pKw, pKb, pKa and NEN6411 I am trying to understand how to account for pH-temperature relations during industrial pH measurements.
Do I have it right?:

The impact temperature has on the pKa of strong acids are in practical terms non existent as the equilibrium at any temperature lays very far to the right.

The impact of temperature on pKa are more significant for weaker acids.

The impact of temperature on pKw is only significant around the point of neutrality. At lower pH's the few hydronium ions produced by H2O + H2O are so few that it bares no meaning on the logarithmic pH scale. In other words, the pH-temperature relationship is important when dealing with neutralization processes, but not so much when having pH setpoints set to anything in the milder acidic region and below.

Compensating for the pH-temperature relationship caused by the temperature dependent pKa is only important when doing measurements on processes where the temperature of the process differs from the temperature of the final product and where the targeted pH level during process control has to match the targeted pH level of the end product (at another temperature). For example: say I want to produce an end  product with a pH at 5 at temperature X. In order to make this product the temperature of my fabrication process has to be something other than temperature X. In this case, temperature compensation to account for changes is pKa becomes important - otherwise the end product won't have the desired pH.

in a different scenario the pH setpoint of a process becomes a means to an end. It's the pH of the process that determines the chemical reactions that takes place in that particular process (perhaps the hydronium ions merely acts as catalysts). In this case, temperature compensation in no way should be applied. This because the pH changes casued by changes in pkA are real and if the process demands say pH 3 for an optimal chemical reaction then it doesn't matter what the pH would have been at another temperature. The important thing is what the temperature is in the process right here, right now.

while pH changes that comes from changes to pKa actually do mean that a solution becomes more or less acidic, changes to pKw doesn't result in a solution becoming more or less acidic even though pH changes. This is because an equal amount of hydroxide ions a produced. So only the point of neutrality is changed.

Do I understand things correctly so far?
If so, then my question now is: What happens when we move into the alkaline region? I mean, pKb is, according to Le Châteliers principle, temperature-dependent as well, right? When looking at this graph https://ibb.co/LRs7G5y (algorithm based on NEN6411 (now
DS/EN ISO 10523:2012)) it seems as if alkaline solutions ALSO becomes more acidic with increases in temperature. Why is this? Does it mean that the equilibrium constant pKb runs in reverse of pKa when temperature increases (less OH-)? Or does it mean that pKw plays are much larger importance once we wander into the alkaline region? I don't, however, see how pKw play any importance on the temperature-pH relationship of alkaline solutions. I mean, if we have an alkaline solution of say pH 12, then any extra/or less hydronium ions produced by changes in pKw would be dwarfed by the proportional much larger amount of hydroxide ions.
Or do I get this wrong?
Last question: Why does the graph https://ibb.co/LRs7G5y indicates that no significant pH-temperature relationship exists below pH 6? Is it because the acid gets so strong at this point that the changes to pKa becomes miniscule relatively speaking?
Help would be much appreciated.
/Lars

The impact temperature has on the pKa of strong acids are in practical terms non existent as the equilibrium at any temperature lays very far to the right.

Practically, yes.

The impact of temperature on pKa are more significant for weaker acids.

Yes.

The impact of temperature on pKw is only significant around the point of neutrality. At lower pH's the few hydronium ions produced by H2O + H2O are so few that it bares no meaning on the logarithmic pH scale. In other words, the pH-temperature relationship is important when dealing with neutralization processes, but not so much when having pH setpoints set to anything in the milder acidic region and below.

$\mathrm{p}K_\mathrm{w}$ changes apply to all $\ce{H3O+}$ and $\ce{OH-}$, not just to those from water autodisscociation. If $\ce{[H3O+]}$ ( [] is the convention for molar conceentration ) is determined by $\ce{[OH-]}$, or if some process depends directly on $\ce{[OH-]}$ and not on $\ce{[H3O+]}$, it is significant.

Compensating for the pH-temperature relationship caused by the temperature dependent pKa is only important when doing measurements on processes where the temperature of the process differs from the temperature of the final product and where the targeted pH level during process control has to match the targeted pH level of the end product (at another temperature).
For example: say I want to produce an end product with a pH at 5 at temperature X. In order to make this product the temperature of my fabrication process has to be something other than temperature X. In this case, temperature compensation to account for changes is pKa becomes important - otherwise the end product won't have the desired pH.

Yes. E.g. for acetic acid, $\mathrm{p}K_\mathrm{a, \pu{25^{\circ}C}}=4.75$. That is the $\mathrm{pH}$ of solution acetic acid / sodium acetate in molar ratio 1:1.  Let imagine for illustrative purposes $\mathrm{p}K_\mathrm{a, \pu{50^{\circ}C}}=4.5$. If you used at $t=\pu{50^{\circ}C}$ for the final process $\mathrm{pH}=4.75$, there would not be reached 1:1 ratio.

in a different scenario the pH setpoint of a process becomes a means to an end. It's the pH of the process that determines the chemical reactions that takes place in that particular process (perhaps the hydronium ions merely acts as catalysts). In this case, temperature compensation in no way should be applied. This because the pH changes casued by changes in pkA are real and if the process demands say pH 3 for an optimal chemical reaction then it doesn't matter what the pH would have been at another temperature. The important thing is what the temperature is in the process right here, right now.

Yes, if I understant the paragraph well.

while pH changes that comes from changes to pKa actually do mean that a solution becomes more or less acidic, changes to pKw doesn't result in a solution becoming more or less acidic even though pH changes. This is because an equal amount of hydroxide ions a produced. So only the point of neutrality is changed.

This depends on what you mean by acidity. If you mean by acidity the activity or concentration of $\ce{H3O+}$, it is affected by $\mathrm{p}K_\mathrm{w}$ changes.

If so, then my question now is: What happens when we move into the alkaline region? I mean, pKb is, according to Le Châteliers principle, temperature-dependent as well, right? When looking at this graph https://ibb.co/LRs7G5y (algorithm based on NEN6411 (now DS/EN ISO 10523:2012)) it seems as if alkaline solutions ALSO becomes more acidic with increases in temperature. Why is this? Does it mean that the equilibrium constant pKb runs in reverse of pKa when temperature increases (less OH-)? Or does it mean that pKw plays are much larger importance once we wander into the alkaline region? I don't, however, see how pKw play any importance on the temperature-pH relationship of alkaline solutions. I mean, if we have an alkaline solution of say pH 12, then any extra/or less hydronium ions produced by changes in pKw would be dwarfed by the proportional much larger amount of hydroxide ions.

For weak acids, the primary parameters is $\mathrm{p}K_\mathrm{a}$. Then for their conjugate bases, $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a}$.
For weak bases, the primary parameters is $\mathrm{p}K_\mathrm{b}$. Then for their conjugate acids, $\mathrm{p}K_\mathrm{a}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b}$.
For solutions of strong acids, $\mathrm{pH} \approx -\log{[\ce{H3O+}]}$,  $\mathrm{pOH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pH}$.
For solutions of strong bases, $\mathrm{pOH} \approx -\log{[\ce{OH-}]}$,
$\mathrm{pH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pOH}$.

The graph you mention is for strong electrolytes, so $\mathrm{p}K_\mathrm{a}$ is not applicable.


Identification: was this gallium or mercury from broken thermometer? I had accidentally broken a thermometer against my computer screen, resulting in the splash of the silvery substance seen below:

I cannot remember whether the thermometer is a mercury-free thermometer, and was wondering if you could help with identification of the substance given the following information:

This picture was taken about 1-2 minutes after the initial spill
The temperature of the room was around 70 degrees F.. at most 80 degrees
The silvery substance clung to the screen as shown.
Small bits of the silvery substance seeped into the computer. However, it was fully functional in the few instances I have used it

Any suggestions or additional follow up questions would be much appreciated. Thanks!

Gallium, and Galinstan have the ability to wet glass, while mercury does not.

So it's most probably gallium (or Galinstan).


Derivation of Ideal Gas Law with Boltzmann's Constant I was reading about the Kinetic Theory of Gases and I came to know about This Particular equation which piqued my interest.

I know about the Ideal Gas Law which gives us $PV=nRT$.
But this one contained Boltzmann's Constant and it goes like this $PV=Nk_BT$ where  $k_B$ is the Boltzmann constant and  T the absolute temperature.
Could I get some insights to how was this form of Ideal Gas Law Derived?
And how was the value of $k_B$ and $R$ determined ? Any help would be appreciated.

$R$ is defined as $k_{B}N_{A}$ where $N_{A}$ is Avogradro's number. Notice that these forms are equivalent once you realize that $N = N_{A}n$.

I think $k_{B}$ is measured experimentally. I learned the derivation via statistical mechanics.


Is is valid to compare reaction quotient to $K_p$ as well as $K_c$? Can I use the reaction quotient compared to $K_p$ to answer the following question?

The solid $\ce{XY}$ decomposes into gaseous $\ce{X}$ and $\ce{Y}$: $\ce{XY(s) <=> X(g) + Y(g)}$ with a $K_p = 4.1$ at $\pu{0 ^\circ C}$. If the reaction is carried out in a $\pu{22.4 L}$ container which initial amounts of $\ce{X}$ and $\ce{Y}$ will result in the formation of solid $\ce{XY}$?

a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$
b. $\pu{2.0 mol}$ of $\ce{X}$, $\pu{2.0 mol}$ of $\ce{Y}$
c. $\pu{1 mol}$ of $\ce{X}$, $\pu{1 mol}$ of $\ce{Y}$
d. none of the above
If I had been given $K_c$, I could put the different answer choices into the equilibrium expression and obtain $Q$. If $Q$ turned out to be smaller than $K_c$, then that would suggest that products would be favored. Contrariwise, if $Q$ was larger than $K_c$, then reactant(s) would be favored. So, I would simply see if any of the choices resulted in a $Q$ larger than $K_c$.
But, I was not given $K_c$. Must I convert $K_p$ to $K_c$ before making the comparison with $Q$?
I suspect the answer is "yes" (i.e. I must do the conversion) because (1) the respective values are very different and (2) I get different answers to the question depending on which value I use. This leads me to believe that I cannot compare $Q$ with $K_p$. Any additional insight into why this might be the case would be appreciated.

