text:
How to specify atomic carbon terms in the coupled and uncoupled representation?
So, we know, that the atomic carbon in the electronic configuration $1s^22s^22p^2$ has the following terms
$${}^1S, {}^1D, {}^3P$$
My question is - how can I correctly specify these terms in the terms of coupled and uncoupled representations?

My attempt
So, in the case of terms, we're only considering the orbital angular momentum, not the spin. Because of that, we can describe the single terms in the coupled representation $\left|L, M_L\right>$ which correspond with the linear combination of microstates, i.e. the uncoupled representations $\left|m_{l1}, m_{l2}\right>$ using Clebsch-Gordan coefficients.
For ${}^1S$ term it's pretty easy, as $L=0$ and $M_L=0$ (as described in this answer):
$$\begin{align}{}^1S: |L = 0, M_L = 0\rangle &= \frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |-1, 1\rangle - \frac{1}{\sqrt 3} |0, 0\rangle\\ &= \frac{1}{\sqrt{3}} \left| 8 \right> + \frac{1}{\sqrt{3}} \left| 11 \right> - \frac{1}{\sqrt{3}}\left| 14 \right >\end{align}$$
In the last expression there are wavefunctions specified with indices from the microstate table below.
But further it gets somewhat more tricky - both ${}^3P$ and ${}^1D$ will contain multiple states.
$P$ corresponds with $L=1$ and so $M_L \in \left\{ -1, 0, 1 \right\}$. I suppose, that its coupled representations are $\left| L=1, M_L=-1\right>, \left| L=1, M_L=0\right>, \left| L=1, M_L=1\right>$.
$$\begin{align}
{}^3P: \left| L=1, M_L=-1\right> &= \frac{1}{\sqrt{2}}\left| -1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, -1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 2 \right> + \frac{1}{\sqrt{2}}\left|5 \right>\\
\left| L=1, M_L=0\right> &= \frac{1}{\sqrt{2}}\left| 1, -1 \right> - \frac{1}{\sqrt{2}}\left| -1, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 3 \right> - \frac{1}{\sqrt{2}}\left|6 \right>\\
\left| L=1, M_L=1\right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> - \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 1 \right> - \frac{1}{\sqrt{2}}\left|4 \right>
\end{align}
$$
${}^1D$ corresponds with $L=2$ and $M_L \in \left\{ -2,-1,0,1,2 \right\}$.
$$\begin{align}
{}^1D:\left| L = 2, M_L = -2 \right> &= \left| -1, -1 \right> = \left| 15\right>\\
\left| L = 2, M_L = -1 \right> &= \frac{1}{\sqrt{2}}\left| 0, -1 \right> + \frac{1}{\sqrt{2}}\left|-1, 0 \right> \\
&= \frac{1}{\sqrt{2}}\left| 10 \right> + \frac{1}{\sqrt{2}}\left| 12 \right>\\
\left| L = 2, M_L = 0 \right> &= \frac{1}{\sqrt{6}}\left|1, -1 \right> + \sqrt{\frac{2}{3}}\left| 0, 0 \right> + \frac{1}{\sqrt{6}}\left| -1, 1 \right> \\
&= \frac{1}{\sqrt{6}}\left| 8 \right> + \sqrt{\frac{2}{3}}\left|14 \right> + \frac{1}{\sqrt{6}}\left| 11 \right> \\
\left| L = 2, M_L = 1 \right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left|7 \right> + \frac{1}{\sqrt{2}}\left|9 \right> \\
\left| L = 2, M_L = 2 \right> &= \left| 1, 1 \right> = \left| 13 \right>
\end{align}
$$
Is this the right approach or do I understand it incorrectly?

Table of microstates

I'll try to illustrate the procedure for two of the nine states that form the term symbol $^3P$. This term symbol has $S = 1$ and $L = 1$, which makes for three possible values of $M_S$ and three possible values of $M_L$, for a total of nine states. These states are "coupled" in both the spin and the spatial dimensions:
$$|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle, |1,0,1,+1\rangle, \cdots$$
The microstates in the table are states which are "uncoupled" in both dimensions:
$$|m_{s1},m_{s2},m_{l1},m_{l2}\rangle = |{+1/2}, {+1/2}, {+1}, {+1}\rangle, \cdots$$
To go from the term symbols ("doubly coupled" - don't use this terminology, I made it up) to the microstates ("doubly uncoupled"), we have to convert both spin and spatial components from coupled to uncoupled representations.
Thankfully, the spin and spatial components are separable, so we can write
$$|S,M_S,L,M_L\rangle = |S,M_S\rangle \cdot |L,M_L\rangle$$
Uncoupling the spin part is simple enough:
$$\begin{align}
|S=1,M_S=+1\rangle &= |\alpha(1)\alpha(2)\rangle \\
|S=1,M_S=0\rangle &= \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \\
|S=1,M_S=-1\rangle &= |\beta(1)\beta(2)\rangle \\
\end{align}$$
where $|\alpha\rangle$ denotes $m_s = +1/2$ and $|\beta\rangle$ denotes $m_s = -1/2$, as usual.
For the spatial parts, you will need to use the Clebsch–Gordan coefficients to expand $|L,M_L\rangle$ in terms of the "uncoupled representations" $|m_{l1},m_{l2}\rangle$. You already seem to understand this. For illustrative purposes we'll use
$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|m_{l1} = +1, m_{l2} = 0\rangle - |m_{l1} = 0, m_{l2} = +1\rangle)$$
and we'll simplify the notation by denoting $m_l = +1, 0, -1$ as $p^+, p^0, p^-$ respectively, so:
$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\}$$

Now, say we wish to find the microstate corresponding to $|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle$. What we need to do is to expand both spin and spatial components in terms of their respective uncoupled representations, then multiply them together again.
$$\begin{align}
&|S=1,M_S=+1\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = |\alpha(1)\alpha(2)\rangle \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{\sqrt 2} \left\{|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle - |p^0(1)\alpha(1)\,p^+(2)\alpha(2)\rangle\right\}
\end{align}$$
Notice how this is a (suitably antisymmetrised) state where one electron is in $p^+$ with spin up, and the other electron is in $p^0$ with spin up. This corresponds exactly to microstate 1.
Although the microstates are uncoupled states, it doesn't mean they aren't antisymmetrised. By virtue of quantum indistinguishability, they have to be antisymmetrised: that's why most of the microstates are linear combinations of "composite states" such as $|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle$ (such a state on its own is not physically permissible, whereas the microstates are physically permissible states).

For an uglier but more instructive case, let's look at $|S,M_S,L,M_L\rangle = |1,0,1,+1\rangle$.
$$\begin{align}
&|S=1,M_S=0\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{2} \left\{ |p^+(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^+(2)\alpha(2)\rangle + |p^+(1)\beta(1)p^0(2)\alpha(2)\rangle -  |p^0(1)\alpha(1)p^+(2)\beta(2)\rangle \right\}
\end{align}$$
Notice the first two terms in this expansion describe an (antisymmetrised) state where one spin-up electron is in $p^+$ and one spin-down electron is in $p^0$. This corresponds to microstate 7. Likewise, the second two terms correspond to microstate 9. All in all, we can write:
$$|S = 1, M_S = 0, L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|7\rangle + |9\rangle)$$

Using this procedure you can work your way through the rest of the states in $^3P$ (and indeed, $^1S$ and $^1D$ too). As a final example, let's look at one of the five $^1D$ states:
$$\begin{align}
&|S = 0, M_S = 0, L = 2, M_L = 0\rangle \\
&\qquad = |S = 0, M_S = 0\rangle \cdot |L = 2, M_L = 0\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle - |\beta(1)\alpha(2)\rangle\right\} \cdot \\
&\qquad\qquad\qquad \left[\frac{1}{\sqrt 6}\left\{|p^+(1)p^-(2)\rangle + |p^-(1)p^+(2)\rangle \right\} + \sqrt{\frac{2}{3}}|p^0(1)p^0(2)\rangle \right] \\
&\qquad = \frac{1}{\sqrt{12}}\left\{ |p^+(1)\alpha(1)p^-(2)\beta(2)\rangle - |p^-(1)\beta(1)p^+(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{12}}\left\{ |p^-(1)\alpha(1)p^+(2)\beta(2)\rangle - |p^+(1)\beta(1)p^-(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{3}}\left\{ |p^0(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^0(2)\alpha(2)\rangle \right\} \\
\end{align}$$
The first line of this expansion corresponds to $(\sqrt{1/6}|8\rangle)$; the second to $(\sqrt{1/6}|11\rangle)$; and the third to $\sqrt{2/3}|14\rangle$. So, your expansion was correct; however, be careful, because the uncoupled representations of the spatial wavefunctions
$$|m_{l_1} = +1, m_{l2} = -1\rangle$$
do not themselves correspond to the microstates. Instead, it is only by multiplying this with the spin wavefunction can you obtain a wavefunction that corresponds to the microstates.

Just one more question - I don't understand, why is there minus in $\left| S=0, M_S=0 \right>$ in the very last example?

@Eenoku that's the singlet spin wavefunction. Atkins MQM 5th ed section 4.12 describes this in detail, and the bottom bit of my previous answer sort of summarises this. But really, any textbook should cover it (sometimes it appears in the context of the helium atom).

No problem. Feel free to ask a new question if you want more detail (I think it’s too long for comments).


text:
Confusion about prioritizing functional groups in case of poly functional organic compounds
According to Wikipedia,
sulfonic acid $(\ce{-SO3H})$ is comparatively stronger than carboxylic acid $(\ce{-COOH})$.
Then why is sulfonic acid lower in priority than carboxylic acid (while naming poly functional organic compounds) in this priority chart? Even my textbook prioritizes $(\ce{-COOH})$ over $(\ce{-SO3H})$.
Or is the strength not the only factor while assigning priority?
Correct me if I'm wrong or missing something. I'm currently learning about IUPAC nomenclature.

Acidity has no relation to nomenclature.

Then why does -cooh get priority over -so3h?

Because IUPAC published a book that says so. There is not much more to it than that...

Did they do it just as such without any justification?

Yes, pretty much.

see Is the precedence of functional groups used in IUPAC naming arbitrary?

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text:
Reason for shorter bond between sulphur and oxygen?
I recently came across this sentence in my textbook:

the bonds between sulphur and oxygen in oxides of sulphur ($\ce{SO2}$ and $\ce{SO3}$) are much shorter than might be expected for a single bond.

I feel it could be due to partial double bond character due to resonance.
And the same textbook gives another explanation:

In these molecules, in addition to normal π bond, a π bond is also
  formed by the sidewise overlap of a filled 2p orbital of oxygen with a
  vacant 3d orbital on the sulphur). This is called pπ - dπ bond and
  results in bringing the two atoms closer and thus accounts for shorter
  bond length of $\ce{S-O}$ bond.

The reason provided in the textbook (as mentioned above) is definitely untrue and incorrect
Because of reasons mentioned in these posts
Why is the bond order in the SO₃ molecule 1.33 and not 2?
.
Hybridization of sulfur in sulfur dioxide
Please provide a explaination for the same (the title question).

