text:
Are there chiral compounds that don't rotate plane-polarized light?
I know that meso compounds have chiral centers but don't rotate plane-polarized light, and I know that there can be non-traditionally chiral compounds (e.g. ones with large substituents that prohibit much rotation around single bonds) that can still rotate light, but I haven't been able to find any molecules that are chiral but don't rotate plane-polarized light (or rotate it impossibly little). I'm imagining something like an enantiomeric pair connected by one single bond with a slight modification to one that makes it have similar optical activity but not actually be identical (so the final molecule is indeed chiral), but I don't know exactly how to make it work.

Achiral compounds don't rotate light.

@IvanNeretin My source for that claim was the first answer to this post on Quora. The reasoning seems OK to me, but I'm not knowledgeable enough about stereochemistry to accurately confirm or dispute it.

@IvanNeretin More precisely (as I suspect the questioner intended) compounds with no chiral centres can be chiral because of restricted rotation or other geometric features eg hexahelicene.

Depends on what you mean by "do they rotate?" What if they don't rotate light by a measurable amount? I'm sure I can contrive a number of compounds where that is the case.

Compounds without chiral centers can be chiral all right.

@Zhe Can you give an example?

@IvanNeretin It seems as if I got my terms mixed up. I've edited my question - hopefully it's more accurate now.

Take octane and replace one of the hydrogen atoms on the 2-carbon with a deuterium.

I haven't been able to find any measured rotations, but I suspect that compounds with a chiral methyl group (-CHDT) as their only source of chirality would have very small optical rotations.

Good question.
There's a phenomenon named cryptochirality[1] (meaning “hidden chirality”), when a compound, though chiral, has practically unmeasurable optical rotation activity.
It can happen to molecules with chiral center(s) bearing very similar substituents. (So, no tricks with bonded slightly modified enantiomeric pairs are needed.)
An example is 5-ethyl-5-propylundecane $\ce{CH3[CH2]5-\overset{$*$}{C}(CH2CH3)(CH2CH2CH3)-[CH2]3CH3}$,[2] don't call it “butyl(ethyl)hexyl(propyl)methane”, found e.g. in beans. Its specific rotation is $[\alpha] < 0.001$.
Another, more common example are fats, i.e. triglycerides, $\ce{RCOOCH2\overset{$*$}{C}H(OCOR')CH2OCOR''}$,[1] if containing e.g. only palmitic, oleic and similar long acyls, optical rotation is not demonstrable.[3]
(Related topic is chirality in polymers, see e.g. Q: Chirality on Carbon of PVC molecule.)

References:

Mislow K. & Bickart P.: An Epistemological Note on Chirality. Israel Journal of Chemistry 15, 1–6 (1976)
Wynberg H., Hekkert G.L., Houbiers J.P.M. & Bosch H.W.: The Optical Activity of Butylethylhexylpropylmethane. Journal of the American Chemical Society 87, 2635–2639 (1965)
Schlenk W.: Synthesis and analysis of optically active triglycerides. Journal of the American Oil Chemists’ Society 42, 945–957 (1965)

Wow this answers a question I've mulled over for years! I had no idea there was a term for this. Thanks!

@NicolauSakerNeto, slightly off-topic, but you might also be interested to know that these cryptochiral molecules are capable of asymmetric induction in the Soai reaction. That's one of the few reactions we know (I think it might be the only one) where a tiny enantiomeric excess in the system (which could theoretically be just stochastic) can be propagated, leading to significant levels of enantioenrichment in the product, so it's somewhat relevant to the origin of homochirality.


text:
Are there gases that spontaneously combust when mixed with oxygen?
Is there a known gas that would combust by there mere action of mixing with oxygen? So, we could put this hypothetical gas in a balloon, and if the balloon popped, it would combust and explode.

Phosphine - PH3

Well, any pyrophoric gas would do. Phosphine is just one of them.

@user69077 Phosphine, silane, borane.

Disilane is way more exciting than silane.

Dichlorosilane will go too

I think we have established that this is a big list question. Note that Wikipedia already has a (fairly short) list of pyrophoric materials. I can readily get over a dozen I've worked with in semiconductor processing, and asking around will yield many more, particularly when including metal organics.

Edit:
Many gasses have this capability, called pyrophoric gasses. These include Phosphine, silane, borane, and more

Note that phosphine is not an element, it is a molecule.

To nitpick phosphine isn't an "element" but rather a "compound."

Thank you, I didn't realize I had put element.

source: Stack Exchange

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

sort: votes


text:
Combustion analysis methodology question
I am questioning the answer key of an exam question my chemistry professor has given, because I think there is a flaw in the methodology of the exam answer.  I have worked out my solution using my assumptions here, and would like to know if I have erred in my reasoning, or if my professor has.

The question asks:

Almost all brake fluids are hygroscopic which means they absorb water from
  the atmosphere. This is why you should change your brake fluid yearly. Assume
  for this exam brake fluid is comprised of only $\ce{C7H16O4}$ (triethylene glycol
  monomethyl ether).
You send a $\pu{0.12235 g}$ sample of your brake fluid, $\ce{C7H16O4}$, for combustion
  analysis and the results give $\pu{0.10674 g}$ of $\ce{H2O}$ and $\pu{0.20199 g}$ $\ce{CO2}$. What is the
  mass percentage of water in your brake fluid?

Because we don't know the purity of the sample of the brake fluid, I started by assuming that the generated $\ce{CO2}$ could be used to determine the amount of extra water, if water was the only impurity.
I started by converting grams to moles:
$$
\pu{0.20199 g \ce{CO2}} \times \frac{\pu{1 mol \ce{CO2}}}{\pu{44.009 g \ce{CO2}}} = \pu{4.5897 \times 10^-3 mol \ce{CO2}}
$$
$$
\pu{0.10674 g \ce{H2O}} \times \frac{\pu{1 mol \ce{H2O}}}{\pu{18.015 g \ce{H2O}}} = \pu{5.9251 \times 10^-3 mol \ce{H2O}}
$$
Then I balanced the equation:
$$
\ce{C7H16O4 + 9O2 -> 7CO2 + 8H2O}
$$
I inputted the known values into a BCA table:
$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & & & &\\ \hline
 C  & & & &\\ \hline
 A  & & & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline
\end{array}
$$
If we assume that $100\%$ of the $\ce{C7H16O4}$ combusted, and that there was extra $\ce{H2O}$ in the reaction mixture, then we can base the rest of this table off of the amount of $\ce{CO2}$ generated, because heating water does not produce $\ce{CO2}$.
So, filling out the necessary components, we get:
$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & & & 0 & 6.7964 \times 10^{-4}\\ \hline
 C  & & & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline
 A  & & & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline
\end{array}
$$
Based on this, we can see that there were $\pu{6.7964 \times 10^{-4} mol \ce{H2O}}$ of extra water before the reaction occurred.
Converting moles to grams:
$$
\pu{6.7964 \times 10^{-4} mol \ce{H2O}} \times \frac{\pu{18.015 g \ce{H2O}}}{\pu{1 mol \ce{H2O}}} = \pu{0.012244 g \ce{H2O}}
$$
Finally, we can divide by the total mass of the contaminated sample:
$$
\frac{\pu{0.012244 g \ce{H2O}}}{\pu{0.12235 g sample}} \times 100\% = \pu{10.007\%\; \ce{H2O}}
$$
However, my professor's exam solution doesn't agree with this. I am skeptical as to whether the given exam solution is correct, because the reasoning of it seems off to me:

Theoretical yield:
  $$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 7.4513 \times 10^{-4} & \ldots & 0 & 0\\ \hline
 C & -7.4513 \times 10^{-4} & \ldots & +5.2159 \times 10^{-3} & +5.9610 \times 10^{-3}\\ \hline
 A & 0 & \ldots & 5.2159 \times 10^{-3} & 5.9610 \times 10^{-3}\\ \hline
\end{array}
$$
  or $\pu{0.22955 g \ce{CO2}}$ and $\pu{0.10739 g \ce{H2O}}$.
Since carbon does not change due to the addition of water, $\Delta \ce{CO2}$ is proportional to the percent $\ce{H2O}$.
Therefore,
  $$
\begin{array}{crCl}
& \pu{0.22955} & \pu{g\; \ce{CO2} theoretical} \\
-& \pu{0.20199} & \pu{g\; \ce{CO2} actual} \\ \hline
& 0.02756 & \pu{g\; \ce{CO2} difference} \\
\end{array}
$$
$$
\frac{\pu{0.02756 g\; \ce{CO2}}}{\pu{0.22955 g\; \ce{CO2}}} \times 100\% = \pu{12.006\%\; \ce{H2O}}
$$

Three things are extremely off-putting about this explanation:

The numerical values of a known-to-be-impure substance are used to calculate theoretical yield
The theoretical yield calculated is of an unrelated substance
I don't think the mass difference of carbon dioxide is proportional to the excess $\ce{H2O}$

Does the professor's answer make sense?

I agree you are right but having a hard time formulating a reason.

@A.K. I spoke with a friend who had the following to say about it:

The professor states that the missing CO2 as a percent of of the amount of CO2 we expect is equal to the percent of water in the sample, but that's not true, because it's not proportional. However, if you multiply by 7/8th, you get 10.5%, which isn't quite right, but closer to right.
When you compare theoretical with actual, actual has an unknown quantity of H2O that's screwing up both sides. You have to find things off the actual CO2 and its share of the result H2O.

Curious - What did the professor say about all this? Kudos to you for working through this.

@MaxW She said:
"I figured out what I did wrong-In my excel macro I made a mistake with the grams of H2O that should have formed - for your exam 0.109186 grams of H2O, not 0.10674 grams. Your exam grade has been adjusted."
I got a 105%.

Methodology
Both methods of computing the percent water are completely accurate and logically sound, and will give the same result when the problem quantities are in the correct ratios.
However, the quantities in the problem lead to a contradiction.

Mass of oxygen based on mass balance
The sample mass is
$$\pu{0.12235 g sample}. \tag{1}$$
The sum of the product masses is
$$\pu{0.10674 g \ce{H2O}} + \pu{0.20199 g \ce{CO2}} = \pu{0.30873 g products}. \tag{2}$$
The mass of the final reactant, $\ce{O2}$, must therefore be
$$\pu{0.30873 g products} - \pu{0.12235 g sample} = \pu{0.18638 g \ce{O2}}. \tag{3}$$
Mass of oxygen based on based on the reaction ratios
There are 9 moles of $\ce{O2}$ per 7 moles of $\ce{CO2}$. We start with no $\ce{CO2}$ and end with no $\ce{O2}$.
We ended with
$$\pu{4.5897 \times 10^{-3} mol \ce{CO2}}. \tag{4}$$
Therefore, we must have started with
$$
\pu{4.5897 \times 10^{-3} mol \ce{CO2}} \times \frac{\pu{9 mol \ce{O2}}}{\pu{7 mol \ce{CO2}}} = \pu{5.9010 \times 10^{-3} mol \ce{O2}}. \tag{5}
$$
We can see this by filling in $\ce{O2}$ in the BCA table:
$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & \ldots & 5.9010 \times 10^{-3} & 0 & 6.7964 \times 10^{-4}\\ \hline
 C & \ldots & -5.9010 \times 10^{-3} & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline
 A & 0 & 0 & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline
\end{array}
$$
$\pu{5.9010 \times 10^{-3} mol \ce{O2}}$ is, in grams,
$$
\pu{5.9010 \times 10^{-3} mol \ce{O2}} \times \frac{\pu{31.998 g}}{\pu{1 mol \ce{O2}}} = \pu{0.18882 g \ce{O2}}. \tag{6}
$$
Conclusion
$(1)$ is clearly not equal to $(6)$. Hence, this is a contradiction.
$$
\pu{0.18882 g \ce{O2}} \neq \pu{0.18638 g \ce{O2}}.
$$
They are $1.3\%$ different.