The reaction of interest is:
$$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$
Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$, no matter what units are used for $P$.For example:
$$P_\ce{X} = \frac{n_\ce{X} \ \pu{mol} \times  \pu{0.082 L atm K-1 mol-1} \times  \pu{273.15 K}}{\pu{22.4 L}} = n_\ce{X} \ \pu{atm}$$
Suppose, initial amounts of $\ce{X}$ and $\ce{X}$ are $x$ and $y$ and amount of $\alpha$ reacted to produce $\alpha$ amount of  $\ce{XY}$ at the equilibrium (to be realistic, $\alpha \gt 0$). Thus, at the equilibrium: $P_\ce{X} = x- \alpha$ and  $P_\ce{Y} = y- \alpha$. Therefore:
$$K_p = P_\ce{X}\cdot P_\ce{Y} = (x- \alpha)(y- \alpha) = \alpha^2 - (x + y) \alpha + xy = 4.1 \tag2$$

If $x = 5$ and $y = 0.5$: From the equation $(2)$, $$ \alpha^2 - 5.5 \alpha + 2.5 = 4.1 \ \Rightarrow \ \alpha^2 - 5.5 \alpha - 1.6 = 0 \\ \therefore \ \alpha = \frac{5.5 \pm \sqrt{5.5^2 + 4 \times 1.6}}{2} = \frac{5.5 \pm 6.05}{2} $$
Thus, $\alpha = 5.78$ (not realistic) or $\alpha \ne -0.28$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 2.0$ and $y = 2.0$: Again, from the equation $(2)$, $$ \alpha^2 - 4 \alpha + 4 = 4.1 \ \Rightarrow \ \alpha^2 - 4 \alpha - 0.1 = 0 \\ \therefore \ \alpha = \frac{4 \pm \sqrt{4^2 + 4 \times 0.1}}{2} = \frac{4 \pm 4.05}{2} $$
Thus, $\alpha = 4.025$ (not realistic) or $\alpha \ne -0.025$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 1.0$ and $y = 1.0$: Similarly, you can prove that there is no reaction.

Hence, answer is "$(d)$. none of the above."


Why do they flush people with water after chemical attack? I recently viewed a video (dated April the 6th 2017), which purports to show the aftermath of a chemical attack (where Sarin was employed).
Could anyone explain why the victims in the video were doused with water? To what degree does such measure help?

A standard protocol after unintended chemical exposure is to quickly douse the victim in large quantities of water to wash away chemicals that could remain and cause further harm, such as powders and other types of solids that could spread over your body as you moved. Also, the water will dilute any chemicals to the point of least harm that soak into clothing (provide a large amount of water is used). However, as Ivan Neretin pointed out in the comment, sarin gas does not do physical damage to the external parts of the body, but enters the body and causes nerve damage extremely quickly. In this type of situation, the water dousing probably did little to nothing to help the victims.


Can precipitation occur in states of matter other than liquids? This answer to Are precipitation and crystallization both analogous between chemistry and meteorology? discusses the history and use of the term "precipitation" in chemistry.
From what I remember in school I was introduced to precipitation in the context of reactions in liquid media and usually water.
In meteorology the term refers to a phenomenon in air where droplets or crystals form from some molecules mixed with others. I don't know if we can call the nitrogen/oxygen mix a solvent and the water vapor the solute, but some of the resulting droplet suspensions are indeed called aerosols.
But my question here is constrained to chemistry. Do we speak of precipitation in media other than typical liquids? Are there gaseous precipitations? Are there precipitations in supercritical situations? Do they happen in glasses or in (presumably pretty hot) solids?

As stated in my previous answer on precipitation, formation of precipitates in supercritical is a common issue. Ask those who do sub/supercritical fluid chromatography. Sometimes you inject a methanol soluble compound into a chromatography column and the mobile phase is supercritical carbon dioxide (often mixed with a small amount of methanol) and the compound precipitates on the head of the column. So there is no restriction on using the term precipitate in fluids other than liquids.
Wikipedia also talks about precipitation in solids as

In solids, precipitation occurs if the concentration of one solid is
above the solubility limit in the host solid, due to e.g. rapid
quenching or ion implantation, and the temperature is high enough that
diffusion can lead to segregation into precipitates. Precipitation in
solids is routinely used to synthesize nanoclusters

The issue of calling a mixture of gases as a solution or not, it purely a matter of semantics. There is no rule or law which restricts that the solvent and the solute must have (strong or a certain type of) interaction in order to be labelled as a solution. Even gases interact with other gas phase molecules. Microwave spectroscopists know that very well. Many authors use the term gaseous solutions. Even one IUPAC's terminology related document uses this term. Heavens don't asunder apart if a trivial gas mixture is termed as a solution. Keep in mind that if you look up IUPAC definition of solvent, it does not include the usage of solution for gases. Only solids and liquid solutions are defined. So it is better to follow IUPAC's nomenclature.
In short, keep the meteorological precipitation away from chemistry.


Are there phosphorescent/fluorescent materials that absorb near infra-red waves and emit higher wavelengths? This SO discussion here lists some chemicals (and interestingly chlorophyll) that absorb visible radiation and emit infra red. I was wondering if there were chemicals that absorbed near infra-red and emitted much larger wavelengths (near microwave).

All molecules emit in the microwave region. They all have rotational energy levels and they absorb and emit all the time as energy flows between these levels.  But this not what you are asking I suspect. If you blast thin Gallium Sulphide wafers with a femtosecond laser then it is possible to generate terahertz radiation.


What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation? The stability order of the following is:

A. III > II > I > IV
B. II > III > I > IV
C. I > III > IV > II
D. I > III > II > IV

Answer:

 D

I came across a question where we had to determine the stability order of this carbocation compared with three others. This cation (I) was given to be the most stable (note that there's complete conjugation in the ring, shown here with a dotted line). Can someone enlighten me on how this is so?
Although this is a homework question, my point is not to just figure out the answer, I wanted to know why the first carbocation has a higher stability as compared to the rest.
This question is different from the other one because the compounds and functional groups are different. I did take a look at that one before I asked mine, as a result of which my question header is more or less the same. This is, however, different.

Questions like these always appear simple, when in fact they are not. First of all it has to be clear that stability is no absolute concept. In this case one could ask which of the cations is most persistent, in other words is least reactive towards nucleophiles. With the exception of the tert-butyl cation, one could also ask the question, which of the remaining substituted cyclopropenium cations has the largest aromatic stabilisation. One would evaluate this traditionally with the hydrogenation energy. This obviously proves difficult, because it is a cation, it has to be neutral first.

Of course you can go ahead and apply approximations like +/-I or +/-M effect, and hyper- or  ordinary conjugation, or negative conjugation, and the there are steric effects to be considered, and dispersion effects, and don't neglect the solvent, and what about the anion. There are many things to be considered, and just naming them makes this problem a lot more complicated. It is our way of looking at these questions that make them appear simple.

While I wanted to use a more rigorous approach and I am generally interested in such weird molecules, I still have not come up with something proper, that suffices my standards (in a reasonable time frame).

First I thought that could be solved with calculating the ions and comparing them via the following isodesmic reaction scheme:

This is a conformational nightmare, a reaction that produces diastereomers, highly strained rings, etc. I therefore dropped this approach.

The other way to evaluate this is to simply add a hydride ion. While this reduces the complexity a little, it is still a conformational nightmare. It is in principle a little bit too much time to invest in such an exercise, and in the end the answer might not be helpful at all.

Coming back to your original question:

  What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?

In extremely simple terms, it's exceptional stable because of the cyclopropenium core structure, because it is aromatic. Larger substituent moieties usually further stabilise such substructures, because they hinder an attack from nucleophiles. With a large enough counter ion (weakly coordinating ions, not necessarily as large as the ones mentioned on Wikipedia), it will be very persistent.

When you look at all the choices you have been given, the first conclusion should be, that they are all comparatively stable. We are not talking about primary, or secondary carbocations, or other clear cases. There really is only very little use in further analysing it.

For the sake of the argument:

Methyl groups can provide some more stabilisation (than hydrogen) because of hyperconjugation.
Phenyl groups are excellent at delocalising a positive charge when it overlaps with its π system. In that case the cyclopropenium ring is too small and the rings are too large, so they have to twist out of plane and therefore conjugation is not ideal. Here is the optimised structure of a DF-BP86/def2-SVP calculation:

Cyclopropanyl groups are rather small. They will usually attach in a way that the hydrogen is in plane with a π system. The group itself behaves like a π donor. (That's a factoid you might want to remember. See my answer on Why does cyclopropane react with bromine? and there are surely others to find. You can search in ron's answers first.)

To demonstrate this a little further, I have calculated the molecule at DF-BP86/def2-SVP level of theory and the π orbitals are below (click to enlarge).

In summary, the cyclopropanyl moiety has a good size and nice π-donating properties.

I still have to reiterate, that I find these questions not very helpful, especially if the limitations of such model views are not thoroughly discussed.


Why is tetrafluoromethane non-polar and fluoroform polar? Consider the Lewis dot structures of the molecules fluoroform, $\ce{CHF3}$, and tetrafluoromethane, $\ce{CF4}$:

My first line of thought is that both of these molecules are symmetrical (the vectors from each individual atom naturally cancel each other out) with no lone pairs of electrons. Now I knew that just because they were symmetrical didn't automatically meant that they were non-polar. If they're symmetrical, they're usually non-polar but not always. This is one of those cases.

My next line of thought was that they both had to be polar and had dipole-dipole interactions. Then I later find out that the $\ce{CHF3}$ is actually the polar molecule with dipole-dipole interaction and $\ce{CF4}$ is the non-polar molecule with London-dispersion forces.
This bothered me a little bit because both of these molecules have a very electronegative atom (fluorine) that is more electronegative than any other element in the compound. This is why I thought they were both polar.

Why is tetrafluoromethane non-polar and fluoroform polar?

Draw the structures in 3D and then you will see why one is polar and the other not.

$\ce{CF4}$:

As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four $\ce{C-F}$ bonds cancel out, leaving no overall dipole.

$\ce{CHF3}$:

Although the molecule has some symmetries, it is not perfectly symmetrical. Since the dipoles of the $\ce{C-F}$ bonds are far larger than the basically non-existent dipole of the $\ce{C-H}$ bond, the dipoles do not cancel out and you are left with a molecule which has a notable net dipole moment of 1.649 D, with the negative end over the fluorines and the positive end over the hydrogen.

As general advice, when determining whether a molecule has a net dipole, always consider the molecule in 3D as opposed to just looking at individual bond dipoles.


Interlocked cyclic compounds Being new to chemistry, there may ought to be some properties I couldn't understand fully. One of them being cyclic compounds.

One particular question about this is, whether these cyclic compounds, say benzene rings 'link' in-between one another like a chain. Having one molecule chain of one benzene molecule, go through the center of another, before closing back up as a ring.

