text:
What is the correct formula for diamminechloridonitrito-n-platinum(II)?
Is the correct formula for diamminechloridonitrito-n-platinum(II) $\ce{[Pt(NH3)2Cl(NO2)]}$ or $\ce{[PtCl(NH3)2(NO2)]}$?
I was taught that the sequence of symbols within the coordination entity should be written in the following order:

The symbol of the central metal atom is written first.
The ligands are written afterwards and should be mentioned in alphabetical order according to the first symbol of their formulae.
For example, for the coordination compound $\ce{[Co(H2O)2(NH3)4]Cl2}$,
$\ce{H2O}$ is cited first and then $\ce{NH3}$. (As H comes before
N)

However, it is given in my textbook that the formula for diamminechloridonitrito-n-platinum(II) is $\ce{[Pt(NH3)2Cl(NO2)]}$, where $\ce{NH3}$ is followed by $\ce{Cl}$.
I assumed that the correct formula would be $\ce{[PtCl(NH3)2(NO2)]}$ ($\ce{Cl}$ followed by $\ce{NH3}$) instead. What am I missing here?

I believe your proposed formula is correct simply because the anionic ligands must be written before the neutral ligands. The alphabetical order applies in case of two ligands of the same type such as two neutral ligands.
Source : https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Nomenclature_of_Coordination_Complexes

Diamine should have two ammonia molecules, so either formula you posted doesn't match. I can't find a structure for that complex, only some exercises from those Indian exams that pay more attention to memorization of garbage rules than understanding of chemistry. Knowing the structure would certainly help... Otherwise, from my point of view, it doesn't matter.

@Martin-マーチン Whoops, my bad! You're right, the formula should have two ammonia molecules. I corrected it. And as for the Indian exams, I totally agree with you! The education system here encourages rote memorization instead of long term understanding. Unfortunately, this problem isn't just limited to chemistry...

in Indian exams it doesn't matter. we just have to memorize the rules for nomenclature and represent the compound the way we want(chlorine can be before or after, it doesn't matter). At our level the best we have to do is to learn to draw structures remember the(non-useful VBT) and just learn to feel their presence in the daily life and their application, presence and that is all

So I did some digging, and here's what I found:
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005,

IR-9.2.3.1 Sequence of symbols within the coordination formula
(i) The central atom symbol(s) is (are) listed first.
(ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in
alphabetical order (see Section IR-4.4.2.2).Thus, CH3CN, MeCN and NCMe would
be ordered under C, M and N respectively, and CO precedes Cl because single letter
symbols precede two letter symbols. The placement of the ligand in the list does not
depend on the charge of the ligand.
(iii) More information is conveyed by formulae that show ligands with the donor atom
nearest the central atom; this procedure is recommended wherever possible, even for
coordinated water.

So in this case, $\ce{Cl}$ should come before $\ce{NH3}$ and the correct formula would be $\ce{[PtCl(NH3)2(NO2)]}$.

But if chlorine is not a ligand, things change. Anyway, good research.


text:
Half sigma bonds

$\ce{N2+}$ has $2$ $\pi$-bonds and a half sigma bond.

First of all, how can a "half" sigma bond exist? Usually when we encounter non integral bond orders, it is mostly because of resonance which creates partial $\ce{\pi-\pi}$ bonds and doesnt involve orbitals participating in sigma bond.
To understand this, I tried to draw the lewis structure of $\ce{N2+}$. While a triple bond and two lone pairs exist in $\ce{N2}$, I couldnt figure out how $\ce{N2+}$ looks like.
Then I wrote down the molecular configuration using MOT. I noticed that the last electron in $\ce{N2+}$ goes into $\ce{\sigma_{2p}}$ orbital.
I suspect that this might have got something to do with partial sigma bonds, but I'm not really able to see through it clearly. Can someone help me out?

In case someone is wondering about the source, it is the 54th question asked in Jee mains 2020 10th January, Shift -I

$\ce{N2}$ has three bonds between the two atoms $\ce{N}$ : one sigma and two pi. One of these bonds has to loose one electron to form $\ce{N_2^+}$. Apparently it is the sigma bond. As a consequence, this bond still exists and becomes a one-electron bond, or a half sigma bond

https://chemistry.stackexchange.com/questions/11007/fractional-bond-order-how-to-visualize

@Maurice your explanation does make sense. However why would the loss of electron be from a stable bond pair? Why cant it be one of the lone electrons? Also why does it have to be a sigma bond and not a pi bond, if at all the loss is not from one of the lone pairs?

@Newbie105. I don't know the answer. I was also surprised to hear that the electron was coming from the sigma bond. Why this bond ? I would have expected another bond...

Yes, in the molecular orbitals of N2, the highest energy occupied orbital is sigma, while the two filled pi orbitals are below it. There are many descriptions online of why this is so and more generally how to determine the orbital energy rankings in diatomics

First of all, how can a "half" sigma bond exist?

Usually, you expect double bonds to be shorter and stronger than the corresponding single bonds, and triple bonds even shorter and stronger. The OP already mentioned bond-orders of 1.5 that occur for conjugate double bond systems, and those have properties in between single and double bonds.
Perhaps the most famous case for a "half" sigma bond is the hydrogen molecule ion, $\ce{H2+}$. It has a bond distance of 100 pm (compare 74 pm for dihydrogen) and a bond energy of 2.77 eV (compare 4.52 eV for dihydrogen). Source: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm
So in terms of comparing the bonding in the hydrogen molecule ion with the hydrogen molecule, it makes sense to talk of a partial bond in terms of bond length and strength. It might make sense to talk of "half" a sigma bonds because the single bond has 2 electrons, but in this case there is only one.
The reason this case is famous, at least in courses introducing quantum chemistry, is that it is a single-electron system, so easier to work with theoretically and computationally than the dinitrogen cation.

Then I wrote down the molecular configuration using MOT. I noticed that the last electron in N2+ goes into σ2p orbital.

Here is the molecular orbital diagram, and a depiction of the highest occupied molecular orbital.

Source: https://www.chemtube3d.com/orbitalsnitrogen/
As you can see, this orbital has two nodes. In comparison, the orbitals corresponding to the pi bonds have only one node. Using arguments similar to the ordering of orbitals in Hueckel theory, you would expect the sigma orbital to be higher in energy.
However, comparing the energies of states in all diatomics, you see that it is more complicated (the difference between the energy of 2s and 2p changes across the 2nd period elements, and with it the amount of mixing between 2s and 2p of same symmetry):

Source: https://chem.libretexts.org/Courses/Heartland_Community_College/HCC%3A_Chem_161/9%3A_Molecular_Geometry_and_Bond_Theory/9.8%3A_M.O._Theory_and_the_Period_2_Diatomic_Molecules

[OP in comments] However why would the loss of electron be from a stable bond pair? Why cant it be one of the lone electrons? Also why does it have to be a sigma bond and not a pi bond, if at all the loss is not from one of the lone pairs?

You are using the valence bond model (or the Lewis structure) to "name" electrons. There is no 1:1 mapping of "lone pairs" to the molecular orbital diagram. It does say that the two combination of 2s orbitals "cancel each other out", i.e. are non-bonding; however, the shape of these orbitals does not match the valence bond model, where sp-orbitals pointing away from the center of the molecule are counted as lone pairs.

Thank you so much for the well detailed and tailored response! I really appreciate the time you have taken to write this answer

"Half" a bond is new to me, but it makes sense if a (full) bond consists of two electrons: one from each atom bonded together. Then half of that would be one electron in the bond, for 50% of a full bond.
It would get interesting in the boranes where you have three-center bonds, or a pi system where you have an electron deficiency. And then, if you have an anion where an electron is added to antibonding orbitals, especially where you have an extended pi system... But then you can use some other numerical system to indicate which bonds are weakened or strengthened.
Those odd-electron molecules do require some extra imagination.

source: Stack Exchange

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

sort: votes


text:
How to draw molecular orbital diagrams for protonated/deprotonated methyl red?

Give a molecular structure-based explanation for the differences in the UV-Vis
spectra in I and II. Use detailed MO schemes for A and B for illustration. (12 marks)

Does the double bond of N=N, C=O and lone pair of oxygen and nitrogen involved in the MO diagram?

The longer the conjugated pi-system, the longer the absorbed wavelength. In the form A, there is a huge conjugated pi system . It is probably II. In the form B, there are two separate conjugated pi system. it corresponds probably to I.

I am able to assign A and B to the their specta, however I am not sure on how to draw detailed MO schemes for A and B in order to show the electron transitions.

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


text:
How does lead toxicity from jewelry happen vs. occupational exposure?
It is known that lead levels in jewelry is a problem, especially for children (https://dtsc.ca.gov/toxics-in-products/lead-in-jewelry/). As a parent with jewelry-loving children I am aware that ingestion or inhalation of dust or any part of the jewelry will be toxic, while simply touching it would not be. However, several sites claim it is dangerous if one touches the jewelry and then touches their eyes, nose, mouth, food, etc and propose hand washing as a solution.
First, I am wondering what form the lead is in when it's found in jewelry metal alloys (not painted) and how exactly it gets to be toxic from there or what happens when you touch it (ions, oxidation?). I can't find any other information online about the form of lead in jewelry. I know it's inorganic lead and was told by friends it's elemental lead*.
Second, I am wondering what is the best way to remove the lead from jewelry exposure from skin? I've read that hand washing with soap and water is insufficient to remove lead from skin according to CDC, NIOSH, and such studies. But these were all focused on PbO from occupational exposure, and not on jewelry exposure (which I presume is not PbO). Is there some difference here that makes regular soap and water effective in this case?
*After reading similar posts on here and here, elemental and metallic lead (what's the difference?) even if ingested does not seem to be as toxic as ionic Pb(II) and covalent compounds of Pb, but there have been and continue to be cases of lead poisonings involving jewelry items, so either it is not elemental lead or elemental lead is more toxic than indicated here.

Lead toxicity mostly happens because people continue to use lead. The detail doesn't matter. There is absolutely not good reason why any jewellery should contain it. Or any modern paint. the best way to avoid it it is not to have any objects that contain it or are painted with it.

I would be very surprised if any jewelry is made of lead today . Even a tin ,lead alloy ( solder )would be unusual.