Well, maybe this is because they are not single bonds?

Please cite the sources for the quotes and images you use, even (or especially) it they are from of own network.

"Which is definitely untrue and incorrect" Why do you think so?

@Karl  “That is thoroughly incorrect. chemistry.stackexchange.com/questions/51168/… chemistry.stackexchange.com/questions/29101/… – Mithoron”

Please explain this (in one or two scentences) in your question. It's not obvious.

Your resonance formulas show double bonds, and single bonds with a charge separation.

If you would consider it more carefully, you would see that the bond order in SO2 and SO3 are, respectively, 1.5 and 4/3. This means that they are shorter than a single bond of sulphur and oxygen.

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text:
Are all alkene geometrical isomers achiral?
I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.

You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.

The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?

@Karl Sadly, that statement isn't true. See my answer below...

@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.

Such questions always remind me of this one, no offence intended: https://xkcd.com/169/

@Karl I forgot about that one :)

While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.


text:
Oxidative Addition - Really concerted?
I guess you are all familiar with the oxidative addition or reductive elimination reaction in organometallic chemistry. If you read further into the topic you may find different pathways like via radicals, nucleophilic substitution, ionic paths or even the famous concerted mechanism.
In general we get a σ-type donation from the, let's take dihydrogen as the approaching ligand, hydrogen's occupied bonding orbital into an empty metal-centered orbital. Some sources say it's a p-orbital but I think the energy gap would be a bit too large so the $\ce{d_{z²}}$-orbital seems a bit more likely to me. On the other hand we get a back-donation from the occupied metal d-orbitals into the σ* of the H-H molecule.

Now while this may be likely for something as symmetric as hydrogen I still have my doubts about this. Firstly the question remains for all reactions if something is really concerted or if two electrons are ever donated at once or if there is a very fast cascade of two single-electron transfers. And related to this, can this actually happen the way it is graphically shown in literature or is the orbital overlap simply too poor for a good transition? You may also ask this for the Dewar-Chatt-Duncanson model or any other insertion reaction. As the metal d-orbitals approach from an angle the overall overlap should be lower than in a head-on σ-type bond. And I think there are already papers, where for the alkene case a pre-coordination via a hydrogen is proposed:

I remember that my professor used to mention a so called 'slip-mechanism' but I cannot find anything related to that. If I remember correctly it was just about this topic. As the overlap is quite poor it seems more likely that the addition process is carried out in two fast single-electron transfer steps where you have a 'slipping' towards one side followed by a slipping towards the other side.
I tried to draw it in chemdraw. This is of course too much but it shows what I mean by 'slipping'. Imagine there is still a weak connection between the orbitals even though it slips to the other side much like in π-bonds (pink dashed lines) but at the moment where the electron is actually transfered it slightly slips and twists to increase the orbital overlap

Has anyone heared about something like that before? That the mechanism may not be a 100% concerted but actually two very fast single electron transfers?

A mechanistic question like this immediately begs two questions: (1) what different outcomes would you expect and (2) can you design an experiment to check for the difference?

(1) it would not change the outcome probably, (2) for ultra-fast reactions like these you'd have to trap the intermediate species (if you don't trust in computational calculations). Doesn't seem very likely to me that you could do that. The question is more related to the fact that we take these simple views of having everything aligned in a square or hexagon (Grignard) and accept that it will happen in a single step so easily. The problem with catalysis is that you need to know about every single step and it's perhaps even microscopic reversibility.

That is at least how I was taught the organic (not the inorganic 'black box' view) of organometallics in catalysis. And the question is if that mattered like for alkynes or alkenes where the angle of the antibonding orbitals is much different it could perhaps have an effect on the actual choices for the reaction conditions or the catalyst itself. To be honest I don't know if there would be any consequence to it but I just imagine that it doesn't remain stiff like it's often depicted but may also be quite dynamic.

So it sounds like you're asking about more than whether this reaction is concerted; you're asking if this addition is symmetric. For example, in inverse-electron-demand Diels-Alder reactions, the reactions are typically concerted but highly asymmetrical in terms of the progression of bond formation. That may be the case here depending on the alkene or alkyne of interest. Again, whether or not that has a discernible impact on the outcome determines how much we should care. Interesting question nevertheless.

That would probably summarize it quite well.


text:
Complex organic molecules
I am studying astronomy and came across the following term in the astrochemistry course called 'complex organic molecules' or also written as COMs. My question is: What is exactly meant with these molecules? Is it just a molecule with more than one carbon atom?

This Wikipedia may be helpful.

tl;dr: two different definitions. Astronomy: multiple carbon atoms in molecule. Chemistry: polymer

Interestingly enough, after reading about COMs here, as well as reading the Wikipedia page and the corresponding arXiv paper, it seems like chemists and astronomers have different definitions of what a complex organic molecule should be!
As far as I knew, in chemistry complex organic molecules were long polymers, such as proteins, which were composed of thousands upon thousands of amino acid units. In the astronomy paper, however, they cite other types of molecules.
$\ce{CH3OH,
CH3CHO, HCOOCH3 and CH3OCH3}$, all cited as "complex" (haha) organic molecules in the paper, would appear to chemists as relatively simple molecules. (I read the paper, because it piqued my interest that something like a protein could be found in space). I then read the Springer article.

The term “complex organic molecules” is used differently in astronomy and chemistry. In astronomy, complex organic molecules are molecules with multiple carbon atoms such as benzene and acetic acid. These molecules have been detected in interstellar space with radio telescopes. In chemistry, “complex organic molecules” refer to polymer-like molecules such as  proteins.

I think the term is highly context dependent. In space molecules are pretty complex in that environment. Some chemists would argue that polymers are complex (others would disagree as they are big but repetitive so don't contain much complexity) Others (like me) would argue that molecules like taxol are complex because it has 4-fused different-sized rings in its core and 10 stereo centres and I don't have enough atoms in my modelling kit to come close to building it.

Don't even think about asking astronomers to define "metals" if this upsets you!

Agreed, astronomers are sloppy in their language... Also the physics and maths is often being treated in a similar way.

@Dani I can't possibly comment. I work with too many of them. Besides I'm a physicist and you should hear how we are with chemistry

I note that you give the example of $\ce{CH3OH}$, but your quotation specifies "multiple carbon atoms". I wonder if this means that there's disagreement about the definition used in astronomy?

@ruakh hmm, the thing is that was directly quoted from ArXiv

Here is a prime example of how much a "complex" is a simple organic molecule for Astrophysicist when the data are coming from interstellar space, which has made a Science article (Ref. 1):

Abstract: The largest noncyclic molecules detected in the interstellar medium (ISM) are organic with a straight-chain carbon backbone. We report an interstellar detection of a branched alkyl molecule, iso-propyl cyanide $(\ce{i-C3H7CN})$, with an abundance 0.4 times that of its straight-chain structural isomer. This detection suggests that branched carbon-chain molecules may be generally abundant in the ISM. Our astrochemical model indicates that both isomers are produced within or upon dust grain ice mantles through the addition of molecular radicals, albeit via differing reaction pathways. The production of iso-propyl cyanide appears to require the addition of a functional group to a nonterminal carbon in the chain. Its detection therefore bodes well for the presence in the ISM of amino acids, for which such side-chain structure is a key characteristic.

In September, 2014, a science reporter, Michael Eyre wrote an article about this discovery in BBC News under the title of "Complex organic molecule found in interstellar space," stating that iso-propyl cyanide has been detected in a star-forming cloud 27,000 light-years from Earth. This star-forming cloud or the star-forming region where the current observations made is Sagittarius B2(N) or Sgr B2(N). According to the authors of Science article, Sgr B2 is the most massive star-forming region in our Galaxy, which is located close to the Galactic Center, and contains two main sites of star formation, Sgr B2(N) and Sgr B2(M). In particular, Sgr B2(N) has been nicknamed Large Molecule Heimat because it is the source of many "complex organic molecules," most of which are found for the first time in space (Ref. 2).
Using the 30-m single-dish radio telescope of the Institut de Radioastronomie Millimétrique (IRAM) in France, the authors have previously discovered the straight-chain organic molecules, ethyl formate $(\ce{C2H5OCHO})$ and normal-propyl cyanide $(\ce{n-C3H7CN})$ in Sgr B2(N) (Ref. 3). To date, more than 165 molecules have been discovered in the interstellar medium or in circumstellar envelopes over the past five decades and approximately 2-4 "new" interstellar or circumstellar molecules are found each year. Among them, “complex” organic molecules with up to 13 atoms have been found, showing that the interstellar chemistry in some regions is efficient enough to achieve a relatively high degree of chemical complexity (For the authors credit, they have clearly stated that these molecules are “complex” for astronomers, not for biologists!). Yet, the discovery of ethyl formate (3 carbon atoms) and n-propyl cyanide (4 carbon atoms) sited in Ref. 3 is claimed to be the first clear detection of these molecules in space, and $\ce{n-C3H7CN}$ be the largest molecule yet detected in this source at the time. It is also noteworthy that the authors detected neither ethyl formate nor n-propyl cyanide toward the more evolved source, Sgr B2(M).
Even though, ammonia $(\ce{NH3})$ and water vapors $(\ce{H2O})$ have been discovered in space respectively in 1968 and 1969, the first "complex organic species" from space is believed to be formaldehyde, which was discovered and reported in 1969 (according to Ref. 2). Since then, a real Gold Rush broke out for "complex organic species," and approximately 165 molecules detected (by 2012) in the interstellar medium, 20 of which are cyanides (Ref. 4).

References:

A. Belloche, R. T. Garrod, H. S. P. Müller, K. M. Menten, "Detection
of a branched alkyl molecule in the interstellar medium:
iso-propyl cyanide," Science 2014, 345(6204),  1584-1587 (DOI: 10.1126/science.1256678).
K. M. Menten, “Perspective from a Younger Generation — The
Astro-spectroscopy of Gisbert Winnewisser,” In The Dense
Interstellar Medium in Galaxies: Proceedings of the 4th
Cologne-Bonn-Zermatt Symposium (Zermatt, 22–26 September, 2003); S.
Pfalzner, C. Kramer, C. Straubmeier, A. Heithausen (Eds.); Springer
Proc. Physics: Vol. 91, Springer-Verlag: Berlin, Germany, pp. 69-82,
2004.
A. Belloche, R. T. Garrod, H. S. P. Müller, K. M. Menten, C. Comito,
P. Schilke, "Increased complexity in interstellar chemistry:
Detection and chemical modeling of ethyl formate and n-propyl
cyanide in Sagittarius B2(N)," Astronomy & Astrophysics 2009,
499, 215-232 (DOI:10.1051/0004-6361/200811550).
M. H. Ordu, H. S. P. Müller, A. Walters, M. Nuñez, F. Lewen, A.
Belloche, K. M. Menten, S. Schlemmer, "The quest for complex
molecules in space: laboratory spectroscopy of n-butyl cyanide,
$\ce{n-C4H9CN}$, in the millimeter wave region and its astronomical
search in Sagittarius B2(N)," Astronomy & Astrophysics 2012,
541, A121 (8 pages) (DOI: 10.1051/0004-6361/201118738).