Unfortunately, I couldn't seem to find any term describing it for further research.

This may be a stupid question or have been asked before, but since I couldn't find my answer anywhere, hopefully someone may either point out why wouldn't this exist in any case or redirect me to an answer.

The compounds you are referring to are called catenanes (see also rotaxanes, as they are a close relative).

(image source)

  one molecule chain of one benzene molecule, go through the center of
  another, before closing back up as a ring

While catenanes involving larger rings can be prepared, having one benzene ring threaded through a second benzene ring is not likely.  If you build a model you can see that there would be severe steric interactions, I've tried to depict this in the following drawing.

The carbon-carbon bond length in benzene is 139 pm.  Therefore, the approximate side to side distance in benzene is $\mathrm{2 \cdot 139~ pm \cdot cos(30°)= 241~pm}$.  The van der Waals radius is defined as "half of the internuclear separation of two non-bonded atoms of the same element on their closest possible approach" (source).  The van der Waals radius for carbon is 170 pm, meaning that the closest approach for two non-bonded carbon atoms is around 340 pm.  This value is significantly larger than the 241 pm side-to-side distance we calculated above for benzene.  This means that a carbon atom cannot pass through the interior of a benzene ring, at least not for carbon atoms with energies typically found in an organic laboratory.


Normal modes of vibration Why Normal modes of vibration are important when compared to translational and rotation modes. And why the frequencies for rotational and translational modes are close to zero?

Rotational frequencies are not close to zero. They are only smaller than vibration frequencies. You may know that frequencies and energies are proportional at the atomic scale. It does not require much energy to make a molecule move or rotate. More energy is needed if you want to deform it and try to stretch its bonds.


Is it possible to apply reaction progress kinetic analysis on non catalytic reactions? I was wondering if anyone familiar with reaction progress kinetic analysis knows if it is possible to apply the method on non catalytic reactions, and if so provide a few references with relevant examples.

Apparently the method works normally with non catalytic reactions according to the example presented in a Mettler Toledo webinar (Titled: Reaction Progress Kinetic Analysis, https://www.mt.com/us/en/home/library/on-demand-webinars/automated-reactors/kinetic.html) where a simple Diels-Alder reaction is analyzed by RPKA.


Why does boiling oatmeal "grow"? When cooking oatmeal (rolled oats, either long-cooking or instant) on the stove or in the microwave, when it reaches the boiling point, if left unchecked it will quickly "grow" and overflow the container in which it is cooking, even if it is a very large container.

A comparable amount of water boiling in the same pot or container would not do this.  It would boil, yes, but it would not "grow" and overflow the container.  I have not seen this happen with other grains.

Why does oatmeal do this?

Oatmeal swells when it absorbs water and is sticky enough to bind into a near-continuous mass. Water that is not actually absorbed into a grain is trapped by swollen soft, sticky goo. But it is still water, not chemically combined, and if you heat it to 100 C, it will volatilize with great expansion, stretching the goo as it does so, but the bubbles can't make it all the way out.

In a way, it's like a slow motion popcorn, or a speeded-up rising of yeast bread. You can do something similar with a marshmallow.

I found that if you exactly repeat a recipe, the microwave timer could be precisely set to swell the oatmeal to fill the bowl, and not overflow. Then you let it cool down and do it again.


Is the index 1 in the substituent name "1-methylethyl" a redundant index? I want to identify the correct IUPAC name for this compound:

I would say this is 2-methyl-5-(methylethyl)-5-(1-methylpropyl)-nonane. However, the solution manual says that it is 2-methyl-5-(1-methylethyl)-5-(1-methylpropyl)-nonane.

I don't understand why the first "1-" index is necessary. It doesn't add any information, as "(2-methylethyl)" would actually be just a propyl.

Am I missing something?

I'd not subtract a point for your "mistake," because there is no point to prove you are wrong. However, suppose the group is different than methyl. Then the position matter at that point. Say its bromo-group. So, 1-bromoethyl- and 2-bromoethyl- are two different things. That's why that rule is for. Can't give any favors just for methyl! :-)


What free software exists to view molecule pictures within a file explorer? I am testing my small api library that generates lots of chemical files such as MOL and SDF. I was wondering if there is a software that can show me the pictures of chemical structures inside each file so that I don't have to open each of these separately. would prefer something free for Windows and Mac.

I found this tool named thumbfish (Windows only). It can display thumbnails for variety of formats and is free


Naming convention for adduct ions In order for a compound to be analysed in mass spectrometry it must first be ionised, forming an adduct ion. In many papers (e.g 1 [PDF], 2) these adduct ions are described in the form:

[M+H]+
[M+Na]+
[M-2H]2-

Where M is the analyte, the other content of the square brackets is the ion, and outside the brackets is the resulting charge. The IUPAC Gold Book also mentions this format for protonation without, as far as I can tell, describing it further.

Is there some reference actually describing this convention - ideally encompassing more complex scenarios such as dimerisation or neutral losses as well - around?

Your convention is generally correct; as more complicated adducts and ions are formed (clusters, adducts, neutral losses, etc) things start to become a free for all where the clearest terminology wins.

As of now, just go with (where A means adduct):

$$\ce{[M + nA]^{n+}}\text{ or }\ce{[M - nA]^{n-}}$$

Protonated molecules can be denoted as $\ce{[MH]+}$ (common) or $\ce{[M + H]+}$ (technical). The $\ce{[M + H]+}$ convention is more precise because if you write $\ce{[MH]-}$ for a deprotonated ion the interpretation becomes muddled (i.e is this a hydride adduct or is this a deprotonated ion?).

One important think to keep in mind that is Molecular Ions ($\ce{M^{+.}, M^{-.}}$) are not the same as $\ce{[M + H]+}$ or $\ce{[M - H]-}$ and that the terminology is not interchangable.

$\ce{M^{+.}}$ or $\ce{M^{-.}}$ retains the elemental composition of the ion and is ionized through the removal or addition of an electron.
$\ce{[M + H]+}$ and $\ce{[M - H]-}$ are protonated and deprotonated molecules, respectively. They are not 'pseudomolecular' ions as there are no such thing as pseudo ionized specties.


How can you build a model of tetrahedral coordination from objects found at home? Our college is switching to teaching online mid-semester in hopes of slowing the spread of Covid-19. All the molecular model kits are still on campus. How could you build a model of tetrahedral coordination (say methane) from materials found at home?

I'm aware of computer visualizations (and will make those available), but I think having a physical model when first encountering three-dimensional structures adds value. I made a model from lawn toys (see below), but these are not common household items.

Inflate balloons, and tie them «at their stem» like a bouquet of flowers.  If you take four of them, not too much inflated, you well demonstrate a situation close to $sp^3$ hybridization. These models equally work well in larger lecture halls by the way, and intentionally using different colors allows many options.

(source)

(screen photo

You need some worked examples?  See videos like this or this.

You need a scientific paper? Well, there are as well, e.g., this (open access) expanding the picture to extended $\pi$-systems:


If there are two longest chains possible in an organic compound and both have the same number of substituents, how do we decide the parent chain? Is the parent chain in the above case determined by atomic mass, alphabetical preference or any other rule?

If two longest chain are possible with same number of side chain in both of them, then we consider sum of locants of the side chain. The longest chain in which the sum of locants of the side chain is least is taken into consideration.


Why are there no dihedral mirror planes in the D3h point group? Consider the molecule $\ce{BH3}$, which belongs to the $D_\mathrm{3h}$ point group. Why are the three mirror planes in this point group labelled as $\sigma_\mathrm v$ instead of $\sigma_\mathrm d$? Don't they bisect the angles between a pair of rotational axis $C_{2}$ axes as shown in the diagram below?

The σ$_v$ planes in a molecule are a direct result of the C$_n$ main axis i.e. they are interchangeable if you perform the C$_n$ operation on them. On the other hand σ$_d$, which are called dihedral planes bisect the dihedral angles between members of the σ$_v$'s set. Therefore molecules with odd C$_n$ such as BH$_3$ wont have any σ$_d$ planes; only n (3 for BH$_3$) σ$_v$'s. In contrast a molecule such as benzene which has a C$_6$ main axis will have 3 σ$_v$'s (the ones including the carbon atoms) and 3 σ$_d$'s which are the ones passing between the carbon atoms. Of course it doesnt really matter which set is called dihedral and which is called vertical, it could be the other way round as well. Ref: Chemical applications of group theory by Cotton.


Why does the O atom of N-phenylnitrosamine donate its lone pair towards H+ when O is more electronegative than N? Nitrogen has lone pair as well which it can donate towards the proton but instead we see the oxygen being protonated despite being more electronegative. Also, this is not like the next part where $\ce{H3O+}$ is formed which is a good leaving group.

If you look at complete mechanism, this wasn't hard to understand why only oxygen get protonated instead of readily available nitrogen, which is supposedly more nucleophilic than oxygen. OP has given N-phenylnitrosamine as the direct product from reaction between aniline and nitrosonium ion, avoiding the first intermediate I (see following reaction mechanism of formation of diazonium ion). As Waylander correctly pointed out in his comment that all of these steps are reversible until you get diazonium ion (counterion $\ce{X-}$ is chloride ($\ce{Cl-}$) in this case):

As I emphasized in the Blue box of the depicted diagram, nitrosamine can be protonated on either $\ce{N}$ (see in blue box) or $\ce{O}$ (see in the total mechanism). As shown in the total mechanism, protonation is slightly favored towards $\ce{O}$ because charged intermediate is resonance stabilized. Nonetheless, this stabilized intermediate is the one directing to the expected product, diazonium salt (see also Waylander comment), making it more favorable. Also note that protonation on $\ce{N}$ make the intermediate going back to original first intermediate I, which cannot give the any other product.

Note: Also keep in mind that direct conversion from intermediate I to resonance stabilized, $\ce{O}$-protonated intermediacte can be achieved by $\ce{N}$ to $\ce{O}$ proton transfer, which is highly likely.


Why are ethers treated as substituents and not as functional groups in IUPAC nomenclature? Why are ethers not considered as functional groups in IUPAC nomenclature?

Let us consider the following example:

Image source

The PIN for this compound is 2-ethoxy-1,1-dimethylcyclohexane. And we get this by first point of difference as ether is also considered as a substituent here.
But ethers do confer specific properties to the compound. They are studied as a separate functional group. Yet they aren’t given the preference in nomenclature in the absence of any other functional group too. Why is this so? Is it just a convention or is there a reason behind it?