Yes there's no reason to contain it, but unfortunately XRF testing done recently by both legal bodies and consumer advocate groups show there's a lot of violating jewelry (and other goods) even to this day. So if there is no way to tell which jewelry has lead, would the best course of action to be to completely avoid all jewelry?

Metallic lead is comparatively inert, but after exposure to air it quickly develops a film of lead compounds, including litharge, $\ce{PbO}$, basic lead sulfate, $\ce{Pb2SO4(OH)2}$, etc. Though these are not very soluble, sweat and bacterial action can convert them to much more soluble compounds.
Lead is also very soft, so touching an article made of lead leaves a residue on the fingers, similar to the way using an aluminum-handled tool leaves the hands covered in gray aluminum and its compounds.
The danger is usually not immediate, but rather the accumulation of toxic lead in the body. However, lead worn next to the skin is steadily abraded. Also, children (and pets) put things in their mouths or even swallow them. Stomach acids do dissolve lead, and there is high mortality in waterfowl from swallowing lead pellets.
This is why lead has no place in play things (or perhaps in most products, since it will almost invariably leach out to poison the environment).
That said, this is an abstract comment, and not to be construed as a way to "safely" use lead in toys.

Thank you! Do you think ISML-containing wipes as tested by NIOSH (https://www.researchgate.net/publication/239522749_Handwipe_method_for_removing_lead_from_skin) could work against every day play things after they are accidentally touched then? Would lead also leave residue easily if it is in an alloy?

That's getting into the area of medical advice...


text:
Is the quantum number $L$ a combination of the azimuthal and magnetic quantum numbers $(\ell, m_\ell)$?
Somehow, I am getting confused about this...
In the $J=L+S$ equation about total angular momentum, which of the four quantum numbers used to describe the electrons and their states is included in the $L$?
And does the $J$ include the first, principal quantum number?

The values of $L$ come from the individual $\ell$ values, according to the Clebsch-Gordan series:
$$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$
where the dots indicate all "in-between" values with a step size of 1. The values of $S$ come from the individual $s$ values, in a similar fashion. Note that it is $s$, not $m_s$.
Finally, the values of $J$ come from $L$ and $S$:
$$J = |L - S|, \ldots, L + S.$$
The principal quantum number $n$ does not play any role in this, since it has no direct relation with angular momentum.

Usually it helps to see an example. Consider the case of the ground-state carbon atom; the $\mathrm{1s}$ and $\mathrm{2s}$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $\mathrm{p^2}$ configuration.
We therefore have two electrons, both having $\ell = 1$ (p-orbital) and $s = 1/2$ (because they are electrons, and all electrons are spin-$1/2$). The permissible values of $L$ are
\begin{align}
L &= |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2 \\
  &= | 1 - 1 |, \ldots, 1 + 1 \\
  &= 0 , \ldots, 2 \\
  &= 0, 1, 2
\end{align}
The permissible values of $S$ are
\begin{align}
S &= |s_1 - s_2|, \ldots , s_1 + s_2 \\
  &= | 1/2 - 1/2 |, \ldots, 1/2 + 1/2 \\
  &= 0 , \ldots, 1 \\
  &= 0, 1
\end{align}
The permissible values of $J$ depend on what $L$ and $S$ are, and must be evaluated on a case-by-case basis. So far, we have identified three possible values of $L$ and two possible values of $S$. Consider the case where $L = S = 0$: here,
\begin{align}
J &= |L - S|, \ldots , L + S \\
  &= |0 - 0|, \ldots, 0 + 0 \\
  &= 0 , \ldots, 0 \\
  &= 0
\end{align}
So when $L = S = 0$, the only permissible value of $J$ is also $0$. If you've seen term symbols before, the corresponding term symbol is denoted as $^1\!S_0$; if you haven't, then all you need to know is that this is a chemist's "shorthand" way of writing $(J, L, S) = (0, 0, 0)$, which is the combination we just found.
You will need to work through this for the other five combinations of $L$ and $S$. This will give you a full series of term symbols, except that there's a catch. The Pauli exclusion principle forbids some of those six combinations: to be precise, the combinations $(L, S) = (2, 1)$, $(1, 0)$, and $(0, 1)$ are not permissible, which leaves only three valid combinations, one of which I've already demonstrated. See Pauli-forbidden term symbols for atomic carbon for more details.

Addendum. As for the $m_\ell$ and $m_s$ quantum numbers: they do have a use, but in a rather more subtle way. I don't expect this very brief summary to be useful, but I'll try; feel free to ignore it at this stage of your learning if it doesn't make much sense.
Much like how (for one electron) $\ell = 1$ implies $m_\ell = -1, 0, 1$, for the entire atom each value of $L$ is associated with a range of $M_L$ values: for example, $L = 1$ implies $M_L = -1, 0, 1$. Therefore, the $L = 1$ term is actually not one single electronic state, but rather a collection of three different values of $M_L$, which is why we call it a "term" rather than a state. Likewise, the $L = 2$ term actually corresponds to five different values of $M_L$. How does this relate to $m_\ell$? Well, it turns out that $M_L$ is the sum of the individual electron $m_\ell$ values. A similar situation exists with $S$, $M_S$, and $m_s$. However, I think that's a topic for another question, if necessary.


text:
How do bonds rotate in pericyclic reactions?
In Pericyclic reactions, it is shown that the terminal bonds rotate (conrotation or disrotation) to allow for allignment of the FMO's, thereby forming the sigma bond

However, a detail that most of my textbooks and references miss out on is how is the double bond rotating about it's axis? As far as I know, pi bonds are restricted in their rotation which is what gives rise to cis-trans isomerism. If this is the case, how are the bonds rotating about their axis to align the orbitals?
One possibility I can think of is that the double bond is broken before/during rotation, but I haven't seen this explicitly mentioned anywhere. Any insights are appreciated.

Well, it is a double bond at the start and a single bond at the end. So as the reaction proceeds, this restricted rotation becomes less and less restricted.

@orthocresol that sounds plausible... Could you provide a more formal answer, preferably with some references/diagrams? I did read this related answer of yours, but I couldn't comprehend it or find anything relevant there.

I don't have a formal answer, sorry, but it seems intuitive to me that as electrons flow out of the old π bonds (to form the new π and σ bonds) the restrictions on rotation decrease. I would guess that it is a continuous process in which the dihedral angle smoothly goes from 0° to 90° as the reaction proceeds. One way to verify this would be computation, but not something I can do now.

One way to maybe think about this: if the FMO's didn't rotate, the reaction would not proceed and the two double bonds would stay rigidly as it is in the π overlap. But because we are heating up the substrate, the molecule will start becoming excited and therefore, bond breakage will be initiated. So it's more like a case of availability of higher degrees of freedom for the molecular components with increased temperature, which allows such rotatory overlaps to occur. This bond rotation of course breaks the π overlap, but it's entropically favorable as heating is going on

Do remember that the pi orbitals on the central carbons will be forming a pi bond as the sp2 becomes more like sp3. Personally, how I was taught is that the drawing above is incorrect. There should be no line bonds drawn if you are representing the p-orbitals.


text:
Need reactants for a rocket
I am doing a project in chemistry at the moment to build a rocket out of materials that can be easily bought and that react strongly to create thrust.
I was wondering if anyone knew a good chemical formula for my rocket fuel? I know lots of good fuels and reactants used in amateur rocketry however commercial fuels have been banned from the project.
Any help would be appreciated.

I remember school days where sodium chlorate NaClO3 has been widely used as total herbicide called Travex . We used as the solid propelent the filtration paper, soaked by concentrated Travex solution and dried. It had to be kept dry, as sodium chlorite is hygroscopic. Not sure if it is still available. Generally, in context of terorism, public access to resources for violent chemical reactions are rather limited. For similar purposes could be used KNO3, but solution must be hot because of solubility. Other option could be classical black gun powder.

See this excellent site about all things rocketry-related for suggestions. There are also other answers on this site so poke around.

https://en.wikipedia.org/wiki/V-2_rocket used oxygen and climate neutral ethanol. Whatever you use, it must burn fast enough to be a good propellant, but slow enough not to burst your rocket. A cardboard rocket for new years day will need different propellant from a high performance steel casing.

Put some potassium nitrate $\ce{KNO3}$ plus powdered charcoal and sulfur in a mortar. Wet it carefully and grind the whole until you get a black homogeneous paste. Let it dry overnight in the mortar. The obtained black gun powder may be detached from the mortar with a wooden spoon. It is a safe rocket fuel. The proportions of the powders are defined by stoichiometry. They must allow the chemical reaction : $$\ce{2 KNO3 + 3 C + S ->  K2S + 3 CO2 + N2}$$
I have done it many times with my students. It's a good exercice of stoichiometry. Never had an accident ! And remember : Never ! Never grind dry powders ! Grind as wet powders as possible ! In case of doubt add more water ! Pastes are even a better choice for grinding purposes !

The chemistry is a safety issue, but the mechanics of rocket guidance is important too. Take it easy with grinding: I was grinding a little mass (maybe 10 grams) of propellant when it ignited (did not explode!), and I jerked back and banged my head.
I made small rockets by winding paper around a pencil, then filling the tube with $KClO_3$ + sulfur + sugar.  Ignited in a copper tube about 2 feet long, they would fly straight out and go about 200 feet, except one that made a right angle turn right out of the tube and sailed right by an observer, hitting a garage door, where the rest of the propellant exploded, leaving a clean spot on the rusty door about 6 inches in diameter.  It probably burned thru the paper tube unsymmetrically, deviating the thrust from a straight flight. The propellant burned well, but had an unexpected impact sensitivity.
With the high energy reactions, scale-up is problematic: you know that at some point, your mix may change from burning fast to exploding very fast. Do things cautiously, small scale at first and do not scale up too fast. Have several layers of safety measures, because you know that you will not stop building bigger rockets until you have a catastrophic failure. Plan to have the catastrophe some considerable distance from you and others.

try 65g of KNO3 per 35g of sugar. Dissolve the mix in the smallest possible amount of hot water on a pan, and boil off the water, once you start seeing large bubbles and it turns into a paste start mixing. Make sure it stops sizzling when you push it down with a spatula. Sounds dangerous but really it's reliable, just do it outside. Coalesce it into a large blob with the spatula while it's still hot. It will harden within a few minutes so you better put it into the motor immediately.
Push a pencil in the middle of the fuel to create a hollow cyllinder, that will make it burn much faster.
Also, it's much easier to set off with an ember than flame


text:
How to thicken Paint
I'm non technical and have a question that you can hopefully help me out with as I've been searching online and so far haven't found any answered
I'm looking for a product that will thicken the following paint
polymarineshop - hypalon paint for avon inflatables
Any help at all would be greatly received
Best wishes
Ant

Is it thinner than it should be ? ( e.g. after using a thinner solvent). Or do you just want it thicker for whatever reasons ? Any thickering may negatively impact the paint properties. I advice rather asking the vendor.

