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Does there exist a molecule of formule XαYβ where X and Y are elements that are also found in molecule XγYδ?
I am not a chemist, but I will be explaining chemistry to some third-graders. I want to give the example of how different things can be made of the same parts, but arranged differently. My "real world" example is a car and a motorcycle. The car has an engine and four wheels, but we can create a different vehicle with an engine and only two wheels: we get a motorcycle. Does there exist a molecule composed of the elements $\ce{X}$ and $\ce{Y}$, where $\ce{X}$ and $\ce{Y}$ can form a different molecule when the quantities of $\ce{X}$ and/or $\ce{Y}$ atoms are different?
I was hoping to use carbon and hydrogen because the children have already heard me use those words, and $\ce{CH4}$ is a gas that will make them laugh when I tell them where it comes from. But the only $\ce{CH2}$ molecule that I could find is this, which is too complicated and unfamiliar to explain. I would prefer molecules composed of the common elements hydrogen, oxygen, carbon. I would like to avoid molecules of elements that are commonly found in their elemental state such as gold, iron, or silver as at this stage I want to reinforce the idea that we cannot see molecules with the unaided eye. I might introduce the idea that a molecule (and macromaterial) can be composed of a single element later, if this lesson goes well.
For purposes of discussion, a molecule can be defined as "a discrete body composed of atoms". There is no need to address quantum chemistry, as the target audience is a third-grade classroom.

How about CH4 and C2H4? And C2H2 for that matter.

Sn and Cl or Pb and Cl. If you're limiting yourself to Period 2 elements nitrogen oxides are your best bet, although there is certainly no NO4

@jerepierre: Thanks, I might be able to work those in. I'll need to think of a different analogy, though.

You could also use formaldehyde ($\ce{CH2O}$) and methanol ($\ce{CH4O}$). You would just have to add something to your idea that gives the oxygen … a driver maybe?

You can do this with single elements and things that are very familiar (at least if you don't demand simple discrete molecules). Graphite and diamond are the same element built with different connectivity and radically different properties. Dioxygen and ozone also work, though ozone may be less familiar. If you must have compounds, ethanol and ether have the same formula but different properties (and are safe enough to show to children). Or oxides of nitrogen (less safe though except under careful conditions).

@matt_black That is a great idea! I can show them the graphite in their pencils (yes, I know that today it is a different compound but for purposes of the class I'll fib) and the black crust of burnt toast, as well as possibly a diamond. Thank you.

$\ce{GeF2}$ and $\ce{GeF4}$
$\ce{GeCl2}$ and $\ce{GeCl4}$
$\ce{GeBr2}$ and $\ce{GeBr4}$
$\ce{GeI2}$ and $\ce{GeI4}$
$\ce{SnF2}$ and $\ce{SnF4}$
$\ce{SnCl2}$ and $\ce{SnCl4}$
$\ce{SnBr2}$ and $\ce{SnBr4}$
$\ce{SnI2}$ and $\ce{SnI4}$
$\ce{PbF2}$ and $\ce{PbF4}$
$\ce{PbCl2}$ and $\ce{PbCl4}$
$\ce{XeF2}$ and $\ce{XeF4}$
$\ce{XeO2}$ and $\ce{XeO4}$
Silicon dihalides ($\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$) have also all been prepared in the gas phase, but are much less stable than the tetrahalides.

Thank you for this invested list. Unfortunately none of those are molecules that form materials which I would expect a third-grader to be familiar with.

@dotancohen I would modify your plan a little and do: CO is to CO2 as bicycle is to  tricycle.

That is a terrific idea. I don't think that the third graders know what carbon monoxide is, but I can bring some H2O and H2O2 for a tricycle and a car. One issue: I don't think that modern children have ever seen a tricycle! At least in Israel, it has been decades since I've seen one myself.

@dotancohen My sons are both still in grade school and they both rode tricycles.  I wanted to give away one of their old tricycles to charity since they ride bikes now, but they said they still want to keep it!  Here (east coast USA) we have carbon monoxide detectors in many homes (too many deaths from generators and heaters) so some would know, but H2O2 would make a better visual demonstration.

How about sulfur and oxygen?
$\ce{SO2}$ is an invisible gas that smells terrible. $\ce{SO4^2-}$ is an anion - it has extra two electrons that it must bond with something. That something can be hydrogen in the example of sulfuric acid, or with all kinds of metals. A favourite example of mine is celestine, $\ce{SrSO4}$:

Source: Rob Lavinsky, iRocks.com – CC BY-SA 3.0, via Wikimedia Commons

The closest I can come up with is $\ce{CH4}$ and $\ce{H2O}$, which obviously have different central elements, but cars and motorcycles have very different bodies with very similar wheels.

I think the other one can be formaldehyde $HCOH$

@shaistha well hey, if X=CO and Y=H, then we have formaldehyde and H3COH alcohol, which are very different, especially their effect on the body.

@Aabaakawad he said that he wants to give the example of how different things can be made of the same parts, but arranged differently.

@shaistha You are right. Some pair of molecules built out of H, C, & O. By the way, formaldehyde is CH2O with no H on the O. HCOH is the hydroxymethylene radical.


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How is (SO4)^2- Charged if S and O all are Neutral Themselves?
As far as I understand $\ce{SO4^2-}$ structure,  sulfur makes two coordinate bonds with two oxygen atoms and two double bonds are formed with oxygen after sulfur expands it's valance shell to 6. At this point calculating formal charge of those two coordinate bonded oxygen gives total formal charge of −2 and the formal charge on $\ce S$ and the double bonded $\ce O$ is zero. Thus the minus 2 charge on $\ce{SO4^2-}$ can be shown but initially all the atoms were neutral. While making bonds electrons were shared only.
How did excess 2 negative charge pop up?
Please help, I am certainly making some mistakes

The ion is not one S atom and 4 O atoms together. The ion on the whole is having two negative charges, Thus it is somewhat like 1Sulphur + 4 Oxygen+2 electrons. It is not as though there was one sulphur and four neutral oxygen atoms, and they became negatively charged during bond formation. The bond must have formed in some reaction, and in that reaction the charge would have been conserved.

Yes, consider sulfuric acid- H2SO4. It is a strong acid and when it completely disassociates, it loses 2 H+ atoms. This leaves a $\ce{SO4^{2-}}$ ion which must be -2 charged due to the conservation of charges.

Yeah I made some mistakes myself. . S and 2 O are not connected with coordinate bonds but normal covalent bonds and extra electrons come from reactions. I was thinking wrong in the first place

Tho sulfur and oxygen are not neutral atoms. The oxidation state of sulfur is 6+, and of each oxygen is 2-. It slightly helps to think about it as being bonded ionically, even though the bond nature is rather covalent. Together it all ends up being 2-.

There are partial charges on the S and the O in SO4[2-], the Sulfate ion.  Sulfate ions are found in sulfuric acid = H2SO4, which ionizes into SO4[2-] + 2 H+
Oxygen is electro-negative. It has a partial negative charge. Think of water = H2O, which can break down into H+ and OH-.
With oxygen pulling the electrons toward itself, the sulfur gets a partial positive charge.
Sulfur can have either a partial positive charge or a partial negative charge.  Hydrogen Sulfide = H2S, which can break down to H+ and HS- [sort of like water].

An S atom has 6 valence electrons and a O atom has 6 valence elections
SO4 => (6x1) + (6x4) = 30 valence electrons
We would need 32 valence electrons for 4 full outer shells(8x4) but only have 30. Thus the charge is 2- for the ion.

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Work in free expansion
I have read that no work is done during free expansion and temperature of an ideal gas undergoing free expansion doesn't change.
Let us assume an ideal system of piston and gas which has vacuum around it. The gas will expand but we say there is no work done by the gas. Why so ? Gas applies a pressure(& hence a force) on the piston and the piston gets displaced. So acc to F.s = work , there is work done by the gas !

The words "free" and "piston" contradict each other.

@ivan I have been taught in school that if we assume a weightless piston and vacuum around, we can regard it as free expansion. Is that incorrect ?

Weightless piston is another story. It is as good as no piston at all.

I think what biogirl is referring to is a massless piston.  If you do a force balance on the piston (using Newton's 2nd law), the force exerted by the gas on one side of the piston is equal to the force exerted by the vacuum on the other side of the piston (i.e., zero force).  So the force that the gas exerts on the massless piston is zero, and the gas does no work on its surroundings.

@ChesterMiller I know this is a silly doubt but can the pistin move freely inside the container?

@ChesterMiller And why did we take it massless

The piston can move freely (without friction) in the container.  It was made massless because the analysis is easier to conceptualize this way.