PS: I checked this similar question on our site but it doesn’t answer my question as to why ethers aren’t functional groups when it comes to naming.

The priority order for functional groups is as follows:

Radicals > Anions > Cations > Zwitterionic compounds > Acids (In the order -COOH, -COO(O)H, S-derivatives, Se-derivatives, Sulphonic acids, Phosphonic acids, etc)> Anhydrides > Esters > Acid halides > Amides > Hydrazides > Imides > Nitriles > Aldehydes > Ketones > Alcohols and Phenols > Amines > Imines > hydrazine > Phosphanes > Alkene > Alkynes > Alkanes> Ethers > peroxides >  Halo, Nitro, Nitroso, Alkoxy > Oxiranes

Functional groups such as Ethers and Nitro-groups are way down the order and will not be the primary functional group more often than not, the reason behind them be named as substituents.

In your question, the alkane group is given priority over the ether as per the above-mentioned priority order for functional groups. Thus the name given is 2-Ethoxy-1,1-dimethylcyclohexane.

Let's say instead of the 2 methyl groups over there, you had a nitro group, then the ether would have become the principal functional group and the nitro group being the substituent. There's no rule saying that ethers are always to be named as substituents as per my knowledge.


Heat of hydrogenation If benzene is more stable than cyclohexane then why is the heat of hydrogenation of benzene negative, ie heat is released during the formation of cyclohexane from benzene ?

All hydrogenation reactions, not just for benzene, are exothermic. You can refer  Why are hydrogenation reactions exothermic .


How to read Robin-Day Classification? I am currently preparing for an exam and am stuck now with the Robin-Day classification. We had these three diagrams and also the explanations how the systems differ but we never talked about what the diagrams are supposed to show or how we have to read them.

The paper suggests that λ here is not (as one might suggest) the wavelenght required to excite an electron but the so called Marcus-reorganization energy and h(ab) is here the electronic coupling.

I understand that you have to treat the system a bit more complicated because you have a system that exchanges electrons without the formation or cleavage of bonds or any major geometric change. And there is also the concept of conductor and non conductive if we see class III compounds as being fully delocalized.

But why does the class I non conducting intervalence compound have these two energy curves, the class II a double well. I suppose because there is no λ in class three that, as there is no localization of the corresponding electrons, this reorganization is not important there. Still I'd be interested in how to actually understand the diagrams and what I can take from them.

Theses are potential energy diagrams, representing the 'stable' forms of both the reactant and the product of IVCT (and their immediate relationship with ligands and surrounding solvent). This is most easily seen in the class I case, where the curves are not merged at all, and instead represent the species as distinct (and the reaction unable to proceed). The reason the curves appear to merge is because the class II and class III cases are where electron transfer is able to occur, and thus the reactant has a reaction path with which to proceed to the product.
There are in fact two reaction paths that a class II mixed-valence species can proceed to the product, one being optical (which is the IVCT transition, caused by absorption of light), and the other being thermal. The optical pathway occurs upon absorption of light at the bottom of the leftmost potential well, where it is excited to the top left of the upmost curve. This corresponds to lambda, which confusingly represents the energy of this optical transition (E=hv=\lambda). The position that the ion is now at on the potential energy curve is representative of a state where the electron has been exchanged, but the molecule is yet to rearrange bond lengths, and the surrounding solvent has not rearranged to accomodate the new distribution of charge. The molecule then relaxes down the adiabatic potential energy curve (not shown here, some diagrams have the same structure as class I in dotted lines, these lines being the adiabatic energy curves), to the rightmost potential energy well, where the molecule has completely rearranged the surrounding solvent and changed bond lengths to accomodate the new charge distribution. The thermal process is less relevant, except to note that the hump in the class II diagram is the thermal activation barrier, where an activation energy equivalent to or greater than this barrier is required for the reaction to proceed via this pathway.

A class III ion is one where there is full mixing of the two potential energy curves, such that they become the structure you see above. Clearly here there is no thermal activation barrier, as there is no hump, and the optical transition energy is actually equivalent to or smaller than 2Hab (twice the coupling constant). The result of the now missing thermal activation barrier is that the ion can freely transition between both states, equilibrating between them, resulting in partially charged ions, and a very delocalised system. Note that optical transitions still occur in class III ions, and are typically more intense than transitions in class II ions. There is just no good way of displaying where the transition occurs as it is in the same position as the 2Hab gap.

I realise this thread was created long ago, so this response likely won't be useful to the original poster, but hopefully anyone else who is interested will have a greater appreciation of how these diagrams work.
I strongly recommend reading this paper https://onlinelibrary.wiley.com/doi/10.1002/9780470166314.ch1 if you can get access to it. It is by far the most understandable of all the papers on the subject, and very comprehensive.


Is the equation below for Gibbs energy only applicable under standard conditions? This is quite confusing to me because it seems as if a lot of people use standard state equations and non standard state ones interchangeably:

$$ΔG^\circ = -nFE^\circ$$

The equation as you wrote it refers to the standard state only. However, if you look at a system that is not at standard state, the relevant relationship looks very similar, dropping the standard state from both sides (the assumption then is that both refer to the same non-standard state that you are looking at at the moment):

$$ \Delta G = -n F E $$

The same goes for relating Gibbs energy to enthalpy and entropy:

Both

$$ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ $$

and

$$ \Delta G = \Delta H - T \Delta S $$

are correct. (Again, G, H, and S should refer to the same state in the lower equation.)

However, there are other relationships concerning Gibbs free energy where you have to be at standard state. For example,

$$ \Delta G^\circ = - RT \ln(K) $$

If you wanted a similar relationship for non-standard state, you have to account for it, e.g.

$$ \Delta G = - RT \ln\left(\frac{K}{Q}\right)$$

usually written as

$$ \Delta G = RT \ln\left(\frac{Q}{K}\right)$$

  This is quite confusing to me because it seems as if a lot of people use standard state equations and non standard state ones interchangeably.

If you have a correct relationship of Gibbs energy, enthalpy or entropy that is true for any state (i.e. written without the $^\circ$), you can always write the same equation specifically for the standard state by adding the $^\circ$ to all the state functions. The reverse (just dropping the requirement for standard state) is not always possible, as shown in the example above.


Meso and achiral does both term means the same? When a molecule has plane of symmetry,  or centre of symmetry it is termed as an  achiral compound (though it contain some chiral centres).  Same goes for the meso compounds due to the internal compensation.  So both terms mean exactly same.  But I am eager to to know the applications of these terms  in different cases.  All meso compounds can be told achiral and as well as all chiral compounds as meso?

From the Gold book:

meso-compound:
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members

Achiral compound:
A compound that can be superimposed on its mirror image.
The difference lies in the presence of the chiral centre. All meso compounds are achiral (due to internal compensation as you have mentioned) but all achiral compunds aren't meso.

Look carefully and you'll be able to spot chiral centres in all the above compounds but you'll also notice that they have a plane of symmetry (internal compensation) which makes them achiral.

Bromocylohexane has a  plane of symmetry which makes it achiral. But it has no chiral centre, that's why it doesn't qualify to be a meso compound.


Does water have a smell? I was washing my cup with hot water (without soap) and upon it nearing my nose, there was some sort of 'smell' (I lack a better word)-

However the 'smell' was different compared to when I was drinking just plain cold water.

So my question is, does water have a smell?

Water in its pure form, i.e. $\ce{H2O}$ does not have a smell, or at least no smell that we can distinguish because the receptors in our nose (and mouth) are continuously exposed to it.

What you smell are dissolved gases and other volatile impurities. The nature of these chemicals will vary mainly by location/source of the water, and might also be a bit influenced by your specific local plumbing.

For example, in Iceland near Myvatn there is a lot of volcanic activity and the water from the tap (in particular the warm water because it can dissolve more) smells like rotten eggs due to the sulfur. Another example is that in many countries the water will smell a bit like chlorine, because chlorine is used to keep the water safe to drink (kills germs).


Can a chemical reaction have two balanced equations? I was recently studying Xenon compounds and read this reaction:

$$\ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 +  24HF + 3O2}$$

However this reaction can also be written as the following:

$$\ce{4XeF4 + 8H2O → 2Xe + 2XeO3 +  16HF + O2}$$

Now both of these are valid but wouldn't that create problems, like in quantitative analysis or in kinetics of the reaction?

You can expand the second equation so it has the same amount of educt on the left side as the first one:
$$\ce{6XeF4 + 12H2O → 3Xe + 3XeO3 +  24HF + 1.5 O2}$$
Difference on the product side, compared to first equation, is
$\ce{-1 Xe + 1 XeO3 - 1.5 O2}$.
This excess of one $\ce{XeO3}$ can go
$$\ce{XeO3 → Xe + 1.5 O2}$$
, exactly the amount of xenon and oxygen that was missing compared to the first equation.
Your equations describe not one reaction, but (at least) two, which don't have to occur in a predefined proportion. It's not directly clear, and would involve an in-depth investigation, what exactly the elementar reactions are.
So you could make an infinite number of chemically correct, but quantiatively different equations for the hydrolysis of xenon flouride. All depends on the reaction conditions, probably.


Jahn Teller Effect on Metal Complexes I've been learning how the Jahn Teller Distortion effects the orbitals in Metal complexes and how the splitting of eg and t2g orbitals happens.
But the book mentions that the effect is strong for electron configurations where the eg orbital has 1 or 3 electrons. I can't figure out what that's the case. Why isn't the effect strong with 2 electrons in the eg orbital as well?

If the orbitals of the metal core is distorted. the amount of splitting is not high enough to promote the pairing of energy. therefore the 2 electrons will go to two separate eg orbitals. let the splitting energy due to this distortion be β1. the electron in the lowers orbital will have an energy loss of β1/2 and the electron in the higher orbital will have an energy gain of β1/2 and therefore the net gain in the energy is 0. which does not give any additional stability to the complex(as there is no release of energy).
hence the distortion does not have a strong effect.


Relationship between pKa and pKb The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$. Does it stand for the following equilibrium:

$$\ce{NH4+ <=> NH3 + H+} \tag{1}$$ With $K_\mathrm{a} = 10^{-38}$ for reaction $(1)$, indicating that the conjugate acid of $\ce{NH3}$ is quite weak?

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$. Does it stand for the following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-} \tag{2}$$ With $K_\mathrm{b} = 10^{-4.74}$ for reaction $(1)$, indicating that $\ce{NH3}$ is a weak base?

Also: $$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 42.74 \neq 14$$

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$.