Diisocyanates are used in some coatings and polymerize in contact with moisture. Flexithane contains a non-reactive polyurethane (already reacted diisocyanate), and so is simply a solution of rubber in a mixture of solvents (SDS: http://www.hmgcoatings.co.uk/msdspdfs/Flexithane.pdf).
The literature claims dry to touch in 5 minutes, ready for a second coat in 20 minutes at 20C. One way to thicken the paint without adding anything that would destroy its performance would be to evaporate some of the solvent. In a can about half-filled with the paint, stir well, with plenty of ventilation (as advised by the SDS) for a few minutes. At £49.66 to £75 per 500 mL, it could be be an expensive experiment if you don't stir well enough and you get lumps or skin in the paint when you try to apply it.


text:
Why do phosphorus and sulfur have variable valences?
I have learnt that phosphorus and sulfur have valences 3, 5 and 2, 4, 6 respectively because of vacant d orbitals. But how?
Phosphorus' configuration is $\mathrm{[Ne]3s^23p^3}$. There are 3 unpaired electrons that can get paired with bonds, so 3 makes sense (similar to nitrogen) and a similar argument for 2 with sulfur (similar to oxygen), but how do the vacant d orbitals help in giving valence 5?
Do 2 electrons from the s orbitals go to the d orbitals and then there are 5 unpaired electrons available for bonding and similarly for sulfur?

You said it yourself. (Arguably, the explanation might be a bit outdated, but that's what they want from you anyway.)

What is the modern explanation?

https://chemistry.stackexchange.com/questions/86275/why-is-there-only-one-bond-length-for-sulfur-hexafluoride https://chemistry.stackexchange.com/a/49844/9961

For a modern answer (with some hand-waving), we can simply say that valence electrons occupy more space as we go from first period to second period to third period. So hydrogen makes one bond, second period elements up to four, and third period elements up to six.
If you want to stick to Lewis structures and hybridization, you can explain $\ce{PCl5}$ with resonance:

Source: https://www.cusd80.com/cms/lib6/AZ01001175/Centricity/Domain/1108/003-center%204-electron%20bond.pdf
If you like molecular orbital diagrams, it is pretty straightforward to expand a $\ce{EX3}$ MO diagram to a $\ce{EX5}$ diagram without using d-orbitals, e.g. http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf. Like in the valence bond model, the bond order will decrease as you increase bound atoms.


text:
CSTR reaction cooling jacket - Transfer Functions
I have a CSTR where an irreversible exothermic liquid reaction takes place. Coolant is flowing in the jacket of the tank to regulate the temperature of the fluid in the reactor and control in this way the concentration of component A. The cooling jacket stays at a uniform temperature. An on-line infrared spectrometer with no lag is the sensor used to measure the concentration of component A. Reaction rate is given by r(t) = k(T(t)) Css.
The goal is to get all transfer functions between the input and output variables.

In a homework question such as yours you must make some attempt at an answer and explain where you are stuck otherwise the question will be closed without anyone providing an answer.

Thanks, the question is to find the transfer functions correlating input and output variables, and then to design a controller


text:
Why don’t we have a list of all compounds yet?
With only three basic particles making up atoms, and their interaction fairly well understood (we can even predict the chemical properties of elements we have not yet discovered) atoms seem to be like Lego bricks: only fitting together in a limited number of ways. So how come, in the age of super computers, we have not yet invented a program that can predict every single possible chemical compound (lets start modestly: of around fifty atoms or less)? Would a “super version” of such a computer program- one that would create a “master list” of all possible compounds and their reactions - theoretically make the field of experimental chemistry obsolete? I realize the vast number of possibilities, but just as with a chess game, the number of actual possible boards is far less than the theoretical number of all possible chess pieces and positions. Many simple compounds could be ruled out as “impossible” and a dead end, thereby considerably shortening the list of possible compounds. Am I missing something obvious here?

Speaking figuratively, we have such a program and it works. The estimated time remaining is $10^{20}$ years. In the meantime, why not do some old-style chemistry?

Possibly related: if you are fine with up to 13 non-H atoms of C, N, O, S and Cl, GDB-13 by Renaud et al. count 977 468 314 molecules.  Have a look at the relevant question on ChemSE, https://chemistry.stackexchange.com/questions/119797/are-there-any-datasets-containing-molecules-with-more-than-38-heavy-atoms/119802#119802 for an entry.  Then teach a computer about the rules about the atoms not yet considered and let it crunch the permutations in silico (because, after all, there probably are not enough atoms around to synthesize all possibilities in gram scale).

There's no such number and the problem is ill posed. A small chunk of polypropylene consists of virtually uncountable amount of "different compounds" - each molecule is different - and so what? Two very big and similar molecules are hardly distinguishable in there. On the other hand a protein molecule and and its misfolded form may be considered "same thing" and yet are very different. Such list would have about as much sense as list of all possible words, lacking their  meaning.

On second thought, adding the criterion of «and, can it be prepared in the lab with reasonable effort?» as in https://pubs.rsc.org/en/content/articlelanding/2021/SC/D0SC04321D#!divAbstract

Or, why haven't all possible books been written?

What if such a list would have a mass greater than the earth?

Look at single stranded RNA. The number of different molecules grows exponentially with the number of nucleotides. So, perhaps sometime in the future, we will be able to describe the behavior for each, but never, ever, of every such RNA. You might be interested in computational complexity, too: some problems require exponential time. Look for keywords like "Turing machine" or "NP-hard". There is a whole universe of things to understand related to your question.

@Karl https://chess.stackexchange.com/a/8334 Chess people have been taking care that there is not an infinite number of possible games, there are rules that no position of figures is allowed more than three times, and if my memory i correct and they didn't change the rules, a game is undecided if no figure is thrown out of the game within 50 moves.

It is similar to the question: “Why do not we generate all the possible text strings, and just collect all the good poems and books from them, so there will be no further need to writers, poets?” What I am saying, practicality is not the only weak point of the question.

I think we have collected enough clever analogies and counterexamples in this comment thread already. If anybody wants to comment further, please consider writing it as an answer; or even better, tie all of these together into another answer with an underlying theme. NP-hard is a good one, @GyroGearloose.

Because the combinatorial complexity is far to big for practical computation
One of the simplest possible cases that illustrates why this is impossible is given by restricting the choice of atoms to just carbon and hydrogen, forbidding any double bonds and disallowing any rings. This becomes equivalent to enumerating the number of possible mathematical trees of a certain structure a calculation interising in graph theory as well as chemistry.
Graphs are not exactly the same as hydrocarbons as there are spatial constraints on the molecules that can exist in 3D space and also because the molecules are in 3D, some variants with the same graph may be chemically different due to chirality (see this answer for more details). But Some calculations based on just 50 carbons put the number of possible variants at >1021 (see this wikipedia page) which is a ridiculously big number.
And this is is with just two elements from the hundred or so that could be used and it is ruling out most of the interesting structural features that could add a great deal of extra complexity. All this complexity arises from the carbon skeleton alone and it is already beyond any computational capacity we can contemplate having (just using one byte per compound would use a great deal more storage than at least one estimate of the total world capacity to store information.
So, if you still think this sounds like an easy task, you haven't done those calculations or appreciated the gargantuan scale of the resulting numbers.

Great answer, thanks. I am not a scientist, but it makes sense to me.
I never thought the process would be easy, by the way - just wanted to know if it was possible, as in: would this be a deterministic process limited only by the scale of computing power and time required?
Much appreciated!

@LuukvanHeerde There is a big difference between what is possible in mathematics and what is possible in a finite universe. This is the sort of mathematical problem that pretty quickly exhausts all the matter in the universe just to store the answer, never mind compute it. OTOH this is the sort of complexity that makes chemistry interesting in practice and not just a branch of math or logic.

Let me add a another point besides the already overwhelming problem of almost innumerable  combinations.
I am afraid that the Lego analogy oversimplifies the underlying problem by a large margin. A Lego brick only cares to which brick it is directly connected, lets say we have bricks A,B,C,D. Lets connect them like this, A-B-C. For A, it doesn't matter that B is connected to C and i could switch C with D without affecting the connection between A-B. This does not hold for atoms in molecules. It matters significantly what type of other bricks are connected to B or even further along the chain.
It is possible to define functional groups that are fairly independent, having similar properties irregardless of the neighbors, but that is always an approximation. The approximation is often good, which is why the rules that one learns in basic organic chemistry work most of the times but there are cases where the approximation is bad and the rules no longer work. It applies even to A and B themselves. The properties of A and B are not conserved when you bring A and B together, the whole idea to identify A and B in the combination A-B is an approximation.
This adds to the complexity to calculating all possible combinations. In principle, you have to recalculate the properties for every new compound. You can often make a good guess for the properties of the new compound based on the constituent parts but that's it.
This poses already a difficult problem but coupled with the insane amount of possible combinations makes it all but impossible to create such a table.


text:
What happens to a solution below its freezing point?
This is the question we did in class-
1000g of 1 molality sucrose solution in water is cooled to -3.534°c. What mass of ice would be seperated out at this temperature? (Kf of H2O= 1.86 K.mol^-1.kg)
We first found the lowering of freezing point (FP) at 1 molality concentration of sucrose which will be 1.86°c. Then from the data of mass of solution to be 1000g we found the mass of solute as 254.84g and mass of solvent as 745.15g. Then we applied the formula of delta(Tf)= Kf* molality, again. (Note- delta(Tf) is the lowering in FP). This time the moles of solute would remain the same so we put that as 254.48/342. The new mass of solvent was taken 'x'(g) so the molality was [(254.84)(1000)]/[(342)(x)]. Delta(Tf) was taken as 3.534°c. Solving the equation we found the value of 'x' and the difference of 745.15-x was taken as the mass of ice seperated out.
Now, I understand the solution till we find mass of solute and solvent. But after that why would we take lowering in FP as 3.543? Why would less ice be obtained if the temperature is even below the FP of solution? I mean shouldn't all the solvent be solidified? Why plug in that formula again?
Can someone please make this clear or give an alternate solution.