Finding the missing mass
The difference in the oxygen mass is also exactly equal to the difference in the calculated water masses from the solutions presented here.
$$
\pu{0.18882 g \ce{O2} - 0.18638 g \ce{O2}} = \pu{0.00244 g \ce{O2}} \tag{1}
$$
Let's expand out the BCA table a little further:
$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 6.5567 \times 10^{-4} & \ldots & 0 & 6.7964 \times 10^{-4}\\ \hline
 C & -6.5567 \times 10^{-4} & \ldots & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline
 A & 0 & \ldots & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline
\end{array}
$$
Now, using the values from the table:
$$
\pu{6.7964 \times 10^{-4} mol \ce{H2O}} \times \frac{\pu{18.015 g}}{\pu{1 mol \ce{H2O}}} = \pu{0.012244 g \ce{H2O}}\tag{2}
$$
$$
\pu{6.5567 \times 10^{-4} mol \ce{C7H16O4}} \times \frac{\pu{164.1995 g}}{\pu{1 mol \ce{C7H16O4}}} = \pu{0.10766 g \ce{C7H16O4}} \tag{3}
$$
$$
\pu{0.12235 g brake fluid} - \pu{0.10766 g \ce{C7H16O4}} = \pu{0.014689 g \ce{H2O}}\tag{4}
$$
Finally, we can see $(5)$ is approximately equal to $(1)$:
$$
\pu{0.014689 g \ce{H2O}} - \pu{0.012244 g \ce{H2O}} = \pu{0.0024451 g \ce{H2O}} \tag{5}
$$
$$
\pu{0.0024451 g \ce{H2O}} \approx \pu{0.00244 g \ce{H2O}}.
$$
To find the amount of water we should have had at the end, we will assume there were $\pu{0.014689 g \ce{H2O}}$ before the reaction, and that the amount of $\ce{C7H16O4}$ is still accurate. In the BCA table, we see the reaction created $\pu{5.2454 \times 10^{-3} mol \ce{H2O}}$, or
$$
\pu{5.2454 \times 10^{-3} mol \ce{H2O}} \times \frac{\pu{18.015 g}}{\pu{1 mol \ce{H2O}}} = \pu{0.094496 g \ce{H2O}}.
$$
The amount of water that should have been recovered at the end is therefore
$$
\pu{0.014689 g} + \pu{0.094496 g} = \pu{0.10919 g \ce{H2O}}.
$$
This gives a mass percentage of $12\%$, as was intended by the professor.

You could go a bit further in the conclusion and point out that the difference in mass of the oxygen is exactly the difference between the different masses of water.

You should accept your own answer. You correctly have pointed out the contradiction with the problem.

@MaxW I have updated the post.

I did this a little differently.  If there were $\pu{4.5897 \times 10^{-3}}$ moles of $\ce{CO2}$ produced, that must mean that there were $1/7$ as many moles of $\ce{C7H16O4}$ present in the original sample, or $$\frac{4.5897\times 10^{-3}}{7}\times 164.1995=\pu{0.10766 g}$$ in the original sample.  The rest of the original sample ($\pu{0.01469 g}$) must have been water.  So the percentage of water must have been $$\frac{0.01469}{0.12235}\times 100\%= 12.006\%.$$  This agrees with your professor's answer.

+1 Very straight forward solution! You could use better mw of 164.1995

Thanks @MaxW  I was just too lazy to work that out.

Good answer, but the professor's approach still bothers me.

Correct me if I'm wrong, but I think you are forgetting about the mass of oxygen that went into the reaction. You have 0.1075 g C7H16O4 plus about 0.1888 g O2.

@LiamWhiteMcShane it's good, he used the number of moles of $\ce{CO2}$ extracted to get the number of moles of triethyleneglycol monomethylether in the sample

For the sake of argument:
Let's assume that the professor is correct.
$$12.006 \% \text{ water} = 0.12235\text{ g} \times .12006 = 0.01469\text{ g}$$
which means that the sample has 0.12235 - 0.01469 grams of $\ce{C7H16O4}$, or 0.10766 grams of $\ce{C7H16O4}$.
So we start off with

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.10766 & \ldots & 0 & 0.01469\\ \hline
C & & \ldots & & \\ \hline
A & & \ldots&  & \\ \hline
\end{array}
$$

However the grams of $\ce{CO2}$ that 0.10766 grams of  $\ce{C7H16O4}$ should produce is:
$$\dfrac{0.10766}{164.1995}\times 7 \times 44.009 = 0.20199 $$
The grams of $\ce{H2O}$ that 0.10766 grams of  $\ce{C7H16O4}$ should produce is:
$$\dfrac{0.10766}{164.1995}\times 8 \times 18.015 = 0.09449 $$
So now we have:

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.10766 & \ldots & 0 & 0.01469\\ \hline
C & & \ldots & 0.20199 & 0.09449 \\ \hline
A & & \ldots&  & \\ \hline
\end{array}
$$

Finishing the table we have

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.10766 & \ldots & 0 & 0.01469\\ \hline
C & & \ldots & 0.20199 & 0.09449 \\ \hline
A & & \ldots& 0.20199  & 0.10918 \\ \hline
\end{array}
$$

Thus the professor's solution is inconsistent with the given amount of water which was  0.10674 g.
Now let's assume that Liam is correct.
$$10.007 \% \text{ water} = 0.12235\text{ g} \times .10007 = 0.01224\text{ g}$$
which means that the sample has 0.12235 - 0.01224 grams of $\ce{C7H16O4}$, or 0.11011 grams of $\ce{C7H16O4}$.
So we start off with

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.11011 & \ldots & 0 & 0.01224\\ \hline
C & & \ldots & & \\ \hline
A & & \ldots&  & \\ \hline
\end{array}
$$

However the grams of $\ce{CO2}$ that 0.11011 grams of  $\ce{C7H16O4}$ should produce is:
$$\dfrac{0.11011}{164.1995}\times 7 \times 44.009 = 0.20658 $$
The grams of $\ce{H2O}$ that 0.11011 grams of  $\ce{C7H16O4}$ should produce is:
$$\dfrac{0.11011}{164.1995}\times 8 \times 18.015 = 0.09664 $$
So now we have:

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.11011 & \ldots & 0 & 0.01224\\ \hline
C & & \ldots & 0.20658 & 0.09664 \\ \hline
A & & \ldots&  & \\ \hline
\end{array}
$$

Finishing the table we have

$$
\begin{array}{|c|c|c|c|c|}
\hline
& \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline
 B & 0.11011 & \ldots & 0 & 0.01224\\ \hline
C & & \ldots & 0.20658 & 0.09664 \\ \hline
A & & \ldots& 0.20658  & 0.10888 \\ \hline
\end{array}
$$

Thus Liam's solution is inconsistent with the given amount of carbon dioxide which was 0.20199 g and the given amount of water which was 0.10674 g.

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


text:
What compound has the strongest optical rotation of polarised light?
A question recently asked here whether compounds know to be chiral can have non-measurable optical rotations: Are there chiral compounds that don't rotate plane-polarized light?.
Apparently they exist and the phenomenon is known as cryptochirality.
This prompted me to think of the opposite end of the spectrum of optical rotation. Which known compound shows the largest known optical rotation?
NB when I say "known compound" I mean something that has actually been synthesised in large enough quantities to be well characterised. So no theoretical-only compounds or esoteric systems.

One aspect which might make this tricky is that many publications are awful at using the correct units for specific rotation, meaning values are often ambiguous.

Various helicenes are said to have rather high specific optical rotations. What about switchable one from “low” rotation to high, is |[α]|=4680° enough? :−)

Maximum observed optical rotation through a polarimeter might come handy, if someone approches the question practically


text:
Why is ammonia removed before the yield is attained in Haber process?
Why does removing ammonia from product mixture before reaching the yield of 20% can increase the amount of NH3 produced per unit time?

$\ce{}$ Haber's process is as follows:
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$
As you can see the reaction is going in gas phase, so removal of $\ce{NH3(g)}$ will push the reaction in forward direction i.e. in product side according to  le chatelier's principle.

I do understand the part regarding the shifting of equilibrium position to increase the yield of ammonia, but I am confused about how it can boost the rate of reaction

From what i understand to , in order to equilibrium to shift forward the rate of product formation will increase which in turn  increase the rate of reaction.
r = 1/2*d[NH3]/dt


text:
Can methyl halide be used in Wurtz reaction?
Yields of product(alkanes) are best for 1°(60%) and least for 3°(10%) alkyl halides. Can methyl halides be used in Wurtz reaction? Would it produce a good yield?

Methyl Halide if used , as it is primary alkyl halide and also can produce only one product ethane which is symmetrical and therefore the yield should  be very much high .


text:
What is the Crystal System of MoS2?
I'm confused about the crystal system of MoS2.
In the case of the monolayer MoS2, it has 3-fold rotation symmetry when viewed from the top of its unit cell (c-axis). Even in the case of bulk (AB stacking) MoS2, it has only 3-fold rotation symmetry from the same direction. So, I expected that it must be the trigonal system. However, 'Materials Projects' doesn't seem the trigonal system.
Here are some database from 'Material Projects'
List of MoS2
https://materialsproject.org/#search/materials/{%22reduced_cell_formula%22%3A%22MoS2%22}
Bulk MoS2
https://materialsproject.org/materials/mp-2815/
Bilayer MoS2
https://materialsproject.org/materials/mp-1025874/
and so on.
Some of them are the trigonal system, and the others are the hexagonal system. I studied the minimum symmetries determine the crystal system, or that is a misconception. But now, it might be wrong by the web data.
Which one is correct and if both are correct, how can I check the 6-fold rotation symmetry for the hexagonal system?
(appended--------------------------------------------------------------------)

This is the figure for the HCP structure. Its symmetry along the c-axis can be called 3/m or bar(6).

This view is the bulk MoS2 along c-axis. In the unit cell, it has two layers of MoS2. However, it has 6/mmm point group. How does this happend?...
https://materialsproject.org/materials/mp-1018809/#

Naturally occurring molybdenum disulfide can occur in either of two crystal systems, hexagonal or "trigonal"; the latter is also known as rhombohedral (see here).
In both cases the coordination geometry around molybdenum has only threefold rotational symmetry, but in the hexagonal form the crystal will add a center of symmetry to become hexagonal.  Such a conversion is seen in other cases such as hexagonal close-packed metals like magnesium and zinc, which actually have only threefold symmetry about any axis perpendicular to the basal plane and passing through an atom, yet the Crystal's are hexagonal.  In general, when you have a unit cell with $D_{3h}$ symmetry -- a mirror plane perpendicular to the threefold exis -- you will get hexagonal crystals in both molybdenum disulfide and hexagonal close-packed metals.
The rhombohedral crystal, with a different arrangement of mirror planes (the point group symmetry is $D_{3d}$ instead of $D_{3h}$, and $D_{3d}$ has no mirror plane perpendicular to the threefold axis), has the required center of symmetry without going from threefold to sixfold rotational symmetry.