This means, the reaction this value is describing is:
$$\ce{NH3 <=> NH2- + H+}$$

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$.

This means, the reaction this value is describing is:
$$\ce{NH3 + H+ <=> NH4+}$$

There is a lot of misuse going on with the term $\mathrm{p}K_\mathrm{a}$ since many people are too lazy to say

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$’s conjugate acid is $9.26$
$$\ce{NH4+ <=> NH3 + H+}$$

rather than

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $9.26$.
$$\ce{NH3 <=> NH2- + H+}$$

but that is really what they should be doing. Thankfully, when looking up values in a table the person compiling the table (hopefully) knew what they were doing and used the correct terminology.
Only for a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$ that describe the same reaction (from opposite viewpoints) is it true that $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$.


Why is temperature constant when an ideal gas expands into a vacuum? An ideal gas in a box has pressure $p$ and temperature $T$. This box is kept in a vacuum, within a large container.

When the box is punctured, what happens to the temperature of the gas as it expands to fill the entire container?

I was taught that the temperature will remain constant. How is this possible? I understand that the pressure $p$ will decrease and the volume $V$ will increase, but for the temperature $T$ to remain constant, the decrease in $p$ must exactly counterbalance the increase in $V$. Is this the case?

This process is called an expansion into a vacuum. As you described, the volume of the gas increases and the pressure drops, such that $pV$ does not change at all. Since $n$ is constant, $T = \frac{pV}{nR}$ is also constant.

I think the fact that $pV$ is constant is not very intuitive if one only uses the ideal gas law. A more formal way of explaining that $T$ is constant would use thermodynamics. If you assume that the large container does not allow any transfer of heat (i.e. it is thermally insulating), then the large container is an isolated system: $\mathrm{d}q = 0$. Since the gas is expanding into a vacuum, $p_{\text{ext}} = 0$ and $\mathrm{d}w = -p_{\mathrm{ext}}\,\mathrm{d}V = 0$. By the First Law, $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w = 0$ and temperature is therefore constant (since the temperature of an ideal gas depends fully on $U$).


MgCl2 acidic or neutral in water? I'm currently taking chemistry 12. On our test we were asked, when given a $0.1~\mathrm{M}$ solution of certain compounds whether the resulting solution when added to water would be acidic, basic, or neutral.

At first, I thought the solution would be neutral, because we learnt in class that alkali earth metals and chloride are spectator ions and therefore don't interact with water and pH.

However, I then realized that $\ce{Mg(OH)2}$ has low solubility in water, and I assume that adding $\ce{MgCl2}$, would dissociate and then precipitate $\ce{Mg(OH)2}$ from the solution, therefore the resulting solution would be acidic. My final answer was acidic.

The teacher marked my answer wrong, but when questioned could only say that $\ce{Mg}$ and $\ce{Cl}$ are both spectators, but was unable to give direct contradictory proof against my answer.

I looked elsewhere online and Wikipedia states that anhydrous $\ce{MgCl2}$ is a Lewis acid, despite being a weak one. But anhydrous seems to mean without water, so I currently have no idea which answer is correct.

I would like an explanation to why it is acidic or neutral and why the contradicting argument (both spectators vs low solubility of $\ce{Mg(OH)2}$ ) is invalid or flawed.

Long story short, the ultimate cause of non-neutral pH in the solutions of some salts is hydrolysis. If a salt is made from a weak acid and strong base, then the acid anion would get partially protonated by water, like $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$, and that $\ce{OH-}$ would be responsible for the overall alkaline pH of the resulting solution. If we have a salt of a strong acid and weak base, it is the other way around.

Which acids or bases are strong and which are weak is determined by looking up their $pK_a$ or $pK_b$ in the reference books. I don't want to get deeper into this because running far ahead of your syllabus might do more harm than good. Just remember that neither $\ce{Mg^2+}$ nor $\ce{Cl-}$ would hydrolyze significantly, so the solution would remain neutral.

The precipitation is not all that important, as hydrolysis of most common salts rarely goes all the way up to metal hydroxide. For example, $\ce{AlCl3}$ would make the solution pretty acidic, but still produce no precipitate. Also, consider $\ce{NH4+}$ which is prone to hydrolysis despite having soluble hydroxide. (One might argue this does not quite qualify as hydrolysis, but anyway.) If you want an example when the precipitation actually does occur, think of $\ce{TiCl4}$.

As for the spectator ions, that's a lame argument indeed. There is no such thing as "spectator ion in general". True, certain ions are spectators in certain reactions, but there is no ion that would always remain a spectator, no matter what the situation is.

As for the Lewis acids, that's a different concept altogether, to the point that you shouldn't automatically think of "Lewis acids" as acids (though many of them are).


SMILES string convertors I'm new to SMILES strings, and have noticed that they're not standardised.
Firstly, what is the reason for different structures of SMILES strings to represent the same compound?  Why not have a standardised format?
Secondly, are there any tools to convert between different SMILES strings for a given molecule?

SMILES doesn't define a particular order in which the atoms show up in the string. Hence the position in the molecule where software starts printing will determine the string. E.g. for CO2:

Ketcher prints C(=O)=O (starts with C and considers one of the O's a branch)
MarvinJS prints O=C=O (starts iteration with O, hence no branching)

Both of these representations are valid, they simply start with different elements. Each tool would recognize SMILES string generated by another tool, so there's no need to convert between strings created by different software.
In order to make different tools generate the same string a so called Canonical SMILES was developed. Not every tool can generate it, but if they do - they'll spit out the same string (in theory).


Formation of O₂, Water, and Electricity from Hydroxide? The Clark Type Oxygen Electrode converts water, oxygen, and electrons into hydroxide via a hydrogen peroxide intermediate:

\[\ce{O2 + 4e^{−} + 2 H_2O -> 4 OH^{−}}\]

Which is the sum of:
\[
\begin{align*}
\ce{2H2O +O2 + 2e- &-> H2O2 + OH- \\
H2O2 + 2e- &-> 2 OH-}
\end{align*}
\]
Can this be reversed?

Yes, the overall reaction can be reversed.  The reverse reaction is called the oxygen evolution reaction or simply water oxidation.  Depending on the pH, you could write it as either

$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$ or
$$\ce{4 OH- -> O2 + 2 H2O + 4e-}$$

The reverse reaction happens when water is electrolyzed to make hydrogen at one electrode and oxygen at the other.  You may have seen that demo at one time or another -- here's a Youtube version.

Oxygen is a thermodynamically powerful oxidant, which is another way of saying that it is easily reduced.  This means that the water oxidation reaction requires energy input, about 1.23 V at the chemical reference state of pH 0.  But it usually takes even more energy to effectively reverse the reaction -- notice they used a 9 V battery in the Youtube demo, so about 8 of those volts are wasted in the demo.  Developing new electrode materials that can carry out the water oxidation reaction with less energy loss is a major area of fundamental research in chemistry today.

You also said the Clark electrode works with $\ce{H2O2}$ as an intermediate.  Depending on the details of the particular materials used, in the water oxidation reaction hydrogen peroxide may or may not be an intermediate.  Working out the fundamental reaction steps that happen at electrode surfaces can be very difficult.


If the hydroxide ion is such a poor leaving group then why does water dissociate? Water dissociates into the hydroxide and hydronium ions but, why does this happen if the hydroxide ion is such a poor leaving group. Alcohols don't undergo $\ce{S_{N}1}$ reactions because the hydroxide ion doesn't dissociate so why does it happen in water?

Only 1 in 10,000,000 water molecules dissociate.

HI, HBr and HCl fully dissociate, and I-, Br-, and Cl- and are much better leaving groups.


How do I make a simple home distillation device? Like this:

I have two medium sized glass flasks. I understand I need a T-shaped adapter from the first flask that will connect the thermometer and the condenser. So, I also need the condenser and a L-type adapter that will connect the last flask with the condenser.

Is the condenser really necessary for a completely beginner set up?

Do I really need the adapters? I'm worried they will not fit. How high are the costs for these parts?

Thanks for any replies. I know some theoretical chemistry stuff, but with this glassware I've got no idea what I'm doing.

EDIT: Thanks guys, I accidentally found a very nice chemical equipment store and a nice lady there sold me all the parts. I only didn't get the thermometer because it was way too expensive (and I can make something similar myself). It cost 80 eur. Flask (1l), adapter, condenser, another adapter, mount for the flask and rubber pipes. The thermometer was extra 35.

This depends on what you need this setup to accomplish, as the amount of solutions that this setup can manipulate vary greatly.

The basic consensus is that if you would want an optimal apparatus you would need:

Round Bottom Flask
Thermometer
A Column
Condenser (with Water In/Water Out nozzles)
Collection Flask


Can nuclear fission or fusion occur together with bond formation in a reaction? Is it possible for a nuclear reaction to occur simultaneously with a chemical reaction involving bond formation?

Can anyone suggest an example of this, please?

It has been done, not with bond formation but rather with bond breaking, and is called the Szilard–Chalmers reaction (credit to Loong for pointing out the name).

When a biphasic mixture of ethyl iodide (with stable $\ce{^{127}I}$) and water is irradiated with neutrons, all the radioactivity from $\ce{^{128}I}$ is found as iodide in the aqueous phase; there is none in the organic phase.  This is because $\ce{^{128}I}$ is initially formed in an excited state that emits a γ-photon, and the recoil from that is more than enough to break the C–I bond.

The energy of a γ-photon is > 10 keV; in comparison, the strength of a typical chemical bond is ~ 1 to 10 eV. The C–I bond strength is roughly 2.5 eV.


Why is ascorbic acid (vitamin c) readily oxidised in the presence of oxygen? I know that vitamin c is notoriously unstable in aerobic conditions, but I'm not sure how to explain why it is - is it due to redox potential, acid strength etc?

According to Juan Du, Joseph J. Cullen and Garry R. Buettnera[1]:

Ascorbic acid ($\ce{AscH2}$, vitamin C) is a water-soluble ketolactone with two ionizable hydroxyl groups. It has two $\mathrm{p}K_\mathrm{a}$’s, $\mathrm{p}K_1$ is 4.2 and $\mathrm{p}K_2$ is 11.6; thus, the ascorbate monoanion, $\ce{AscH−}$, is the dominant form at physiological pH. Ascorbate is an excellent reducing agent and readily undergoes two consecutive, one-electron oxidations to form ascorbate radical ($\ce{Asc•−}$) and dehydroascorbic acid (DHA). The ascorbate radical is relatively unreactive due to resonance stabilization of the unpaired electron; it readily dismutes to ascorbate and DHA ($K_\mathrm{obs} =\pu{2×10^5 M−1 s−1}$ at $\mathrm{pH} = 7.0$).