Solutions are weird. Let that sink in. Solutions are not like pure compounds. Pure water freezes at one temperature. Solutions are not like that.

@Ivan Neretin- Ya realized that very well!

Good. Is your question solved, then?

No not really. Ok if we say it doesn't have a single FP and accept all that weirdness, what the value of x tells us is the amount of water there needed to be so that it would freeze at -3.543°c so isn't the value of x itself the amount of ice formed? Why subtract it from original mass of solvent?

I'm sorry, but this is not clear at all. You say that $x$ is the amount of water, then immediately say that $x$ is the amount of ice. Which of these is your point of view? Make up your mind.

I am saying x should be the amount of ice that's formed. So why isn't it that way in the answer we were given?

Cryoscopy aside, that's not how problem solving works at all. We see a lot of unknowns, and we are unsure what to do, so we choose one unknown, denote it $x$, and then we stick to that definition. You can't change it halfway through the process. In the beginning we chose that our $x$ is the amount of water in the solution. So be it.

Accept some experimental facts and then proceed to calculations.
a)First of all, as you know very well that 1 m (molal) sucrose solution will not freeze at 0 C.
b) Additionally, you have also calculated that 1 m sucrose solution has 254.84 g sucrose and  745.15g water. This is the solution before cooling.
c) Now the question mentions that that the solution has been cooled to -3.534 C. Again, experimentally, it may be observed that some pure water freezes out and some liquid containing sucrose remains there.
When (c) happens, there will be a change in molality of sucrose. Now you have a biphasic system, solid pure ice and sucrose solution, you now would start your calculations.
Under the experimental conditions $\Delta$T= 3.534, this temperature change corresponds to a sucrose molality of = 3.534= K$_f$.m.
Therefore the molality of the remaining liquid sucrose solution at -3.5 C must be 1.88 m! The sucrose solution has become concentrated now pure water has frozen out.
From 1.88 m you calculate the amount of water in the liquid. You also know the original amount of water you started with before cooling. This should help you calculate the amount of ice.

Okay this actually makes sense. The whole solution freezes slowly over a range of temperature! As it constantly maintains the molality required for FP at that temperature and the rest solidifies. Thanks.

Yeah. Can't upvote cuz I don't have enough reputations


text:
Si(OH)4 vs. H4SiO4 vs. SiO2.2H2O
Can someone please explain to me what is the differences between $\ce{Si(OH)4}$, $\ce{H4SiO4}$, and $\ce{SiO2.2H2O}$? I am currently studying sandstone acidizing geochemical models and this question came to me. It seems that $\ce{SiO2.2H2O}$ and $\ce{Si(OH)4}$ are basically the same thing and are considered minerals that may precipitate during the process, but $\ce{H4SiO4}$ is the aqueous form.
References

Quinn, M. A., Lake, L. W., & Schechter, R. S. (2000, February 1). Designing Effective Sandstone Acidizing Treatments Through Geochemical Modeling. Society of Petroleum Engineers. doi:10.2118/60846-PA
Walsh, M. P., Lake, L. W., & Schechter, R. S. (1982, September 1). A Description of Chemical Precipitation Mechanisms and Their Role in Formation Damage During Stimulation by Hydrofluoric Acid. Society of Petroleum Engineers. doi:10.2118/10625-PA
Sevougian, S. D., Lake, L. W., & Schechter, R. S. (1995, February 1). A New Geochemical Simulator To Design More Effective Sandstone Acidizing Treatments. Society of Petroleum Engineers. doi:10.2118/24780-PA

There is no difference.

First, my answer here is relevant: What is chemical formula of Bauxite?, and also follow links inside.
Regarding your question specifically, there is not much difference. However, there are some conventions that come into play here.
The three formulae describe compounds that stoichiometrically have four atoms of oxygen and hydrogen for each atom of silicon.
$\ce{SiO2.2H2O}$
This notation simply says that for each component of "silica" there is a single component of "water". This notation is often used for solids, and never for dissolved species in aqueous solution. In some cases, this can denote that this entire thing has a crystalline structure, in which there are actual molecules of $\ce{H2O}$ held together with "molecules" of $\ce{SiO2}$. This is often called water of crystallisation, but this is not the case here because hydrated silica (which is what this thing is) is not crystalline, but is amorphous.
$\ce{Si(OH)4}$
This is a slightly more simplified way of writing the formula for hydrated silica, withour referring to specific chemical components (such as water or silica). In some cases this indicates that the crystal structure contains hydroxide groups ($\ce{OH-}$), but again, as hydrated silica is amorphous this is not the case here.
$\ce{H4SiO4}$
This is the formula for "silicic acid", which when in solid form, is exactly the same as "hydrated silica". However, this notation is often used when discussing hydrated silica in aqueous solution chemistry, as you correctly inferred. This does not necessarily mean that it is an actual acid, at least in the sense we think of when discussing things like sulfuric or phosphoric acid. We have a bottle of commercially acquired "silicic acid" in our lab. It's not an acid - it's just some hydrated silica, but written in the "acid"-type notation.

Something to remember here, is that some of these types of notation imply that the hydrogen occupies a specific structural site in the crystal lattice. The first implies molecular water, the second implies hydroxide groups, the third implies hydrogen protons.
This is misleading, because in fact hydrated silica can contain all three structural types of hydrogen simultaneously, and figuring out exactly which and their relative abundance is often of little importance in most studies. Adding to the confusion is that the compound is not really crystalline (like opal). And finally to confuse you even more, a ratio of 2-to-1 of water to silica isn't required, and it can vary. In fact, sometimes the formula is written as $\ce{SiO2.nH2O}$ because we simply don't know (and don't care) exactly how much water is in there. Finally, to add even more confusion, some groups use one notation while others use another for exactly the same thing. The same author can even use two different notations in two different papers!

Awesome! Thank you for this detailed explanation! It helped me so much!
The thing is I don't know how introducing such species, affects equilibrium calculations; For example, I am using Reaktoro which uses Gibbs energy minimization for equilibrium calculations but it only has SiO2(aq) in one of its databases. I don't know exactly what is SiO2(aq)! Is it in equilibrium with H4SiO4? or they are basically the same thing? If they are not the same (I think this way), Should we include thermodynamic properties for H4SiO? How could it affect equilibrium calculations?


text:
Chemical equilibrium of inorganic complexes
After mixing Cobalt(II) Chloride hexahydrate with ammonium chloride a violet complex forms, what would be the complex formed?
It was the precursor for producing hexamine cobalt chloride so I didn't characterize it.
I expect many compounds like hexaqua cobalt chloride and pentamine aqua cobalt chlride or amine chloro coblat complex?

The ammine complexes of cobalt form in stages and have different colors. Which complex forms depends on the relative concentration of ammonia to the cobalt(II) ions as well as the pH.

Well thanks, amount of ammonia is approximately three times larger than that of cobalt, I think ammonium tetrachloro cobaltate (NH4)2[CoCl4] would form also but still not sure about it.


text:
Are there safe inks that allow one to eat a fish after painting it?
Just came across a specific type of art called Gyotaku where specific types of ink are used to paint over one side of a fish and then a paper (rice paper, shoji paper, ...) is placed on top in order to retrieve the fish impression. After the impression, the ink is removed with water.
As catching the fish just to print doesn't seem worthy, I am wondering if there there safe types of ink that let's one eat the fish afterwards? Will washing it fully remove the ink?
Is it safe to eat the fish if one uses acrylic paints?
What kind of inks should one avoid?

Edit
Here's a video of an artist painting a oni, washing the ink out and then eating it. It is not clear if the ink is absorbed by the fish before he eats it or not. He states that he uses water-based acrylic and that it is water soluble and washable.

I Googled "edible ink" and found this item, which is a set of edible ink-containing pens.
The keyword is "edible." I'd stay away from anything not specifically labeled "edible."
Finally, it is safe to eat fish adorned with ink specifically certified to be safe to eat.
Disclaimer: There might still be some issues with the fish, so keep your eye/nose on that.

I think one should not suggest anything edible in a chemistry forum especially if they are dyes.

@M.Farooq Well, there are many chemically synthesized dyes that are officially marked as edible, coded E1nn as food colourants.

@ToddMinehardt I am not sure if that type of ink would not be suitable for gyotaku as it may not pass to the paper. I have added a video in the section Edit where one can see how does the Gyotaku process looks like and a comment on the type of ink the artist uses: water-based acrylic.

@Poutnik, yes, everyone has been eating and drinking synthetic dyes in this century. No idea what good they have done to mankind by adding colors to food and drinks.

@M.Farooq While I agree with you about the questionable general purpose, they are here and some of them eventually useful for the OP purpose. Or, natural products may be used as well. But I would avoid acrylic paints.


text:
Why doesn't NASA or SpaceX use ozone as an oxidizer for rocket fuels?
Presumably, it would be expensive to use ozone ($\ce{O3}$) as an oxidizer instead of $\ce{O2}$, but would the extra oomph be worth it? Does $\ce{O2}$ provide as much thrust/energy/heat as can be provided, given the liquid hydrogen propellant?

I figured Chemistry was better than Physics, Engineering or Astronomy for this question.

... This is unrelated to your current question and I apologise, but I notice now I've answered a couple of your previous questions, and never got any sort of reply. This seems to be a bit of a pattern with your questions on Chemistry, not just the ones I answered. You're obviously not new to SE, but could I encourage you to either accept answers if they helped you, or if they didn't, try and engage with people who comment/answer your questions? Thanks!

@orthocresol Agreed—it's just commmon courtesy.  It is discouraging when you put a lot of work into giving a good answer, and the poster doesn't bother to acknowledge it.

All that would happen is a massive explosion, before even getting to the rocket. Ozone is not as temperamental as FOOF, but it cannot be tamed.