More precisely, let's see hcp structure. Its point group is 6/mmm or D_6h. By the way, can you think space or point group as Hermann–Mauguin notation because I prefer this? So, 6/m in hexagonal, it means it has 6/m symmetry along the c-axis. However, when I rotated six times and reflected atoms, I couldn't see the overlappings. Where should I insert the mirror perpendicular to the c-axis?

Any mirror perpendicular to the threefold axis makes the symmetry you need to get a hexagonal crystal.  You have to look for it.  In the hcp case it would be along the basal plane.  In molybdenum disulfide try a plane of molybdenum atoms.

I've found the figure for the symmetry along the c-axis and posted it. I should change the word I mentioned. The symmetry of HCP structure along the c-axis is one mirror after 3-fold rotation or inversion with 6-fold rotation one after another (bar(6)m2).
https://en.wikipedia.org/wiki/Crystallographic_point_group
So, bar(6) means anyway it has six-rotation. Andhttps://en.wikipedia.org/wiki/Crystal_system

And, then I can understand. en.wikipedia.org/wiki/Crystal_system
But, one case! Can you check this structure? https://materialsproject.org/materials/mp-1018809/#
This is bulk MoS2. And Materials Project noted this has 6/mmm. I posted this too.


text:
C-4 oxidation of octanoic acid to 4-oxooctanoic acid
I am leading a project in which I am attempting to synthesize 4-oxooctanoic acid in a (mostly) biological pathway.
Currently, I have reached a dead-end. I have a way to create octanoic acid through bacterial growth, I could not seem to find a way to proceed to the 4-oxo step with enzymes so I turned to chemistry.
Although this seems to be impossible due to the structure of octanoic acid. There appears to be no way to specifically target the fourth carbon to obtain the 4-oxooctanoic acid.
My question is: Am I correct in assuming that converting octanoic acid to 4-oxooctanoic acid is impossible (at least via simple oxidation)? Or would there be a better way to go about doing this?
Thanks in advance
PS: Here are pubchem links to the two structures involved.
https://pubchem.ncbi.nlm.nih.gov/compound/octanoic_acid#section=2D-Structure
https://pubchem.ncbi.nlm.nih.gov/compound/4-oxooctanoic_acid#section=Top

I would start from both functional groups and then attach the tail. That said, bacteria might well be more versatile in their business. What's a simple oxidation, come to think of it? Nothing is simple in biochemistry (at least, that's my impression of it).

Very curious, it seems that your end product 4-oxo-octanoic acid can be easily made by direct hydrolysis of "gamma-Octalactone" (used in perfume and flavoring industry). The naturally occurring gamma-Octalactone compound is readily available, but it does seem very pricey. Would there be a way to 1) oxidize the octanoic acid non-specifically. 2) cyclic esterify to form reactant mixture plus gamma-Octalactone. 3) purify away unwanted reactants (distillation perhaps). 4) hydrolyze the purified gamma-Octalactone into the desired 4-oxo-octanoic acid product. http://www.leffingwell.com/glacton.htm

It has been a while since I have thought about organic chemistry. I thought the "cyclic esterification step 2" might help with a purification process. The question piqued my interest. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5636794/

If I was tasked with making 4-oxooctanoic acid I would not start from octanoic acid! There is no easy way of making 4-oxooctanoic acid directly from octanoic acid. It is however a relatively trivial exercise from other starting materials.
My suggestion is to react the commercially available ethyl succinyl chloride (ethyl 4-chloro-4-oxobutyrate) with n-butylmagnesium bromide (also commercially available) modified with manganese(II)chloride/lithium chloride as described in Tetrahedron Lett. 1992, 33 (31), 4439–4442. This will give the ethyl ester of the required 4-oxooctanoic acid. A simple base hydrolysis gives 4-oxooctanoic acid. You could probably use the widely available n-butyllithium in place of the Grignard.


text:
Molpro - Simple explanation of OCC, CORE, FROZEN and CLOSED orbitals
I'm beginning with quantum chemistry and I'm supposed to do some ab initio calculations with Molpro. The problem is, I don't understand the orbital specification very well.
Let's say, I have the following computation:
 {mrci;
  occ,4,2,2,0,4,2,2,0;
  core,1,0,0,0,1,0,0,0;
  wf,13,5,1;
  state,3;
  option,nstati=5;
  option,maxit=200;
  option,maxiti=200;
 }

I don't understand much the flags OCC, CORE, CLOSED and FROZEN. I've read about them in the documentation, but I have troubles understanding it. Moreover, they possess different meaning in different methods, i.e. MCSCF and MRCI.
Could you, please, explain it in a basic way and graphically, if possible? I'm not looking for the deep understanding of methods now (I'll study them later), but I need to be able to use Molpro for some simple (di/tri-atomic) systems.

In MCSCF we have:

occ are occupied orbitals, it refers to all orbitals that may be occupied in any configuration. For a FCI calculation this equals the total number of orbitals (per IRREP). If occ is smaller than that, the difference will be the number of virtual (unoccupied) orbitals. Virtual orbitals are never occupied in all the configurations.
closed orbitals are doubly occupied in all configurations. This means occ - closed equals the number of active orbitals. closed orbitals are optimized in an MCSCF calculation.
frozen is similar to closed, but they are not optimized in MCSCF, i.e. their MO coefficients will be kept fixed.

In MRCI there are:

occ and closed being similar to MCSCF. They are used to define the reference space of configurations.
Multi-reference means additional external configurations are created by excitations from the reference space. Here excitations from closed orbitals are considered, but not from core orbitals. In other words: closed orbitals are only doubly occupied in the reference space but may have different occupations in the external configurations, while core orbitals are doubly occupied in both, the reference space and external configurations.
frozen is not used here, as orbitals are not optimized anyway.

The "lower" set of orbitals is always included in the "higher" set, e.g. occ includes all closed. Thus we always have core/frozen <= closed <= occ <= total number of orbitals.
Here is a graphic for the MRCI case. The curved arrows to the left indicate considered (black) and neglected (red) external excitations into the virtual orbital space. For MCSCF replace core with frozen and ignore the curved arrows.

One more note: The picture shows only excitations into the virtual space. But in principle you could also have excitations from closed to active orbitals.

Text only answer for now, I will try to make a graphic later.

Great! Thank you very much! And I'll be grateful for the pictures.

@Eenoku I added an orbital diagram to show the different orbital spaces.


text:
Lewis acids as ligands
Usually Lewis bases as my textbook says:

2.3 Ligands
  The ligands are the ions or molecules bound to the central atom/ion in the coordination entity. This is better visualized as the combination of a Lewis acid (the central atom/ion) with a number of Lewis bases (ligands). The atom of Lewis base that forms the bond to the Lewis acid (central atom/ion) is called donor atom ...

According to me its not compulsory for a ligand to be a Lewis base, as $\ce{BF3}$ is a Lewis acid and can act as a ligand by accepting electrons from a metal in vacant orbital of boron. Is it true?

In that situation, the thing that the boron is binding to is the ligand.

I can't understand. Please clarify

@Zhe Actually OP is right. Ligands can have both acceptor and donor properties like CO, or indeed be only acceptors like BF3.

Can H+ also act as ligand for electron rich metals. If it can its a pure acid . AlCl3 might also be such a ligand due to vacant d orbital of Al

@Harshjain You're going to far there. While H+ and AlCl3 are Lewis acids (and don't even mention this "d-orbital" nonsense) it's a matter of formalism when something can be called Z-Ligand.

@Mithoron Thanks for the reference on "Z-ligands." I've not encountered those before.

Yes it can happen.  This paper offers an example, in which $\ce{BF3}$ or $\ce{B(C6F5)3}$ adds to a low oxidation state iridium complex.  There is really not a lot of mystery behind it.  The iridium in its low oxidation state has electrons to offer and the boron with its vacant orbital can take them up.


text:
Classification in Stereoisomerism
I was trying to understand stereoisomerism. I am having trouble creating a 'flow chart' of Stereoisomerism. While there are many sources available over the internet and books, I don't find a standard answer. Some sources mention conformers and geometrical(cis/trans) under diastereomers while others seperate them completely. According to definition of diastereomers available on Wikipedia,
Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.
While this definition includes geometrical isomers very well into it, not much clarity arises regarding conformers as it used the word configurations rather than conformation.
Also I was finding difficulty in placing Mesoisomers in the chart.
It would be really nice if you could draw the flow chart and post it here.
Thank you!

Geometrical isomers are not all stereoisomers...

https://goldbook.iupac.org/html/S/S05983.html

The problem with looking for information about stereochemistry on the internet is that many sources use outdated terminology, imprecise definitions or have downright erroneous information!
Here's a flow chart that I put together that I believe to be accurate, yet hopefully straightforward:

When using the flow chart, it is really important to remember that each decision on the flow chart is related to comparing two molecules, ie. it doesn't make sense to ask "what is compound A", but it does make sense to ask "how are compounds A and B related"? A molecule may be the enantiomer of another molecule, but a diastereomer of another.
Implicit in the flow chart is that stereoisomers have different 3-dimensional arrangements of their atoms, while having the same connectivity of atoms to each other.
The term "geometrical isomer" is now deprecated and its use is "strongly discouraged". It is an older term for cis-trans isomers, which are just a subset of diastereomers.
Back to your original post - the part in bold about diastereomers is problematic because the definition of diastereomers does not relate directly to stereogenic centers. It emerges that compounds with two or more chirality centers (the correct term for what are often referred to as stereogenic centers or stereocenters) can often exist as diastereomers, but chiral centers are not a pre-requisite. For example, cis- and trans-2-butene are diastereomers according to the definition.
A meso-compound is an achiral compound (the mirror image is super-posable), which is a member of a set of diastereomers, where at least one of those diastereomers is chiral. A meso-compound most commonly arises when the compound has two or more chiral centers, yet is not chiral due to an internal plane of symmetry. Meso-compounds don't fit neatly on the flow chart, which is based on comparing two molecules, but are a particular type of diastereomer.


text:
what should be the onset potential using cyclic voltammetry?
I am working for PEM fuel cell cathode material. I am confused regarding the onset potential determination for my catalyst (CV attached, fig 1 is complete CV while fig 2 is its zoom image for clear understanding of change in reduction curve direction) and also why the reduction starts from 0.8e-4 A instead of 0 A. I performed this CV in calomel electrode under oxygen in 0.5M H2SO4.
i) If I use Argon and overlap both the graphs then probably the curves start to separate at around 0.58V.
ii) I have read in some discussions to consider the onset potential at which current density approaches 0.1mA/cm2 while in other discussion, I have read to consider 20microA/cm2 as onset potential. If I plot current density instead of current only, then in first case (0.1mA/cm2), it would be 0.785V while in second case (20microA/cm2), it would be 0.775V
iii) If I only consider the point from where curve starts to change the direction of reduction peak, it would be 0.64V or 0.56V
iv) if I draw tangents then this would be around 0.5V.
So please help me in i) analyzing onset potential ii) guide me why there is positive current in the reduction peak iii) i also don't know what does the background correction mean and should I need to do it for my CV? All potentials are measured in SCE.