Also,

Ascorbate readily undergoes pH-dependent autoxidation producing hydrogen peroxide ($\ce{H2O2}$).

So, in presence of oxygen, this process is accelerated due to formation of hydrogen peroxide. It is also known that this reaction may happen even faster with a metal as a catalyst[2]

References:

Du, J.; Cullen, J. J.; Buettner, G. R. Ascorbic acid: Chemistry, biology and the treatment of cancer. Biochim Biophys Acta. 2012, 1826 (2), 443–457. DOI: 10.1016/j.bbcan.2012.06.003.

Jiang, D.; Li, X.; Liu, L.; Yagnik, G. B.; Zhou, F. Reaction Rates and Mechanism of the Ascorbic Acid Oxidation by Molecular Oxygen Facilitated by Cu(II)-Containing Amyloid-β Complexes and Aggregates. J. Phys. Chem. B 2010, 114 (14), 4896–4903. DOI: 10.1021/jp9095375.


Syntactic vs Semantic validity of molecular graphs While reading this paper on molecular graph generators using deep learning algorithms, I encountered the terms syntactically and semantically valid molecular graphs:

  $\dots$ to ensure the generation of syntactically and semantically valid
  graphs.

The output molecular graph should at least satisfy the following constraints:

Valency: Number of incident edges to an atom shall not exceed the valency of the atom on that node
Connectivity: The output graph should be a single connected graph.

I was wondering what semantic and syntactic validity means in this context? Which type are each of these constraints?

Some concepts from computer science:

  Syntax is the structure or form of expressions, statements, and
  program units but Semantics is the meaning of those expressions,
  statements, and program units.

But I can't make an exact connection to molecular graph's case. Also,

  In computer science, the syntax of a computer language is the set of
  rules that defines the combinations of symbols that are considered to
  be a correctly structured document or fragment in that language

Then, valency and connectivity, unlike the linked paper in the answer below, should be syntactic constraints.

Generative machine learning methods in chemistry are typically tasked with generating new molecular structures. Think of it as 'dreaming up molecules.'

We see these kinds of ML models with predictive text programs.. if I'm typing, my phone attempts to guess the word or complete my sentence. (NB this doesn't work so well with molecule names or scientific terms.)

There are a wide variety of ways to represent a molecule for such a generative ML method. One of the common ways is via SMILES in which the molecular graph is encoded as a text string:

c1ccccc1  benzene
CC(=O)O   acetic acid

In this case, syntax can be a problem. SMILES uses parenthesis to represent branching, and they must match. Numbers are used to represent ring closures, which must match as well.

Consequently, many generative ML method generate syntactically incorrect molecules (ignoring questions of chemical valence, etc.)

c1ccccc2 error
CC(=OO)) error

There are solutions. Noel O'Boyle and Andrew Dalke suggested DeepSMILES which encodes a new syntax that's resistant to such issues.

As you mention in your answer errors in connectivity and/or valence represent semantics - that is what makes a chemically viable molecule (i.e., carbon does not take 5 bonds, isolated oxygen atoms aren't stable, etc.)


What should be the nomenclature of this organic compound? I am confused with my teacher's explanation for the nomenclature of this organic compound.

He said that we should start numbering from the sulphonic acid at the bottom of the structure. But one of the students asked if we should start numbering from the other sulphonic acid and with this method, the double bond will be on number $1$ (as shown in the image below). But my teacher didn't give a reason for that.

So isn't the student saying correct ?

Although this isn't disputed by anybody here, let's start by noting that in general, suffixes have greater "priority" over double and triple bonds in numbering. The rules are given in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(e) saturation/unsaturation

This means that the sulfonic acid suffixes should obtain the locants 1 and 2, which are the lowest possible for two adjacent groups. There are two possible ways of doing this.

With the sulfonic acid groups sorted, we can now turn to the thorny issue of the alkene. It is more complicated than I thought, actually. First, let's consider how to refer to the double bonds in the names. Again, the answer is provided by the Blue Book.

P-31.1.1.1
[...] Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

On the left, this is a "typical" double bond where the locants (2 and 3) differ by 1, and so only the lower locant is cited.
On the right, the locants (1 and 6) differ by  more than 1, so both locants are cited with the higher one in parentheses.

Lastly, we have to make a choice between these two. We turn to the last rule for today:

P-31.1.4.2 If there is a choice of names and numbering, the following criteria are considered in order until a decision is reached:
(1) a minimum number of compound locants. A compound locant is used for a double bond if the locants of the atoms at each end of the bond do not differ by a value of one. When a compound locant is required, the higher locant is cited in parentheses. [...]

Since the option on the left-hand side has no compound locants, and the one on the right has one compound locant, the left choice is preferred. Therefore, the preferred IUPAC name for the molecule is cyclohex-2-ene-1,2-disulfonic acid.


Most stable resonating structure of 1-nitro-4-nitrosobenzene The most stable resonating structure of 1-nitro-4-nitrosobenzene is:

In my view since in each case we have 4 charges and except in case 1 all have complete octets, so 1 is neglected. However, I am facing difficulty in eliminating other two in order to get the desired answer. I noticed that in case 3 it is in conjugation (alternating single and double bonds).

The answer is not that difficult. You should have a certain number of rules to follow in determining the most stable resonance (not "resonating") structure.

It probably starts with check all for octets (which you correctly did). 1 has ten electrons on the nitroso nitrogen, so that rules it out.

It probably then goes on to ask you to check for the least number of formal charges (which, judging by your description, you probably did).

The next step should probably say something about the negative charges being on the most electronegative elements. Electronegative atoms like O, F... (and N to a slightly lesser extent) are quite happy to take on a negative charge whereas carbon will not quite be so pleased.

Here, resonance forms 2 and 4 have the negative charge on carbon, whereas structure 3 has it on oxygen. 3 is therefore the most stable resonance structure.

Incidentally, all of the resonance forms 2 through 4 have complete conjugation throughout the molecule so that's not a factor. The $\mathrm{sp^2}$-hybridised anionic carbon can take part in conjugation equally well. Conjugation does not require alternating single and double bonds - it only requires an extended, continuous, $\pi$ system. Generally this means that all atoms along the chain must have unhybridised p orbitals parallel to each other.

Side note: "Most stable" resonance structure is a slightly sloppy description. None of these resonance structures actually exist and the true structure is a linear combination of all of these resonance structures. It should more properly be called the "greatest resonance contributor" or "greatest contributor to the resonance hybrid".


Chemical test for ferrous ions I have a reference sheet for inorganic analysis which says that the confirmatory test for ferrous ions is as follows:

"Make a solution of the given salt (whose cation is to identified) and add potassium ferrocyanide [ potassium hexacyanidoferrate(II) ] solution to it.

If a dark blue precipitate is formed, then the presence of ferrous ions in the given salt is confirmed."

Chemically, what is the composition of this dark blue precipitate? Or is there a flaw in the test that I have specified?

If yes, then what is the real test and the chemical formula of the dark blue precipitate?

The test is better suited for $\ce{Fe^3+}$.
$$\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$$

The compound $\ce{Fe4[Fe(CN)6]3}$ has a specific name. Its called Prussian Blue.

If you perform the test for $\ce{Fe^2+}$, the precipitate won't be that dark. First partial conversion of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ takes place. Then the above reaction takes place to give the prussian blue precipitate.

There are specific tests for $\ce{Fe^2+}$ as well. You can use Dipyridyl to get a red colouration which is due to the formation of $\ce{[Fe(dipy)]^2+}$.
Or, you can use Dimethylglyoxime to get red colouration again which is due to $\ce{[Fe(DMG)2]}$.


What is the difference between Half Equivalence point and Equivalence point? OKay I learned in class that equivalence point is the time of neutralization, where the # of moles of acids and base present is in equivalence/same amount.

Now I was also told that Half equivalence point is when [HA] = [A$^-$] , the concentration of a weak acid = concentration of conjugate base.

Problem:
I'm getting a bit confused as to the differences. I've read around and am finding definitions that don't really shed any light on the differences between them. What is the difference between a half equivalence and an equivalence point?

Based on what i have been told, they sound the same to me.

The equivalence point is where the amount of moles of acid and base are equal, resulting a solution of only salt and water. If you are titrating an acid against a base, the half equivalence point will be the point at which half the acid has been neutralised by the base. For instance, if you have 1 mole of acid and you add 0.5 mole of base, exactly half of the acid will have been neutralised. The solution remaining will be half salt and half acid.


Conc. vs Time Graph for Second Order Reaction From my understanding plotting the conc. over time for a 0-order reaction yield a negative linear slope. Plotting conc vs time for a first order reaction intuitively yields an exponential type slope. however, my teacher never covered the looks of a conc vs time graph for a second order reaction. What may this look like. In conjunction, what does it look like when we graph its rate over concentration?

Let's take a single-component second-order reaction:

$$\ce{2A -> B}$$

If it is a 2nd-order reaction, the rate will be proportional to the concentration of $A$, squared:

$$\frac{dA}{dt} = -kA^2$$

That equation can be integrated:

$$\frac{dA}{A^2} == -k dt$$

$$-(A^{-1}-A_0^{-1}) = -k(t-t_0)$$

And solving for $A$:

$$A = \frac{1}{k(t-t_0) + A_0^{-1}}$$

That is the equation of a hyberbola.  It goes to zero more slowly than a the expression for a first-order reaction, i.e. an exponential decay.  This makes sense because as $A$ gets more and more diluted as the reaction occurs, collisions between $A$ molecules become rarer and rarer.

See this web page for more information.


For isomers, what does the term "connectivity" mean? I would like an explanation for the term "connectivity" in the question "Do the compounds have the same connectivity? I do not understand what "connectivity", a vague term, means.

Connectivity describes how the atoms are connected to each other. Consider a daisy chain (image courtesy of Wikipedia).