Related: Why is the work done by a gas during free expansion zero and not infinity?

Little bit tricky question. As you say, the work is force acting along the path. But there is no force acting, as there is weightless piston with vacuum on the opposite side.
To visualize the process better, imagine an inflatable balloon in outer space, when you puncture it. All gas molecules continue to fly in the direction and speed they had in the moment of accident (therefore the temperature is the same) and they exert no work (no collision thereafter).

Is there a reason for which we considered a weightless piston

I have seen some answers on this similar question they all point to same thing that pressure outside is zero so from $W = p.\Delta V$, work done will be zero or as said in the answer from @ssavec that there is no force acting on the piston, the point is that you cannot have a balloon in space, it will burst due to inner pressure.
My physics teacher gave another perspective for this situation which in my opinion is much more relevant. It goes like this:
Since the gas is in a vessel with massless piston, then it will definitely have some pressure in normal conditions and even in vacuum, but the twist here is that, this pressure once pushes the piston and after that because piston had no resistance, it went out with some constant velocity. There was no direct force and displacement relation but just a sudden push from gas which made piston leave immediately.

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Entropy of universe for non spontaneous processes
When spontaneous processes are accompanied with an increase in entropy of universe and remains unchanged in a reversible process, what happens to entropy when one process is non spontaneous? Would entropy decrease??

Yes. Don't take the second law to mean that these non-spontaneous reactions can never happen. They can happen. They are just statistically speaking extremely unlikely to happen.

What are you referring to when you use the term "non-spontaneous?"  Please define this term so that we are all on the same page.

@chester miller, this is a question from Brown's Chemistry The central science 13th ed. Page 827. The question is just as simple as that. And the answer is no.

@orthocresol  Please see my comment on Costa's answer, below.  And just to paint a picture of how large 10^(2 x 10^23) is, consider that there are estimated to be ~10^80 atoms in the observable universe.  If there were one of our universes for every atom, then the total number of atoms in those 10^80 universes would be "only" 10^160.  Now suppose you do it again, where each atom represents 10^80 universes. Then you've only made it to 10^240 atoms.  You would have to repeat this process (of expanding atoms into universes) 8.2 x 10^20 times before you would reach 10^(2 x 10^23) atoms!

Yes. That is why non-spontaneous reactions do not occur unless you provide some artificial interference. If you were to cause a non-spontaneous reaction to occur in this way, in that single reaction, entropy would decrease.
However, in doing so, the entropy of the universe will still increase or stay the same BECAUSE, in the process of creating such an artificial interference, entropy is created.
For example, if you add energy for a reaction to occur, the energy must be obtained from... well, a non-spontaneous process, where an increase in entropy occurs.
Or more simplistically, if your room is very messy (high entropy), and you clean your room (lower entropy), it may seem like the entropy has just decreased. Actually, the entropy has still increased because in order to clean your room, you do work, which involves multiple other increases in entropy: digesting food which came from a food-processing plant, which was driven to this plant by a truck that adds entropy by burning gasoline, and so on.

Start by remembering that when we say that a process is spontaneous we mean under certain conditions e.g. at room temperature. The reverse of spontaneous processes are non-spontaneous processes. Non-spontaneous processes do not occur naturally according to classical thermodynamics and it is statistically extremely improbable that they will occur naturally according to Boltzmann thermodynamics.
But non-spontaneous processes can be forced to occur with the continuous expenditure of energy. For example, the expansion of a gas in a piston can be reversed using a heavy weight on the piston. A chemical example of a non-spontaneous process is provided by the electrolysis of water.  In both these cases, the values of the thermodynamic parameters for the non-spontaneous process are equal to those for the spontaneous process but of opposite sign. However, this ignores the source of the consequence of using the energy we use to bring these processes about. As Frank put it (above), 'the process of creating such an artificial interference, entropy is created'. For example, in the electrolysis of water, the entropy change of the H2/O2/H2O system is negative, but because the use of electricity causes the water being electrolysed to warm up, this involves thermal enerfgy transfer to its surroundings, causing a positive entropy change and  the overall entropy of the universe increases. Spontaneous or non-spontaneous, the entropy of the universe always increases.


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How to control the purity and quality of fatty acids or their CoA thioesters?
Background: We are looking into getting a series of cis-mono-unsaturated very-long-chain (C20-C32) fatty acids (with different, specific double-bond positions) — and/or the corresponding acyl-Coenzyme A thioesters — synthesised by a company. These compounds are to be used in primary research for a biochemical enzyme activity assay.
Question: Would MS or 1H-NMR be preferable for quality control/identification of such (custom-synthesised) compounds? Or would it not matter for these compounds at hand? What would be the advantages/disadvantages of these two options we are being offered?

Short answer: Both methods are (technically) suitable but it depends on what impurities you want to detect. But in your case, I would probably suggest mass spectroscopy in most cases.
Long answer:
It depends on how many different compounds you’re getting, how much you trust the company, which impurities you are expecting and how important it is to have exactly the one compound you need.
Mass spectroscopy, ideally coupled with HPLC beforehand, will give a specific retention time for a specific compound, and of course you will get the compound’s mass. That alone is not sufficient for a priori identification! When one detects purity by mass spectroscopy, one would usually have a standard run of one’s desired compounds and compare it to the unknown sample. To identify, compare the retention time (ideally standardised, if the project goes on for more than, say, a year) and the compound masses.
The different fatty acids will have similar, but not identical retention times. The cis-double bond should allow for easy detection via UV absorption (good for quantitising known impurities) and the mass will, of course, differ only by the number of carbon atoms. Impurities that are not fatty acids will have completely different retention times, different UV spectra and different masses which should allow an easy attribution of ‘good peak, bad peak’. If it is not of paramount importance to have only exactly one fatty acid, one can even rest happily with quantitising the amounts of the different ones, too.
Note that it will take some fine-tuning of the HPLC gradient, since most gradients are designed to fit a wide range of polarities with average accuracies, while what you want is a high accuracy in a low polarity range.
Everything I said about HPLC-MS of fatty acids also holds true for the CoA-thioesters, although it is possible that the HPLC solvent hydrolyses the thioester. Work-arounds would be buffered solvents and similar.
$\ce{^1H}$-NMR is best used on pure substances, but can also be used to detect impurities. It should allow for an easy identification on which bond is a double bond by the characteristic shifts and/or multiplets of the two vinylic protons (shift range $6-7~\mathrm{ppm}$). However, you will most likely get unseperable signals at around $1-2~\mathrm{ppm}$ — the aliphatic region where all the saturated carbons’ protons will resonate. Thus it will not be easy to detect saturated impurities, say for example the corresponding unsaturated fatty acid. However, it will be easy to detect other impurities stemming from the synthesis such as remaining reactands (which will depend on the synthetic route).
If you go to the CoA-thioesters, it will be a lot harder to estimate purity just by an NMR spectrum, because CoA has a lot more signals in the $2-5~\mathrm{ppm}$ range. But it is still possible.
NMR would allow the quantification of impurities if you can identify them just by integrating the proton signals and dividing them by the amount of protons giving identical signals.
On a time scale, a typical HPLC-MS run will be around $40~\mathrm{min}$ while a typical proton NMR can be recorded within $5~\mathrm{min}$.
I would suggest NMR in situations where you have a really high number of samples that need to be analysed very timely, where it does not matter if the saturated acid is present or where identifying the stereochemistry and regiochemistry of the double bond is of prime importance. In any case where saturated fatty acids would impede your results, I would strongly suggest HPLC-MS. In either case, I would suggest acquiring pure samples (at least of the fatty acids) first to run a series of standards.

Thanks for this detailed answer. I guess part of the problem we face is that no good standards are commercially available, which is in fact the reason we are looking into custom synthesis. I suppose we would be most concerned with impurities that would matter in terms of biochemistry, especially other compounds that could act as enzyme substrates -- in other words, saturated vs unsaturated acyl-CoA and what proportion of acyl-CoA have the "right" double-bond position.

@PhiS It would boil down to using the same compounds as standards … You could hope to find a preparative HPLC somewhere to separate samples (at least as far as you can detect them) and then using the separated samples as standards. Pure samples can be further analysed by NMR to unambiguously determine correct double bond positioning.

authors: امنة فارس عباس


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Name of round piece of glassware with thin neck?
Found in the lab, it looks like a round bottom flask with an extremely thin neck. Effectively a glass sphere about 60 mm in diameter with what looks like a capillary tube as the neck. Not sure whether it is a standard COTS item or custom glass.