It is way too reactive (like aggressive) if stored at in high concentrations.  Maybe the conditions performing an ozonolysis are not familiar to you.  However, descriptions about ozone itself and reactions against e.g., metals (think about tubings, fittings, framework of a rocket) quickly outweigh the little «extra umpf» to reach a higher temperature of combustion.  Because rockets' propulsion is based on  momentum ($p = m \cdot v$), overall, it is cheaper to generate more gaseous volume per time.

Liquid ozone has tendency for spontaneous explosion. especially at contact with anything ozone can oxidize. As an astronaut or a satellite, I would not like to sit on a basin of liquid ozone. Additionally, it is serious environmental pollutant even at as low concentration as 100 ppb v/v.

@theorist Yes, I agree but most of the OPs are one-timers. They post their questions and don't even bother to check back if they have received any answer/comments. My point is "Why bother asking if you don't come back to check the answers?" I once gave a detailed answer to a question and thought of having a good discussion with the OP but he hasn't checked back yet. It's been 3 years :(

Ozone's boiling point is −112 °C ("normal" oxygen's is −183 °C).

What's the advantage in Joule/kg, if any, when burning RP-1 with O3 instead of O2?

Read 'Ignition', linked in Matt's answer. Recommend any technical friends read Ignition too, even if they're not into rockets. It's a great read, you don't need to know anything about rockets to appreciate the humour, of which there is a foretaste in the quotes Matt has chosen to include in his answer.

"I figured Chemistry was better than Physics, Engineering or Astronomy for this question" – There is also [space.se], although as always the same rule applies: take the tour, read the help center, especially the on-topic, off-topic, and "How to Ask" pages, read the FAQ, search the site to make sure your question is not a duplicate, browse the meta site, browse the questions to get a feel for what is and isn't considered a good question, and make sure you have done your research.

@orthocresol Stack Exchange sites aren't forums. There's no chit-chat. And accepting some particular answer (or any answer) is not mandatory. We expect people to pay back the community by their contributions, not by "acknowledging" things. If you're not okay with that, this may not be the site for you.

@AsteroidsWithWings Do you realize that you are a new user who is on this site just for a few days; and you are talking to a long-term moderator and high-reputation user of this site? maybe you should take your time and see how this site works before making any comments.

@Loong I've been on Stack Exchange for 9 years with over 400k accumulated between accounts (twice as long as the moderator in question with almost eight times as much reputation network-wide). I know what I'm talking about, and I'm also quoting the basic, standard guidance given to all users since the dawn of time. I'd expect a moderator to know all this already. But thanks for your sarcy comment, Mr "I've been here for all of 36 days".

Sorry!  I DO read the answers to my many questions, (O.K., occasionally I forget for a while)... And I usually upvote the answer(s)... I HAVE green-checked the best answer on a number of questions....

BUT, when I first joined Stack Exchange, I read in the 'instructions' somewhere NOT to just leave a comment or answer saying 'Yay, thanks!'... But, instead, to upvote helpful comments and answers, and (possibly), green-check the best one....       I did, in my early days here, leave a few comments saying 'Yay, thanks!' anyway, and someone said NOT to keep doing that...

As is usual with rocket fuels, the problems of ozone are practicality not performance
Almost every answer for why a specific rocket fuel component is used or not will end up referring to John D Clarke's magnificent and sparklingly written book: Ignition: An informal history of liquid rocket propellants (a rare technical book worth reading for the brilliant and humorous style in addition to the technical content).
His summary of why ozone, despite its apparent functional advantages, is not more widely used is fairly simple: the practical problems outweigh the apparent advantages.
He points out the advantages:

What makes it attractive as a propellant is that (1) its liquid density is considerably higher than that of liquid oxygen, and (2) when a mole of it decomposes to oxygen during combustion it gives off 34 kilocalories of energy, which will boost your performance correspondingly. Sänger was interested in it in the 30’s, and the interest has endured to the present. In the face of considerable disillusionment.

But every available way of creating a liquid with a high proportion of ozone is dangerous. Ozone is extremely toxic and unstable:

For it has its drawbacks. The least of these is that it’s at least as toxic as fluorine. ... Much more important is the fact that it’s unstable—murderously so. At the slightest provocation and sometimes for no apparent reason, it may revert explosively to oxygen. And this reversion is catalyzed by water, chlorine, metal oxides, alkalis—and by, apparently, certain substances which have not been identified. Compared to ozone, hydrogen peroxide has the sensitivity of a heavyweight wrestler.

Work was done on solutions of ozone in liquid oxygen which is somehow more stable. But has the disadvantage that ozone/oxygen mixtures separate into two phases the ozone rich version of which is difficult to prevent in feed tubes after firing and is extremely unstable. Another mixture considered to make handling easier was with liquid fluorine (!!!).
Ultimately he concludes the known work on ozone mixtures of any sort:

For ozone still explodes. Some investigators believe that the explosions are initiated by traces of organic peroxides in the stuff, which come from traces, say, of oil in the oxygen it was made of. Other workers are convinced that it’s just the nature of ozone to explode, and still others are sure that original sin has something to do with it. So although ozone research has been continuing in a desultory fashion, there are very few true believers left, who are still convinced that ozone will somehow, someday, come into its own. I’m not one of them.

Maybe there are theoretical advantages, but they are outweighed by the practical and safety problems. In a profession used to testing things like FOOF (fluorine dioxide) and chlorine trifluoride, this is some admission to make.

Ignition! and Things I Won't Work With have some of the greatest chemistry quotes of all time.

In this context, does "mixture considered to make handling easier" mean "mixture which was evaluated for the purpose of making handling easier" or "mixture which, in people's opinions, made handling easier?" (If it's the latter, then it's awfully interesting that people tried it and it appeared to work.)

@TannerSwett Opinions don't count in rocket fuel research. You have to test. In this case they tested ozone/fluorine mixtures and they were better than pure ozone, despite the seemingly insane safety implications. But then, this is a profession that took chlorine trifluoride seriously as an oxidiser.

I think it should be apparent by now that "we" use the most powerful fuels and oxidizers which are consistent with practicality.  If you -- an amateur -- has thought of it, experts have thought of it; if they aren't using it, there's a good reason why they aren't using it.

@RonJohn: That's more or less the difference between science and engineering. I never understood why "rocket science" is used as a placeholder for "insanely difficult", because it is essentially high school level maths and newtonian mechanics, and undergrad organic chemistry and fluid mechanics. Rocket engineering is where it gets hard and rocket economics is where it get really hard. (Actually, I consider economics part of engineering: "an engineer is someone who can do with a dollar what any fool can do with two". [Source unknown])

@JörgWMittag orbital mechanics is only "high school level math" after "egghead" mathematicians figure out all the complicated maneuvers and orbits.

@JörgWMittag Shrug. It's not undergrad OC. Just because any highschool kid can balance the equations doesn't mean you've understood how the reaction works. Especially with ozone. ;-) That's where engineering becomes difficult: If you have no scientist with whom you can work up your experiment and find out why it went wrong.

https://library.sciencemadness.org/library/books/ignition.pdf is readable

@Karl: Yeah my dad had that problem in college chemistry. He would produce balanced and energy-balanced equations but if you actually set it up, a different set of reactions would take place.

Liquid ozone is rather stable (as in "can be stored") as an up to 70% solution in LOx, if you add some stabilisers. That would give quite a bit of additional boost in a rocket, but it's surely going to explode already in the rocket's fuel pumps and tubing, instead of burning in the engine's nozzle.
Rocket engines are tricky enough to build, you don't want to add the requirement that all material is of a kind that does not catalyse a premature decomposition of ozone. It's usually a runaway process: Decomposition increases temperature, temperature increases decomposition rate, .... boom.

Apart from practical considerations, what are the actual numbers with regard to the difference in (potential) energy? Is there only a 1% difference between using oxygen and ozone?

@PeterMortensen That's a very simple piece of math. Just look up the formation enthalpies of water and ozone.

Back-of-envelope math says +25%.

I wonder if it's worth the pressure-fed tank rather than the fuel pump.

Does it matter if the ozone decomposes just before use? It seems rather convenient to have oxygen radicals, after all. That speeds up the reaction, which is important as the part of the flame outside the combustion bell produces little trust. The chief problem appears to keep the reaction front stable; it shouldn't move up the pipes into the main tank.

"rather stable" is somewhat misleading. The stability of ozone in liquid oxygen is stable relative to pure ozone liquid as is the solution in liquid fluorine (!!!) but it not on the spectrum of things any sane chemist would call "stable". And it is very hard to prevent even less stable mixtures emerging in fuel pipes as oxygen/ozone mixtures can separate into two phases under common conditions where one phase is far less stable than the normal solution and this makes large quantities of nitroglycerin look benign.

@matt_black de:wp links this patent https://patents.google.com/patent/US3400024 , they sound very optimistic. ;-) Have you got a reference for the phase separation? The dipole moment is a respectable .5 D , which is a lot more than e.g. CO (0.1 D).

@karl The airco experience of ozone stabilisation with those and other stabilisers was mentioned in one of the chapters in Ignition which is also where the phase separation information came from. There is a lot more detail there than in my answer.

@MSalters: Oxygen radicals usually result in an engine-rich combustion.

@MSalters "Does it matter if the ozone decomposes just before use?"... not really. as long as your plumbing is up to the job of having its contents jump in temperature by about 730K, with very little notice. (this may occur in your main oxidizer tank, if the Ozone is in a bad mood)


text:
Can cycloaddition reactions be antarafacial?
Cycloaddition reactions of the form [4+2] and [2+2] are always suprafacial, as the geometry is too constrained to allow for antarafacial interaction between the FMO's. However, for cycloaddition of longer chains, say [6+6] cycloaddition, can the cycloaddition be antarafacial?
Here's a small scheme I came up with for antarafacial cycloadditions, and the resulting product:

Is this reaction possible? I haven't been able to encounter any literature online for reactions of this type, which is why I'm doubtful. I just took [6+6] as an example, but if there are references of other conjugated systems undergoing antarafacial cycloaddition, they would be helpful.

This is the ChemDraw bit which I use, if you're curious.