"Complex" is a relative term and I would say the situation in Astronomy is more complicated than that alluded to by JavaScriptCoder, who states that methanol (amongst others) is considered complex in Astronomy.
A distinction can be made between the Interstellar Medium, where it is indeed reasonably remarkable to find a substance as complex as methanol, a Nebula (which contains a greater concentration of matter but is still by Earth standards a hard vacuum) and a Condensed Body such as a planet, moon or comet, where it is not particularly remarkable to find substances with a molecular weight several times that of methanol.
The definition of "complex" on lifeless condensed bodies seems to be at a molecular weight a little over 100. Even more complex organic molecules called tholins are responsible for the colour of some planets. The first image in the article https://en.wikipedia.org/wiki/Tholin lists "complex organics" as having a molecular weight of 100-350 (Daltons), but then lists "negative organic ions" (and presumably tholins) as having higher molecular weights still. This article makes a distinction between "simple" organic molecules with molecular weights below 50 and "complex" organic molecules with molecular weights above 200.
Astronomers are also interested in the function of "complex molecules" in the origin of life. There is a lot of interest in the presence of polycyclic aromatic hydrocarbons (which have been detected in nebulae and condensed bodies) and their potential role in mediating the synthesis of RNA / DNA type substances https://en.wikipedia.org/wiki/PAH_world_hypothesis
People who study planets are called planetary scientists. Perhaps they are not astronomers, or perhaps they are a type of astronomer. What is clear, though, is that "complex" means "at least fairly complex for that particular environment or context" so there is no fixed lower limit for what is considered complex. It is also clear that what astronomers consider "complex" is nowhere near as complex as what chemists consider "complex."


text:
What is meant by forward reaction?
While reading my chemistry textbook I came across the phrase:

forward reaction is favoured by decrease in temperature in exothermic reactions.

I was tripped off by the term 'forward reaction', can someone explain in layman terms what is meant by this?

The forward reaction is the reaction as written, towards products, in the direction of the reaction arrow. Naturally, the reverse or backwards reaction is in the opposite direction.

This phrase comes from Le Châtelier's Principle. Given a reversible reversible reaction, you may (arbitrarily) choose which chemicals to consider the reactants/products, since the reaction goes from both the reactants to the products and from the products to the reactants. Simply put, the forward reaction starts with what you consider to be the reactants. The reverse reaction starts with the products and ends with the reactants. If you flip the equation these names change.

So in all, if the word 'reaction' was used instead of 'forward reaction' it wouldn't have made much a difference?

Not all reactions are reversible. That is key. But essentially, yes.

So the reaction favoured ahead in case of a reversible reaction is forward reaction?

In the reaction $\ce{A <=> B}$, if B is favored, the forward reaction is favored.

It's arbitrary and may be a cultural misunderstanding. Most English speaking cultures regard left to right on a page as moving forward and reverse is going from right to left. Thus for a reversible reaction such as producing ammonia from hydrogen and nitrogen:
$$\text{(reactants)}\qquad\ce{H2 + N2 <=> 2NH3}\qquad\text{(product)}\tag{reversible reaction}$$
the forward reaction is forming product(s) on the right from reactant(s) on the left
$$\text{(reactants)}\qquad\ce{H2 + N2 \color{blue}{->} 2NH3}\qquad\text{(product)}\tag{forward reaction}$$
where as the reverse reaction is forming reactant(s) from product(s)
$$\text{(reactants)}\qquad\ce{H2 + N2 \color{red}{<-} 2NH3}\qquad\text{(product)}\tag{reverse reaction}$$

If I were to produce hydrogen and nitrogen from ammonia, the reaction would look more like:
$$\text{(reactant)}\qquad\ce{2NH3 <=> H2 + N2}\qquad\text{(product)}\tag{reversible reaction}$$
the the forward reaction is:
$$\text{(reactants)}\qquad\ce{2NH3 \color{blue}{->} H2 + N2 }\qquad\text{(product)}\tag{forward reaction}$$
and the reverse reaction is:
$$\text{(reactants)}\qquad\ce{NH3 \color{red}{<-} H2 + N2}\qquad\text{(product)}\tag{reverse reaction}$$
The reactions are the same but because the order of the reversible reaction flipped, which reaction is considered the forward and and reverse has also flipped

On my screen, the annotations on the right side of each equation run over into the links in the side bar...


text:
Does a C=C double bond donate electron or withdraw it?
I have learned that a $\ce{C=C}$ double bond is electron-rich because of the p-p overlap and hence can easily donate an electron pair: I agree with this statement.
However, in a $\ce{C=C}$ double bond, the carbon is $\ce{sp^2}$ hybridized and I've also learned that more '$\ce{s}$-type' character increases electronegativity. Does this mean that the carbon atom - being more electronegative - tries to withdraw electrons?
For example consider the double bond connected to a positively charged Carbon, then will it donate an electron and stabilize it? or will it (being more electronegative) withdraw an electron?

As you advance in chemistry, you tend to find that different factors often produce competing effects with different impacts in different situations, producing the complexity we observe in chemistry.

You need to specifically give us the context. Your question as of now is too general and unspecific. Where is the position of the double bond? Is it adjacent to a positively charged carbon (i.e. allylic to it)? Or is it adjacent to a negatively charged carbon?

oh ok. I will edit my question then

An allylic double bond can be said to be both electron-donating and electron-withdrawing, depending on the context. The simpler way to explain this is based on the idea of resonance. Both the allyl cation and allyl anion are stabilised by the presence of this double bond.

Source

Source
The behaviour of allylic double bonds can also be rationalised based on molecular orbital (MO) theory. A molecule containing a $\pi$ bond has both an antibonding $\pi^*$ MO and a bonding $\pi$ MO. Due to the lower strength of a $\pi$ interaction, compared to $\sigma$ interactions, the $\pi$ MO is not too lowered in energy and the $\pi^*$ MO is not too raised in energy, relative to the energy level of the $\ce {C}$ p orbitals. The consequence of this is that, to attain greater stabilisation of the molecule, this filled $\pi$ MO can interact with empty orbitals at adjacent sites in the molecule and this empty $\pi^*$ MO can interact with filled orbitals at adjacent sites. The allyl cation and allyl anion are just two examples to illustrate the above concept.
Other systems illustrating these effects would be the conjugated enone, as well as allyl halides. For the case of the enone, the effect seems to be more of an electron donation as the $\ce {C=C}$ $\pi$ bonding MO interacts with the $\ce {C=O}$ $\pi^*$ MO. For the case of tha allyl halide, it also seems to be that there is electron donation of the allylic double bond to the partially positively-charged $\ce {C}$ bonded to the halogen atom as the $\ce {C=C}$ $\pi$ bonding MO interacts with the $\ce {C-X}$ $\sigma^*$ MO.
From these examples, we can see that the electronic effects of the allylic double bond is highly dependent on the system we are considering. We can also seem to qualitatively derive the rule of thumb that the allylic double bond donates or withdraws based on the "needs" of the adjacent site. If it is electron-deficient, there would be donation from the double bond. If there is a site of high electron density, there would be withdrawal towards the double bond.

I guess it depends on the way you quantify "electron donation/withdrawal" (because it is a useful concept, but not really a measurable observable). I'm sure someone will find a counterexample, but if you consider methane and ethylene as Brønsted acids, you find that methane is a much weaker acid. Thus, ethylene is better at stabilizing the negative charge. (This is in line with your sp2 argument.)

True. But then again I would argue that in lots of compounds having C=C double bond, the pi electron cloud donates some negative charge in order to stabilize it. And that goes against my sp2 argument.


text:
What is the driving force for Claisen rearrangements?

I came across this mechanism (3,3 sigmatropic rearrangement). Why does the allyl vinyl ether have the ability to rearrange itself giving a new product? What drives the reaction?
Is it the instability of allyl vinyl ether (if yes, state why)?
Thank you in advance!

Try adding up typical bond strengths. (Most of the bonds are the same, which saves some time.)

The formation of the carbonyl group is a gain in thermodynamic stability over the vinyl ether. In your example, the monosubstituted alkene is a "wash". In any case, the alkyl substitution pattern of the non-enol ether double bond is not a problem.
The heat of this Claisen rearrangement, first conducted by Hurd and Pollack1 in 1938, can be calculated by determining the heat of formation (ΔHfo) of allyl vinyl ether (1) and 5-hexenal (2). Heats of formation of allyl vinyl ether (1), hexanal (3), 1-hexene (4) and hexane (5) are known (NIST and Chemeo). These heats of formation are shown in black. However, the ΔHfo of 5-hexenal (2) is not accessible in tables but it can be calculated. The heat of hydrogenation of monoalkyl substituted alkenes is ~125 kJ/mol (~30 kcal/mol). The reduction of 1-hexene (4) to hexane (5) is -124.6 kJ/mol, which is computed from the respective heats of formation of 4 and 5. Applying this heat of hydrogenation to the reaction 2 ---> 3 gives a ΔHfo of -128.2 kJ/mol for 5-hexenal (2). Accordingly, the Claisen rearrangement of allyl vinyl ether (1) is exothermic by -100.6 kJ/mol.

1. C. D. Hurd and M. A. Pollack, J. Am. Chem. Soc., 1938, 60, 1905.


text:
Acid value of linseed oil
For this we weigh $\pu{1.015 g}$ of linseed oil which is dissolved in alcohol. 3 drops of phenolphthalein (pH indicator) are added and a solution of potassium hydroxide of concentration $C = \pu{9.5e-3 mol L-1}$ is poured. The indicator turns to $V = \pu{7.4 mL}$
I calculate like that:
$$n_\mathrm{acid} = n_\mathrm{OH} = C_\ce{OH}\cdot V_\ce{OH} = 9.5\cdot 10^{-3} \cdot 7.4\cdot 10^{-3} = \pu{1.444e-4 mol}$$
$$\pu{1.444e-4 mol} → \pu{1.015 g}$$
$$\pu{1.423e-4 mol} ← \pu{1 g}$$
$$\implies I_a = m_\ce{KOH} = 1.423\cdot 10^{-4} \cdot 56 = \pu{7.967 mg}$$
But the teacher responds with $I_a = 3.8$. How to solve?

Hi vodanh, welcome to Chem.SE! We hope you enjoy your stay, please visit the [help] and take the [tour] if you haven't already. We have MathJax and mhchem to help you prettify your formulae! I have added an edit that is currently awaiting peer review.