Might be a pyknometer -quite a bit bigger than what I've used before, though

Or possibly a hydrometer, if one believes the wiki page https://en.wikipedia.org/wiki/Relative_density

We used to have glassware like that in our lab.  It is basically an "unfinished" piece of glassware. In a nutshell, it was used to handle (collect, transport) gases.
It was easy and inexpensive for the glassblower to make several of those bulbs at a time; he'd make a variety of sizes.  We'd keep them in a drawer and then before we used them we'd do some glassblowing ourselves and add the expensive part - a stopcock with a ground glass joint.  Then, having already measured the volume, we could attach one to a vacuum line (no vacuum) with the stopcock open and feed a gas into the vacuum line, close the stopcock and remove the bulb.  Knowing the pressure, temperature and volume, we knew how many moles of gas were in the bulb.  We could then attach the bulb to a reaction flask and open the stopcock, so that a known amount of gas was used in a reaction.
Here is a simple picture, imagine the tube on the bottom is the vac line.

These bulbs were also used to trap gaseous reaction products by attaching the bulb with its stopcock and joint to a vac line along with a reaction flask.  The bulb was cooled to the desired temperature and when the reaction produced gases, they could be collected in the cooled bulb.

Given the stuff we do in the lab, that sounds correct

That piece of glassware is quite possibly a Florence flask. It has a round bottom and long neck, and is used to boil liquids.
To be fair, I have not seen a Florence flask with such a narrow and long neck, so there was probably a very specific need in the design of the glassware.

I never knew that was the name for such a flask!


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The weight of a 2cc glass pharmaceutical flame sealed ampoule?
This is one of the flame sealed type whose neck is broken before use. There seems to be lots of info on dimensions but none with regard to weight.

Break one and weigh. Note that there is not much sense in weighing to a greater accuracy than grams because every ampoule will have a different weight; these weights will differ at least by tens of milligrams if not by hundreds.

Problem is that I don't have one, and I do not want to buy a big batch just to weigh one.

Why do you need its weight then?

To attach to a microbalance that is mass limited and to weight in gas to determine gas density. Volume would be found by calibrating with gases of known density at a given temperature

I grabbed 10 empty, unsealed ones and threw them on the analytical balance and got mean: 2.4071 g, s.d.: 0.01264 g for Wheaton amber glass pre-scored vials (part number 176796). The variation between vials isn't too bad, but I don't know if they can be filled and sealed consistently enough to determine gas density like that to any useful precision and other brands may be better or worse.

$+1$ for ‘I don't know if they can be filled and sealed consistently enough to determine gas density like that to any useful precision’

Think buoyancy...


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What do positive, negative and zero overlap of atomic orbitals mean?
I am confused about positive, negative and zero overlaps.
Do they represent the extent of overlap or are they related to the bonding and anti-bonding orbitals as described in Molecular Orbital Theory?

Where do you get these? Mention the resource or the book where you got that & elaborate it.

Page 7 @ http://www.chem.tamu.edu/rgroup/dunbar/Teaching/CHEM362/Lecture_7.pdf

I think by positive overlap, constructive interference between the atomic orbitals may've been referred;  negative overlap to anti-bonding  that is,  destructive  interference & zero-overlap to  zero-bonding`.

Related: What are overlap integrals?

The terms in the lecture slides you refer to describe the situation in a certain, very specific subset of molecular orbital theory: the idea of linear combining single atomic orbitals to form bonding and antibonding orbitals. While one can arrive at the same bonding/antibonding orbitals with elaborate molecular orbital theory through the localisation of orbitals approach, the simplified view these lecture slides (and many others) present is not consistent with molecular orbital theory. (However, a subset of the result can be obtained by a proper symmetry analysis.)
Let’s consider a rather simple case such as $\ce{HCN}$. It is obvious that the carbon atom’s atomic orbitals will in some way interact with the atomic orbitals of both hydrogen and nitrogen to form bonds. In the method presented in the lecture slides, you would look at certain pairs of orbitals and then determine what type of interaction would be between. I will only present a few representative examples. For consistency, I will define the molecule axis to be my coordinate system’s $z$ axis.

The $2\mathrm p_z$ orbital of carbon and the $2\mathrm p_z$ orbital of nitrogen.
These two p-type orbitals are aligned in the same way. Without taking their phases into account, we can see that the $-z$ lobe of the nitrogen orbital and the $+z$ lobe of the carbon orbital can interact. Depending on whether the phases match or do not match, we would arrive at a bonding and an antibonding orbital, respectively. The resulting bonding orbitals would be a $\sigma$ and $\sigma^*$ orbital.
The $2\mathrm p_x$ orbital of nitrogen and the $2\mathrm p_x$ orbital of carbon.
These orbitals are parallel and still well aligned. However, at any one time both lobes of one will interact with both lobes of the other. There is a nodal plane that completely contains the bond axis. If the $+x$ lobe of both orbitals feature the same phase, the interaction is bonding. Otherwise it is again antibonding. The bonding orbitals are $\pi$ and $\pi^*$, respectively.
The $2\mathrm p_x$ orbital of carbon and the $1\mathrm s$ orbital of hydrogen.
Now, we are trying to mix an orbital that only has one phase (the s orbital) and one that has two phases (the p orbital). This would be okay if only one of the two phases of the p orbital were pointing towards the s orbital; however, because the p orbital is perpendicular to the bond axis, both its phases would have to interact at the same time. Thus, we have one interaction on the $+x$ half of the coordinate system which is bonding and at the same time an interaction on the $-x$ half which is antibonding. Mathematically, the bonding and antibonding contributions cancel out perfectly. Thus, there is no interaction between the two orbitals. This is zero overlap.
The $2\mathrm p_x$ orbital of carbon and the $2\mathrm p_y$ orbital of nitrogen.
Now, we are attempting to mix two different p orbitals. However, they are not oriented in the same direction. The $+x$ direction of one orbital and the $+y$ direction of the other have the same phase while the $-y$ and $-x$ directions have the other. If we start at $+x$ and go around in a circle, we will start in an area of no interaction (the $\mathrm p_y$ orbital has a nodal plane in the $xz$ plane). Then, turning towards $+y$ we reach an area of bonding overlap. After crossing $45°$, the bonding overlap decreases until we reach zero again at $+y$. From there to $-x$, we experience an increasing and decreasing antibonding interaction. We then rinse and repeat what we just saw for the remaining $180°$.
It would probably not be impossible to consider this interaction a curly bond that rotates around the central axis by $90°$ between the two atoms. However, traditional analysis would note that at the same time there is something like an antibonding second part displaced by $90°$ to this. Therefore, the analysis would consider these two orbitals nonbonding with zero overlap.

In a proper MO or LCAO approach, molecular orbitals will stretch across the entire molecule. Thus, the analysis of whether an orbital is bonding or antibonding with respect to a certain bond reduces itself to the question of ‘does this orbital feature a nodal plane along the bond axis or does it not?’. From the $+,-,0$ picture we had above, we arrive at a black and white picture. This is because nonbonding interactions are typically separated out before the molecular orbitals are even calculated by a symmetry analysis. If you do that, you would realise that the $2\mathrm px$ orbital and the $1\mathrm s$ mentioned in case three are representative of two different irreducible representations. (Strictly speaking, $2\mathrm p_x$ and $2\mathrm p_y$ both belong to the same degenerate irreducible representation.) Only orbitals that belong to the same irreducible representation can be combined in bonding or antibonding orbitals.

The atoms combine by colliding with each other. But what does this mean on atomic level. This situation refers to the process in which the two atoms comes so close to each other that they penetrate each other’s orbital and form a new hybridized orbital where the bonding pair of electrons reside.  This hybridized orbital has lower energy than the atomic orbital and hence are stable. It is in the minimum energy state. This partial penetration of orbital is known as orbital overlap.
The extent of overlap depends on the two participating atoms, their size and the valence electrons. In general, greater the overlap, stronger is the bond formed between the two atoms. Thus, according to the orbital overlap concept, atoms combine together by overlapping their orbital and thus forming a lower energy state where their valence electrons with opposite spin, pair up to form covalent bond.
The importance of orbital overlap was emphasized by Linus Pauling while explaining the molecular bond angles observed through experimentation and is the basis for the concept of orbital hybridization.
Directional Properties of bond
The molecular bond angles were explained through the directional properties of bond. The molecule of hydrogen is formed by overlap of 1s orbital in head on collision.
Overlapping of Atomic Orbital
When two atoms come in contact with each other to form a bond, their overlap can be positive, negative or even zero depending upon the phase and sign of the two interacting orbital.
Positive Overlapping of Atomic Orbital – When the phase of two interacting orbital is same, then the overlap is positive and in this case, the bond is formed. The phase of the two interacting orbital (+ or -) comes from the sign of orbital wave function and is not related to the charge in any sense.
Negative Overlapping of Atomic Orbital – When the phase of two interacting atomic orbital is opposite, then the overlap is negative and in this case, the bond is not formed.
Zero Overlapping of Atomic Orbital – When the orientation of two interacting atomic orbital is such that there is no overlapping of the orbital, that is known as zero overlapping.