@orthocresol I use the online 'trial' version, which is good enough for reactions but doesn't have orbitals. Thanks for the edit though :)


text:
Relation of Energy required to change an orbit by electron of H atom with spectrum
We have the value for expression of line spectrum of hydrogen.
$\Delta E = 13.6 \cdot  Z^2 \cdot \left(
 \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Amount of energy required by electron to change its orbit.
Now for wave number $\pu{1.09677e7} \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
Does this mean the wavelength of light that it can absorb.
If yes , then why do we compare it with this .How is this compared with it.
Since this shows the light emitted by spectrum and only certain colours show up.
My main question is that does any of these formula have any relation with the fact that you know , we see certain colours emitted when seen on a black screen and not all the colours.
Form the picture that I have shown.

The differences in formulas must be in parentheses.

Wavenumber is number of waves that fits the chosen length unit. Conventionally, it is usually 1 cm, so the wavenumber unit is 1/cm.

My main question is that does any of these formula have any relation with the fact that you know , we see certain colours emitted when seen on a black screen and not all the colours@Poutnik

Both ( equivalent ) formulas, after recalculation to wavelengths from energy differences or wavenumbers, predict particular wavelengths to be either emitted by hot and excited hydrogen atoms, either absorbed by hydrogen atoms being excited.
Hot atoms emit just particular wavelengths, not continuous spectrum like condensed matter. So their emission spectrum is linewise on black background.
OTOH, relatively cold  atoms absorb at the same wavelengths they emit when excited. So their absorption spectrum is linewise as well, with the same lines as the emmision spectrum, but this time dark lines on bright background.
Both phenomena are the opposite direction of the common process, how atoms deal with discrete energy levels of their electrons. Emission of photons to reach lower energy, or gaining energy by absorption of photons. These photons belong to light of the very particular wavelength.

Comment feedback:
Incandescent bulb filaments have as condensed matter continuous spectrum. If their light passes through gas containing free hydrogen atoms ( i.e. not only molecular hydrogen ), particular wavelengths are absorbed, leaving dark lines in the spectrum.
Excited hydrogen atoms then either reemit at the same wavelength in any direction, or convert it during collisions in kinetic energy.

Yes.You understood my question.Like for example we a bulb.Now the filament of becomes hot when switch is on.So when that light is passes through prism.Only certain colours appear on screen.Whereas , if we pass light through hydrogen gas , it first absorbs and then emits different wavelength of light.

So,What is the relation of the formula with this?

Does the formula tell us that there will be some black spaces in the formula ? Is the wavelength of light that it absorb same as wavelength of light it emits ?

I hope you get my question.

Read what I have written. Then read again. Then again, until you get it.

https://en.wikipedia.org/wiki/Emission_spectrum   https://en.wikipedia.org/wiki/Absorption_spectroscopy

Ok.Yes @Poutnik.I have got it now.Thank you.


text:
How can I know which molecule is more stable when bond order it's equal?
So, I was wondering, I have these molecules $\ce{O^+2}$, $\ce{O^-2}$, $\ce{O2}$ and $\ce{C^+2}$.
When I try to find their bond order and their magnetic nature I get this:
$\ce{O^+2} = \frac{1}{2} \left( 10 - 5 \right) = 2.5$ - Paramagnetic
$\ce{O^-2} = \frac{1}{2} \left( 10 - 7 \right) = 1.5$ - Paramagnetic
$\ce{O2} = \frac{1}{2} \left( 10 - 6 \right) = 2$ - Paramagnetic
$\ce{C^+_2} = \frac{1}{2} \left( 7 - 4 \right) = 1.5$ - Paramagnetic
You can note that $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same bond order and are also paramagnetic, but how can I know which one is more stable so I can arrange them in stability order if $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same orders?
Thanks!

For future reference: In addition to MathJax, the bodies of questions, answers and comments in ChemSE may use mhchem which eases the input and improves the rendering of chemical formulae and equations.  Because it is somewhat special not all engines understand and render properly, MathJax and mhchem shall not be used in the titles.

To recognize the more stable molecule you may extend the description with bond orders is the one considering how the mathematical concept to mix atomic orbitals by LCAO and visualizing the results in Molecular orbital diagrams and compute the overall energy of such a molecule.  Eventually, you compare the total energy of $\ce{O^-_2}$ with $\ce{C^+_2}$ using the same external reference.
As about the hydrogen molecule shown below:

Two important details about symmetric diatomic moleculs to retain are:

For $n$ atomic orbitals, LCAO yields $n$ molecular orbitals.
The split between the lowered bonding molecular orbital and the higher antibonding orbital is not symmetric.  The difference between the energy level of the initial atomic orbitals and the the antibinding molecular orbital is higher, than the difference between the initial atomic orbitals and the binding molecular orbital.

So you draw the diagram including the energetic levels, and populate the levels with electrons from the bottom up.  If there multiple molecular orbitals of same energy, you fill the orbitals first with one, and later with the second electron. You then compute the the overall energy of the molecule and later compare the results about the molecules in question against a reference in common, e.g. vacuum.
Note that the notation $\sigma$ and $\sigma{}^*$ is about the rotation symmetry around the bond between the molecule (contrasting to $\pi$ and $\pi{}^*$ as perpendicular to this), and not if the molecular orbital in question is conceptually formed from $s$, or $p$ atomic orbitals.  Equally, the energetic consecution of levels of the molecular orbitals may change, e.g. between nitrogen and oxygen:

(using material from here)

Thank you! I did some research and now I see that C energy levels are a little higher than O levels due to electronegativity values.

So I ended up like this O2+>O2>C2+>O2-
From higher to lower stability, would you agree?

@Adolf O2+ more stable than O2? I see why you ended up here, but that energy should be counted for electron number and is differently shifted compared to vacuum.

Thanks for commenting @Alchimista, Indeed its also weird to me that O2 its more stable than O2+ but when calculating bond order O2 its a 2 while O2+ its a 2.5, since they have the same number of protons in the nucleus I think this is because O2 has 2 unpaired electrons while O2+ only has one therefore making it more stable.
What do you think?

@Adolf well there is one term missing. No ion is more stable than a neutral species in term of its absolute energy vs some and the same reference. What you are at the moment is more or less that a second ionisation potential is higher than the first one, and it is not a surprise, isn't?


text:
Introducing n-propyl group to benzene using cyclopropane
I am studying Friedel–Crafts alkylation reactions and my textbook states the following proposition [1, p. 1090]:

Another very good means of introducing $n$-propyl group is to use cyclopropane as the alkylating agent.

I do understand the mechanism of Friedel–Crafts alkylation, but I am not able to write the mechanism for this particular reaction. Any hints are appreciated.
Reference

Shahi, R. Essential Organic Chemistry for JEE Main & Advanced; Arihant  Prakashan: New Delhi. ISBN: 978-93-13192-10-7.

Please -- not only for ChemSE, but as a general applicable technique -- not only quote, but reference the source of a quote.  Cyclopropane has considerable ring tension (building it once with a molecular kit, I literally broke once one of the bonds ...) but so far I thought alkanes always had to be halogenated, then permitting $\ce{AlCl3}$ to yield an electrophilic reagent.

This paper uses cyclopropanes to alkylate electron rich benzenes with Lewis acid catalysis but as it is behind a paywall I can only read the abstract. It may discuss the mechanism  https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.201500856

@Waylander From the SI of the publication you indicate (annotated NMR spectra), accessible without subscription, it looks like cyclopropane always bears an acceptor decoration (1,1-dicarboxylate) and an electron donor (e.g., 2-phenyl group).  By the title of the papers citing this work (equally accessible), this donor-acceptor pattern seems to repeat, too.  It is not so obvious that it is about cyclopropane itself (boiling at $\pu{-33 ^\circ{}C}$).

This was a very interesting question, and I found the answer (or the reference thereto) in Thomas's "Anhydrous Aluminum Chloride in Organic Chemistry" (1941). On page 501 is referenced the very thorough work of Ipatieff et al. on the very subject in question, "alkylation by naphthenes".
First, we can say that because there is no isomerization of the chain (n-propyl benzene is formed, and not isopropyl) that neither chloropropane nor propylene are intermediates. Friedel-Crafts alkylation with either of those leads to at least some isopropylbenzene formation, because of the better stability of the secondary carbocation.
Second, and most convincing, Ipatieff et al. state that "The presence of hydrogen chloride is essential for the alkylation of aromatics with olefins or naphthenes when aluminum chloride is used as a catalyst.". This leads them to propose a hydrogen chloride-aluminum chloride complex, "hydrogen aluminum tetrachloride" or "hydrogen tetrachloroaluminate", $\ce{HAlCl_4}$, which then reacts with cyclopropane to form n-propyl aluminum tetrachloride. This intermediate is the active alkylation agent, and goes on to produce only n-propyl benzenes.
$\ce{HCl + AlCl_3 -> AlCl_3\cdot HCl or HAlCl_4}$
$\ce{(CH_2)_3 + HAlCl_4 -> CH_3CH_2CH_2ClAlCl_3}$
$\ce{CH_3CH_2CH_2ClAlCl_3 + C_6H_6 -> C_6H_5CH_2CH_2CH_3 + HAlCl_4}$
So far as I can tell, this theory has never been refuted and stands as the current state of the art.
Ipatieff, Pines, and Schmerling, J. Org. Chem., 5, 253 (1940)

The book also states that the reaction is highly favorable. So much that it could occur even below 0°C.


text:
Do ions still "conduct electricity" if a physical connection made by aqueous electrolytes between the anode and the cathode is nonexistent?
I always had the misconception that when people say ions conduct electricity, what they actually mean is a connection is made between the cathode and the anode by the ions and electrons can flow from the cathode to the anode through this connection.
However, I recently learned that ions conduct electricity because chemical reactions are happening around the electrodes.
Taking the example of the electrolysis of sodium chloride water solution. At the cathode, $\ce{Na+}$ captures electron ($\ce{Na+ + e- -> Na}$); at the anode, $\ce{Cl-}$ releases electron ($\ce{2Cl- -> Cl2 + 2e-}$). And this is because the battery takes electrons from the anode and feeds electrons to the cathode, and positively charged anode attracts anions and negatively charged cathode attracts cations. That is to say electrons do not flow from cathode to anode but are given to and taken from the ions.
I still feel confused after knowing this explanation because it seems to me that the anode and the cathode do not need to be connected by aqueous electrolytes to "conduct electricity". Suppose the cathode and the anode are placed into different solutions in different containers, anode still attracts anions and cathode still attracts cations. So that there should still be electricity flowing through the circuit.
Some people say that a "closed circuit" is required for ions to conduct electricity. If that is the case then what am I missing here?