Unfortunately, the proportion you are using is not correct. Acid value $\mathrm{AN}$ determination procedure is not much different from an ordinary titration with indicator with the only difference that the mass of titrant is normalized to the mass of oil $m_\mathrm{oil}$:
$$\mathrm{AN} = \frac{T\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}$$
where $ΔV_\mathrm{eq}$ is equivalent volume (ideally you have to use a difference with an external standard ("spiking solution"), but you probably didn't do that); $T$ is a so-called titer, or basically mass concentration of the titrant:
$$T = CM$$
so that
$$
\begin{align}
\mathrm{AN} &= \frac{C(\ce{KOH})\cdot M(\ce{KOH})\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}\\
  &= \frac{\pu{9.5e-3 mol L-1}\cdot\pu{56.106 g mol-1}\cdot\pu{7.4e-3 L}}{\pu{1.015 g}}\\
  &= \pu{3.89 mg g-1}
\end{align}
$$
Also, a couple of side notes:

Always carry units across all steps and never omit them.
I recommend not to do intermediate calculations if you don't have to. Instead, first find an algebraic equation and only then plug in all the numbers.
Mind significant figures. If the mass of oil is given with 4 significant figures, it's a good idea to use a molecular mass of the titrant with at least 4 sigfigs too.


text:
Potential energy curve for dissociation of cation of carbon dioxide
Can anyone suggest me some literature where I can find the potential energy curve or the electronic states for singly ionized $\ce{CO2}$ ($\ce{CO2+}$) dissociating into singly charged oxygen ($\ce{O+}$) and carbon monoxide ($\ce{CO}$).

Check out my edit, you can get more info: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here?r=SearchResults&s=24|10.4115 https://chemistry.meta.stackexchange.com/questions/261/is-mathjax-in-titles-a-problem-and-why?r=SearchResults&s=21|12.8023

Few of the literatures are here:
http://iopscience.iop.org/article/10.1088/0953-4075/47/12/124025/meta
https://pubs.acs.org/doi/pdf/10.1021/jp907351s
https://aip.scitation.org/doi/10.1063/1.444413


text:
What would happen if you burned paper with a laser, in a vacuum?
If you hit a piece of paper with a powerful laser inside a vacuum, what would happen? It sounds like it should not "burn" and have a flame spread over the paper, as there is, of course, no oxygen.
Would it perhaps turn the paper into a thin sheet of charcoal where the laser makes contact? If so, perhaps one could make a cnc-type machine within a vacuum chamber that automatically cuts out stencils.

One doesn't need vacuum to cut paper with laser. It's been about a decade since affordable laser paper cutting machines appeared on the market, I remember there was a little spike of interest in that area, but then 3D printing took the mind of the masses.

https://chemistry.stackexchange.com/questions/49827/gases-produced-by-pyrolysis-of-cellulose


text:
Pale blue species forming during electrolysis of NaHCO3
I have a solution of 3500 ml $\ce{H2O}$ + 454 g $\ce{NaHCO3}$ in electrolysis at a potential of 12.5V using a lead anode and copper cathode. The surface areas aren't measured but they are enormous. I made the electrodes myself from lead and copper foil.
The idea was to make NaOH from $\ce{NaHCO3}$ (an inexpensive source  $\ce{NaHCO3}$ $$0.86/454g  and NaOH $7/500g) meanwhile producing the PbO2 layer on the lead anode.
Everything seemed to be going as expected until I see there's a pale blue species being produced. What could it be? I'm thinking H2O2, water having the potential to completely become peroxide, so I used the 5V output instead and the blue species vanished.
I've read electrolysis of salts in basic solution usually result in the salt reaching its highest state of oxidation at the anode.
Eg.:
$\ce{NaCl + H2O ->Cl- + OH- -> HClO + OH- ->ClO2- + H2}$
eventually becoming $\ce{ClO4-}$
(correct me if I'm wrong but I believe that's the way that reaction carries out),
or
$\ce{(NH4)2SO4 -> (NH4)2S2O8}$
EDIT
It's just copper. Apparently in a solution of NaOH, there is some oxidation of a Cu cathode. I've seen it occur many times afterward. Hydrogen peroxide cannot be made without special laboratory equipment.

Wasn't there a copper wire?

copper cathode so nope, and remember the blue color comes back when it's reacted with something else. I also reacted it with ammonia with no tetraamine formation.

Well, if there was copper then however you did it, it's much more probable then any peroxides or other highly oxidated species (which aren't colorful BTW).

Copper oxidation

The blue color was caused by copper being stripped from the wire. btw, when your hair turns orange, it is because it is being stripped of color, among other other things (chemical bonds), which will cause it to be brittle and also absorb any hair dye to the extreme that it feels waxy...I'd suggest you cut off a piece of hair for a test before dunking your whole head in bleach (chlorine will turn it greenish).
P.S. Use nitric acid to dissolve Cu.

I actually use copper chloride, The fumes from Nitric acid are pretty nasty. I made some nitric acid with Calcium Nitrate and sulfuric acid. You could imagine the fun I had drilling the plaster out of my flask. lol. I just realized how old this question is. This is before I made reagent grade H2SO4 from magnesium sulfate.  Time flies.


text:
How to make equal density pool dye without diffusion
I have a leak I have been struggling to find in my in-ground pool.  I am losing about 1/4" of water per day and I believe the issue is one of the fittings in the bottom of the pool.  There are two main drain fittings and one deep pot that has two pipe plugs that if removed allows groundwater to enter into the pool to prevent the forces of buoyancy from floating the shell out of the ground like a boat when the pool is drained.
I want to make a dye that is has a specific gravity equal to water that has minimal to no diffusion.  It has to be safe because it will be introduced to the pool water.
The only thing I have found that is almost the same gravity of freshwater is beer, or milk.  The problem with those is the diffusion rate.  I was hoping to find something that I could mix with dye that had a specific gravity (I hope that is right) as near to freshwater as possible that wouldn't diffuse when introduced to freshwater.
My intention, hope, is that I would place a blob of dye over these plugged fittings and if there was a leak around the fitting it would eventually suck that dye into it.  It could take 15 minutes or more for the dye to be pulled into the leak, so it has to have a very slow diffusion rate.

1/4 '' is not much, could it be due to evaporation?

I have performed a bucket test as well, and the water level decreases at a higher rate than the water in the bucket does.
I filled the pool to a known level (a pencil mark) this morning at 6am.  At 5pm, eleven hours later, the level in the pool has decreased by a quarter-inch.
I have also set a clear pitcher on one of the pool steps and filled it with water to match the water level of the pool to conduct another bucket test.

Another idea would to introduce something that has a slightly higher specific gravity (or density) of freshwater that would diffuse widely and cover the entire bottom of the pool.  My theory in this would be the dye would then be pulled into the leak and there would be a clear hole in the dye to indicate the approximate position of the leak.

In my kitchen I filled a clear glass bowl with tap water.  I then mixed red and blue dye into a small bowl of lactulose free whole milk to make a purple mixture.  I used a hypodermic needle to slowly introduce the mixture near the bottom of the bowl to observe the diffusion rate.  The mixture slowly diffused across the entire bottom of the bowl, but settled, leaving a thin purple sediment like layer across the bottom of the bowl.  So, the specific gravity/density of this mixture is slightly greater than water (desired), but the diffusion rate is too high (undesired).

If the DIY aspect is not crucial why not to go for a pool dye? It can be some fun especially if children are around. Lot of us here have ideas of what they are but to avoid any safety concern I would go for dyes sold as pool dyes. Saw they aren't expensive.

It really is more of the aspect of DIY, but the pool dye the plumbing company used had a pretty high diffusion rate.  It only lingered for about 30-45 seconds.  I needed to come up with something that wouldn't diffuse for five or more minutes, which the whole milk solution did that.

Whole milk with food coloring solved my problem.  Whole milk has a specific gravity of 1.01, which allowed it to settle to the bottom around the main drains.  The diffusion rate was extremely slow over regular pool dye, which allowed the dyed milk to linger around the drains for an extended period of time (5+ minutes) to allow for the detection of the slow leak.


text:
How to derive a state symmetry in D2h point group from its wavefunction?
Let's have an atomic carbon with the following electron configuration:
$$
1s^2 2s^2 2p^2
$$
One of it's levels is ${}^1S_0$, which is corresponding with the following state:
$$
\begin{align}
\left| S=0, M_S=0 \right> &= \frac{1}{\sqrt{2}}\Big( \alpha(1)\beta(2) - \alpha(2)\beta(1) \Big)\\
\left| L=0, M_L=0 \right> &= \frac{1}{\sqrt{3}}\Big( \left| p^+(1)p^-(2) \right> - \left| p^0(1)p^0(2) \right> + \left| p^-(1)p^+(2) \right> \Big)\\
\left| S=0, M_S=0, L=0, M_L=0 \right> &= \frac{1}{\sqrt{6}} \Big( \left| p^+(1)\alpha(1)p^-(2)\beta(2) \right>- \left| p^+(1)\beta(1)p^-(2)\alpha(2) \right> \\&- \left| p^0(1)\alpha(1)p^0(2)\beta(2) \right> + \left| p^0(1)\beta(1)p^0(2)\alpha(2) \right> \\&+ \left| p^-(1)\alpha(1)p^+(2)\beta(2) \right> - \left| p^-(1)\beta(1)p^+(2)\alpha(2) \right> \Big)
\end{align}
$$
Now I know, that ${}^1S_0$ state is supposed to be totally symmetric, i.e. $A_g$ in the $D_{2h}$ point group. But is it possible to derive it "rigorously" from the wavefunction?

Too late for an answer right now, but the idea is you need to perform all the symmetry operations of the D2h point group on the wavefunction. For example $\sigma_{xy}$ reflects points (x,y,z) into new points (x,y,-z), so if you replace all instances of z in the wf by -z, and you find that the wf remains unchanged, that corresponds to a character of +1. If the sign of the wf is flipped, that’s a -1. You can ignore the spin part because symmetry operations don’t act on them.

Only the spatial part is important here so why not use the symmetry of the spherical harmonics with the appropriate L and ml  values ? The right-hand most columns in most point group tables give these functions.

@porphyrin I understand, what you're saying, but I don't see the connection somehow... Should I "redraw" every microstate wavefunction as p-orbitals?

No I don't think so: if, for example, L = 0 there is no orbital angular momentum and the wavefunction must be totally symmetric, so $A_g$.


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

I read that -NHCOCH3 is ortho para direction. By making the resonance structure, option d is meta directing. I am confused.

Related: https://chemistry.stackexchange.com/questions/31531/friedel-crafts-reaction-of-phenol

AlCl3 reacting with an amide may be difficult compared to amine since lonepair in an amide is in conjugation with carbonyl type group as well as benzene.So FC reaction of benzamide maybe possible

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


text:
How To Scan "improper dihedrals" in Gaussian 09?
I am trying to parameterize Amber force field for a novel molecule, but I couldn't obtain improper dihedrals. Because if I scan impropers like dihedrals; later or sooner the molecule is overlapping and Gaussian exits with errors.
Let's say it is the $\ce{BH3}$ molecule. I tried many things for example with/without  symmetry=none keyword, with O 4 2 3 1 S 10 18.000000 or   D 4 2 3 1 S 10 18.0 (out of plane), but I couldn't get the results for a 360° improper scan.
I got this error at step 6:
 Old curvilinear step not converged, using linear step
 Error imposing constraints
 Error termination via Lnk1e in C:\G09W\l103.exe

What should I do?
Here is my gjf file:
%nprocshared=4
# opt=modredundant am1 geom=connectivity

Title Card Required

0 1
 B                  0.00000000    0.00000000    0.00000000
 H                  0.00000000    1.10873800    0.00000000
 H                  0.96019500   -0.55436900    0.00000000
 H                 -0.96019500   -0.55436900    0.00000000

 1 2 1.0 3 1.0 4 1.0
 2
 3
 4

D 4 2 3 1 S 10 18.000000

Please define what you understand as an improper dihedral angle, or out of plane angle (or what is the definition for them in the program you are calculating them for). I could not find a common definition for these, and it appears to me that they are 'special' for force fields. I have a very hard time understanding what you are actually trying to accomplish.