That is very simplistic and also part wrong. It is not true that more overlap means stronger bonding. There is always a careful balance, as electron election repulsion is also increasing.

These are the lines from the high school Chemistry Textbook of India

@Martin ... So you say that its not completely true. Why?.. The extent of overlap decides the strength of covalent bond, is'nt it?

No it doesn't. More overlap means more repulsion. It is very dependent on the system studied, which one rises faster than the other, and where the equilibrium is. In the H2 molecule, overlap goes to 1 when distance goes to 0, but the bond would be weakest there.


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What is the purpose of froth in washing clothes?
I stay at a rented place and there is no washing machine. I, now, wash clothes in a bucket. When I add detergent, sometimes there is froth, sometimes there is not. I wonder if this had any impact on the washing process. Though I am aware of micelles and all that, but I wonder if froth formation is nearly redundant.

The froth has little or no effect on the detergent action. In fact detergent manufacturers have to add anti-foaming agents to stop excessive foam generation in automatic washing machines.
Froth/foam is generated because surfactants in the detergent adsorb at the air water interface and stop the water film that makes up the walls of the bubbles from collapsing. Note that the surfactant does not cause bubbles to form - bubble formation is always energetically disfavoured and the foam is only kinetically stabilised. The foam is caused by agitation entraining air into the surfactant solution so the amount of foam will depend on the intensity of the agitation as well as the type and concentration of the surfactant.
It is conceivable that in some cases the foam may assist with dispersal of mineral soils. This is the basis of the industrial froth flotation process. However this plays at most a very minor role in domestic washing machines. Not that it's directly relevant, but I mention it out of interest, foam is thought to play a role in cleaning teeth i.e. in toothpaste. It entrains particles of food so they can be spat out along with foam.

"kinetically stabilised" went over my head: I presume the link I looked up and added is the relevant one?

@WetSavannaAnimalakaRodVance: yes, though that's a very general article. Why foams are kinetically stabilised would be a whole new question.

FYI, detergent manufacturers want foam in some cases:  (A) in consumer-visible applications (e.g., dish liquid, or hand "soap") where consumers perceive the foam as evidence of the product's cleaning ability, and (B) in applications where it is desirable to spread the product over a large area without using too much water.  The foam bulks it up, and makes it easier to spread.

I recall a marine aquarium I used to have used a "foam fractionator"/"protein skimmer" as part of the filter. Some organic molecules are attracted to the air/water interface.  Now if I happened to spill that on my shirt when emptying it, the cloth is now dirty with gunk that specifically will come out with froth.

And then they do exactly the opposite to maintain the foam aka "head" on a glass of beer!

I like the technical answer. Here's a practical one: Froth indicates that you have more than enough detergent to do the job. Use too little detergent for an especially oily mess and all of the available detergent is "consumed" by the oil with none left over to make froth. That's why after using dishwater for a while, the plates don't "feel" clean anymore. One might say that any froth is a waste of detergent for cleaning purposes, but others say just a little froth is a small price to pay for the reassurance it gives. A lot of froth is simply a lot of waste.


text:
Does PO3- exist as a free ion or in a cyclic/linear polymeric form?
My book says that $\ce{PO3-}$ exists as cyclic or linear polymeric structure but doesn't elaborate on this. I Googled this and found no result either.
Also, what would be the structure of $\ce{PO3-}$ cyclic or linear polymeric structure.

https://en.wikipedia.org/wiki/Sodium_hexametaphosphate for example

Also, there is nothing unusual about that: $\ce{SiO3^2-}$, too, exists as a variety of polymeric structures, but never as a free ion.


text:
Why is honey more viscous than water?
Why is honey more viscous than water? And, if we boil honey to just below its boiling point, will it become equally viscous as water at normal temperature? How does changing temperature affect viscosity?

Viscosity generally decreases with increasing temperature so honey will become less viscous at higher temperatures. Regarding the actual values, look at this article. It appears that even at quite high temperatures honey remains a couple of times more viscous than water.
I think the reason for higher viscosity in honey is the high sugar () content - hydroxyl groups in sugars can form strong, hydrogen-bonded interactions with water and each other, and the sugar molecules being quite long, they are difficult to shear.


text:
Difference between precipitation and crystallisation
What is the main difference between precipitation and crystallization.
Is precipitation a chemical change and crystallization a physical change.
How can we say that the resulting solute is precipitate or a crystal?
Is it necessary that when we want precipitates to form then its ksp should be less than its Qsp or the precipitates form even after mixing two unsaturated aqueous salt solutions?

The two words are often used interchangeably (perhaps incorrectly), though there is a slight nuance.
IUPAC defines the following:

Crystallisation: The formation of a crystalline solid from a solution, melt vapour or a different solid phase, generally by the lowering of the temperature or by evaporation of a solvent.
Precipitation: The sedimentation of a solid material (a precipitate) from a liquid solution in which the material is present in amounts greater than its solubility in the liquid.

In essence, a crystallisation is the formation of a crystalline solid (i.e. not amorphous), for instance leaving a beaker of $\ce{NaCl}$ in water out for a long time will eventually grow crystals as the water evaporates.
Precipitation can be thought of a something 'crashing out', for instance if you have certain kinds of amines, you can add $\ce{HCl}$ in diethyl ether to precipitate out the $\ce{HCl}$ salt of the amine as they are not soluble.
There are of course in-between cases. Attempts at 'crystallisation' can yield amorphous solids if the conditions are not right, and, equally, precipitation can sometimes give crystalline solids.
The choice of words is therefore always ambiguous, but in general the IUPAC nomenclature should be followed where appropriate (eg. you can never crystallise to give an amorphous solid).
To answer the other part of your question about chemical vs. physical changes, you are correct that crystallisation is always a physical change (different isomorphs of the same material: amorphous, crystalline needles etc). Precipitation can be either, for instance in the $\ce{PPh3}$ case described by Jan, or in the $\ce{HCl}$ amine salt I previously described.

I don't think the distinction between chemical and physical is quite right. Yes, sometimes we add HCl to cause a salt to crystallise, but this is a chemical change followed by crystallisation which we do because some types of chemical (e.g. salts of amines) are easier to crystallise than the raw base.

Both are physical changes. In neither does the substance crystallising/precipitating truely change, in both it is more or less the same after as before. (Technically, ions in solution are something totally different than ions in a crystal lattice, but especially for organic chemicals the difference is neglegible if present at all.)
The terms are not really clearly defined and there is a big grey area in the middle where both can apply. But in general, one would speak of crystallisation if it is a slow process that mainly happens at the bottom of the solution and yields something with a defined crystal structure. Precipitation would be a process which can happen throughout the entire solution rather quickly and tends to result in something cloudy, ill-defined.
So in the case of adding a chloride solution to one of silver nitrate, when silver chloride is formed immediately and almost quantitatively as a cloundy substance everywhere where there are chloride ions, one would not call it crystallisation but precipitation. On the other hand, if I dissolve copper sulphate in water and let it stand at cold temperatures for a long time to generate large single crystals, I would speak of crystallisation and deem precipitation to be wrong.
An example for the grey area would be the removal of triphenyl phosphane from a solution in dichloromethane or ether by adding hexanes. The precipitate forms rather fast, especially if the more polar solvent is evaporated, but at the same time it is rather crystalline looking (I have no clue if they are actually crystals or just look nice). I have heard both crystallisation and precipitation in this context.

This is not a definition, but mostly we use it as follows:

Crystallization - Product forms as nice crystals, generally because you reduce the solubility slowly.
Precipitation - Your product forms microcrystals, because the solubility was reduced too fast and crystals couldn't grow idiomorphic
Crash out - Either your product is very insoluble, or you were way too impatient to let it cool slowly. Your product will form as a a fine powder, a clump of microcrystals or something along those lines.

Crystallisation is formed from a super saturated solution while precipitation is formed due to reaction of 2 substances to form sparingly soluble product

Well, I guess this could count as a difference but who says precipitation can only occur because of reactions?