All the daisies are similar enough much like carbon atoms are similar enough between them; we cannot distinguish. Counting ten daisies in that chain, I could also represent it as (with the ends of the chain reconnecting):
$$\ce{-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$$
This compound would by cyclodecane.
Now imagine you replaced two daisies with violets. Of course, violets are immediately distinguishable from daisies like oxygen atoms are from carbon atoms. We now have a number of different options how to replace daisies with violets. They could be next to each other, separated by one, two three or four daisies. That would give rise to the following structures:

1-peroxecane
$\displaystyle\ce{-O-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,3-dioxecane
$\displaystyle\ce{-O-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,4-dioxecane
$\displaystyle\ce{-O-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,5-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-{}}$

1,6-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-{}}$

In connectivity terms, in the first case and the first case only the two violets are connected to each other. In the second case and the second case only, there is a single daisy connected to two violets. These should be the obvious cases of different connectivities, although each different violet placement gives rise to a new connectivity.
These compounds are different; especially the first (which is a peroxide) and the second (which is the formyl acetal of a dialcohol) have very different chemical (and physical) properties from the others (which are simply cyclic diethers).
Added difficulty arises for issues that I have purposefully ignored in my daisy chain analogy: it does not consider three-dimensionality. Of course, atoms in reality are connected in a three-dimensional manner. You want to be sure that the connectivity is the same in all three dimensions (but not necessarily in the same orientations). Taking butan-2-ol as an example, we speak about identical connectivity whether we are dealing with (R) or (S)-butan-2-ol; because the same carbons atom are still connected to the same other atoms (e.g. carbon 2: Connected to a methyl group, an ethyl group, a hydrogen and a hydroxy group). But a different connectivity would be butan-1-ol, where carbon 2 is suddenly connected to a hydroxymethyl group and a second hydrogen (in addition to the ethyl group and the hydrogen which remained unchanged).


Is this the correct mechanism of the formation of the nitronium (NO2+) ion from sodium nitrate and sulfuric acid? I am currently working on a mechanism for forming paracetamol (or N-(4-hydroxyphenyl)-ethanamide). The typical reaction to nitrate the phenol (starting product) is $\ce{NaNo3}$ with a catalyst of sulfuric acid. The sulfuric acid is required as a catalyst to form the nitronium cation - an electrophile - to attack the aromatic phenol. Unlike the nitration of toluene (for TNT) or benzene, it is possible to use $\ce{NaNO3}$ because the oxygen in the hydroxy group is activating.

My proposed mechanism for the formation of the nitronium is this:
$$\ce{NaNO3 + H2SO4 + 2H2O <=> Na+ +NO3- + SO4- + 2H3O+}$$
I think this is a likely equilibrium between the starting ions. Sulfuric acid is a strong acid, and therefore to use a mechanism with it undissociated I think unlikely - the reaction would be slower, due to lower concentrations of the undissociated sulfuric acid. I know that $\ce{NaNO3}$ is fairly soluble, so would dissociate into ions.

I suggest attack of the $\ce{NO3-}$ by $\ce{H3O+}$, protonating one of the oxygens:
$$\ce{NO3- + H3O+ -> HNO3 + H2O}$$
$\ce{HNO3}$ has a Ka of ${24 * 10}$ - 24 moles of $\ce{HNO3-}$ present for every $\ce{NO3}$ we can attribute this to the stability of the $\ce{NO3-}$ ion, but it also is fairly low compared to the Ka of sulfuric acid - ${1*10^3}$. By adding sulfuric acid, a strong acid, it increases the concentration of $\ce{H3O+}$, therefore shifting the equilibrium to the right. This favors the reaction.

In the $\ce{HNO3}$ molecule the oxygen-hydrogen bonds are all different lengths - $\ce{N-O}$ is 119.9pm, or 121.1pm, and the $\ce{N-O-H}$ bond is 140.6pm. The $\ce{N-O-H}$ bond is clearly weaker than the other two bonds. This is attributed to the resonance effect - a valence electron from each oxygen interacts with a lone pair on nitrogen favorably, shortening and strengthening the bonds. Protonating the oxygen in $\ce{NO3-}$ thus weakens the bond. Protonating the oxygen a second time:
$$\ce{HNO3 + H3O+ <=> NO2+ + H2O + H2O}$$
Would probably break that bond.

The hydronium protonates the oxygen already protonated - resulting in an unstable intermediate. The positive charge pulls electron density from the $\ce{N-O}$ sigma bond onto the oxygen (inductive effect). This breaks the bond, and stabilizes the positive charge on the oxygen, and it becomes $\ce{H2O}$

However, the hydronium may protonate one of the other two oxygens. Why is the one already protonated more basic? The other two oxygens have a higher electron density, so would be more nucleophilic - better for the electrophilic $\ce{H3O+}$ to attack. A fraction would surely react in the way I have described, but I would think a much higher proportion would react on other oxygens.

Is this mechanism likely/correct? Could you explain how the second protonation occurs?

The electrons on the two other oxygens are delocalized making them much less basic.


Block on the Periodic Table? What are the blocks of the periodic table? What is the purpose of naming the elements per block group? What do they signify?

Below is a picture of the 4 blocks in the Periodic Table.  The elements in a block all use the same orbital to hold their valence or outer shell electrons.  For example, all of the elements in block 1 have their valence electrons in an $\ce{s}$ orbital, those in block 2 have their valence electrons in a $\ce{p}$ orbital, block 3 the $\ce{d}$ orbitals and block 4 the $\ce{f}$ orbitals.


Compare the ionization energies of C and Cl? I saw a test question today that basically boiled down to comparing the ionization energies of $\ce{C}$ and $\ce{Cl}$. I know that in the periodic table, ionization energy generally increases as we move right across a row (because nuclear charge increases), and decreases as we move down a column (because atomic radius increases, so valence electrons feel less attraction to the nucleus). Note that $\ce{Cl}$ is in the row below $\ce{C}$, but is also to the right of $\ce{C}$, so it is not obvious how to compare the two. So, without looking up the ionization energies and using only the periodic table, how can I answer this question?

It's the other way around,

moving left to right across a row increases the IP due to increasing nuclear charge, and
moving down a column decreases the IP, due to increasing shielding of the nuclear charge by core electrons.

Chlorine is only one row below carbon, but it is three columns to the right.  As the following chart illustrates, moving to the right by one position generally has a similar or larger effect than moving down one position.

(image source)

Therefore, in this case moving further to the right by 3 positions (and increasing the IP) strongly outweighs moving down by only one position (and decreasing the IP).  Hence, in this case the IP of chlorine would be predicted to be greater than the IP of carbon.  Indeed, the IP of chlorine is reported to be 13.0 eV, while that of carbon is only 11.3 eV.


Which of the following carbocations is more stable? Which of the following carbocations is more stable?

I know two major factors in determining carbocation stability: resonance and the number of carbon atoms neighboring the positive carbon. Does the double bond in the  substituent of the first molecule participate in resonance?

The second one will be more stable.
We usually first check for resonance as it is more effective.
But here it has to be explained on basis of hyper conjugation because the resonance due to the ring is same in both. But the second one will have more hyper conjugating structures as greater number of alpha hydrogens are present.
More about hyper conjugation concept:
https://en.wikipedia.org/wiki/Hyperconjugation

Edit :  No the double bond wont participate in resonance.


Ionization energy of beryllium Suppose I have $\ce{Be^3+}$. What would be its 4th ionization energy?

By trying to solve the issue I saw that its a "hydrogen-like" atom – means that beryllium left with only $1$ electron in his valance shell. Hence, I can use the Bohr atom model to solve the problem.

Is my guess right? If so, what values I need to plug in the equation and why so?

In terms of Bohr model ionization potential $E_\mathrm{i}$ is the work $A_\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity:

$$E_\mathrm{i} = \frac{A_\mathrm{i}}{e}$$
$$A_\mathrm{i} = h\nu = \frac{hc}{\lambda}$$

Unknown wavelength $\lambda$ can be determined from the Rydberg formula:

$$\frac{1}{\lambda} = R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

so that final equation for ionization energy looks like this:

$$E_\mathrm{i} = \frac{hc}{e}R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Since we are determining 4th ionization energy of beryllium ($Z = 4$), $n_1 = 1$ and $n_2 = \infty$:

$$E_\mathrm{i}^\mathrm{IV} = \frac{\pu{6.63e-34 m^2 kg s-1}\cdot\pu{3e8 m s-1}}{\pu{1.602e-19 C}}\cdot\pu{10973732 m-1}\cdot 4^2\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = \pu{217.86 eV}$$

This value is in a good agreement with the one listed in CRC Handbook of Chemistry and Physics [1, p. 10-204]: $E_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.


Why do potassium, rubidium, cesium all have lower standard potentials than sodium? Why do potassium, rubidium, cesium all have lower standard potentials than sodium?

Sodium has a higher charge density than any of them, so its enthalpy of hydration should be larger, right? The argument applies for why lithium has a more negative standard reduction potential than the other three, but why doesn't this argument apply for sodium?

I think lithium is the outlier in this series due to the enthalpy of hydration. The trend from sodium to caesium is linear and consistent with ionization energies, meaning the enthalpies of hydration are not big enough to account for the difference in ionization entropies, thus the trend.


How can glycine act as a buffer at pH 6.00 and why? How can glycine act as a buffer at pH 6.00 and why? Please use
chemical equations to show how it can act as a buffer at this pH
specifically?

My work:
At pH 6.00 glycine acts as a buffer, because at pH 6.00 the condition is acidic and glycine is an amino acid, so its zwitterion forms an equilibrium, which means there is a positive charge on the N of the amino group and a negative charge on the carboxylate. So there are no sudden pH changes and we know buffer solutions resist sudden pH changes, so at pH 6.00 glycine acts as a buffer solution.
Equation:

What I need explained:
HOW does glycine act as a buffer at pH 6.00 with chemical equations, meaning to show an example of how it specifically buffers another species at pH 6.00 with the proper chemical equations.

Your equation is not correct for pH 6; it should be for pH 1. For pH 6, the equation should be:

If an acid is added, then the equation becomes:

and if a base is added, then the equation is:


Why is aluminium carbide bent? The geometry of aluminium carbide is attached. Why isn't it completely linear (in accordance to VSEPR), but bent on the Al atoms?

That structure is incorrect. Several online chemical distributors have similar structures posted in their entries for aluminum carbide; so does PubChem! Some sources show the structure as linear; some show it as bent.

That structure basically assumes that aluminum forms primarily covalent bonds with an "incomplete octet" in VSEPR structures, rather like boron. But aluminum atoms are a lot larger and more metallic than boron atoms, and its compounds will have comparatively more ionic character. $\rm Al_4C_3$ isn't a molecular compound.

Cotton & Wilkinson's Advanced Inorganic Chemistry says that aluminum carbide "reacts instantly with water to produce methane, and X-ray studies have shown it to contain discrete carbon atoms (C-C = 3.16 Å); for these reasons it is sometimes considered to be a "methanide", that is, a salt containing $\rm C^{4-}$, but this is probably an oversimplification."