I want to add Dac's answer about $\ce{NH3}$ impropers here I think it will help you:

The very simple answer to your question, is that improper dihedrals are needed when they are necessary. (Not very helpful, huh :) ) Let me elaborate some. In my experience, there are times when improper dihedrals are not needed because proper dihedrals are sufficient for modeling the energetics and the conformations of the atoms in question. (Note that in Amber proper and improper dihedral angles are defined by the same functional form.) Other times the person who has done the parameterization deems it necessary to use impropers. So ultimately, to use or not use impropers often comes down to the individual tastes of the person who created the parameters. I personally feel that impropers are need in selected case to correctly model out-of-plane bending, and there are very few, if any, force fields that do not use them.
Some else who knows better can correct me if I am wrong, but I believe that parmchk reports provides any improper dihedrals if it recognizes that they are present in the molecule's topology. A value of 10.5 kcal/mol is reasonable for an improper dihedral. If you are uncertain, the best way to determine if you need the impropers, or not, is to run some test simulations with and without them - then choose the setup that gives the best results. (However, if the goal is to force the improper dihedrals to be 100% planar, then adding an additional force constant to do so is "probably" okay. What you will miss by doing this is the proper modeling of out-of-plane bending, but I am not sure how well Parm99 models this anyway.)

Yes, this is David's opinion from the amber forum.
Additionally, if you can not scan improper dihedrals using ab inito method, you can use genetic algorythm with sander.
To do this first use in your *.frcmod file:
I-J-K-L  1          180.0         0.0

and draw chart with CPPTRAJ, then change energy and repeat it until  you get a reasonable result. Actually you can find for this some software, too.
Or you can use ATB server for gromacs. If your molecule doesn't have metal ion, upload pdb format to ATB server and then it will give you force all force constants with improper dihedrals. If you use it in amber, just change the energy from joule to kcal ($\pu{1 kcal} = \pu{4.18 J}$) and multiply it by 0.5 and then use it in your *.frcmod file.

After I asked this question, I found answer. I am putting the solution here so maybe somebody else use this.
The solution is this: if gaussian exit from the job (for example) at step 6; change starting geometry and take it 6 step back. How? For example in my question every step size is 18 degree and step 6 is 90 degree. Open your *.gjf or *.com file select the same dihedral and decrease angle 90 by degree (don't forget 90 degree is just for this example, you wil change it according to your step and the size of step) *sometimes you may have to fix one group . And then start job again.
you can see my "wrong gjf file" in the question. And here is the my "right gjf file": I only changed the starting dihedral angles.
%nprocshared=4
# opt=modredundant am1 geom=connectivity

Title Card Required

0 1
 B                 -0.54526231    0.17208210    0.44752404
 H                  0.54967689    0.02475247    0.03303383
 H                 -1.22032311    1.04666245    0.03303383
 H                 -0.96514071   -0.55516861   -1.21043678

 1 2 1.0 3 1.0 4 1.0
 2
 3
 4

D 1 2 3 4 S 9 18.000000

NOTE: it seems I defined different dihedrals at first glance.
in question I defined:
D 4 2 3 1 S 10 18.000000 and here I defined:
D 1 2 3 4 S 9 18.000000 but out of the symmetry and all H bonds are equal two dihedrals are the same indeed. Here is the article about this:  https://pdfs.semanticscholar.org/ab92/5f43824e59dd9e3be3474ebbac57a64998c3.pdf (DOI 10.1002/jcc.21425  - "Symmetrization of the AMBER and CHARMM
Force Fields", 2009, EDYTA M. et all)
in this article in figure 5 shows only 3 different improper can exist for  I–J–K–L but in my question I=L=J and thus actually 1234 dihedral and 2314 dihedral are the same and gave the same results for the same technics.


text:
How do I make a 0.5 M citrate buffer at pH 4.5?
I need to make a citrate-sodium citrate buffer at 0.5 M concentration and pH 4.5. I calculated the molarities of the citrate and disodium citrate based on the Henderson Hasselbalch equation, using a pₐ of 4.76 . To cross-check my calculations, I tried a couple of online calculators, but the results are way off from mine. I think the formulae should change based on where I use monosodium, disodium or trisodum citrate. What is the right way to go about making this calculation?

The relevant pKa for the trisodium citrate/disodium hydrogen citrate pair is 6.40.

The best would be either adjust the pH of citric acid by NaOH with pH meter and note the ratio, either to use published data for preparation of citrate buffer.

There are many ways to make simple pH buffers like this one. As @Poutnik noted, preparation via titration works fine and in very commonly used. To illustrate how this is done, several possibilities are presented. Assume the desired buffer characteristics are pH = 4.5, concentration = 0.5 M, final buffer volume = 1 L, and the following reagent grade chemicals are available: citric acid, sodium dihydrogen citrate, disodium hydrogen citrate and 2 M NaOH solution. Also assume a laboratory quality pH meter is available and that the pH meter has been properly calibrated with buffers of 4.00 and 7.00, bracketing the desired pH of 4.5.  Before proceeding, it is useful to examine citric acid’s $\alpha$ dissociation fractions.
Brief refresher on $\alpha$ dissociation fractions.
Citric acid, denoted $\ce{H_3Cit}$, is a triprotic weak acid. The successively less protonated citrate species are $\ce{H_2Cit^-}$, $\ce{HCit^{2-}}$, and $\ce{Cit^{3-}}$. Since all the citrate species are in the same volume of solution, molar concentrations (denoted by the usual square brackets, e.g., $[\ce{H_3Cit}]$) are proportional to numbers of moles. Define $D$ as the sum of $[\ce{H_3Cit}] + [\ce{H_2Cit^-}] + [\ce{HCit^{2-}}] + [\ce{Cit^{3-}}]$. Then $\alpha_0$ is $[\ce{H_3Cit}]/D$, $\alpha_1$ is $[\ce{H_2Cit^-}]/D$, $\alpha_2$ is $[\ce{HCit^{2-}}]/D$, and $\alpha_3$ is $[\ce{Cit^{3-}}]/D$. The sum of the $\alpha$ values equals unity, as expected: every citrate species must be somewhere in the solution.
At any given solution pH, all four citrate species are present (though possibly at negligible concentration), as determined by the pH, the three acid dissociation equilibria and their associated equilibrium constants: $\mathrm{p}K_\mathrm{a1}$, $\mathrm{p}K_\mathrm{a2}$, and $\mathrm{p}K_\mathrm{a3}$. For citric acid, the $\mathrm{p}K_\mathrm{a}$ values are 2.13, 4.77, and 6.40. At any given solution pH, the $\alpha$ fractions are simply the proportion of total citrate that is in the relevant ionic or molecular species.
The screenshot below shows the $\alpha$ fractions up to pH = 8 (it is boring above pH = 8):

For a desired citrate buffer at pH = 4.5, the figure shows that only $\ce{H_2Cit^-}$ (green $\alpha_1$ curve) and $\ce{HCit^{2-}}$ (blue $\alpha_2$ curve) are present in significant concentrations. Furthermore, $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] > 1$.
Method 1
Weigh out an amount of citric acid equal to approximately 0.5 moles, i.e., about 91 g. Dissolve that, with stirring, in about 0.5 L of water in a large beaker. Be careful to not drop a magnetic stir bar into the beaker!  If 375 mL of 2 M NaOH solution were to be added, the pH would be about 4.76, which is a little too high. The reason is that 0.75 moles of $\ce{OH^-}$ would neutralize the first citric acid proton and lead to a roughly equimolar ratio between $\ce{H_2Cit^-}$ and $\ce{HCit^{2-}}$. This is too much added NaOH solution because $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 1.82$, as is easily determined using the Henderson-Hasselbalch equation or the $\mathrm{p}K_\mathrm{a2}$ expression.
Add 250 mL of 2 M NaOH and then, with constant stirring and monitoring of the pH, slowly add more 2 M NaOH solution until the pH increases to 4.5. Finally, the last step is to add water until the final buffer volume is 1 L.
Method 2
Start by assuming $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 2$, so the pH will be a little below the target value of 4.5, and weigh out about 71 g of $\ce{NaH_2Cit}$, i.e., about 1/3 mole. Likewise, weigh out 39 g of $\ce{Na_{2}HCit}$, i.e., about 1/6 mole. The total number of moles of citrate species is about 0.5. Dissolve the $\ce{NaH_2Cit} $ and $\ce{Na_{2}HCit} $, with stirring, in about 0.8 L of water in a large beaker. Then, with constant stirring and monitoring of the pH, slowly add 2 M NaOH solution until the pH increases to 4.5. The last step is to add water until the final buffer volume is 1 L.
What happens if a mistake is made and the pH is too high? Simply add a little more of the most acidic species, making sure it dissolves, and then titrate back up with NaOH solution. It can be helpful to have a lower concentration NaOH solution available, e.g., 0.1 M, either for this corrective procedure or for tweaking the pH.
Other possibilities include 1) citric acid plus trisodium citrate or 2) citric acid plus disodium hydrogen citrate or 3) sodium dihydrogen citrate plus trisodium citrate, but these are not considered further.

The problem is that 0.5M buffer will have pH strongly affected by activity coefficients.

No, it has its worth. Keep it.


text:
Why does heating glucose in the excess of HI produce n-hexane?
Alcohols react with acid halides such as $\ce{HI}$ to create alkyl halides, as such:
$$\ce{C2H5OH + HX → C2H5X}$$
However, this doesn't seem to be the case with glucose, as the entire chain is reduced to an alkane instead. Is this due to the steric hindrance factor of the large chain or perhaps the aldehydic group attached to it?
An explanation would be appreciated.

Iodine is somewhat special among halogens, in that it falls off easily.

@IvanNeretin Even though Iodine is a great leaving group, can it be replaced with a hydrogen?