No, this is incorrect. The terms are not defined like this.


text:
Elucidating reaction mechanism and purpose
Please help me understand synthesis sequence shown below:
Benzylbromide is used for protection, probably from action of $\ce{CuBr2}$, but it's reaction mechanism is unknown for me. The bromination of the methyl group is some kind of radical reaction? What would it do to the rest of the molecule weren't it for benzylbromide? Next, 18-crown-6 is clearly used as catalyst for halogen exchange — but how does it function? Does it enclose bromine making $\ce{Br-}$ a better leaving group or $\ce{F-}$ better nucleophile? Finally, how does $\ce{CF3COOH}$ manage to remove that protecting group? Maybe it protonate benzilic $\ce{O}$ leaving the stabilized carbocation behind? But this seems quite unlikely since this would probably attacked fluorine or carbonyl.

Please use the \ce{...} command for chemical formulae. $\ce{CuBr2}$, for example, automatically gives you $\ce{CuBr2}$. Crown ethers are generally used to solvate the cations not the anions; the reaction on the second arrow can be thought of as a simple $\mathrm{S_N2}$ reaction. Also, on the first arrow you have first and second steps, meaning benzyl bromide is applied first (likely with a base), that product is isolated and then subjected to $\ce{CuBr2}$. The latter must be the reason for bromination, but I can’t give you a mechanism.

Thanks for suggestion Jan, it is indeed more practical. I see, so this crown ethers sequester from $\ce{F-}$ their $\ce{K+}$ countercations which make them better nucleophiles?

I would guess so, 18-crown-6 is known for its particularly high affinity for $\ce{K+}$ if I am not wrong. Each of the crown ethers has their own "favourite" cation, for 12-crown-4 it's $\ce{Li+}$, 15-crown-5 is $\ce{Na+}$, and so on. On top of that, if you are using an organic solvent, I think the addition of the crown ether is what makes KF soluble (an important practical consideration).

Not sure about the particular case of KF. I know the solubility thing is important for some other salts.

@orthocresol I think you’re right there. However, it is impossible to know whether the crown is needed without knowing the reaction solvent.

I have no idea what the addition of benzyl bromide accomplishes. There doesn't seem to be any need for a protecting group in any of the subsequent reactions. The rest of the reactions are relatively easy to explain.
This paper reports essentially 100% yield for bromination of 4-hydroxyacetophenone with $\ce{CuBr2}$, refluxed in chloroform-ethyl acetate for 15-20 mins. There is no mention of any need for a protecting group for the phenol. No mechanism is given but you are correct in your assumption that radicals will be involved.
The second reaction is a regular $S_N2$ reaction. The transition state of the reaction is highly stabilized by the adjacent carbonyl group and so the reactivity is dramatically increased compared to a regular primary halide. 18-crown-6 is used to solvate the $\ce{K+}$ cations, for which it has a very high affinity. This leaves the fluoride free to react, making it far more nucleophilic than it would otherwise be. The reaction will go to completion easily because bromide is a far more stable leaving group than fluoride.
The final reaction is a standard ether cleavage, catalysed by a strong acid. The oxygen is protonated by the acid and leaves to reform the phenol and the stabilized benzyl cation.


text:
Can a brine water electrolysis be looped?
Electrolysis of brine yields hydrogen and chlorine gases and aqueous $\ce{NaOH}$. So is it possible to combine the gases to form $\ce{HCl}$ gas, and mix it back with the aqueous $\ce{NaOH}$, so it returns to brine water?

Of course. But the reaction of $\ce{H2}$ and $\ce{Cl2}$ would be explosive. The combination is often called Chlorknallgas — chlorine explosive gas — in German.

Yes I have seen the individual reactions on Youtube, but I just never saw anyone loop the whole process.

I have to know, is this the start of the most ill advised perpetual motion/over unity machine attempt ever?

No, I never thought that this can be an over unity machine from the start. This may be of other uses though.

Well there would be no reason to loop the process other than demonstrate that it's possible …

there are actually very good rea sons to loop this process. if you can use these salts to store and transport hydrogen this could be an important component in a renewable energy system.


text:
Is xenon electropositive in compounds of xenon?
Do you agree with the below passage in my text?

Electronegativity is defined as the ability for something in a covalent framework to attract electrons also in covalent bonds.
Electropositivity is the inverse of electronegativity or the ability of something to donate/release electrons in a covalent framework.
How can a highly oxidized element be very electropositive? A highly oxidized element by definition has a minimal amount of electrons to be donated.

Hmm, there is some sort of electronegativity defined for noble gases. Seems like good ol' Pauling scale, and suggests that Ne is more electronegative than F.

@Dissenter You seem to be mixing up several unrelated concepts. Electronegativity is a property of an element in general, not of its particular oxidation state.

@IvanNeretin - I welcome an answer from you on this topic; I remember we chatted about this but a drawn out, written, and permanent answer here on this site would be helpful too! I think it's good to have competing viewpoints (as me and my professor already do).

How can a highly oxidized element be electropositive?

An element can be electropositive, the highly oxidized element less so.
There are a number of things that don't sound quite right in the passage you quote.  For example, the Pauling electronegativity for xenon has been assigned (2.6, if you'd like to see the methodology used to caculate the electronegativity of a noble gas, see this earlier answer). Also, depending on the compound, hypervalent (or hypercoordinate) bonding, rather than use of d orbitals, might be used to explain the bonding in some noble gas compounds

thank you for the helpful answer! "An element can be electropositive, the highly oxidized element less so." - Haha, I agree, and I think you know what I meant :p. Will edit wording though.


text:
Constrain bond angle In Gaussian molecular structure optimization
I am computing the optimized molecular structure of an organic molecule in the Gaussian 09 computational chemistry package.
As I have it now, I do not think that the optimized structure that I am getting from Gaussian is correct, in part due to the symmetry of some of the bonds.  To try and correct this, I want to fix two of the bond angles in the molecule in order to help/force it to produce a more believable structure.  Even though I want to fix these bond angles, I want to leave everything else variable and open to optimization, including those bond lengths.
How can I specify this in the input file?
Fixing the bond angle to a set value seems like an extreme solution to this problem.  Perhaps I can just constrain the angle within a certain range, is this possible?
After a lot of work and playing with settings, I have figured out how to fix bond angles, but now my difficulty is fixing them to 180 degrees (my desired value), for linear bonds. It seems that this is very hard to do.
Fixing the bond angle to 179 degrees (or 179.9 degrees), results in an error saying "Error in internal coordinate system." A lot of people online say this is because of linear/nearly-linear bonds in the molecule being optimized. Some people also say this might be avoided if I try to optimize in Cartesian coordinates, instead of internal coordinates, using the cartesian parameter after opt. However, I do not think I can use ModRedundant and cartesian parameters at the same time.
How can I fix a bond angle to be linear, but still optimize the rest of the structure in Cartesian coordinates? Is it even possible to fix angles to 180 degrees?

Is this an organic or inorganic structure and what method do you use? But fixing an angle to sth that you believe might not be what is "right".

And what method do you use for that organic molecule, which leads to conpletely unbelievable results?

I don't think it is the method/basis set's fault.  Try many different choices all give me the same problem.

Fixing the bond angle to a set value seems like an extreme solution to this problem.

It is. To be honest, fixing parameters in a calculation is almost always no solution at all. The resulting geometry has little to no meaning and all derived properties are for a purely hypothetical case. If the located geometry is no stationary point, what does that even mean? A wrong solution is worse than no solution at all. You have to be very careful with what you fix and when you fix, and you should always confirm what you have found with a full optimisation and a frequency calculation. (On various levels of theory.)
I think Greg's advice is also very important. Symmetry can be a very powerful tool if utilised correctly, but you should make sure that the implied symmetry is correct. You also need to check whether your methodology matches the system you are computing.
All that being said, there are certainly a few very productive uses for partial optimisations:

finding higher or unusual local minima
scanning the potential energy surface
modifying converged geometries
fixing transitional modes
etc.