The reason the current starts at 0.8x10^-4 A is probably because you have an equilibration time built in to your procedure (say of a few seconds),or even if you do not,there are parasitic currents always occurring, especially if you are starting at 0.8 V (vs SCE). Given that I don't know much more about your system or how you arrived at the onset potential calculations, I cannot comment. Could you please provide more details?

Thanks for your response. I dont know what information do you require so i am mentioning all the details: the compound is C/N/S/transition metal,  medium: 0.5 M H2SO4, 3mm glassy carbon electrode, scan rate = 0.01V/s, quiet time = 2sec, sensitivity=1e-4A/V.


text:
Why does nucleophilic aromatic substitution occur at C-2 and C-4 of pyridine?
In the following reaction of 2,3,4-tribromopyridine with sodium methoxide in methanol,[1] why are only the C-2 and C-4 bromines replaced with methoxy groups?

Why doesn't the bromine at C-3 also get substituted?

Gibson, K. J.; D'Alarcao, M.; Leonard, N. J. Rearrangements of azabiphenylenes. The impact of nitrogen number and position. J. Org. Chem. 1985, 50 (14), 2462–2468 DOI: 10.1021/jo00214a012.

Related: Is there any selectivity between the C-2 and C-4 positions of pyridine in SNAr reactions?

Nucleophilic aromatic substitution on pyridines (or related heterocycles such as quinolines or pyrimidines) regioselectively occurs at the 2- and 4- positions. These can be thought of as being "ortho" or "para" to the nitrogen.
In the mechanism for nucleophilic aromatic substitution, the initial step involves attack of the nucleophile (in this case, methoxide anion) on the π-system of the aromatic ring. Because this step generates a high-energy anionic intermediate which has its aromaticity broken, it is also typically the rate-determining step. In general, the stability of this intermediate is what dictates whether a nucleophilic aromatic substitution is possible or not.
If we consider the resonance forms for this intermediate resulting from attack at each of the three positions, we can understand why attack at C-2 and C-4 is favoured.

When the nucleophile attacks at C-2 or C-4, the resultant intermediate possesses one resonance form where the negative formal charge is located on nitrogen. On the other hand, this is not possible when the nucleophile attacks at C-3.
In general, resonance forms involving negative charges on electronegative atoms such as N and O are much more important contributors than resonance forms with negative charges on carbon. Thus, the anionic intermediate resulting from attack at C-2 or C-4 is better stabilised by resonance, allowing the nucleophilic substitution to take place.


text:
finding out fractions of ions(different valencies/charges) of an element in non-stoichiometric compounds

Analysis shows that nickel oxide has the formula $Ni_{0.98}O_{1.00}$. What fractions of nickel exist as $Ni^{2+} $and $Ni^{3+}$?

What i did:
Let the number of $Ni^{2+}$ and $Ni^{3+}$ atoms be $x$ and $y$ respectively
one  atom/ion of $Ni^{2+}$ satisfies one  $O^{2-}$ atom
$2 \over 3$ atom/ion of $Ni^{3+}$ one  $O^{2-}$  atom
and then i wrote down the equations:
$x + $$3 \over 2 $$y =1$  and  $x+y=0.98$
Then i solved the equations to get the answers:
$x=0.94 $ $and$ $y=0.04$
Is what i did correct? If not then guide me how to solve the question

i am getting used to mathjax and markdown, so please if my question is tough to read someone edit it

i don't know what is wrong with this question that nobody would answer it. I've given question and also said what my approach was. please someone answer

What you have done seems to be correct. But you might want to convert those numbers to percentiles. I can tell you another method based on charge conservation.
Let us take  100 oxygen atoms and 98 Nickel atoms into our consideration.
Total negative charge$= 200$
Total positive charge$= 2x+3y$ where $x$ and $y$ are what you have assumed to be in the question . Now Net charge =0.Equate the positive and negative charges.
Also $x+y=98$. Solve them and you will get the same number. This of course has an high degree of
similarity to what you have done

yes, i agree with your last point. What i have done is also in a sense "charge conservation"

yes , of course


text:
Which is more soluble in water ammonia or Sodium dodecyl sulfate
I've tried looking this up and can't seem to find the solubility of SDS if any one knows the answer I would really appreciate it.
Which is more soluble in water ammonia or Sodium dodecyl sulfate.
Thanks,

en:wp actually doesn't have a number, but https://de.wikipedia.org/wiki/Natriumlaurylsulfat

SDS water solubility is 100g/L according to this source https://pubchem.ncbi.nlm.nih.gov/compound/Sodium_dodecyl_sulfate#section=Solubility
Ammonia's maximum concentration in water is 880g/L at just above 0C data here


text:
Extrapolating from a calorimetry lab to find a new delta T with different volumes and grams of substance
So, we just finished an AP chem lab where the question was basically:

Given the results of the lab were that $\pu{5 grams}$ of $\ce{CaCl2}$ and $\pu{45ml}$ of water produced a $\pu{13.33 ^\circ C}$ change, find the new temperature change of $\pu{10 grams}$ of $\ce{CaCl2}$ in $\pu{40ml}$ of water.

Our teacher used a ratio strategy where basically $$\frac{\pu{5 grams}}{\pu{10 grams}}=\frac{\pu{13.33 ^\circ C}}{x}$$
Then, since x turns out to be $\pu{26.66 ^\circ C}$;
$$\frac{\pu{26.66 ^\circ C}}{45ml}=\frac{x}{50ml}$$
solving for $x$, the final temperature change is $\pu{29.6 ^\circ C}$.
My question is basically whether this is true, because I find the denominator in the first fraction of the second ratio sketchy - its saying $\pu{10 grams}$ for 35 ml of water, but, the $\pu{26.66 ^\circ C}$ is derived from finding $\pu{10 grams}$ with same amount of water. Thus, shouldn't it be 50, since in the first ratio we only changed amount of grams of solution, so water is still 40 ml, and now $\ce{CaCl2}$ is  $\pu{10 grams}$. $\pu{40ml + 10 grams = 50 grams}$.  Thus, shouldn't the first fraction be $$\frac{\pu{26.66 ^\circ C}}{50ml}$$
The final answer happens to be $\pu{26.66 ^\circ C}$ (which is pretty and convenient), but  I think that's just because we chose 5 grams to 10 grams which is a nice number to double and the volume decreases by 5, so everything cancels out nicely.
Kinda of a big segue and separate question but related:
What's wrong with the following logic to solve the problem?
$$\pu{50 g}\times \pu{4.184 J g-1 ^\circ C-1} \times \pu{13.33 ^\circ C}=\pu{2789 J}$$
Basically, using q=mcat (m is mass of water, 45, plus mass of solution, 5) to find q of 5 grams $\ce{CaCl2}$ and 45 ml water, and then, assuming if we double the mass we double the q, work backwards? (If my logic is wrong, I think it occurs in this assumption)
thus, since we need 10 grams and 40 ml in the end, the mass is still 50 grams, but the 2789 is multiplied by 2 (since mass is doubled) and 13.33C becomes an X.
$$\pu{50 g} \times \pu{4.184 J g-1 C-1} \times x = \pu{5578 J}\\
x = \pu{26.66 ^\circ C}$$
Basically, I think the answer is $\pu{26.66 ^\circ C}$ and my teacher thinks it's $\pu{29.66 ^\circ C}$.
I'm honestly very confused about q... Google does not seem to help answer this question, if amount of water in a solution/reaction (ie water and $\ce{CaCl2}$) changes, but mass of reactant stays the same, does q change?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{ }$ for chemical formulas/equations, $ $ for math term/equations, and $\pu{ }$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

Yes, the second ratio is sketchy. The first one is also not quite right except to a first approximation where you say the heat capacity is independent of composition.

The ratio strategy will definitely work, but I think going back to $q=m\cdot C\cdot \Delta T$ is easier to understand.
You've already figured out a bunch of these constants: $m = 50$, $C = 4.184$ (we assume that the heat capacity of the water doesn't change too much when adding the solid), and $\Delta T = 13.33$.
$$\begin{align}
q & =m\cdot C\cdot \Delta T \\
& = 50\cdot4.184\cdot13.33 \\
& = \ce{2789 J}
\end{align}$$
The heat produced by mixing $\ce{5 g}$ of $\ce{CaCl_2}$ is $\ce{2789 J}$, therefore,
the heat of reaction ($\ce{\Delta H_{rxn}}$) in $\ce{J/g}$ is $\ce{557.7 J/g}$.
When we mix $\ce{10 g}$ of $\ce{CaCl_2}$, the total heat produced will be $\ce{10 g \cdot 557.7 J/g = 5577 J}$.
Now we need to determine the heat change had we mixed in $\ce{10 g}$ of $\ce{CaCl_2}$. We can rearrange the specific heat formula to get $\ce{\Delta T = \frac{q}{m\cdot C}}$. We know $q = 5577$, $m=50$, and $C=4.184$.
$$\begin{align}
\Delta T &= \frac{q}{m\cdot C} \\ & \\
&=\frac{5577}{50\cdot4.184} \\ & \\
&= 26.66\, ^\circ \ce{C}
\end{align}$$
The reason the ratio works is the same logic as the gas laws:
$P_1\ V_1 = C$, $P_2\ V_2 = C$, therefore $P_1\ V_1 = P_2\ V_2$.
$q_1 / T_1 = m\cdot C$ and $q_2 / T_2 = m\cdot C$.
Therefore, $q_1 / T_1 = q_2 / T_2$ by the transitive property of equality

Thanks for the more mathy way to justify my ideas! Just to clarify, you agree with my proposed change to the ratio and not my teacher's original ratio?

What makes you think that the heat capacity of a 20% CaCL2 solution is the same as that of water?

Specific heats of calcium chloride solutions as function of temperature and composition.  https://www.oxy.com/OurBusinesses/Chemicals/Products/Documents/CalciumChloride/173-01791.pdf

@ChesterMiller The heat capacity is certainly not the same, I agree, but for a high school AP Chem class, the heat capacity difference is somewhat advanced.

@catdog If you want to use the teacher's ratio, it'll have to be between 45 mL and 40 mL, but I insist on it being incorrect because the total mass of the solution is changing temperature, therefore the ratio will be between 50 mL and 50 mL (which changes nothing).

@ChesterMiller I guess I just thought it was a negligible amount at first... using the document you provided, I guess I convert the .16 to the units I need, but I don't know where to go from there - do I just do q=mcat only with masses of CaCl2, but which heat capacity would I use?

You first calculate the mass fraction of CaCl2.  The heat capacity you use is the average between the values at the initial and final temperatures (determined from the graph in the document).  You may need to do this by trial and error, since you don't start out knowing the final temperature.


text:
How does Borax Solution act as a buffer?
What type of buffer is it? If it's a Weak Acid-Conjugate Base type of buffer, what is the weak acid and conjugate base?