Leaving is irrelevant; falling off is another story.


text:
In dS≥dQ/T, material equilibrium means that there is not material changes at all or that a material change can be reversible and at equilibrium?
I am reading Levine's Physical Chemistry, I know how to arrive to that equation, but I have questions.
First, $\mathrm{d}S=\mathrm{d}Q/T$ for material equilibrium, but if there is a heat flow I understand that there is a reversible process going on, what kind of process can it be? Only a reversible thermal or mechanical process or also a reversible material change?
Second, the book says
$$\mathrm{d}S > \frac{\mathrm{d}Q_\mathrm{irrev}}{T} \qquad\text{and}\qquad \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T}.$$
Then states the equation $\mathrm{d}S ≥ \mathrm{d}Q/T$ and says $\mathrm{d}S$ equals $\mathrm{d}Q/T$ for a reversible process and $\mathrm{d}S$ is greater than $\mathrm{d}Q/T$ for an irreversible chemical reaction or phase change.
I imagine that the heat flow is different if the process is reversible or irreversible and that the equation $\mathrm{d}S≥\mathrm{d}Q/T$ takes into account that the $\mathrm{d}Q$ is different, is this correct? I think I am a little confused just because there are no more subscripts in $\mathrm{d}S≥\mathrm{d}Q/T$.
Also, I think that the heat flow due to a irreversible process is greater than the heat flow due to a reversible process that goes to the same states, why is that?

Here is a tutorial to help you understand the change in entropy for a system experiencing an irreversible process and application of the Clausius inequality:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

In the inequality, the T is not the system temperature (which can vary from location to location within the system for an irreversible process).  The T here is the temperature at the interface between the system and the surroundings where the heat transfer is occurring.


text:
What are Technetium's physical stress characteristics?
I've looked at periodic and data tables from several sources, but none give mechanical / engineering property data on Technetium.  Why is this?

Tc is unstable ans is certainly not a constructing material, but it doesn't mean there is no data for it. What mechanical properties are you interested in, exactly?

Elastic moduli of polycrystal Technetium: https://www.sciencedirect.com/science/article/pii/0022508870900391


text:
How Can I Induce Corrosion Electrochemically?
I'm very interested in measuring corrosion in metal alloys, particularly in salt solutions. I understand metals corrode over time, however, I was wondering if it's possible to electrochemically "speed up" or induce further corrosion in saline solutions? How would I go about this electrochemically? I've considered applying a specific current / potential and utilising it to induce oxidation, however, I have not found much in the way of literature actively seeking to induce / speed up corrosion, most are content to investigate over longer time periods.
Many thanks.

Corrosion is a more or less involuntary/undesired reaction. If you apply a current, the corrosion is defined by the current you apply, so you are controlling the system, not just observing it. In general, electrochemical forcing puts your system into a highly unnatural regime, but can certainly provide interesting information.
Corrosion can be affected (in ways that you can determine and document) by temperature, by contact with other metals, by the corrosivity of the solution (concentration and specific ions), and by inhibitors.
Some time ago, we were interested in stress-corrosion cracking (failure) of stainless steels. Hydrogen, (slowly generated in a natural environment) would embrittle the steel. By increasing the corrosivity of the solution (chlorides) and adding an activator (sulfides), we were able to speed up the hydrogen evolution/absorption so that steel specimens under tensile stress would fail within days or weeks, rather than months or years in nature. This enabled choosing alloy compositions that were most resistant to stress-corrosion cracking.

Alloy composition has little to do  with hydrogen stress cracking ; it depends on strength level  and to some degree on stress and microstructure ( once hydrogen is in the matrix). Austenitic stainless ( non-magnetic ) , and ferritic stainless are more or less immune. Martensitic stainless  is susceptible  under the same conditions as carbon and low alloy steels. Sulfides cause H entry , not evolution, that is why the international standard ( MR 01-75) is based on the level of hydrogen sulfide.

Yes you can!
I would suggest one of two methods for you the Electrochemical one known as Electrolysis https://en.wikipedia.org/wiki/Electrolysis which is somewhat scalable to your need (Just make sure you don't fill your room with chlorine gas) or Galvanic Corrosion which is passive and safe but slower https://en.wikipedia.org/wiki/Galvanic_corrosion

How would you go about inducing electrolysis with aluminium? Setting copper as the cathode, immersing the aluminium in salt water and then applying a potential?

Electrolysis of aluminum is usually called "Anodizing" and is done in sulfuric acid; there must be whole books on the subject.

Since I can't comment I'll add this new response:

It needs to be DC
Rust will fall of the negative electrode and form at the positive IIRC
If the current is too high you'll form chlorine gas so you might want to use baking soda instead of salt. At the minimum do it outside.
If you want Aluminum oxide then make sure you only immerse aluminium on that side
Don't try to get Iron Oxide from stainless steel as that can create certain toxic heavy metals.

Read more than what I say, it's been quite some time since I did this and some information might be incorrect.

There is a field of chemistry/metallurgy that uses applied potential to measure corrosion response of metals in various solutions : lookup Pourbaix Diagram.  There are many books on the topic. ( My excuse for not posting this a year ago is that hydrogen stress cracking is my favorite subject and I got distracted.)


text:
Why does d/l-2,3-dibromobutane on reaction with NaI/acetone give cis-2-butene?
This is what I have read/know about the dehalogenation of vicinal dihalides by NaI/acetone:
The two halides which are leaving must be on opposite sides (from the mechanism).
In case they are present in the same side, we must rotate the C-C bond first and then proceed with the reaction.
In both cases we end by with a trans alkene.
But now I came across this:

meso-2,3-dibromobutane on reaction with NaI/acetone gives trans-2-butene.
d/l-2,3-dibromobutane on reaction with NaI/acetone gives
cis-2-butene.

I have no problem regarding the first statement. But regarding the second, from what I have learnt, the product again must be trans-2-butene.
I drew the Fisher projection and tried to figure out the reaction. For the 1st reaction, meso-2,3-dibromobutane must undergo C-C bond rotation, (performing two "swaps") and then we would end up with d/l-2,3-dibromobutane leading both 1st and 2nd reactions to trans-2-butene.
Where is my mistake? Is my understanding of the mechanism of the dehalogenation of NaI/acetone wrong? Any good sources to read regarding the same (I couldn't find it in any of my books) are appreciated, too.

Build a 3d model of the two compounds: it is far more obvious what is happening. Fisher projections don't help nearly as much as a 3D model.

I just noticed that you give your questions very, very non-descriptive titles. We would appreciate it very much if you could avoid 'buzzwords' like question, doubt, problem, ... , and give your question a titel that summarises the actual question.

Sure will do that in the future.

The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the formation of vicinal di-iodide, the compound now undergoes bond rotations if required and make the two iodine atoms orient themselves anti-periplanar for a E2 elimination.
So, essentially not only elimination, but substitution also occurs prior to elimination, and thus meso-2,3-dibromobutane never actually gets converted into (d/l)-2,3-dibromobutane during the reaction. Also, going by Sawhorse Projection formula is particularly useful in this case rather than Fisher projection. The former gives you a better visualisation of the syn or anti-periplanarity.
Here is the mechanism of the reaction with the meso-compound as starting material:

Similarly, you can draw the mechanism with the optically active 2,3-dibromobutane as your initial compound:

Now you can easily see how the corresponding alkenes are formed depending on the substrates.

This answer would be slightly better if you did not use wedge and dashed bonds in the products.

I believe the wedge/dash bonds are wrong according to IUPAC guidelines. // Please do not abuse MathJax for styling elements.

I will take account of that from further on.

Your stereo configuration of the structure of starting material in last scheme is actually $(\mathrm{2}R,\mathrm{3}R)$-, not $(\mathrm{2}R,\mathrm{3}S)$- as indicated. It is worth note that $(\mathrm{2}R,\mathrm{3}S)$- is the  stereo configuration of meso- compound in first scheme.

Oh yeah !! My bad.. Sorry..


text:
Reaction of orthocresol with butyrolactone in presence of AlCl3 and heat
What will be the major product of the following reaction?
Please also help me with the mechanism. I'm not sure whether C- acylation or O- acylation will occur here. And if O- acylation does occur, I think the heat will cause a Fries' rearrangement. What will be the major product of this rearrangement?
Also, if simple Friedel Crafts acylation occurs, will AlCl3 affect the reaction due to formation of -O-AlCl3 (Lewis acid-base interaction), something like what happens in the case of Friedel Crafts reactions with aniline? (I know aniline cannot undergo FC reactions due to deactivating effect of -NH-AlCl3 interactions)
I'm unable to upload a pic of the question due to some technical problem, but what I mean is reaction of ortho cresol and butyrolactone in AlCl3, along with heat.

When you write "a cyclic ester" do you mean a lactone?

Waylander it is a lactone.Recently a question on product formation between butyro lactone and ortho cresol in presence of aluminium chloride was asked in national exam conducted nation wide.How ever phenols may not undergo freidal crafts reaction.

@Chakravarthy Kalyan, Yes that is the question I am referring to. The official answer given has the carbonyl acylation at the position para with respect to methyl group. Is that right? I was thinking the acylation should be done at para position with respect to hydroxy group.

It's possible that different conditions provide different products.

I can take a guess since no one else has provided an accepted answer.
First, I don't think lactones will take part in FC-acylation on an arene, but rather prefer alkylation as mentioned in this paper. In the case of o-cresol and heat, this could be a possible explanation for your question why the O-acylation takes place instead of the C-acylation.
Second, when the Fries rearrangement occurs, and the acylium ion is released in the intermediate, the phenolic oxygen is complexed with the aluminum chloride (second to last step here). If you imagine an ortho-methyl group, as well as the complexed aluminum chloride, it's plausible substitution to the remaining ortho-site is not sterically favorable. This would force it to substitute elsewhere on the ring.
As for why "elsewhere" prefers para to the methyl group and not the oxygen—going by what you wrote in your comment—I'm not sure. The para alumino-oxy substitution should produce the more stabilizing resonance structure for the arenium intermediate. At least, in one other reaction of electrophilic aromatic substitution with ortho-cresol, namely its nitration, the 3/5-nitro products are not found. Maybe it's a solvent matter. Wikipedia states a non-polar solvent prefers ortho in the Fries rearrangement; perhaps if the ortho site is blocked, it will settle for para to a methyl substituent.
Edit
In a summary of the comments, the m-hydroxy-ketone was called into question as the correct product. I mentioned none of the papers I found of similar reactions furnished anything but the para or ortho product. Here are the papers I looked through:
paper 1 - first paragraph: o-tolyl acetates yield predominantly p-hydroxy-ketones.
paper 2 - tables 2 and 3
paper 3 - table 1 (page 378)
paper 4 - 5th paragraph
A few others: book p 223, paper, paper

Thanks for your answer. But I still don't understand this... does the complexation of phenolic oxygen with AlCl3 convert it to a deactivating group? Why then do Friedel Crafts reactions and Fries rearrangements occur in phenols? And if it doesn't convert it to a deactivating group, does it just lower the directive power of the -OH group such that -Me becomes more dominant?