There are a few possibilities how to make partial optimisations in gaussian. All have some caveats and all may lead to undesired results. There are certainly restrictions to all of them. I'll give a short overview the most common ones with a few practical examples.
You have already noticed, that fixing a bond angle to 180 degrees is a challenge. This is due to the fact, that it is a requirement to have linearly independent variables, which is simply not possible with threes point on a line. However, there are certain ways to work around the problem. Spoiler alert: It is only possible with z-matrices or very rigid symmetry (where you fix the symmetry, not the bond angle).
Partial optimisations with z-matrices
Let's start with something very simple. A full optimisation of methanol without any symmetry restrictions.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt
int(ultrafinegrid)
sym(none)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc4         2 hch4          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 hoc6          2 dih6

hc2           1.1
hc3           1.1
hch3        109.5
hc4           1.1
hch4        109.5
dih4        120.0
oc5           1.4
och5        109.5
dih5       -120.0
ho6           0.9
hoc6        109.5
dih6        180.0

This will result in the following optimised geometry:

 c
 o    1 oc2
 h    1 hc3         2 hco3
 h    1 hc4         2 hco4          3 dih4
 h    1 hc5         2 hco5          4 dih5
 h    2 ho6         1 hoc6          3 dih6

oc2           1.414851
hc3           1.109697
hco3        107.476
hc4           1.118853
hco4        113.586
dih4       -118.329
hc5           1.118853
hco5        113.586
dih5       -123.342
ho6           0.974680
hoc6        107.408
dih6        179.999

Fixing an angle in a z-matrix is straight forward. You need to specify that you are doing a partial optimisation (popt) and move the fixed parameter to the constants section. As an example, I am fixing the COH angle, and also invoke symmetry.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 coh           2 dih6
  Variables:
hc2           1.100000
hc3           1.100000
hch3        109.5
dih4        120.000
oc5           1.400000
och5        109.5
dih5       -120.000
ho6           0.900000
dih6        180.000
  Constants:
coh          90.0

This will give you the following list of optimised parameters:

       hc2         1.1104
       hc3         1.1173
      hch3       107.0973
      dih4       116.0195
       oc5         1.4471
      och5       108.92
      dih5      -121.9694
       ho6         0.9796
      dih6       180.0092
       coh        90.0

Now we will try the same, but we will make the bond angle 180 degrees. Here we need to introduce dummy atoms:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
xx    5 xxo6        1 xxoc6         2 dih6
 h    5 ho7         6 xxoc6         1 dih7
  Variables
hc2           1.100000
hc3           1.100000
hch3        109.500
dih4        120.000
oc5           1.400000
och5        109.500
dih5       -120.000
dih6        180.000
ho7           0.950000
  Constants
xxo6          1.000000
xxoc6        90.000
dih7        180.000

Which will give you once again the following set of parameters:

       hc2         1.1209
       hc3         1.1211
      hch3       106.3981
      dih4       113.09
       oc5         1.3644
      och5       112.3772
      dih5      -123.4351
      dih6       180.0
       ho7         0.9465
      xxo6         1.0
      xxoc6       90.0
      dih7       180.0

And the following Cartesian coordinates:

    6
scf done:  -115.581115
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.120932
 H     1.075500     0.000000    -0.316498
 H    -0.421786    -0.989342    -0.316498
 O    -0.695163     1.052866    -0.519431
 H    -1.177401     1.783245    -0.879763

Quick and Dirty fixing with Cartesian coordinates
Appending -1 to an element in a row in the Cartesian coordinates block will fix it. This is a very quick and dirty approach, which may lead to a brute force computation. You are basically restricting the algorithm to rotate the molecule, which leads to less effective calculations. Fixing Cartesian coordinates also fixes the bond length, but it is the only way to force a 180 degree angle with this choice of input.
Here is an example input:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)
symm(loose)

title card required

0 1
 C  -1    0.000000     0.000000     0.000000
 H        0.000000     0.000000     1.300000
 H        1.221600     0.000000    -0.444626
 H       -0.479083    -1.123738    -0.444626
 O  -1   -0.695162     1.052869    -0.519426
 H  -1   -1.177400     1.783250    -0.879755

This will result in the following optimised geometry:

    6
scf done:  -115.581115
 C    -0.663620    -0.000005     0.000000
 H    -1.090591     0.003267    -1.036538
 H    -1.090556     0.896080     0.521076
 H    -1.090648    -0.899281     0.515456
 O     0.700780    -0.000004     0.000001
 H     1.647272    -0.000004     0.000001

Note that this calculation takes about five times longer.
Fixing parameters with modredundant
This option is described quite straight forward in the Gaussian online manual. But let's make an example where we try to restrict the COH angle to 90 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(modredundant)
int(ultrafinegrid)

title card required

0 1
 C     0.000000     0.000000     0.000000
 O    -0.659966     1.143095    -0.466667
 H    -0.000000     0.000000     1.089000
 H     1.026720     0.000002    -0.362997
 H    -0.513360    -0.889165    -0.363000
 H    -0.502133     0.869719    -1.359507

A 5 6 1 F

With this option it is impossible to fix the angle to be 180 degrees, since this is not an allowed value. You also cannot use dummy atoms, as these do not exist in the redundant coordinate system. It might be a good idea, to kill this variable completely, to ensure that the  program will not fail trying to construct it, or when an optimisation is getting close to that value.
Alternatively the angle can be restricted via a linear bend. This is the best option, if you know that the angle is 180.0 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(Modredundant)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L 1 5 6 2 F

The last line freezes the linear angle between atoms 1, 5, and 6. Atom 2 needs to be specified to determine the plane in which the linear bend lies, or is orthogonal to.
Fixing linear angles with general internal coordinates (GIC)
A new way of handling the coordinate system was introduced with Gaussian 16: generalised internal coordinates. It is in many ways superior to earlier versions, however, also a little bit harder to learn. Nevertheless, it is absolutely worthwhile trying.
Here is a sample input for oxoketene $\ce{O=C=C=O}$. In this case I specified the complete set of coordinates necessary.

#p PM6
opt geom(AddGIC)
scf(xqc,MaxCycle=250)
int(ultrafinegrid)
symmetry(none)

Oxoketene (constrained)

0 1
O        0.000000000      0.000000000     -2.210000000
C        0.000000000      0.000000000     -0.770000000
C        0.000000000      0.000000000      0.770000000
O        0.000000000      0.000000000      2.210000000

BondC1O1(Active)=R(2,1)
BondC2O2(Active)=R(3,4)
BondC1C2(Active)=R(2,3)
LinBend1O1C1C2(Freeze)=L(1,2,3,0,-1)
LinBend2O1C1C2(Freeze)=L(1,2,3,0,-2)
LinBend1C1C2O2(Freeze)=L(2,3,4,0,-1)
LinBend2C1C2O2(Freeze)=L(2,3,4,0,-2)

The format is
L(     i,      j,      k,      l,                                 M)
L(Atom 1, Atom 2, Atom 3, <Atom 4 | Orthogonal to Plane>, component)

The linear bend is therefore defined between i, j, k as Atom 1, Atom 2, Atom 3 and must not be dummy atoms (and with GIC these are hardly necessary anymore). The fourth value l, can be another atom to determine the orthogonality/ plane to, or it can be defined as a plane where l = 0, -1, -2, -3 standing for automatic, $yz$, $xz$, $xy$, respectively. The last value M gives the component, of which two are necessary, and must be M = -1, -2.
The methanol example is as follows:

#p PM6
DenFit
opt(AddGIC)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L(1, 5, 6, 2, -1) Freeze
L(1, 5, 6, 2, -2) Freeze

As mentioned, Z-matrix input is one solution.
You also can constrain angle (and other vairable) of Cartesian structures using opt(ModRedundant) Gaussian manual
If your problem is related to symmetry, you should be careful about two other things:

Do you enforce a (probably incorrect) symmetry? Gaussian may picks up a wrong symmetry due to a bad input geometry. You can check that from your output file, and see what symmetry are you constrained to. You can switch off symmetry (both for optimization and SCF part) or can modify your input that Gaussian can find the right symmetry.
Do you have the right method to represent your wavefunction? It is often found mistake that you you a single determinant method to a multideterminant problem or you don't realize the presence of static or dynamic Jahn-Teller effect. If your structure is very different from the expected, you may have this problem.

About related question:
Not that I am aware off, but generally it makes not much sense anyway. If you want to fix e.g. the HOH angle in water between 80 and 90 deg, it will be 90 anyways - same as you fix it at 90 deg. I.e. it will stick to one extreme, that you already can figure out from chemical intuition or previous, unconstrained runs.

Thanks for all answers it really helped me. By the way, all links which has given above were died so here is the new gaussian manual link for z-matrix:
http://gaussian.com/zmat/
I generated my z-matrix file by using Avagadro (from extensions->> Gaussian menu) software here is the download link: https://sourceforge.net/projects/avogadro/files/latest/download I hope it will work for somebody.

I believe one can accomplish this by giving the geometry input in z-matrix form, and using the 'constants' and 'variables' keywords. This page from the Gaussian 09 manual shows in detail how one could use these keywords.


text:
Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acid
Should the C=C double bond below be labelled as (E) or (Z)?

When using the Cahn–Ingold–Prelog (CIP) priority system and the ‘Sequence Rules’, double bonds are split into two bonds and the doubly bonded atoms are duplicated.
Thus

is treated as

where (O) and (C) are duplicate representations of the atoms at the other end of the double bond;
whereas

is treated as

Therefore, the name of the structure that is given in the question is (2E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid.