Forget ions for a moment; chemistry is complicated. It is said that electrons conduct electricity in metals. What if I connect two pieces of wire to the positive and negative electrodes, but without a connection between the pieces? Will the electrons still flow from the negative electrode to the attached piece of wire? Why not?

@IvanNeretin Electrodes are mostly made of metal so attaching wires do not change anything. Electrons will flow from the negative electrode to the wire, until the voltage between the two wires is the same as the voltage of the battery, just like how electrons can flow to the cathode. The reason why ions are different from metals in this case is that ions can give and take electrons, therefore creating non-stopping current until electrolytes are depleted (or not?).

Well, same thing here: the electrons will flow from the negative electrode to the solution, until the voltage between the two solutions is the same as the voltage of the battery. Ions may react all they want; what they can't do is to make the charge disappear.

The question risen on the title   Do ions still "conduct electricity" if a physical connection made by aqueous electrolytes between the anode and the cathode is nonexistent?   does not make sense to me. If there is no electrolyte between the electrodes, where are the ions then and why to call them the anode and cathode ??

Put electrolytes away for now and consider ionized gases. How they conduct electricity ?

@Poutnik Yes, the title can be misleading, but the emphasis is on "a physical connection" instead of "electrolytes". Ionized gases are essentially plasma and have free electrons, so they conduct electricity the same way metals do. And the question I'm having is not around conductivity either.

Well, it is about conductivity, as without connection, there is one. Placing metal into electrolyte causes transient effect until charge transfer sets such an electrode potential where there is zero net electrode reaction. This is independent on external potential of the open circuit, as electrolyte potential is "floating", similarly as potential of one wire of ungrounded transformer output.

If a cathode (negative pole) is placed alone in an ionic solution (without any anode in the solution), it will attract cations from the solution, and discharge maybe one or two or one thousand of these cations. This will last maybe 1 nanosecond, and then it will stop. The reason is that this number of atoms is extremely small, so small that the neutral atoms which are deposited on the cathode cannot be seen or weighed. And then the cathodic discharge will stop, because the whole solution has lost positive charges (cations), but has not lost any negative charges (anions). As a consequence, the whole solution becomes negatively charged. This negative charge increases with the time and prevents the cathode from bringing new electrons necessary for discharging further cations. The electrolysis will stop. The electrode is said to be "polarized".

Thank you @IvanNeretin and Maurice , I see what I'm overlooking here. So what happens is that the solution can become charged to create an electric field that cancels out the electrodes' ability to attract ions, therefore preventing chemical reactions, unless the solution is able to remain neutrally charged. I can now see how the analogy between metal and electrolytes applies here. When connected to a battery, metal is essentially also creating an electric field to cancel out the battery's ability to attract and repel electrons.

I want to further expound upon your newly developed misconception as written in your comments although Maurice has already explained it well.

I see what I'm overlooking here. So what happens is that the solution
can become charged to create an electric field that cancels out the
electrodes' ability to attract ions, therefore preventing chemical
reactions, unless the solution is able to remain neutrally charged.

No the bulk solution has to remain neutral all the time. There can be no charge build up in the electrolytic cell. Everything takes place at the interface of the dipping electrode and the solution next to it.
By definition, whenever electricity will pass through an electrically conducting solution, chemical reactions will take place. "Something" has to oxidize at the anode and "something" has to reduce at the cathode. That something could be the solvent, neutral molecules, ions or anything in the solution. Something has to lose electrons and something has to gain electrons.
I think you are looking for the right words, because it seems your confusion arises from the semantics. Ions carry electric charge, and rate of movement of charges constitutes current. Therefore, we say that nature of current in a solution is ionic current.

Suppose the cathode and the anode are placed into different solutions
in different containers, anode still attracts anions and cathode still
attracts cations. So that there should still be electricity flowing
through the circuit.

Yes, do you recall a capacitor which are two disconnected plates of metal. When it is connected to a battery, quickly a small amount of current flow. It it is exactly like this scenario. It may be called as capacitive current.
Now remember that capacitors in a circuit do not allow direct current to pass through it. Similarly two separated electrodes sitting in two beakers will not allow direct current to pass in the external circuit.

Some people say that a "closed circuit" is required for ions to
conduct electricity. If that is the case then what am I missing here?

Now this is a different situation. When you want the direct current to flow, then yes, the circuit has to be closed. Think of a battery as a recycling pump. Imagine a pump in your house, which fills up a water tank placed on the roof of your house, and this tank is connected to your faucets, and all that water you use is pumped back into the water tank. Dirty analogy but good for a battery.

Thank you for going out of your way to help me better understand. I found your answer very informative. You said Maurice explained it well but I was mostly rephrasing Maurice's answer by saying that "the solution can become charged". What Maurice said is, "As a consequence, the whole solution becomes negatively charged". You pointed out that "There can be no charge build up in the electrolytic cell". Is this a contradiction or I'm missing something here?

The solution can never be charged by dipping an electrode and there can be no charge build up! It always remains electrically neutral all the time. Maurice may have written this incorrectly.

@ M. Farooq. You are right. The solution never gets charged. I should have said that this "charged solution" is only a hypothesis stated to show that no chemistry happens around a cathode dipped alone in a solution,

@ Maurice, There might be a small capacitive current but no redox process.


text:
pKa = pH for strong acid — strong base?
I have learnt that for a weak acid — strong base titration, $\mathrm{p}K_\mathrm{a} = \mathrm{pH}$ at the half equivalence point.
However, the same conclusion is not drawn when discussing strong acid — strong base titrations. Why does the above only hold for weak acid — strong base and not strong acid — strong base?
Similarly, why is $\mathrm{pOH} = \mathrm{p}K_\mathrm{b}$ at half-equivalence point only true for strong acid weak base and not for strong acid strong base?

What do you know about buffers?

I know that a buffer solution is one which resists changes to pH when small amounts of acid/alkali are added and I know the two ways in which a buffer solution can be formed. I also know that a buffer is most effective when the concentration of the ions is very high so that additional ions added (H+ or OH-) won't have a large relative impact...is that useful?

How about pKa = pH vs buffers then and why strong acid won't work?

https://chemistry.stackexchange.com/questions/133853/why-does-the-ph-before-the-equivalence-point-of-a-titration-depend-on-the-initia

The reason is that strong acids have $\mathrm{p}K_\mathrm{a}$ values that are poorly known. These $\mathrm{p}K_\mathrm{a}$s cannot be determined directly, because the exact concentration of the ions is difficult to know with precision: the electrodes do not react with the concentration of $\ce{H^+}$ or $\ce{H3O^+}$ ions. They react with the activity of $\ce{H+}$ or $\ce{H3O+}$, which may be quite different from the concentration in concentrated solutions.
So, the $\mathrm{p}K_\mathrm{a}$ values of strong acids can only be determined indirectly, for example by extrapolation of values obtained in organic solvents, and mixtures of organic solvents plus water. This extrapolation is never precise enough.

Thanks Maruice - I still don't get. Surely the Ka of strong acids is almost entirely to the right (complete dissociation) - why don't we know it? Also, is the topic of buffers important at all, as mentioned by Mithoron above in the comments? Thank you!

@ vgupt. I don't know what Mithoron wanted to say with his short question !

So, can the pKa value of a strong acid not be determined from it titration curve?

are you there? Would appreciate any help, if possible!

No. The pH cannot be determined in a concentrated solution. Or, if you prefer : the measured pH values has no relation with the concentration of the ion $\ce{H^+}$.

Just to explain my purpose. If you measure the pH of a 0.01 M solution of HCl, you obtain 2. If now you dissolve enough NaCl to saturate this solution, you measure pH 1.1, as if the solution were more concentrated. This shows that the pH is the log of the concentration calculated NOT by unit of volume, but by unit of volume of "free water". Free water is the water that is not adsorbed around the dissolved ions. In a concentrated solution, there is practical no free water any more. The concentration calculated by volume of free water  seems to increase. That is the same for pure strong acids.


text:
Why does Ethylene Glycol have higher boiling point than Propylene Glycol?
Looking at the two alcohols, both have hydrogen bonding and are singly bonded throughout the molecule. However, propylene glycol has a larger electron cloud which to my knowledge would increase its London dispersion forces and thus its boiling point. I already know both molecules exhibit hydrogen bonding since they're both alcohols.
However, it seems that ethylene glycol has a higher boiling point despite propylene glycol seemingly having stronger IMFs. I believe this has something to do with molecular shape but I can't seem to find the definitive answer explaining why ethylene glycol has a higher boiling point.
I've used https://pubchem.ncbi.nlm.nih.gov/compound/Propylene-glycol#section=RTECS-Number and https://pubchem.ncbi.nlm.nih.gov/compound/1_2-Ethanediol#section=Odor for comparison of the two alcohols.

"larger electron cloud which to my knowledge would increase its IMF"  That sounds very wooly. IMF is a pointless catch-all term. Its about hydrogen bonds!! If you have them, everything else is largely irrelevant.

I would have other considerations but the fact that propylene glycol has a boiling point about 20 °C higher than that of etandiol should solve your issue.

I'm not sure which sources say that propylene glycol has a higher boiling point, but if you've found such a source that'd be of great help!

According to Merck, ethylene glycol has a b.p. of 195-198 °C at 1 bar, and propylene glycol has a b.p. of 187 °C at 1 bar.

This is all about linearity of the molecule and steric hindrance. Consider the linear structure of ethylene glycol versus the branched structure of propylene glycol. EG can "stack" more easily and therefore engage in more intermolecular bonding than PG. More attraction means more energy required to vaporize. On the other hand, 1,3 propane diol, which is the linear isomer of propylene glycol, has a higher boiling boiling point than EG at around 213C. To further illustrate - 1,4 butanediol (linear) has a higher boiling point than 1,3 butanediol (branched).