Borax reacts in hydrochloric acid in the preparation of the buffer to give the boric acid through the following reaction $\ce{Na2B4O7.5H2O + 2HCl -> 4H3BO3 + 2NaCl + 5H2O}$
http://www.novedu.ru/anotes/refer/buffers.htm#bor1
The above link gives you the procedures to making the buffer. The reason why the borax solution is able to keep the system at around pH of 9 because the boric acid that is produced from this reaction has a pKa of 9.46. When the concentrations of the acid and its conjugate base, that is, the borate ions is exactly 1:1, the pH will be most difficult to change since any addition of hydrogen ions will only bond with the borate to form the undissociated boric acid.
So to answer you question, boric acid is the weak acid, and borate ions are the conjugate base.


text:
Bonding in Zeise's salt
I am not sure I completely understand the bonding in coordination compounds formed by donation of pi electron cloud by the ligands to the central atom.
For example, consider Ziese's salt:

I came across a question which asked the total number of sigma bonds in the salt. I expected the bond between ethene and platinum to be a sigma bond too because it appears as if it is formed by overlap of the orbitals along the axis and hence my answer was 9. But the answer given was 8 and the answer key didn't consider that bond as a sigma bond.
So is the bond between platinum and ethene a $\pi$ bond? If yes, then how?
What orbitals are actually involved in that bond?

I suppose they assumed that a bond is something that is between two atoms. The bond between platinum and ethene isn't.

The bond between $\ce {Pt} $ and ethene is not purely a $\sigma $ bond and also not a pure $\pi $ bond. The bonding  actually consists of a combination of both bonding and back-bonding effect.

If you look into the Molecular Orbital Diagram of Ethene, the picture will be more clear. For the formation of bond between the metal and the ligand, electrons from filled $\pi $-bonding MO of ethene ($\pi_b $) is donated to the vacant orbitals of Platinum. And electrons from the filled orbitals of $\ce {Pt} $ is back-donated to the vacant $\pi^*$ orbitals of ethene. Thus the bonding would have been a $\sigma $ bond if the electrons are donated from a $\sigma_b $  or $\sigma_{nb} $ orbitals. But here the bonding mainly is $\pi $ donation and $\pi$ acceptance ( known as Synergic effect).

Thus, it would be better not to consider the bond as $\sigma $ bond, but may be considered more of a $\pi $ type but also not a pure one due to the back-bonding.


text:
What is the entropy of Cu(NO3)2?
I have searched for the entropy of copper nitrate in various tables across the internet* and all I could find was its Hf. Ullmann also didn't give an any better clue.
If it was not measured and released, how can I then calculate it?
*http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm

You can't.$;!$

@IvanNeretin I can't but someone else sure can.

Sure, and that's already covered in the answer by porphyrin. It is just that these calculations require quite a bit of data which are even harder to come by than just entropy.

You can find the entropy in two ways, however, both need some experimental data. The first is to find tables of how the heat capacity varies with temperature and then use $$S=\int_o^T\frac{C_p}{T}dT+ \frac{\Delta H_x}{T_x} + etc.$$ where the subscript x allows for any phase changes, such as melting or one solid phase changing to another and etc. if there is more than one.
The second way is to use spectroscopic data, and calculate vibrational and rotational energy levels and use the relevant partition functions to calculate the internal entropy and the Sakur -Tetrode Equation for the translation.
There are examples of both methods in 'Basic Chemical Thermodynamics' by E. B. Smith, publ OUP.


text:
How to get ionic radii for coordination number 12?
I am studying $\ce{ABX3}$ perovskites, and I would like to calculate Goldschmidt tolerance factors for them. The $\ce{A}$ sites in these materials have a coordination number $12$.
The Shannon's paper (R. D. Shannon, Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides. Acta. Cryst. A 32, 751 (1976)) provides values for ionic radii for the coordination number $12$ only for a few elements, but doesn't do it for most other elements.
Are there any sources, where I can get ionic radii for $\text{C.N.} = 12$ for all elements?

Goldschmit has been shown to be a poor model to use with perovskites.

How many ionic compounds give coordination number 12?  Close-packed structures forcing like charges together don't work well for ionic compounds.  Our experimental database might be a bit limited.

The commonly used method of obtaining ionic radii for higher coordination numbers (C.N.) is to extrapolate values from the Shannon's scale [1] using the relationship between ionic radius and coordination number proposed by Zachariasen [2]:

... the bond lengths $D(N_1)$ and $D(N_2)$ for cation coordination numbers $N_1$ and $N_2$ were related as follows
$$D(N_2)=D(N_1)(N_2A_1/N_1A_2)^{1/n} \tag{1}$$
where $n + 1$ is the exponent in the Born repulsion term of the lattice energy and $A_1/A_2$ is the ratio of the Madelung constants. The quantity $A_1/A_2$ was taken to be $0.972$ for $N_2/N_1 = 12/9 = 8/6 = 4/3$ and $0.927$ for $N_2/N_1 = 12/8 = 9/6 = 6/4 = 3/2$.

Jia [3] used Zachariasen's formula and published effective ionic radii for all lanthanides with C.N. $6$ to $12$.
Using the same approach, selected ionic radii for the 12-coordinated $\ce{A}$-sites in oxo-perovskites $\ce{ABO3}$ [4] as well as halide perovskites $\ce{ABX3}$ [5] have been determined.
References

Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst. A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Zachariasen, W. H. Bond Lengths in Oxygen and Halogen Compounds of d and f Elements. Journal of the Less Common Metals 1978, 62, 1–7. https://doi.org/10.1016/0022-5088(78)90010-3.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Li, C.; Soh, K. C. K.; Wu, P. Formability of $\ce{ABO3}$ Perovskites. Journal of Alloys and Compounds 2004, 372 (1), 40–48. https://doi.org/10.1016/j.jallcom.2003.10.017.
Li, C.; Lu, X.; Ding, W.; Feng, L.; Gao, Y.; Guo, Z. Formability of $\ce{ABX3}$ ($\ce{X = F, Cl, Br, I}$) Halide Perovskites. Acta Cryst. B 2008, 64 (6), 702–707. https://doi.org/10.1107/S0108768108032734.


text:
Effective diffusion coefficient of a species in a heterogeneous mixture
I want to calculate the effective diffusion coefficient of hydrogen radical in a mixed-phase consisting of metal oxide, metal hydride and metal. I consider the interfaces to be diffuse.
How can I calculate the effective diffusivity or effective diffusion coefficient of a species (ion or radical) in a heterogeneous mixture consisting of 3 different components? Can I weight average the diffusivities or is it more complicated than that?

I’m general, with great difficulty.

This greatly depends on macro structure. If the phases were say arranged in a sheet-like structure (like a diffusion front) then the low diffusivity phase would be the dominating determinate of the diffusion rate. For this macrostructure, if we assume steady state with no change in phase or length from diffusion or time, then mass transport of the bulk can be approximated by:
$$\frac{\mathrm dn}{\mathrm dt} = \frac{A \times \nabla C}{\sum\frac{L_i}{D_i}}$$
If the high diffusivity phase is continuous then that could be the dominating determinate of the diffusion rate. That does not mean it is easily calculable, just noting that dominating factors can change. If no phase is continuous or other phases are continuous then this also becomes convoluted. For non-sheet-like structures calculating diffusion requires a computer model of the grain structure and finite element analysis to resolve an approximate solution.


text:
Is there radioactivity at absolute zero?
Theoretically, will a radioactive material still be radioactive at absolute zero? What would happen at the lowest realistic temperatures we have ever achieved?
Will radioactivity stop at absolute zero, since it is a nuclear phenomenon and nuclear motion slows down as we approach absolute zero (and theoretically stopping entirely at absolute zero)?

Your premise is that the  nuclear decay is supposed to be caused by their kinetic energy? Nuclei clashing with each other or what, because this seems rather unclear?

@Mithoron I would think radioactivity and nuclear phenomena are borderline, but would work here in a lot of cases.

@Tyberius Yeah, maybe "unclear" is better reason here.

Thermal vibration in molecules (which could be described as "nuclear vibration") is irrelevant to radioactivity which depends on internal quantum processes in the nucleus that have no meaningful dependence on temperature. Or any notable dependence on any other chemical process.

Theoretically, a radioactive material will still be radioactive at absolute zero, and its rate of decay will be $100.00\%$ of that at room temperature. Practically, at the lowest achievable temperatures we observe the same thing: radioactivity is still there, not affected the slightest bit.
Nuclear motion does not slow down as we approach absolute zero, because there is no such thing as nuclear motion in the first place. In a way, all nuclear motion has stopped already at room temperature. Each nucleus just sits there in the ground state and does not know what happens in the chemical world above. From its point of view, the room temperature is the same as absolute zero. To reach its first excited state, it would need energies a great deal greater than that.
Say, you heat your radioactive sample until it melts. Then you heat it a few more thousand degrees, until all materials, including tungsten, melt and then evaporate. Then you heat it some more, until even the strongest chemical bonds are broken and there are no more molecules, just atoms. Then you heat it about ten times more, until atoms lose much of their valent electrons and you have a highly ionized plasma. Then you heat it about a hundred times more, until all atoms lose all their electrons and you have something like a stellar plasma. Then you heat it some more, just in case. Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort.
Short of that, you could just as well have asked if the radioactivity in a sample stops when you paint it blue.

I meant nuclear vibration

Yes, I got that. What I said still applies. Then again, you might point out that atoms do  move, and so do nuclei, being part of them. That's right, but that's irrelevant.

Do you mean nuclear internal degrees of freedom when you say that nucleus sits in the ground state? Or do you assert that there're no phonons at room temperature?

I mean internal processes, of course. Phonons do exist (and then again, some things are liquid or even gaseous at room temperature, so their atoms move even more), but that's irrelevant to radioactivity.

This has been covered on Physics SE a few times.

Would have been surprising if it hadn't.

I'd just like to point out that hyper-heating a sample really seems to get as far away from proving the point as possible. Would be better to say something about why it would retain its radioactive properties at absolute zero. Sure, chemical processes stop there, but then why don't the quantum properties also stop? What are the rules in play that make radioactivity not stop at absolute zero?

@KlomDark It sort of does prove the point. What Ivan is saying is that the temperatures required to start affecting nuclear processes is very high. He is saying that absolute zero is effectively the same as room temperature on the scale of nuclear processes. Absolute zero does not mean zero energy in the ground state. Ivan is saying that temperatures required to bring nuclear states out of ground state are very very high. So absolute zero is effectively the same as room temperature.

Dividing the temperature of the core of the Sun by room temperature yields a pretty big number, but how does that compare to dividing room temperature by zero?

@SolomonSlow Who said you should divide? Who said it should make any sense? In my book, you should compare things like $e^{-{E\over RT}}$, and those behave quite differently.

This is a bit errant. The nucleus is not internally excited, but it is moving as a complete unit as its attendant atom vibrates. Indeed, if it could not do so, I could not throw a block of matter with nuclei in it, as they could not move through space as part of the block. Motion of the atom in thermal vibration is still motion.

The reason it can move as a whole without being internally excited is in fact the same reason that cooling something down will not affect the radioactivity.

I know little about nuclear physics, but a little intuition suggests your answer is wrong. A system at 0K is incapable of work, because it has zero entropy. Since nuclear decay can be used to perform work, this means a radioactive material is not at 0 K. qed.

Ok, but the implications of what I laid out are interesting: it implies (obvious perhaps in hindsight) that a radioactive material has residual entropy, due to the structure of the nucleus. Care to comment? Congratulations on being right, by the way!

"Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort." Why should the energy of the nucleus show any dependence on the electronic structure?