To add to the confusion, this book states that the phenolic -OH group does not interact with Lewis acids like AlCl3 to the extent of affecting reactions. https://books.google.co.in/books?id=YzbjDQAAQBAJ&pg=PA1037&lpg=PA1037&dq=friedel+crafts+acylation+of+phenol+chemistry3&source=bl&ots=R4oqD_Dwhy&sig=Z4-7Vk_6Kzl16bS2OSwVEmOtzFY&hl=en&sa=X&ved=2ahUKEwjV3eOP9fLfAhVQfysKHTzVDcsQ6AEwH3oECAYQAQ#v=onepage&q=friedel%20crafts%20acylation%20of%20phenol%20chemistry3&f=false

I don't expect the phenol to initiate with aluminum chloride. O-acylation isn't a FC mechanism. O-acylation is catalyzed by an acid or base, and is kinetically favored under heat. I don't have literature on it, but my assumption would be that aluminum chloride acts as a lewis acid to coordinate to the carbonyl group of the lactone. The phenol attacks the labilized carbonyl to form an orthoester. The phenol hydrogen shifts to the neighboring oxygen, which is subsequently released, and the ester is formed. From there, a Fries rearrangement takes place, as described here: https://bit.ly/2VXs2yq

As for why methyl takes precedence in being para directing, I do not know. Are you sure the para-methyl product is correct? I would have also expected it to be para to the hydroxyl group. For example, this paper ( https://pubs.rsc.org/en/content/articlepdf/1958/jr/jr9580002926) describes o-tolyl acetates with AlCl$_3$ as resulting in a para-hydroxy arrangement, and other papers of similar reactions have the same result.

As the mechanism in the link shows, does Fries rearrangement always shift the attached acyl group to the ortho and para positions (with respect to phenolic hydroxy group)? Or does the product also depend upon other substituents on the aromatic ring? Are you aware of any such examples where Fries rearrangement shifts the acyl group to meta with respect to -OH due to other substituents of the phenyl ring?

I recently gave a (quite popular) national level examination in India for admission in undergraduate courses in national universities. This question was in my question paper. The official answer key released by the government authority has the para-methyl product as major product. I have challenged the answer key and am hoping for a correction, but I wanted to be sure.

The substituents on the aromatic ring do matter (that is the topic of the rsc paper, where incidentally the para product is majority). The solvent and catalyst also play a role, at the very least in determining the relative proportion of products. I did try to find the reaction online and went through 4-5 papers of similar reactions. I never found mention of a meta product -- always ortho or para. It's certainly not my expertise though, so I am hesitant to rule out the answer key on my authority and after only a cursory investigation.

Could you please post links to the other 4-5 papers on similar reactions, which you have mentioned above? I'd like to take a look at them too.

I edited my answer with them.

It will follow the same mechanism as friedel crafts acylation.Just break the C-O bond of that cyclic Ester and add it to para (with respect to OH group)  and the H removed will be added to the last O- and form OH group.

My doubt is, will the presence of AlCl3 as a medium here, affect the reaction, by interacting with the hydroxy group to form -O-AlCl3? My teachers are giving this explanation and saying that the -O-AlCl3 group will have lesser directive influence than the methyl group on the benzene ring, thus acylation will be done at the para position with respect to methyl group.

@Ankit Kumar Misra phenol is an ambidentate nucleophile.Nucleophillic attack can take place can take place at C C -ACYLATION  and O - Acylation.O acylation takes place via nucleophillic acyl substitution. IF 1 equivalent of AlCl3 is used then it will not function as a catalyst.In the paper given  excess AlCl3 is  not given.Product may not be formed.

@Chakravarthy Kalyan, so O-acylation product will be formed. And if we then assume excess AlCl3 and heat, what will happen next? I think a Fries rearrangement. What will be the product then? Which position of the benzene ring will the acyl group migrate to?

@Ankit Kumar Misra heat or photo has not been given , thetefore fries rearrangement or photo fries rearrangement would not take place most likely.If these conditions ate given then acyl cation will participate in eas para to O assuming that one lone pair is in coordinate covalent bond with Al and other is not.This makes it to have greater directing then methyl group

@Chakravarthy Kalyan: Okay, thanks for the clarification. By the way, heat was given in the question, so I think Fries rearrangement should be considered here too. I think that means the product with acylation at the position para to -OH should be the major product.


text:
What is the refractive index of poly(methacrylamide)?
I've been looking for the index of refraction for poly(methacrylamide). I searched several vendors and in https://polymerdatabase.com/ with no success. Is there any database where I can find this information?
The refraction index of the monomer would be useful too.

As reported in  Gooch JW (Jan W. Encyclopedic Dictionary of Polymers. Springer; 2011. doi:10.1007/978-1-4419-6247-8. Poly(N-methyl methacrylamide) has a refractive index of 1.5398


text:
How does the pH of solutions of salts in which the cation is acidic and anion basic, depend only on the relation of their constants?
My book is not very explanative about it, I'm trying to get the equation:
$$[\ce{H^+}]=\sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$
but I didn't succeed, can someone explain the theory behind and how to get the formula comparing the two ionization constants?
Translation of the actual page of the textbook (Cacace Fondamenti di stechiometria (in Italian)):

Salts which cation has acid features and anions has basic features. A salt like ammonium acetate, $\ce{NH4Ac}$, release after the following (complete) dissociation:
$$\ce{NH4Ac -> NH4+ +Ac-}$$
A cation will behave as an acid and an anion will behave as a base.
In cases like this one, acid-base equilibrium in solution depends only by the strength of the cation compared to the anion. It can be proved

(but it does not, that's why I'm here)

that in first approximation , the concentration of $\ce{H+}$ in those solution is indipendent from the concentration of the salt and it depends only from the comparison between ionization constants of the weak acid and the weak base from which the salt has been formed:
$$[\ce{H^+}] = \sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$
From which:
$$\mathrm{pH} = \frac{1}{2}(14 + \mathrm{p}K_\mathrm{a} - \mathrm{p}K_\mathrm{b})$$

It seems to be unclear. I would like to understand the mathematics of the equation that brings the comparison of the two constant became like the one I already wrote. Why is independent from the salt concentration?

I suspect this is the formula for finding the pH of the solution of the highly-soluble salt of weak acid and weak base that undergoes hydrolysis. If the question gets reopened, I'm going to post an answer with how this formula is derived.

@andselisk go ahead :D // This may be related: Calculating pH of diprotic and amphoteric solutions

@Martin-マーチン Thanks for reopening the question:)

Disclaimer: as we will see later, the formula
$$[\ce{H+}] = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{1}\tag{1}$$
is only applicable to highly soluble salts with equimolar rests of monovalent weak acid and weak base such as $\ce{NH4OAc}$, $\ce{NH4HCO3}$, $\ce{NH4HSO3}$, $\ce{NH4NO2}$ and cannot be used for polyvalent electrolytes, poorly soluble salts and when hydrolysis is irreversible (e.g. $\ce{Al2S3}$).
Let's derive formula \eqref{1} in for the salt of the general $\ce{BHA}$ formula. As we agreed, the salt is highly soluble (also true for your textbook example, ammonium acetate), so the main approximation here is that we assume this salt dissociates completely in solution:
$$\ce{BHA → BH+ + A-} \label{R1}\tag{R1}$$
Each ion participates in hydrolytic equilibrium (\eqref{R2}, \eqref{R3}), plus there is autoprotolysis of water, \eqref{R4}. Subtracting \eqref{R4} from \eqref{R2} and \eqref{R3} we get a brutto equation for hydrolysis \eqref{R5}:
$$
\begin{align}
&+\quad&\ce{BH+ + H2O &<=> B + H3O+} \label{R2}\tag{R2}\\
&+\quad&\ce{A- + H2O &<=> HA + HO-} \label{R3}\tag{R3}\\
&-\quad&\ce{2 H2O &<=>[$K_\mathrm{w}$] H3O+ + HO-} \label{R4}\tag{R4}\\
\hline
&&\ce{BH+ + A- &<=>[$K_\mathrm{h}$] B + HA}, \label{R5}\tag{R5}\\
\end{align}
$$
where $K_\mathrm{h}$ is hydrolysis constant. According to the law of mass action, it can be written for \eqref{R5} as such:
$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} \label{2}\tag{2}$$
Now let's multiply both divider and denominator in \eqref{2} by $[\ce{H3O+}][\ce{HO-}]$ so that
$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]}\qquad \right|\cdot \frac{[\ce{H3O+}][\ce{HO-}]}{[\ce{H3O+}][\ce{HO-}]}\\
 &= \frac{\color{blue}{[\ce{B}]}\color{red}{[\ce{HA}]}[\ce{H3O+}][\ce{HO-}]}{\color{blue}{[\ce{BH+}]}\color{red}{[\ce{A-}][\ce{H3O+}]}\color{blue}{[\ce{HO-}]}}\\
 &= \frac{[\ce{H3O+}][\ce{HO-}]}{\color{red}{K_\mathrm{a}}\color{blue}{K_\mathrm{b}}}\\
 &= \frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}\label{3}\tag{3}
\end{align}
$$
where $K_\mathrm{a}$ and $K_\mathrm{b}$ are dissociation constants of the conjugated acid and conjugated base, respectively:
$$
\begin{align}
\ce{HA + H2O &<=>[$K_\mathrm{a}$] A- + H3O+} \label{R6}\tag{R6}\\
\ce{B + H2O &<=>[$K_\mathrm{b}$] BH+ + HO-} \label{R7}\tag{R7}\\
\end{align}
$$
Since the dissociation of the salt is complete (see \eqref{R1}), its initial concentration $c_0$ equals to the concentrations of $\ce{BH+}$ and $\ce{A-}$ in solution:
$$c(\ce{BH+}) = c(\ce{A-}) = c_0 \label{4}\tag{4}$$
and the corresponding equilibrium concentrations of the ions participating in hydrolysis are
$$[\ce{BH+}] = [\ce{A-}] = (1-α)c_0, \label{5}\tag{5}$$
where $α$ is the degree of hydrolysis showing the ratio between the amount of hydrolyzed salt $n_\mathrm{h}$ to the total amount of salt $n_0$ ($α = n_\mathrm{h}/n_0$; $α\in [0;1]$). The equilibrium concentrations of the weak acid and base formed during hydrolysis are
$$[\ce{B}] = [\ce{HA}] = αc_0 \label{6}\tag{6}$$
and eq. \eqref{2} for $K_\mathrm{h}$ can be rewritten using degree of hydrolysis:
$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} = \frac{α^2c_0^2}{(1-α)^2c_0^2} = \frac{α^2}{(1-α)^2} \label{7}\tag{7}$$
Note that hydrolysis constant doesn't depend on salt's concentration. Since we are interested in $\ce{pH}$, it is convenient to express $[\ce{H3O+}]$ from $K_\mathrm{a}$:
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]};$$
$$[\ce{H3O+}] = K_\mathrm{a}\frac{[\ce{HA}]}{[\ce{A-}]} = K_\mathrm{a}\frac{αc_0}{(1-α)c_0} = K_\mathrm{a}\frac{α}{1-α} \label{8}\tag{8}$$
From \eqref{7} $α/(1-α) = \sqrt{K_\mathrm{h}}$ so that \eqref{8} finally becomes the first equation you asked about:
$$[\ce{H3O+}] = K_\mathrm{a}\sqrt{K_\mathrm{h}} = K_\mathrm{a}\sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}} = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{9}\tag{9}$$
An equation for $\mathrm{pH}$ is obtained trivially by taking negative decimal logarithm on both parts of \eqref{9}.

What's the rationale for assuming that $\alpha$ is the same in both cases?

@Zhe What do you mean? There is only one case: a single hypothetical salt (see 2nd paragraph) with its own degree of hydrolysis.