You should note that in the first step (turning $\ce{C(=O)OH}$ into $\ce{C(-O-(C))(-(O))(-O)}$ and comparing that to $\ce{C(-O)(-O)(-O)}$ the two are still equivalent. You need to step down one step further to see that the highest priority O of the carboxyl group is bonded to a dummy C whose priority is higher than that of the true H on the carboxylic hydrate. That might help clear remaining confusion.

The compound doesn't exist as drawn, it would dehydrate to the symmetric diacid.  That being said, I'd guess Z as the extra hydrogen atoms on the imaginery acid hydrate increase it's priority over the acid.

no changed my mind - according to this The acid is higer priority.  looks like your old professor was right.. The old dog.  http://www.vanderbilt.edu/AnS/Chemistry/Rizzo/chem220a/priority.pdf

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].

@WooTangClan You should edit your answer instead of putting edits in the comments.

Whether or not the orthoacid exists is irrelevant!

Simple. The rule is that at the tie..(carbon)...there are three ligands of equivalent priority...however, since their not both acids, the rule is that sp2 hybridization beats sp3.
Also, you'll never isolate that geminal (or triplet) alcohol, someone called it an acid hydrate..(even hydrates of aldehydes are difficult to isolate). Only as an orthoester can such structures be isolated.

authors: កូនទីបី កំពូលខូច


text:
What type of reaction is the reaction between an acid and a carbonate?
Consider the following reaction:
$$\ce{2HCl + Na2CO3 -> 2NaCl + H2O + CO2}$$
Is this a double displacement reaction or a decomposition reaction?

Does it have to be one or the other?

The first step is a proton transfer (acid-base reaction), from HCl to carbonate, to produce hydrogen carbonate ion. The second step is also a proton transfer, from HCl to hydrogen carbonate to produce carbonic acid. The third step (or maybe better described as a two-step acid-catalyzed process) is the dehydration of carbonic acid to carbon dioxide and water.

The answer is both. A very hand-waving explanation is the following process:
$$\ce{2HCl + Na2CO3 -> 2NaCl + H2CO3 -> 2NaCl + H2O + CO2}$$
The double displacement of $\ce{H+}$ and $\ce{Na}$ occurs between the $\ce{Na2CO3}$ and $\ce{HCl}$.  Since the $\ce{H2CO3}$ product is unstable it will decompose at the high concentration to produce water and carbon dioxide.
A more thermodynamic explanation is that first the reactants dissociate into their conjugate acids and bases.  The $\ce{H+}$, $\ce{NaCO3-}$, and eventual $\ce{HCO3-}$ ions will interact in various proportions to form a equillibrium. The equillibrium between $\ce{CO2}$ gas and water with the aqueous $\ce{H2CO3}$ is important:
$$\ce{H2CO3_{(aq)} <-> H2O_{(l)} + CO2_{,(g)} }$$
The equillibrium will be dictated by the vapor pressure of $\ce{CO2}$:and the concentration of carbonic acid:
$$K_\mathrm{eq} = \frac{[p_{\ce{CO2}}]_\mathrm{eq}}{[\ce{H2CO3}]_\mathrm{eq}} \qquad Q = \frac{[p_{\ce{CO2}}]}{[\ce{H2CO3}]}$$
If the concentration of $\ce{H2CO3}$ is too high we will get: $Q<K$ and thus  more $\ce{CO2}$ must be produced to acheive equillibrium.

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


text:
Why does O2NCl only have two resonance structures?
Why is it that $\ce{O2NCl}$ has only two resonance structures? More specifically, why is the Lewis drawing where $\ce{N}$ is double bonded to $\ce{Cl}$ not considered a resonance structure along with the other two where $\ce{N}$ is double bonded to an oxygen?

$\ce{O2NCl}$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if weighted accordingly.
This last expression in italics is important. Not all resonance structures are equal. People teaching resonance structures usually use expressions like most probable resonance structures. To determine whether a resonance structure is more probable than another one, check the following list:

does it give all main group atoms a valence electron octet?
are formal charges minimised?
are formal charges distributed in accordance with electronegativity (i.e. do electronegative atoms have more formal negative charge than electropositive ones)?

The list is ordered from most important to not so important although traditional teaching sometimes weighs maximised amount of bonds higher and thus arrives at hyperoctet structures.
Now let’s check the structures of $\ce{O2NCl}$ that you are talking about:

We see that all structures fulfill rule 1. However, rule 2 clearly shows a difference: The two structures on the left contain $+1$ and $-1$ formal charges each while the rightmost structure has $+2$ and $-2$. We don’t need to check rule three, we can see that the rightmost structure is unfavourable and thus less close to the truth. Finally, the formal charges are distributed correctly on all three structures, because oxygen is more electronegative.
Therefore, the left two structures are equally probable while the rightmost one is not so; oftentimes in exams, only the left two will be considered ‘correct’. But as I said, the true picture (or something really close to the truth) can only be determined by quantum chemical calculations.

There's nothing wrong with your suggestion. If $\ce{N}$ was double bonded to $\ce{Cl}$, then both $\ce{N}$ and $\ce{Cl}$ would have positive charges and both $\ce{O}$ atoms would have negative charges. That structure satisfies valence bond theory and octet rules, but it looks a bit strange being doubly zwitterionic. The structure with the $\ce{N=O}$ double bond $\ce{N}$ positive and the other oxygen is negative looks the most plausible.


text:
Best way to prepare reference crystal structure with a given force field
I was wondering if somebody could give me some insight into best ways to produce crystal structures, with a given force field, at 0 K. I am interested specifically in glycine, and have obtained the experimental structure from the CSD. I have seen in the literature people refer to annealing, where I equilibrate my crystal at a large temperature, and then anneal. However, I am concerned about the effects of such things as annealing rate here. I have also seen people mention performing energy minimization. However, my concern is typical energy minimization routines, at least to my knowledge, do not adjust the simulation volume according to a specified pressure. Thus, if my virial is positive for my experimental atomic coordinates using my force field, I don't think the minimizer would expand the volume to bring the pressure back to 1 atm. Could anyone give me some guidance. I am trying to prepare glycine crystal structures, at 0 K, using the GAFF force field with the CNDO charges.

I'd also have a look whether there are multiple crystal structures available - if so (one would believer there are), compare the different conformations manually, they might give you some insight into which conformations are allowed/preferred.

Glycine is indeed polymorphic, if thats what you were referring to. I don't think I am quite getting your point. Could you explain further?

It is possible to perform a simulation with constant pressure or constant volume boundary conditions, so it is not a concern. However you should realize that prediction / simulation of accurate crystal structure is neither simple nor the purpose of most MM techniques.

I'm not entirely sure what you're asking. Are you attempting to match the geometry of a molecule from a crystal structure or the entire crystal packing?
Computational prediction of crystal structures is very much an unsolved problem. There have been multiple "blind tests" from the Cambridge Crystal Database, with varying levels of success. In general, the most accurate results come from dispersion-corrected DFT, not force field methods.
In some cases, to speed results and screening, teams will prepare "tailored force fields" by deriving optimized parameters for a particular compound on-the-fly from the DFT calculations. In no way are these intended for general purpose force field methods.
Also, most methods typically have to optimize both the unit cell parameters and the geometry of the compounds in them. Your note about expanding the volume is important - if you are working in this field, you may wish to use such heuristics to optimize the cell parameters, but typical minimizers will focus on either atomic coordinates or unit cells but not both.


text:
Calculating reaction enthalpy given other reactions
$$\begin{alignat}{2}\ce{X(s) + 1/2O2(g) &-> XO}&&\qquad\Delta H=-895.5\ \mathrm{kJ} \\
\ce{XCO3(s) &-> XO(s) + CO2(g)}&&\qquad\Delta H= +484.3\ \mathrm{kJ}\\
\ce{X(s) + 1/2O2(g) + CO2 &-> XCO3(s)}&&\qquad\Delta H = \;?\ \mathrm{kJ}
\end{alignat}$$
I wish to know how to do it. I have seen some questions where there's 3+ equations and at the end a final equation that I would need to determine the $\Delta H$ for the final equation. I have looked online on how to solve this but the tutorials are very confusing and I can't understand them. Can someone explain how to do this and other similar problems like this?
One of the tutorials talks about making sure they cancel out and flipping the equations? Can someone please explain this?

Well, that's pretty much about it: you can cancel things out and flip the equations, just like you would do in algebra, say, with $x=x^2+x-1$.

@IvanNeretin  Ok but why would you need to flip it? Like can you explain step by step how to solve these types of problems? I'm totally lost here.