Many things have some residual entropy, due to nuclear or other reasons. I don't see how is that relevant. 2. The energy of the nucleus does not depend on the electronic structure.

There are a few examples of substances exhibiting interesting properties near 0 K due to residual entropy, or despite zero entropy (liquid $He^{II}$). It's been a potential source of confusion for chemists for instance during the application of the 3rd law.  It's not relevant to radioactivity, but it is a thought exercise. For instance, another answer suggests applying pressure to contain nuclear decay. But at 0 K, P goes to zero.

Also, the sidebar links to https://chemistry.stackexchange.com/questions/20449/does-radioactivity-affect-chemical-reactions?rq=1 which provides indications as to how it might be useful to control nuclear decay for chemical studies. If temperature were a factor, that would be useful. It isn't.

Yes, it will continue to be radioactive. This is because the instability of the nucleus is due to the balance of forces internal to it, not to the motion of it as a whole object. It's just like how that throwing a red ball doesn't stop it from being colored red anymore while it is in motion. Likewise, no chemical process can stabilize a radioactive nucleus because chemistry deals with how the atoms containing the nuclei fit together due to interactions of their electron clouds, and this is another factor external to the nucleus itself and so does not affect that instability originating from within.
That said, in theory it is possible to stabilize atomic nuclei with the application of pressure, as it is a question of balance of forces and thus with enough force you can squeeze on the nucleus and keep it together (roughly speaking - from a quantum mechanics POV it's more like "crowd out all the other energy states so that it cannot decay without entering one that's already occupied by somebody else".). But the pressure required here is profound: it is comparable to those in a neutron star, $10^{30}\ \mathrm{Pa}$, around which any more would cause it and the other nuclei in the heavily compressed matter to "melt" into a stew of neutrons and (some) protons. Within a neutron star, there is a layer near the surface - shortly before you get to the neutron stew - of nuclei too heavy to exist at ordinary pressure without being extremely radioactive, but here are stable and persistent.
(There is also one other way you could "stabilize" a radioactive material and that is if you send it off moving through space close to the speed of light. There, time dilation will expand the apparent half-life of the constituent nuclei relative to you on the ground. But you will, of course, not be able to use it for anything while it's flying away from you at extreme speed. No internal properties of the material have been changed, so in its own frame it is still as unstable as ever, but the geometry of spacetime allows you to experience its evolution as being slowed down. This can be observed realistically with unstable single subatomic particles like muons, and these observations provide a good test of Einstein's special theory of relativity.)

Another way of lookung at this problem is to identify nuclei, at least heavier ones, as having their own internal temperature.  Dynamic processes take place within the nuclei, and the energy exchanged in these processes corresponds to roughly $10^{11}$ Kelvins when translated to a degree scale -- a temperature locked within the nuclei and not generally seen by us outside observers.  Compared with it, the temperatures we do access, driven by long-range but inherently weaker interactions, are just a tiny drop in a huge bucket, scarcely impacting the equilibrium within the nucleus.
We can alter the equilibrium by altering the externally applied temperature on this scale.  Say we use adiabatic compression.  Inside the Earth gravity produces pressures in the megabar range, between $10^{11}$ and $10^{12}$ pascals.  But the corresponding temperature there is only a few thousand Kelvins, not enough to materially alter what goes on within the internally much hotter nuclei.  To get the right order of magnitude temperature effect requires those neutron star pressures described by The Sympathasizer.

Radioactive materials by definition cannot reach absolute zero since internal energy in/out is non-equilibrium. (First Law of Thermodynamics)
The stellar events that create these internal nuclear energies and their inherent quantum properties can't be overcome by simply reducing electron energy to ground state.
The only way to stop the decay of radioactive materials is conversion to pure energy (#), or implementation into a system of extreme gravity that nuclei escape velocities cannot be exceeded by decay or emission velocities (*).

(#) Reaction by conventional nuclear fission/fusion/fission devices results in less than 1 gram per 2200 grams of Uranium or Plutonium isotopes converted into pure energy.
(*) It has been observed that in these types of extreme systems, matter and energy is converted to gamma ray emissions that violently exceed Black Hole, Quasar, Pulsar and Magnetar gravity.

I think the premise you have that absolute zero is somehow decreasing energy within a system is true to a point.  We are removing energy from the system at the macro level but the conditions of an atom for example blend to a single state as we move near absolute zero.
Strings, or whatever current theory says is at the heart of matter spread out to a resonating mass with but a single state.
Its at this point that many of us think is the key to readjusting proton counts in the effort to neutralize radiative material.  Making matter stable or configure din such a way as to be more controllable, creating elements that are stable at higher counts of the periodic table.


text:
Doubt from Thermodynamics
If we have to make graph for reversible adiabatic process involving an ideal gas in a P - T curve then how to proceed ?
My try : As we know the relation between P and T in Adiabatic process , differentiated that and tried to find the relation but it went too long and also difficult.

Is this correct

What did you want to achieve with derivatives? I don't think you need them at all.

What you did was correct, but all you really needed to do was write $$T=CP^{\frac{\gamma-1}{\gamma}}$$

Can you explain me how it's graph will be drawn and what is it's slope

It's a straight line on a log-log plot with a positive slope of $(\gamma - 1)/\gamma$


text:
Equalizing Pressure Question
I have a question about the gas collection over water lab:
We learned that when the water level inside the graduated cylinder is higher than the outside water level, the gases inside the graduated cylinder have a lower pressure than atmospheric pressure.
Then by adjusting water level of the graduated cylinder to match the water level outside would equalize pressure. However shouldn't the volume of the gas decrease if the pressure is increased? Shouldn't the water level go up as you push the cylinder down beneath water level or go down as you lift it through the water? Does Boyle´s law (PV = PV) not apply here? Why?

Why don't you set up a model calculation to quantify exactly what happens?  Treat the gas as ideal, and use the hydrostatic equation to quantify the pressure at the top of the water surface within the cylinder?

Inside the graduated cylinder there is some water and the air bubble above it. The key point is that at a depth inside the graduated cylinder, at the same height as the water line outside, the pressure is the same. That means that the pressure on the outside, which is 1 atmosphere, is equal to the pressure inside, which is the pressure exerted by the column of water that is higher than the outside water level, plus the pressure of the air bubble.
The pressure exerted exerted by a column of liquid is $P = \rho g h$ where $\rho$ is the density of the liquid. But that means that for a small displacement on the order of a centimeter, $P \approx 100\ \mathrm{Pa}$. One atmosphere on the other hand is roughly 1000 times that magnitude, so you've managed to decrease the pressure of the gas bubble by about 0.1%. Boyle's law does apply and implies that you've increased the volume by the same amount.
The takeaway here is that the change in pressure inside the graduated cylinder is quite small, enough to pull or push the liquid up and down as you move the graduated cylinder.
Also, you shouldn't be using a graduate cylinder. There's a special glass tube called a eudiometer that you're supposed to use for these types of measurements.


text:
Why is -OCH3 more strongly activating than -CH3 in electrophilic aromatic substitution?
Facts I know:
1) More the electron density in benzene ring, the faster the reaction.
2) Lone pair on -OCH3 group undergoes resonance and makes ortho-para positions electron rich, and so does -CH3 by hyperconjugation.
3) -OCH3 group shows -I (electron withdrawing) and -CH3 shows +I (electron releasing).
Now, -OCH3 and -CH3 both undergo resonance and hyperconjugation respectively and both make ortho/para positions equally electron rich.
So the factor that should come in play is I effect. As the -CH3 group releases electron into benzene ring by +I effect, and -OCH3 withdraws electron by -I, what should be more activating is the -CH3 group.
Where am i wrong?

Not all resonance effects are made equal...

CH3 don't show resonance and a stable resonance structure is more contributing than hyperconjugation one

Since we are talking about the rates of these reactions, we really want to examine the structures of the transition states in the rate determining step in these reactions, rather than the starting materials. For electrophilic aromatic substitution, there is a high energy carbocation, the Wheland intermediate, formed when the benzene derivative attacks the electrophile. Formation of this intermediate is normally the rate determining step. Since this carbocation is high in energy, the transition state leading to it is going to be similar in structure, according to the Hammond Postulate. So, looking at the influence of varying substituents on the stability of these Wheland intermediates is a good approach to justifying differences in reaction rates (not to mention selectivities).
If we take the resonance theory approach, we can draw many resonance structures for eg. the para-substitution intermediate, putting the positive charge ortho and para to the "E" substituent we just added. Looking at the methyl-substituted benzene substrate (toluene), resonance theory would allow us to draw an additional "no bond" resonance structure, where we break the C-H bond and have a proton associated with a neutral methylidene cyclohexadiene. This is the resonance theory approach to hyperconjugation. This resonance contributor can justify why toluene is more reactive than benzene. However, how important is the contribution of a resonance structure where we have one less σ-bond and a positive charge on a tiny hydrogen atom?

Looking at anisole (methoxybenzene) the same way, resonance theory would lead us to draw a resonance structure where the non-bonding (lone) pair of electrons on oxygen is used to make an additional π-bond and the oxygen ends up with the formal positive charge. It may not feel nice to put a formal positive charge on the oxygen, but this resonance contributor has a lot going for it - we have gained an additional π-bond relative to the other resonance structures and all non-hydrogen atoms satisfy the octet rule. The "loss" of the lone pair is no problem - as a non-bonding pair of electrons they were already relatively high in energy. For these reasons, this is an excellent contributor to the overall structure of this intermediate - probably the most important.

Taken together, and using resonance theory, the above can help justify the differences in reactivity. Another approach would be to use molecular orbital theory. Hyperconjugation would be viewed as overlap of the C-H σ-orbital with the adjacent carbocation π-system. The efficiency of this overlap is poor because there is a big energy difference between a low energy C-H σ-bond and π-system of the carbocation. In contrast the energy difference between the oxygen non-bonding (lone) pair and the π-system is relatively small and overlap is therefore efficient.

The last paragraph describes a Valence Bond approach to hyperconjugation, i.e. Localised orbitals. You do not have those on a molecular orbital approach. Here the π system would smear into the CH bonds, delocalisation is the key descriptor.


text:
According to the Cahn-Ingold-Prelog system assign a priority to each of the following groups. (Highest priority is 1)
1-CH(CH3)CH2F
2-CHCl2
3-CH2Br
4-C(CH3)3
this is the order i did using the system but i am having a hard time understanding it and was wondering if anyone could tell me if my answer is correct, and if it is not, explain to me please

The keys are:

going outwards, only look one bond at a time
look for the first point of difference
to break ties follow the chains of highest priority first

So, going out from the point of attachment, the first atom in each chain is a carbon, so we expand from there to the next three atoms in each case (in red). Draw these out in priority order (ie. highest to lowest atomic number), top to bottom. Now compare the highest priority atom on each chain, ie. C vs Cl vs Br vs C. Bromine has the highest atomic number, so this makes that whole substituent number 1 priority, chlorine is next, so that whole group is number 2 priority, the remaining two groups are a tie, so we go down their list of attached atoms, looking for the first point of difference. The second atom on each is a carbon - ie. a tie. The final atom is a hydrogen on the first group, but a carbon on the final group, so the latter has the higher overall priority. In this particular example, there is no need to go further out along the chains to determine priority order for the substituents.