This is the structure that's posted on Wikipedia:

It says that "Aluminium carbide has an unusual crystal structure that consists of two types of layers. It is based on AlC$_4$ tetrahedra of two types and thus two types of carbon atoms. One is surrounded by a deformed octahedron of 6 Al atoms at a distance of 217 pm. The other is surrounded by 4 Al atoms at 190–194 pm and a fifth Al atom at 221 pm." This is attributed to Greenwood's Chemistry of the Elements, but I don't see it in the edition I have.


Why does dielectric constant of crystal fluctuates in case of Frenkel defect in a crystal? So I was studying Frenkel defect in crystal and came across this line that said

  The closeness of like charges tends to increase the dielectric constant of the crystal.

I'm not able to understand it properly because in a dielectric an number of dipoles align themselves in a direction parallel to electrical field and thus they create a counter field to reduce the electric field that created or aligned it. Now, if we think like that here, dipole moment is charge multiplied by distance, and distance has definitely reduced between a positive and negative ion here, so dipole moment should decrease and consequently dielectric constant should decrease but why is the opposite happening here.

The explanation provided in the text you quote is of the "hand waving" sort. Disrupting the crystal structure via Frenkel defects will have a rather complicated effect. That the effect is complicated is illustrated (in a rather rudimentary fashion) in the wikipedia, from which I derived the following figure:

Clearly the dislodged cations are shifted into closer proximity of other cations, but also of anions, and the orientational effects are not simple.

So the text you quote does not provide an entirely satisfactory explanation, I agree.

Now consider your line of argument:

  Now, if we think like that here, dipole moment is charge multiplied by
  distance, and distance has definitely reduced between a positive and
  negative ion here, so dipole moment should decrease and consequently
  dielectric constant should decrease but why is the opposite happening
  here.

The scenario you describe in the paragraph makes sense. You can write the dipole moment $p$ for two point charges of opposite polarity $q$ and separated by distance $d$ as

$$p = qd $$

Note that, as the wikipedia explains, the value of $p$ is independent of the choice of reference point, provided the overall charge of the system is zero.

Obviously decreasing $d$ decreases $p$.

The problem is that this does not describe the situation layed out in the text you quote:

  The closeness of like charges tends to increase the dielectric constant of the crystal.

Consider what it means to increase the closeness of like charges. Presumably if the crystal remains constant in volume and overall charge, it requires that opposite charges move apart.

But this is still a hand-waving explanation. You need to attempt some calculations with a model to get closer to a convincing explanation.


Will washing child seat belt straps with baking soda and hydrogen peroxide make them unsafe to use? My friend is afraid to use the child seat for her next child.  She was told washing them with the solution weakened the fabric,  The straps weren't submerged, but they did get throughly wet.

I have seen some companies that refurbish child seats, perhaps a new strap kit would put your friend's mind at ease.

If the straps were not soaked for hours in concentrated H2O2 and I saw no damage and could not cause them to break with my strength (healthy adult) I would have used it with my kids.  The baking powder or soda should have no effect.

We bought a used seat for the next size up and have no idea of how it was washed.  Would do the same again.

Stay at home, spend the coin, take out a loan or live with the risk, those are the choices. There are other risks in life like not being sure that the excavator operator is of sound mind and will not swing the bucket into your car at the stop sign, does this stop you leaving the house?

EDIT:
It is possible the dangers of washing are circulated by the seat refurbishing and manufacturing industries to increase sales and not safety.  A car seatbelt can last for decades and sees a lot more use and has no washing warnings.


Purifying cysteine containing peptides I would like to purify my peptide having cysteine residue by using HPLC. It is known that thiol containing peptides undergo oxidation and form disulfide linkages. In order to prevent it I used proper cocktail solution during cleavage from the resin. However, I am still worried what if my peptide undergo oxidation during HPLC purification? How can I prevent it?

I searched in research gate, but I couldn't find suitable answers.

I dont know about HPLC column tolerance to various reducing agents, this would need to be checked. However, in biochemistry we routinely add reducing agents to avoid disulphide-bridge formation in the proteins.
Overall there are three main reducing agents used, to avoid this problem. The choice of which one to use depends on your needs. They include:

TCEP: Considered the golden standard, it is relatively odourless compared to the other reducing agents. Once you have dissolved it, it should also remain very stable over time. Typically used at a 0.4 mM concentration.

DTT: Cheaper than TCEP, although I have heard that it is not as stable over time

2-Mercaptoethanol: The absolute cheapest, stinks like crazy. Typically to a final of 5 mM.


Why is dipositive dilithium more stable than neutral dilithium? According to J.D Lee, compounds with fraction bond number are unstable. I calculated that the bond order of $\ce{Li2^2+}$ is 0.5 while that of $\ce{Li2}$ is 1. Hence, $\ce{Li2^2+}$ must be less stable than $\ce{Li2}$ due to half bond character.

But, in reality, $\ce{Li2}$ is more stable than $\ce{Li2^2+}$. Why is it so?

The comment asks "Why is Li2+ more stable than Li2". Li2 has a relatively low bond energy (gas phase) of 27 kcal/mol (Cotton and Wilkinson, Inorganic Chemistry). The energy cost to make one mole of Li gas is 37 kcal (CRC Handbook). The ionization energy of Li is 5.39 eV = 124 kcal/mol.

So, breaking a Li2 gas molecule into 2 Li (gas) costs 27 kcal/mol.

Breaking a Li2+ gas molecule into Li + Li+ involves not the separation of two uncharged atom, but the separation of a very small Li+ ion from a Li atom which provides some charge accomodation. The heat of formation of Li+ in water is 66 kcal/mol (presumably from aquation with 4 waters); the hydration bonding is worth about (124 + 37 + 66)/4 = 54 kcal/mole each. So I would estimate the bond between Li+ and Li to be perhaps 40 kcal/mol, at least a significant fraction of 54 kcal/mol, and probably more than 27 kcal/mol.

The important point is that the unusually small size of the Li+ ion makes it able to polarize a neutral atom to form a strong bond (and come in closer!), whereas the neutral atoms have only uncharged molecular orbitals to spread their electrons over. It is highly unlikely that larger atoms (Na, K) would show similar stability for an ion-molecule compared to a neutral molecule. Magnesium might be a similar exception, however, since both Li and Mg have small radii; it might be interesting to compare stabilities of Mg2 and Mg2+.


Indicator error in acid-base titration If we value a 20 mL solution of 0.25 M acetic acid with a 0.1M solution of NaOH

The end point of the titration is reached when 50 mL of NaOH is added from the burette. At this point, the pH of the solution is 8.81

Since the pH at the end point is 8.81, phenolphthalein, whose pKa is 9.7, has been used as an indicator.

I need to calculate the error associated with using this indicator, because as the turn does not exactly coincide with the pH at the end point, there must be some error.

For future titration practices, you should never use the buret until the end. Most high quality burets (Class A) are limited to 50 mL (max) or even 25 mL. If it is a made up exercise by a teacher, they should not promote this approach either. You will have far more error from the incorrect volume measurements rather than the choice of your indicator. Also be careful with reporting concentrations: 0.1 M NaOH is not the same as 0.10 M NaOH.

Anyway, one cannot determine the indicator error from the given data. Practically, you would plot a pH titration curve on a graph paper. Determine the equivalence point (= the true point at which the titration is stoichiometrically complete) from the graph. Lets say, you get 40.56 mL (last digit estimated). With an indicator you get an end-point at 40.60 mL. The end-point is the point where the indicator "hints" you the end of the titration. Now the excess concentration of acetic acid calculated via the end-point is your indicator error.


Why is crystalline aspirin washed with chilled water in the purification process? Take as example this
or even better this:

  Wash the crystals with 2-3 ml of chilled water.

Why does the aspirin need to be cooled down? And why do we wash the aspirin with chilled water?

Because aspirin is less soluble in cold water. In fact, aspirin is not very soluble in water at all, which is why you are supposed to take it with lots of water. You are not "chilling the aspirin", you are preventing it from dissolving too much so that you have a better yield of your product. You wash it to remove impurities that do dissolve in water.


Exceptions to motif: nucleophile attacks electrophile I hear this motif - nucleophile attacks electrophile - all the time with regard to organic chemistry. Are there any exceptions?

I think this statement is always true in general because it's a restatement of Coulomb's law. But then are there any organic reactions that cannot be characterized as nucleophile attacking electrophile?

What I learned: Radical termination does not involve nucleophiles attacking electrophiles.

I believe that you are asking if there are any reactions that are essentially "non-polar". There are two classes that immediately come to mind: (1) radical reactions and (2) pericyclic reactions.

A common radical reaction studied in introductory organic chemistry is the radical halogenation of alkanes. Instead of cationic and anionic reactants/intermediates/products, there are radical intermediates that do not fall into the nucleophile/electrophile classification.

Pericyclic reactions involve the flow of pi-electrons and do not require any charged reactants/intermediates/products. Reaction 1 below is a Diels-Alder reaction and reaction 2 is an electrocyclization. Depending on your definition of nucleophile and electrophile, a Diels-Alder reaction might fit the motif of "nucleophile attacks electrophile." That is, the HOMO of the diene (1,3-butadiene in the example) interacts with the LUMO of the "dienophile" (ethylene). If you define a nucleophile as the reaction component providing the HOMO, then a diene is essentially a nucleophile in the Diels-Alder reaction (and vice-versa for the "dienophile"). However, since these reactions don't use the same types of nucleophiles as say SN2 reaction, they are often considered to fall outside of the usual "nucleophile attacks electrophile" motif. That's the way they were first introduced to me, at least.

I should add that many organometallic mediated reactions fall outside of the "nucleophile attacks electrophile" motif as well. My answer was confined to organic reactions.


Are all reactions with ΔG < 0 spontaneous? Given the formula $\Delta G = \Delta H - T\Delta S$. It is commonly written that all spontaneous reactions necessarily have $\Delta G < 0$. I was curious if the converse is true: Are all reactions with $\Delta G < 0$ spontaneous. If not, why?

No, Gibbs energy is only the relevant state function in the special case of constant temperature and pressure.

In the wording of IUPAC:

  partial derivative of Gibbs energy with respect to extent of reaction at constant pressure and temperature [is negative] for spontaneous reactions

In the special case of constant temperature and volume, Helmholtz energy is the relevant state function.

Most generally, a process is spontaneous if the entropy of the universe (system + surroundings) increases.