Welcome to Chemistry SE! The addition of a reputable reference may help to improve how it is received. Also, commenting on the involvement of EG's terminal hydroxyls to forming intermolecular H bonds would help as well.


text:
Silicon–Boron Covalent Bond
This is a semiconductor-related question, but in order to understand the acceptor level energy state of boron in bilicon, I figured I'd ask here for hints.
Boron is a p-type material when introduced in a semiconductor like silicon. Boron forms three covalent bonds with silicon, leaving one silicon atom frustrated, not forming a bond. This doping process introduces the idea of the hole, that is, absence of electron.
However, my chemistry knowledge is limited, and I don’t understand what kind of bond exists when an adjacent silicon valence electron jumps in this electron vacancy. Is this new electron now shared by both silicon and boron, or does it belong to silicon, since silicon complies to the octet rule, whereas boron does not?
Why does the acceptor level reside above the silicon valence band edge and what does this level exactly represent? Does it represent the energy needed for an adjacent silicon valence electron to jump into this vacancy?

Image source: National Schools' Observatory

Crystals aren’t molecules. Bloch functions extend throughout the crystal.

well this doesn't really help me further. but I see your point. still, not clear what this acceptor level represents to me

Using your diagram, the boron-silicon bond with only the silicon electron has only half of the possible electron content.  A "hole" as you said. The hole is not confined to the one $\ce{B-Si}$ bond, but can be in any of the 4 positions. It's not like the boron sits there and the silicon pops its electron into the potential bond; the bond is there and it's only half filled.
Now this isn't a molecule, but you can focus on the boron part of the crystal and say that all four adjacent silicons are bonded to the boron, but you have only 7 electrons where the "molecular" orbitals could contain 8.
Electrons don't really "belong" to one atom or another; they comprise some sort of sea of electrons, and the sea above this boron is a little lower than elsewhere. I think of the sea of electrons in a copper-boron alloy as being very fluid, and having a level surface everywhere, but the sea in a p-type semiconductor is more like mud (quicksand?) and can have some dips that could catch and puddle some water (~electrons). The sea is still controlled mostly by the silicon, just modified by boron (submerged holes).

yea, all that makes sense, however, why does the acceptor level reside above the valence band and not below is or inside it? this level is the energy a nonfree electron inside the "disturbed"/imperfect/doped silicon crystal will have , when it is located in one of these 4 incomplete bonds. an energy level that mathematically derives from the the modified hydrogen model? are my assumptions so far correct?

The band gap in semiconductors is described in terms of levels. https://en.wikipedia.org/wiki/Band_gap however, levels don't have energy, electrons have energy. The levels are what we think we put them in to indicate their energy.

In the picture, a, toward the left, represents pure silicon. If we remove an electron from one $Si-Si$ bond, there will be a little dip in the filling of the valence band because we have taken one electron away. When we add boron to make the $B-Si$ bond, we get that same little dip in the valence band. However, when we add another electron into that bond (from some electron excitation, or from an electron donor somewhere else in the material), that electron is somewhat stable, but not as tightly bound - it is either in a peak of the valence band or just into the conduction band.
The picture shows the transition from atomic orbitals (on the right), which is what the OP shows in the diagram in the question, to a continuity of structure and a broadening of energy levels (on the left). Drawing individual atoms keeps us at the right end of the picture, whereas discussing semiconductors moves us to the left side of the picture. Holding those two sides of the picture (and the middle, too, which represents metallic conduction) in your mind at the same time is somewhat like wave-particle duality.
So, in answer to the comment on my first answer, the electron in the $B-Si$ bond is lower than the top of the valence band, but the second electron in the bond is more energetic than the top of the $Si-Si$ valence band, and may go into the conduction band, or more easily be excited into it. The levels don't conduct electricity, the electrons (or holes) do.

that makes more sense. i am aware of how  the bands are formed, but the energy vs atomic radius picture is new to me. in the case of SI-B, this means that the interatomic distance between B and SI is smaller or bigger than from Si-Si?

The overall silicon structure remains, so the Si-B-Si distance is pretty much the same (maybe a little shrunken). But the individual electron energy at the "hole" is affected even more by the nucleus substitution (-9) plus the electron shielding (also -9, but directed). Focus on the electron energy, but switch from atomic orbital levels to lattice energy bands and gaps. Imagine the situation at one end of the graph, then at the other end; then try to switch back and forth as smoothly as you can.


text:
Which quantum chemistry approximation to use?
I'm just getting started with quantum chemistry using some simple test problems.  I have a few of the software packages installed (ASE, Psi4, GPAW etc) and everything is working - no issues with the software itself - but there is a bewildering variety of methods, from semi-empirical (PM7) to DFT (B3LYP) to coupled cluster (CCSD(T)); plus options for each such as basis functions, and nothing in the documentation gives any specific advice on what to use when.
Is there concrete guidance on which method is appropriate for which type of problems?
What I'm mostly interested is biological systems: molecules in water at room temperature, and both covalent and non-covalent interactions; most of the atoms are CHNOSP.  An example would be taking a system of a few hundred atoms and optimizing the coordinates (either locally or trying to find the global minimum).  A slightly more ambitious project would be to optimize the coordinates a few hundred atoms within a much larger system of thousands of atoms (where the other atoms have partial charges and produce electrostatic forces, but are otherwise fixed).

The simulation of chemical systems is literally the subject of a whole field inside chemistry, so taking an introductory course would be much more helpful (and you could start asking actual questions). Also the size of your system and resources automatically exclude most if not all quantum chemistry methods.

This is more computational than quantum.

A large part of computational chemistry is about benchmarking methods. I.e. testing how good different methods/basis sets are for a problem. If you have a specific problem in mind try searching the literature for it and check what methods were already tested/used. The best practical method for a problem is hard to predict and most of the time determined empirically. People simply test a bunch of methods and compare it to known results. Another "filter" is also what you can actually do with your hardware. The theoretically best method is often simply not feasible especially for large systems.

Related: https://chemistry.stackexchange.com/q/27302/4945 The question is probably better suited on [mattermodeling.se].

A good guide comes from this series of slides by Jan Řezáč:

It seems that semiempirical methods (of which PM6-D3H4 in MOPAC is probably most accurate for the types of systems encountered in biology) are the only methods able to handle structure optimization of a whole protein (or whole domain of a larger protein).  DFT can handle an order of magnitude fewer atoms, and MP2 or CCSD(T) are mainly used for creating benchmark data sets to test the accuracy of methods, such as S66.

I get a virus warning for the pdf file. From when are these slides? There are multiple forests of methods missing from that scheme...

@Martin-マーチン What software gives the warning? Virustotal thinks it's okay, https://www.virustotal.com/gui/url/3eeca80d622347a44810f051f266c89146d97910b4e56c08959c595fa28d73e6/detection
Slides are from 2014.  Could you fill in the methods that are missing?


text:
Reaction rate constants for the inhibition of cholinesterases by various carbamates
Darvesh et al. [1] have conducted a study on the anticholinesterase activity of various carbamates derived from phenothiazine. The authors measured the inhibition rate constants of rivastigmine for acetylcholinesterase and butyrylcholinesterase as $\pu{1.13E3 M^-1 min^-1}$ and $\pu{129E3 M^-1 min^-1},$ respectively.
I want to see how the anticholinesterase activity of the different compounds studied by the authors compares to that of more traditional carbamates. What are the AChE and BuChE inhibition rate constants for pyridostigmine, neostigmine, TL-599, their uncharged analogs, and carbaryl?
Reference

Darvesh, S.; Darvesh, K. V.; McDonald, R. S.; Mataija, D.; Walsh, R.; Mothana, S.; Lockridge, O.; Martin, E. Carbamates with Differential Mechanism of Inhibition Toward Acetylcholinesterase and Butyrylcholinesterase. J. Med. Chem. 2008, 51 (14), 4200–4212. DOI: 10.1021/jm8002075.


text:
What does Pauli’s exclusion principle mean in atomic or fundamental way?
It means is that no electron can have same n , l and $m_l$ but can have two different spin quantum number.
I want to know why is this rule valid?Means there must be some other things happening also Inside that atom which forces this rule to be true.
I got to know from web that electron dual behaviours and electromagnetic forces result in the formulation of this law but what is the real reason?

And what reason would you like? We don't know what experimental data leading to quantum mechanic really mean - that's why there are many interpretations (not only Copenhagen!). Pauli’s exclusion principle is like one of most fundamental properties of matter - it's the other stuff that stems from this.

https://physics.stackexchange.com/questions/22263/what-is-degeneracy-pressure

http://eng-web1.eng.famu.fsu.edu/~dommelen/quantum/style_a/cboxdp.html

Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways).
Fermions exhibit an associated statistical property when you have a collection of the otherwise indistinguishable particles. The statistical property follows from how the wavefunction of the collection behaves under symmetry (exchange) operations. You have two choices when you swap particles in such a wavefunction: retain the sign or switch signs. That is one fundamental distinguishing property between fermions and bosons (all particles can be classified as one or the other). It happens that nature allows both cases. We chose to call one fermions (to honor Enrico Fermi who co-discovered them), and the particles we call electrons are fermions (you can look into the details of the Standard Model to work out exactly how changes to the properties of the electron would change the universe). Fermions have half-integer values of the spin quantum number. Bosons have integer values. If you have multiple indistinguishable fermions, the QM wavefunction changes sign (is antisymmetric) under exchange of any two of the particles. For the exchange to alter the sign the two exchanged fermions must differ in some property, making the wavefunction antisymmetric, for instance if intrinsic angular momenta (spins) have opposing quantization (quantum numbers $\pm \frac12$).
The consequence of this is explained well in the Wikipedia

If two fermions were in the same state (for example the same orbital with the same spin in the same atom), interchanging them would change nothing and the total wave function would be unchanged. The only way the total wave function can both change sign as required for fermions and also remain unchanged is that this function must be zero everywhere, which means that the state cannot exist. This reasoning does not apply to bosons because the sign does not change.

Comments are not for extended discussion; this conversation has been moved to chat.

I did enter the chat to say "indeed I either misread something in recent days or fall myself down this doubt. Indeed we can't say much more than particles come in two types as far Pauli principle is considered, the fermions obey it and the bosons don't. Other point is to say that degeneracy pressure is an intrinsic (T unrelated) property originating from the indeterminacy principle and the Pauli one. Doubts of this type nevertheless signal that I am far from a deep comprehension or at least is always better to refresh such complicated concepts" send button was unreactive so I copy it here.