So, the final priority order is: (1) -CH2Br, (2) -CHCl2, (3) -C(CH3)3, (4) -CH(CH3)CH2F


text:
Function of 'organic' vs regular pipe cleaner
Recently, I needed to buy some pipe cleaner, as sometimes even with care hair and other detritus gets lodged so deep down the shower drain that it is not possible to get it out. I always try to avoid/reduce use of bleach and other hard chemicals or products that incorporate them, as in the end it goes down the drain is not good for the environment.
However, unfortunately the 'bio' product of the same brand of cleaner I had used before (didn't know it existed before) was not very satisfactory so far. Is there maybe someone here who could offer details of the chemical function of these types of cleaners, and either knows a method to improve the use of the bio/organic one or recommend another product or compound that is easy on the environment but cleans drains well? For the proper use, the instructions say that for the 'organic' one (yes sorry for that chemically improper use but ...) you should add hot or warm water to start the reaction, though my worry is also that this added water will just flush the product down the drain depending on the size of the drain.
I know some of these ingredients and how they likely function, but not all of them and I am not sure what the combined effects might be. To my understanding the 'non-organic' one relies on the reactive power of chlor(ide/ine) and resulting reaction products, whereas with the organic one I am not really sure.
Specifically, it is about these two products:
rorax and rorax bio (at least on the German market)
The ingredients of rorax (said to be chlorine based in marketing) are given as water, sodium hydroxide, sodium hypochlorite, myristamin oxide, sodium silicate, fatty acids, C6-12-, natrium salts as per http://detvo.werner-mertz.de/?lang=de&country=DE&prod_id=109655
The ingredients of rorax bio are given as water, sodium chloride, urea, sodim laureth sulfate, sodium thioglycolate, sodium hydroxide, Amide polyglycol ether, parfum, colorant
according to http://detvo.werner-mertz.de/?lang=de&country=DE&prod_id=114331

Congratulations on being environmentally sensitive!
The problem with sink drains is the device that lifts and lowers the stopper. It is a rod that protrudes into the drainpipe above the U-trap; it catches hair and provides a bacterial nest and eventually slows water drainage.
Older drain cleaners were based on alkali to dissolve hair; some used aluminum to foam and break up the hairy mass. Bleach plus alkali works well too, and kills bacteria. The newer, less aggressive formulas can use thioglycolate to dissolve hair (also used on human legs!).
In the overall scheme of life, with all the other bodily wastes going down the drain, a little drain cleaner may not be a big problem, and where it is important, the government usually has a statement or a regulation.
Even without regulations tho, it would be kind to the environment to limit our use of toxic chemicals. Using a lot of ineffective product may be worse than using a little of a more aggressive product. I would suggest two approaches:
1) Pour just a little of the effective product down the drain, but only when it drains slowly (you can measure the drain speed by filling the bowl to a certain line, then timing how long it takes to empty). This little bit (10-20 mL?) will saturate the clog because it is higher than the bottom of the U-trap. Give it some time (many minutes, even hours); see if it has opened up. If not, repeat. This may go against product recommendations to use a lot of product, but I suspect that the bottom of the U-trap is not blocked (any water flow usually cleans that out). This will reduce your environmental impact by using your brain.
2) Use a mechanical method. a) A stiff copper wire with a tiny hook at the end is useful for digging out hair, and often opens up the drain with just a lot of hot water. b) Cut a hollow rubber ball in half to make a tiny plunger. Place it over the drain (with the stopper lifted a little) and press it up and down to push the clog all the way down the drain.

Thanks for your answer. Just some comments. Maybe I should have added that in this case the most problematic is the shower drain, not the sink, so there is no 'stopper' in this case. If it was a sink, I would likely just disassemble the U-trap and clean it that way. The product recommendation in the one I use (there are more concentrated ones) is to use 250 ml of the 1 L bottle at a time, then wait between 5 mins to overnight. I always let it sit overnight. However due to the shape (very flat/narrow) of the drain, it is even harder to get to.

Perhaps you would feel better just adding half of the recommended amount (125 mL) and see if that works. Or whatever minimal amount of whichever product can do the job with the least amount. This is also a money-saver!


text:
What is LSD-1, …, LSD-24, LSD-26, …?
LSD (lysergic acid diethylamide) was formerly often referred to as ‘LSD-25’.
If it really is a (Albert Hofmann / Sandoz)'s code name (and not a combination of ‘LSD’ and usual LSD (Delysid) dosage numeral ‘25’ μg; see also Quora: What does the "25" in "LSD-25" mean?), are some other substances from the series LSD-1, LSD-2, … etc., being at a time probably a part of industrial secret, currently already known?
UPDATE
Summary
Identify at least one another Sandoz LSD-x, where x ≥ 1 and x ≠ 25 (with good sources or at least proven insider knowledge).

It was Hofmann's 25th synthesis of the compound-it has nothing to do with dosage.

@ToddMinehardt … but it wouldn't make much sense to append serial number to the compound name abbreviation. I share the hypothesis, that in the series, ‘LSD’ means ‘Lysergsäurederivat' rather than ‘Lysergsäurediethylamid’.

The number "25" indeed comes from the fact that it was the 25th compound of the lysergic acid amide series or lysergamides. I found an article* by Frank J. Ayd, Jr. & Barry Blackwell who accounted Alexander Hoffman's investigation in his LSD synthesis:

A number of pharmacological experiments were carried out by Professor
Ernst Rothlin with this new compound, which was given the laboratory
code name LSD-25 because it was the twenty-fifth compound of the
lysergic acid amide series.

You can check the other compounds of the series in the Wikipedia article linked above. Most of them have their own specific names and not LSD-1, LSD-2 etc.

I also did find something in an e-book(Ref) which says that the number has something to do with date:

Dr. Hoffmann described the results of accidental ingestion of
LSD. It was in Switzerland in 1943, five years after Dr. Hoffmann and
an associate has synthesized lysergic acid diethylamide tartarate
(from ergot). The substance was labeled "25" because of the date: "May
2, 1938" (the date when LSD was synthesized(?)).

Ref.: LSD-25: a Factual Account: Layman's Guide to the Pharmacology, Physiology, Psychology, and Sociology of LSD, Louise G. Richards, U.S. Bureau of Narcotics and Dangerous Drugs, 1969
ಠ_ಠ

*I was researching about something else i.e "serendipity in Chemistry". While I was surfing the Internet, I stumbled upon this article and from that, I came to know about LSD, searched for the sources of their information and from there I got access to the above article and then proceeded to compile this answer without any prior intentions. What a beautiful irony!

Dr. Albert Hoffman has prepared lysergic acid diethylamide (LSD) for the first time in 1938 as part of a systematic chemical and pharmacological investigation of partially synthetic amides of Iysergic acid in the Sandoz pharmaceutical-chemical research laboratories in Basle. Sandoz was a Swiss chemical company founded in 1886 where its pharmaceutical-chemical division was headed at the time by Professor Arthur Stoll. Dr. Stoll was the first to isolate lysergic acid based active substances such as ergotamine and ergobasine from ergot, a fungus found in tainted rye.According to this article, within a few years researchers at Sandoz had determined the chemical structure of these various biologically active compounds in ergot, all of which shared a common nucleus. This chemical starting point was called lysergic acid, or, in German, Lysergsaure.

Hofmann developed a synthetic process to build the ergot compounds from their component chemicals. Using this method, he recreated ergot’s active ingredients as well as novel but similar compounds that, based on the potency of the ergot compounds, could reasonably be expected to have medical uses.
In a sense Hofmann was playing God, combining lysergic acid with various other organic molecules just to see what happened. He created 24 of these lysergic acid combinations. Then he created the 25th, reacting lysergic acid with diethylamine, a derivative of ammonia. The compound was abbreviated as LSD-25 for the purposes of laboratory testing.

Hofmann was hoping one of these compounds that could stimulate circulation and respiration. Lysergic acid diethylamide (LSD) and four other Hofmann developed compounds are listed below:

Accordingly, as  Todd Minehardt pointed out, the name LSD-25 has nothing to do with the dosage used in therapy. It is strictly for the purposes of laboratory testing, which originally had Sandoz to decide “The new substance, however, aroused no special interest in our pharmacologists and physicians; testing was therefore discontinued.” However, Hofmann's accidental discovery of hallucination by LSD-25 in April 16, 1943 has changed everything forever.


text:
Which are the radicals containing phosphorus in ATP?
The ATP molecule contains 3 $\ce P$ atoms linked through esteric bonds. The IUPAC name of triphosphate adenosine is "[[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy-hydroxyphosphoryl] phosphono hydrogen phosphate" as PubChem states. This name presents 3 groups containing phosphorus in the end: hidroxyphosphoryl, phosphono and hydrogen phosphate. I'm not sure about how are these groups placed in the molecule because I can't seem to find the hydroxyphosphoryl. To ilustrate what I mean, I've drawn it. This is how I found the other groups, but I'm unsure it's the right way to look at the end of the molecule. In blue: methoxy, In green: hydrogen phosphate, In red: phosphono. The other black part does not match the hydroxyphosphoryl group.

You're going backwards...

@Mithoron The one in red is still phospono I guess. I'm unsure about the rest. If I take what I drawn in green as hydoxyphosphoryl, the rest doesn't match phosphate, and vice-versa. I don't understand how must I take them.

@Mithoron, can you give me a short answer on this: I've searched for phosphoryl some more and found 2 answers and I don't know which is the correct one: $\ce{P^{3+}O}$ (which would seem correct here) and $\ce{P^{+}O3^{2-}}$ (which is shown on Wikipedia). I'm really confused.

You have almost answered yourself completely.

'Phosphoryl’ in current IUPAC Nomenclature of organic chemistry, preferred names (2013):

P-67.1.4.1.1.2 Fundamental acyl groups for substituent derived from mononuclear noncarbon oxoacids
(…)
$\ce{-P(O)<}$
  phosphoryl

Example compound with “hydroxyphosphoryl” group:
$\ce{C2H5-P(O)(OH)-CH2COOH}$
[ethyl(hydroxy)phosphoryl]acetic acid
(software generated, in the book there is an example without hydroxy parens: O-({[(2R)-2,3-bis(octadecanoyloxy)propoxy]hydroxyphosphoryl})-L-serine)
‘Phosphono´:

P-67.1.4.1.1.1 Retained names for substituent groups derived from mononuclear noncarbon oxoacids
(…)
$\ce{-P(O)(OH)2}$
  phosphono (preselected prefix)

(There's no “phosphonyl” if someone wonders, there's however different group with similar name, phosphonoyl $\ce{-PH(O)\bond{-}}$.)
Example:
$\ce{(HO)2P(O)-CH2COOH}$
phosphonoacetic acid
And, I think that the correct layout for the name provided should be rather:
{[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy}(hydroxy)phosphoryl phosphono hydrogen phosphate
This name is somewhat overcomplicated. ‘Triphosphate’-based name could be used, see

P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement analogues as functional parents for naming organic compounds
(…)
P-67.2 Di- and polynuclear noncarbon oxoacids
(…)
P-67.2.1 Preselected names
(…)
$\ce{(HO)2P(O)-O-P(O)(OH)-O-P(O)(OH)2}$
  triphosporic acid (preselected name)

and

P-106 Nucleotides
(…)
P-106.2 Nucleotide diphosphates and triphosphates
Diphosphate, triphosphate, etc. esters of nucleosides are named by citing a phrase such as diphosphate, after the name of the nucleoside. The presence of hydrogen atoms on the diphosphate, triphosphate, etc. component of the molecule is indicated by the words ‘hydrogen’, ‘dihydrogen’, etc. Parentheses are used to avoid ambiguity.
Examples:

  uridine 5’-(tetrahydrogen triphosphate)

(but this is natural compounds section, which is more permissive)
Seemingly apparent reason why ‘triphosphate’ is currently not used in software generated systematic names is that there's an ambiguity in binding the residue to the first (last), or the middle phosphorus' oxygen (positional isomers), and no numbering system for non-carbonic acids is defined.
However it is defined (P-67.2.2.1), with esters examples (P-67.2.5.2). The numbering is from-to central atoms including the oxygen connectors, so the terminal phosphorus atoms get numbers 1, 3, 5; so (generalized) names for the two isomeric alkyl triphosphates (alkyl on terminal or middle phosphate unit, respectively), are

O1-alkyl tetrahydrogen triphosphate
O3-alkyl tetrahydrogen triphosphate

which might be confusing, especially in the latter case. So this can be the main reason, why these are not used.
Let's use it anyway (with a minor improvement – indicated hydrogen):
O1-[(2R,3S,4R,5R)-5-(6-amino-9H-purin-9-yl)-3,4-dihydroxyoxolan-2-yl] tetrahydrogen triphosphate.
It should be OK to omit ‘tetrahydrogen’, as phosphates are ionized in physiological conditions.
(Not speaking of yet better names like adenosine 5’-triphosphate or ATP.)