Sorry, I think I'm not familiar with thinking about it this way. I was basically at the same solution as this except that I did not get to a point where the concentrations could effectively cancel to 1 because that requires an assumption that the relative concentrations of acid/conjugate acid and base/conjugate base are very close to 1. Perhaps that's the first order approximation I needed, but I didn't find a convincing reason why this should just be true.

@Zhe Oh, yes, as I mentioned, the salt must be very soluble to assume that. Otherwise you end up with a single equation with 4 variables or something. And if there also were a multi-step hydrolysis, then the math would've really gotten very wild.

I noticed that the OP's question said "to first order," so I was trying to figure out if there was an assumption we could make (aside from what you did) to basically reduce that ratio to $1 + \epsilon$ where $\epsilon \ll 1$.

@Zhe Sure, one can show that for equimolar weak acid and base salts $$\frac{[\ce{H3O+}]}{[\ce{OH-}]} = \frac{K_a}{K_b},$$ express $[\ce{H3O+}]$ as $$[\ce{H3O+}]=\frac{K_aK_w}{K_b[\ce{H3O+}]}$$ from there and you arrive at the same equation. I just thought that involving degree of hydrolysis is a more elegant and general way to illustrate the derivation of that formula. Also, I didn't invent this method of calculation; it is listed in many old Soviet books on analytical and physical chemistry that I used to read when I studied chemistry (yes, they are heavy on math, too).

@andselisk Thank you for your time, it was truly clear. i'll try to make it myself and  truly understanding it. Best regards

@GabrieleSortino You are very welcome:)


text:
Why does this region fit NMR spectroscopy technique?
I was reading a paper and came across a sentence (highlighted one) about NMR spectroscopy.

Several Leishmania Rabs, like Rab1, Rab5a, Rab5b, and Rab7, have been functionally characterized. LdRab5a is a GTPase that has been shown to
  bind to GTP and hydrolyze it to GDP. It specifically regulates
  fluid-phase endocytosis. Although the G1–G5 loop sequences
  are well conserved in LdRab5, it has unique
  interswitch region with substitutions and a 20-residue insertion
  with respect to the human Rab5. The unique insertion
  feature of LdRab5a prompted us to choose NMR spectroscopy as the most suitable technique for determining
  its structure.
It is pertinent to mention that NMR solution
  structure has not been determined for any Rab protein
  from any organism. This would also provide the opportunity
  for exploring the residue-specific dynamics for LdRab5a in
  particular and reveal the general implications for the Rab
  family of proteins. Hence, we have undertaken the determination
  of solution structure and characterization of backbone
  relaxation dynamics for LdRab5a by using NMR
  spectroscopy.
  However, stabilization of LdRab5a through systematic
  and specific mutations and deletions was necessary to arrive
  at a construct that was stable enough for recording long
  multidimensional NMR experiments.

Why do we need to have a unique interswitch region so as to use the NMR spectroscopy technique?

Well you can't analyse switching with crystallography, 'cause you get only one structure. NMR can provide insight into dynamic behaviour.

You don't need to have this. The rest of the structure was already clear, it was just about this "interswitch" (I have no idea what that is exactly). They had the background and equipment to do the analysis, and it proved sucessful, I guess. Read the rest of the paper, and the references within, and you will likely find that the idea wasn't a very long shot.

Here is a link to the paper (not yet available on PubMedCentral)

I don't think describing switching behavior was (directly) the goal in this case. Also, in principle you can capture (crystallize) a protein in different conformations - not easy and not a dynamic description but this can provide structures.

The insert in the Ld ortholog occurs in a beta sheet and suggests the presence of additional strands in that sheet compared to the human variant. The structure of the human protein has been determined by crystallography. It would therefore appear that it should be possible to determine the structure of the Ld protein the same way (via crystallography), provided a crystallizable construct is identified.
Based on the available reference provided, the insertion feature from LdRab5a appears to make the protein unstable in vitro and difficult to crystallize, leaving NMR as the only good alternative for a structure determination. However, it seems the insertion feature also rendered the protein too unstable to be able to complete long NMR experiments, so the authors chose to remove the insert in order to stabilize the protein and perform NMR.


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will two weak acids lower their degree of dissociation when put together in solution compared to when put separately
When two weak acids are put in aqueous solutions separately they will have some value of degree of dissociation,  now when both are put together will there be a decrease in their dissociations, I think they should lower each other dissociation due to common ion effect (H+ from both) , the problem is that in my textbook they give resultant H+ion concentration by adding individual concentration , those values being same as if when put in separately .. can anyone please explain me what's correct mathematically ?

You are correct. Obviously the acid with the smaller pKa would contribute more to the H+ concentration than the other acid, but the other acid would contribute some too.

Thanks , atleast I know that I was correct , but how to calculate the new degrees of dissociation ? when we take suppose a weak acid with a strong one , I can calculate degree of dissociation by assuming that conc. of h+from strong acid part I constant (strong acid ) but I cannot do it here , because both are depreciating each others dissociation


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Apparent contradiction in the name of group 2 elements
From my understanding (I might be very wrong), the "earth" in alkaline earth metals means non-metallic, insoluble in water, and resistant to heating, therefore the oxides and hydroxides of group 2 metals should be insoluble in water by definition of "earth".
However, doesn't the term "alkaline" by definition mean bases that are soluble in water? How can the two definitions be co-applicable?

Earth means just a soil.

ok thank you, i did some research and about three sites said earth meant insoluble in water so i was sort of confused.

Alkaline simply means a pH above 7. The term for bases that are soluble in water is alkali, in which the alkali metals are appropriately named because the hydroxides that they make are soluble in water ,for example sodium hydroxide.

This may give you some context: https://chemistry.stackexchange.com/questions/66546/why-are-group-1-elements-called-alkali-metals-and-group-2-elements-are-called-al

"Insoluble (in water)" is a relative term, especially for ionic compounds, and so is "alkaline".  That relativeness gives us enough wiggle room for two of our most common bases to slip through.
Roughly speaking, it takes a thousand volumes of water to react with and then dissolve one volume of calcium oxide, so the solubility of lime in water would not be evident from casual observation.  Yet if you put an indicator into the water it reacts in a way that indicates the water has turned alkaline.  The solubility of lime in water is hard to see with the eye, but it overwhelms the tiny intrinsic autoionization of water, so lime gives evidence of being both an "earth" and "alkaline".  Ditto (despite the still lower solubility) for magnesia.
Lime and magnesia, then, are the quintessential alkaline earths, far exceeding anything else with similar properties in their common occurrence.  They have become the namesake of all the "alkaline earth metals" in Group 2.

Your name identifies you but does not define you or totally describe you.  In a like manner the names of chemicals of groups of chemicals identify them and in some cases do specify some general properties because of the periodicity in electronic structure.  The common names of chemicals arose from the familiar elements and in many cases were extended to similar elements when they were discovered.  The origin of names is fascinating in its own right but let's treat each element as a unique entity as it deserves to be.

This definition has a historical origin, as these names have been given in the $18$th and $19$th century.
In the $18$th and $19$th century, oxides and hydroxides were not always separated. Hydroxides were often considered as hydrated oxides, like salts were known sometimes as anhydrous and sometimes hydrated. Metallic oxides and hydroxides were not considered really different.
In these old times, the oxides (or hydroxides) were known and studied before the elements were discovered. So first alkaline oxides (or hydroxides) were found to be very soluble and strongly alkaline. Then earths were similar to alumina (aluminum oxide) which is insoluble in water and neither basic nor acidic. But aluminum oxide (or hydroxide) neutralizes acids somewhat like an alkaline hydroxide. So there is a common character to alkaline hydroxides and earths : the reaction with acids, making salts.
Well, now, chalk, magnesia and baryta (oxides or hydroxides) are half way between alkaline hydroxides and earths : they are a little soluble, and this behavior is between soluble alkaline hydroxide and insoluble earths. Furthermore, chalk, magnesia and baryta react with acids like alkaline hydroxide and earths. So these oxides and hydroxides are intermediate between alkaline compounds and earths. This is why they are called "alkaline earths". Later on, the same name have been given to the metal, when the corresponding metal has been produced, usually by electrolysis.

An alkaline substance doesn't always have to dissolve in water(or at least very well), for example, copper(II) hydroxide, as shown in Wikipedia is negligible.
But then there are bases like sodium hydroxide and potassium hydroxide which are quite soluble.
Wikipedia says:

An alkali can also be defined as a base that dissolves in water.
Alkalis are normally water-soluble, although some like barium carbonate are only soluble when reacting with an acidic aqueous solution.

More information below:
Sodium Hydroxide
Copper (II) Hydroxide
Alkalis
Bases

Copper hydroxyde is not an alkaline substance

It is considered a weak base. Also it's hydroxide.

Yes. Copper hydroxide is a weak base and a hydroxide. But it is not an alkali, even if Wikipedia states it. And barium carbonate is NOT an alkali, despite Wikipedia.


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Binding aqueous solution and oil solution without an emulsion
I am trying to solve this mystery...There is this company called 100% pure who makes lipsticks using fruit extracts as their pigments in their formulations. Fruits pigments are water soluble - so, the listed ingredients are the usual oils and butters, wax and all the fruits extracts.
Even if there is wax in the formulation it isn't an emulsion - it's a solid lipstick. How can they bind the oil phase with the aqueous phase without the emulsion? Also if there is a water component you need a preservative which they don't seem to use other than rosemary leaf extract :)
Same thing with a lipstick I bought last week that contains the usual oils and butters, wax, a tincture which has alcool and water in it and honey. It is not an emulsion either - I tried to make a lipstick with honey once and it separated...
I hope you can explain this chemical trick to me because I am truly wondering !?!?
Thanks :-)

Just because it is an emulsion does not mean it will not separate. Often phase separation is just a question of time. Such emulsions are called "metastable".

Thanks for the answers that have been provided - it does open some paths and I will keep researching with this in mind. I am sorry if my question was not that well formulated, I am not a chemist - I studied in Botanics and I am French, also my English is not perfect :-) ! As for the butters used in those formulations, they are plant based not animal - therefore will the plant based butters will act the same way as an animal butter for an emulsion of water in fat ?

Your question is neither sufficiently specific nor detailed to provide a proper answer. However you provide a clue as to how it is possible to disperse the more water-loving components in a fatty matrix without the use of emulsifiers listed explicitly as such in the ingredients list. Assuming that 1) the product is bona fide - the composition is as listed; and 2) the pigments are only of limited fat-solubility; then evidently one or more of the constituents has to provide the required emulsifying activity.
My guess is it is butter, which is an emulsion of water in fat. I would guess the water-loving ingredients are dispersed in the butter before blending in the remaining fatty components, but that is an educated guess.
As for the absence of preservatives, I would guess the low water content would limit bacterial growth. There are available certainly guidelines regarding such matters.

My explanation comes from this video  https://youtu.be/zC99ih56S_8 in this he explains that the water and oil is mixed without emulsion because there is less air in the water due to low pressure hence the whole explanation.