apple cider vinegar to freshly pressed fruit juice preserve it Would adding apple cider vinegar to freshly cold pressed fruit juice be able to preserve the fruit juice and extend its shelf life? Would the juice change into vinegar with time?

Apple cider vinegar is commonly used as a food preservative so I'm sure you would see some extended shelf life.

The acetic acid (the "active ingredient" of vinegar, usually present at around $\pu{5 - 20\%}$) in the vinegar would not "initiate" the production of more vinegar from the juice.  The production of acetic acid comes from a two-step fermentation process. First, alcohol is produced from the sugar in the juice by adding yeast and the appropriate bacteria. Then, acetic acid forming bacteria are added to oxidize the alcohol to acetic acid.

So, simply adding the apple cider vinegar to your juice will likely increase it's shelf life, but will not result in the production of more vinegar.


Energetic stability in relation to α-hydrogens I am new to organic chemistry and was asked to rank the following materials according to energetic stability. I think I have understood that the way to do this is to count the amount of α-hydrogens: the more α-hydrogens, the more stable the molecule. However, the definition of an α-hydrogen is "a hydrogen that is attached to an α-carbon", where the α-carbon is the first carbon next to a functional group. In this case, it is not clear to me where the α-carbon is located: the only candidate for a functional group is the CH2/CH3 (depending on the case) at the very end of the chain, but then would the α-carbon be the very first one or the one next to it?

Any further general insight into the ranking of molecules according to energetic stability would also be very much appreciated… Many thanks!

Apparently the intent of the question is to assess the stability of the three alkenes. Evaluation of the heats of formations provides useful information. The NIST site provides heats of formation and hydrogenation all of which are shown in black in the diagram. Computed values appear in red. Because hydrogenation of A, B and C all provide ethylcyclopentane, the greater the heat of hydrogenation the less stable the alkene. Alternatively, the less negative the heat of formation, the less stable. Thus, the order of stability is A < C < B.