Wow! You really gave me more than I asked for. Now I understand. I was confused by the fact that someone put a question here and in the answer you can see that the phosphoryl group is $\ce{—PO3^2-}$. Guess that's wrong

@StefanOctavian Right, that's wrong. I've checked inorganic nomenclature, and it's the same, trivalent $\ce{PO}$ group.


text:
In 1 dimension, the curvature of Ψ = K.E. of wave. Is this the case for 2 dimensions and 3 dimensions also?
In 1 dimension, the curvature of Ψ = K.E. of wave. Is this the case for 2 dimensions and 3 dimensions also? Moreover, is curvature of Ψ = K.E. of wave = 2nd derivative of Ψ?

If you drop the usage of the word "curvature" altogether, the rest would be true.

I think "curvature" is still used in three dimensions, at least, I've seen it before when discussing KE + PE contributions to the energy...

Ivan Neretin slide 19 of this powerpoint, https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-007-electromagnetic-energy-from-motors-to-lasers-spring-2011/lecture-notes/MIT6_007S11_lec40.pdf, states clearly that curvature of wavefunction = K.E. of wave


text:
Calculating the number of energy states in a particular momentum direction
Given is a 2D scenario of a particle within a container:

The circumference shows all the possible 2D spatial directions for a given momentum value of the particle. One of those momentum directions is $p_e$.
From what I understand, the number of states within that $p_e$ is given by:
$$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2}$$
Where $p_{e,x}$ and $p_{e,y}$ are the projections of $p_e$ in the $x$ and $y$ coordinates.
Since $p_e$ has a spatial direction, this means that the container has a length in that same spatial direction. Let's call that container length in that same spatial direction $L_e$, illustrated as follows:

Does this $L_e$ have any relationship with the number of states in momentum $p_e$? For example, is the following equation valid?
$$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2} = \frac{L_e p_e}{h}$$
EDIT: I forgot to state to use periodic boundary condition in such a case, since rigid boundaries gives zero momentum to particles. So my question is: is it possible to calculate the number of states in a single momentum that has just one direction in p-space? If so, how is this derived from the formula:
$$n=\frac{V \cdot p^3}{h^3}$$

Don't all eigenstates of a particle in a box have zero momentum?

Apologies, I should have stated to use periodic boundary conditions


text:
Reverse Image Searching a ChemDraw Figure
Is there any image-recognition software out there that can read in a high-resolution image of a ChemDraw-generated figure and back out the structure (such as the SMILES or INChI)? I have a bunch of review papers where there are tables upon tables of ChemDraw images of molecules and would really like to put them into some useable database.

You can try this http://chemdataextractor.org/
If you manage to obtain something, I'd kindly ask you to post the code. Thanks.

The best tool I have come across so far to do this is this one: https://cactus.nci.nih.gov/osra/
I was also reading about this other one, but can't find the link to the tool itself.

There is a new tool for this: pic2mol.com. It's a cloud-based optical structure recognition service and an add-in for ChemDraw. There is a free trial. If you have ChemOffice, you can give it a try.
Disclaimer: I am affiliated with pic2mol.

I guess it you have a chemdraw object, you could convert it to SMILES, which was the objective of the question. While this is probably not the most convenient way of doing it, I accept this answer as a possible way to do this, and therefore regard this post as an answer (on the technicalities of what is an answer).


text:
where is sulphate going here? electrolysis of magnesium sulphate
I am electrolyzing a water + magnesium sulphate solution using copper for both electrodes. I am getting a light-blue precipitate forming which I believe to be a mixture of copper hydroxide and magnesium hydroxide, which are both poorly soluble. The solution is remaining clear which tells me that basically no copper sulphate is forming.
My question is, where is the sulphur from the sulphate going?
The mixture is not becoming acidic as if sulphuric acid was forming, and I do not detect any smell of sulphur is if some sulphur oxides were being released as gas.

It stays in solution?

I wasn't aware that could happen. Could the sulphate ion just sit in solution by itself? I thought it would have to be in equilibrium with a positive ion as the copper sulfate would be?

Are you sure the solution is not becoming acidic? Precipitation of Mg(OH)2 indicates that the MgSO4 has been split into Mg(OH)2 + H2SO4.  If no mixing occurs, the cathode region will become alkaline while the anode region will become acidic (and will dissolve some copper).

I'm pretty sure. I don't have pH indicators but when I add a base to the mixture, there is no reaction. And when I add an acid(vinegar), almost all of the precipitate disappears as it reacts with the acid to form soluble salts (magnesium and copper acetate in this case). If any acid was formed, wouldn't it just immediately react with the hydroxides and it would go back to mostly magnesium sulphate again? There is a really huge amount of white precipitate that forms very quickly when electrolysis starts.

In electrolysis of aqueous solutions sulfate ions just sit there.  Protic hydrogen (or the metal in the cation if this metal is not highly electropositive like magnesium) is reduced more easily at the cathode, and water is more easily oxidized at the anode.
Sulfate ion (actually, bisulfate ion) can be oxidized electrolytically at the anode if you use sulfuric acid instead of water as the solvent.  See the preparation methods for potassium peroxydisulfate.

What is "protic hydrogen" ?


text:
Can all oxidizing agents be used interchangeably in organic chemistry reactions?

Butanol can be oxidized with pyridinium chlorochromate (PCC) yielding butanal. If the case is that butanol can be oxidized with PCC, can it be oxidized with CrO3, H2SO4, HNO3, OsO4, and other oxidizing agents? If yes, why? If no, why?
Thanks.

Simple answer no.
For a simple transformation there may be many conditions that will get the same result (in varying yields), but that does not mean that all oxidising agents could be used. Nitric acid is a powerful and indiscriminate reagent better suited to oxidation of inorganic species as it is more likely to degrade your organic substrate than produce clean product. In the example you quote CrO3, if used with acid, will take butan-1-ol to butanoic acid rather than butanal. Solvent makes a difference too; PCC will give the aldehyde in CH2Cl2 but the acid in DMF. Temperature is also important; the Swern reaction for alcohol to aldehyde is routinely run at below -60C as above that temperature side products proliferate.
Selecting the right conditions for a transformation goes to the heart of synthetic organic chemistry and a knowledge of reagents is a basic requirement for any synthetic organic chemist.


text:
Is there an intuitive reason for the proportionality constant 2/3 in pv=nRT=(2/3)U?
Why is there a $\frac23$ term?
I fully understand that it makes sense for pressure to be the volume weighted average kinetic energy. And the equation above shows that this is so (for an ideal gas). I also fully understand how to prove this with momentum and force. I'm wondering is there an intuitive reason why the for one cubic meter, the the pressure should be two thirds of the total internal kinetic energy, not some other value. Or is it just because the math says so?

The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), the particles have equal probability of hitting a face in the x, y or z direction.  The pressure is exerted equally among the three orthogonal directions, hence a factor of 1/3.
Wikipedia also shows the steps in deriving the kinetic theory.

The expression you've written can be obtained from the equipartition theorem
$$U = n_\mathrm{d.f.} \times \frac{k_\mathrm B T}2$$
stating that the energy is evenly distributed over all available degrees of freedom $n_\mathrm{d.f.}$, where each degree of freedom contributes an amount of energy equal to $k_\mathrm B T/2$. For a monoatomic gas of $N$ particles, each atom has only translational degrees of freedom (a total of three per particle, corresponding to three orthogonal spatial coordinates), and $n_\mathrm{d.f.} = 3N$, so that
$$U = 3N\frac{k_\mathrm B T}2$$
Combining this with the ideal gas expression $N k_\mathrm B T= pV$ leads to a constant of proportionality of 2/3.


text:
Which is stronger ? Picric acid or phoshoric acid
I would like  to have clarity as to which acid is the stronger acid; picric acid or phosphoric acid.
Because different  sources are giving me different  answers

Where are you getting your data?
Picric acid is the stronger, pKa 0.38 data here
The first pKa for phosphoric acid is 2.15 data here. Remember that phosphoric acid is a tribasic acid so there are 3 dissociation constants, one for each acidic proton.


text:
Stability order of methy benzyl cations?

I know that meta position shown no resonance effect and hyperconjugation effect . So According to the order given I guess Hyperconjugating effect is taken into account as -CH3 has no Resonance effects.
But are the hydrogen in methyl group alpha hydrogens ?
As per I know that hyperconjugation is only shown if alpha hydrogen is present .
I am not getting the order . If only inductive effect was considered the order should have been 1>2>3 but its 1>3>2  please explain this order to me .

First understand that resonance stabilized carbocation is much more stable than that of inductive stabilized one, so out of the three ones, one can conclude that I,III>II. Now the ambiguity arises in choosing between I and III, now we turn to inductive effect for help, and here in this case, is simple because in I the methyl group is closer to the cation than in III, so you can conclude I>III, based on the two conclusions, one can conclude the order.

How can CH3 show resonance. Can you please elaborate

The effect is called hyper conjugation, look it up on Google

Hyperconjugation is a different thing and its very less effective than resonance


text:
Why lattice energy of NaCl is more than CsCl?
Why is the lattice energy of Nacl more than that of CsCl?
If size of Na is smaller than that of Cs, then according to Fajan's rule, Na should polarise Chlorine more and therefore, NaCl must have more covalent character than CsCl (and higher covalent character means low lattice energy).

CsCl is very much an ionic compound, and NaCl (if Fajan's rule applies) is a tiny bit more covalent. Makes no difference.

Diamond has quite a high covalent character, but I believe its lattice energy is quite high ...

Lattice Energy is directly proportional to the Charge on ion and inversely proportional to radius of atom . Now in between NaCl and CsCl the radius of Cs+ is way more than Na+ , even though CsCl is more ionic but still radius is more and hence the lattice energy is less .
It can also be explained using
Kapustinskii Equation