@surelyourejoking I'm assuming that's not proper formatting, but that's how I think of chemical equations in my head, as equations. But can you please explain how to solve these types of problems step by step?

Flip the second equation, add it to the first, cancel out whatever you can. See where this gets you.

@IvanNeretin Ok so under what conditions would I need to flip an equation? Is there anyway to tell when I will need to flip an equation? I mean thank you for telling me what to do to solve it but I want to understand the topic so I can apply it to other problems

Well, you want to arrive at the third equation. It has $\ce{XCO3}$ on the right. Your first equation has no $\ce{XCO3}$ whatsoever, so it can't help you with this one. Your second equation, however, does have some $\ce{XCO3}$, but it is on the wrong side. What shall we do, really?

I was being pedantic, because it is often dangerous / bad practice to use the equality symbol. However, in this particular context, I concede that it may be of some pedagogic value, because as Ivan Neretin points out, you are just manipulating simultaneous equations.

Ok so for future reference if anyone needs help with this you look at the last equation and then at the other ones, if the compounds(ex: CO2) of the other equations don't appear to be on the same side of the final equation you flip that equation and as a result the sign. After you have done this add everything up and it will give you the answer

While the method used to obtain a result is definitely useful, the reasoning behind that method is much more important. Please don't rote learn methods like this, the answer given by Yomen is better as it also captivates the reasoning.

As enthalpy is a state function, so regardless the multiple stages of a reaction, the total enthalpy change for the reaction is the sum of all changes.This is Hess's law.
So, you can see clearly that the third reaction results from adding  the first reaction to the reverse  of the second reaction, and so is the enthalpy change.


text:
Structure diagrams indicate plausible structures?
Structure diagrams have a hexagonal, "chicken wire" shape, which leads to more elegant representations of some molecules than others.
For example, the following diagram must deviate from the usual convention such as the horizontal connection above the nitrogen.

The following hydrocarbon would have the same problem.
  H H
  | |
H-C-C-H
  | |
H-C-C-H
  | |
  H H

Does this indicate anything about these molecules, for example they are rare or unstable, or is it just a weakness of this type of diagram?

I do not really understand what the question is here. A 2D representation of a molecule is obviously missing the third and can hence only be incomplete. A structural formula is one representation, but nothing more.

Reworded. It's more about every intersection naturally having a maximum of three connections than the 2D nature of them.

I still don't get it. There are so many different possibilities to draw such compounds - they are all just drawings, i.e. representations some of them more accurate than others in certain respects, some less, some are even completely wrong. Even a ball and stick model cannot accurately depict a molecular structure. And a structure, or its representation does not say anything about the stability of a compound.

"For example, the following diagram must deviate from the usual convention." Sorry but I don't quite get it. What is this "usual convention" that you are talking about? "The following hydrocarbon would have the same problem." What is the "problem"?

Structural diagrammes illustrate connectivity in molecules but they are often very limited in their ability to represent the real 3D shape of complex structures.
A drawing on a 2D surface can never properly represent a 3D structure. This is as true for maps as it is for chemical structures (which is why all map projections contain distortion, e.g. the Mercator projection seriously distorts areas).
Chemical structure drawings also omit lots of detail from molecules like the relative size of atoms and the relative lengths of bonds. What they focus on is the connectivity of the molecule, showing us what connects to what. For many structures this gives a poor representation of the 3D shape, especially (as in the example in the question) where there are multiple ring systems. The flat diagram itself doesn't tell you much about the stability of the molecule. That is one reason why chemists need modelling kits that get the 3D geometry right. Even then, the kits are imperfect. Structural kits sometimes just do the connectivity and bonding (in effect a 3D version of the 2D connectivity picture). But, sometimes, you need to understand the spatial interactions that take into account the size of atoms. There are 3D space filling kits for that sort of modelling.
Having said all this, most modelling is done via computers not by creating hand-built models which are relatively expensive and limited. Computers can allow for extra calculations to represent the interactions of bonds and spatial factors to give plausible 3D geometries and can sometimes do detailed calculations of energy levels. Many of the software packages (e.g. Chemdoodle) that allow you to draw 2D structures will automatically attempt to generate plausible 3D structures based on those to give you better intuition about the geometry of the molecule.


text:
Hydrogen displaces lead?
there was a reaction in my book. It is as follows:
$\ce{PbO + H2-> Pb + H2O}$
My question is Hydrogen is below lead in the reactivity series. Then how is this reaction possible?

I looked up the enthalpy of formation of PbO and it's -219, so that reaction would be endothermic.  http://chemister.ru/Database/properties-en.php?dbid=1&id=4082

The activity series you're referring to is constructed for ambient conditions where the temperature is $25°\text{C}$ and water is a liquid.  But, of course, thermodynamic properties can depend on temperature and especially on phase.  At higher temperatures where water is ordinarily a gas, it has a lot more entropy than the liquid.  That makes oxidation of hydrogen to water more thermodynamically favorable than what appears in your ambient temperature series.
For high temperature reactions where water mixes with hydrogen in the gas phase, the entropy effect pushes hydrogen up the series so it roughly matches iron.

The enthalpy of formation of H2O is significantly larger than the enthalpy of formation of PbO, so this reaction is exothermic.  Reactivity series are not perfect predictors.  Results depend on what an element is being reacted with.

you contradicted your initial comment that the reaction is endo-thermic.

Look at it from a redox reaction perspective. I'm not sure what the parameters the reaction is taking place under but the Pb is going from Pb2+ to just Pb. This does make sense because lead oxide is commonly used as an oxidizing agent and hydrogen gas is considered to be a powerful reducing agent.
Hope this helps
Just verified this with my advisor. It's an odd reaction but it does work.
All the best

Lead oxide which is commonly used as an oxidizing agent is another lead oxide, $\ce{PbO2}$.


text:
How is a molecular orbital a 'quantum superposition' of the atomic orbitals?
Let me quote from Feynman's lectures the concept of superposition:

(1)The probability of an event in an ideal experiment is given by the square of the absolute value of a complex number $ϕ$ which is called the probability amplitude:
  $$\begin{equation}
\begin{aligned}
P&=\text{probability},\\
\phi&=\text{probability amplitude},\\
P&={|\phi|}^2.
\end{aligned}
\end{equation}$$
  (2)When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately. There is interference:
  $$ϕ=ϕ_1+ϕ_2.$$

The second point is the core of quantum superposition.
Also, in Dirac's language from The principles of Quantum Mechanics,

[...]It requires us  to assume that between these states there exists peculiar relationships such that whenever the system is definitely in one state we can consider it as being partly in each of two or more other states.[...]

These all imply that when a system is in a state of superposition, it is 'partly' in both of the states that contribute to the superposition( Here 'partly' means possibility which implies the system has the possibility to exist in both states that contribute in the superposition).
For instance, let's take $\ce{H_2^+}$ to illustrate the concept of superposition. Firstly, the words of Feynman:

A positively ionized hydrogen molecule consists of two protons with one electron worming its way around them. If the two protons are very far apart, what states would we expect for this system? The answer is pretty clear: The electron will stay close to one proton and form a hydrogen atom in its lowest state, and the other proton will remain alone as a positive ion. So, if the two protons are far apart, we can visualize one physical state in which the electron is “attached” to one of the protons. There is, clearly, another state symmetric to that one in which the electron is near the other proton, and the first proton is the one that is an ion. We will take these two as our base states, and we’ll call them $|1⟩$ and $|2⟩.$ They are sketched in Fig.

]1
Now, the general state of $\ce{H_2^+}$, say, $|\psi_{\ce{H_2^+}}\rangle$ is in quantum superposition of the two base-states $|1\rangle$ & $|2\rangle.$
Now using the point no.(2) of Feynman, we can prove the superposition of the states as :
What should be the amplitude for finding the electron in, say, $x'$ from the center of the molecule $\ce{H_2^+}?$  (We would use each-coordinate as our new base-state, that is, $|x\rangle$ is a base-state; this means we would now treat continuum of base states).
We would've to find $\langle x'|\psi_{\ce{H_2^+}}\rangle.$ This can happen in two ways & thus applying the point no.(2), we get $$\begin{equation}\langle x'|\psi_{\ce{H_2^+}}\rangle\\=\int \langle x'|x\rangle \langle x|\psi_{\ce{H_2^+}}\rangle dx\\= N\left(\int \langle x'|x\rangle \langle x|1\rangle dx +\int \langle x'|x\rangle \langle x|2\rangle dx\right)\end{equation} ;$$ where $ N= \text{normalisation constant}.$ From this, we can easily shew, $$|\psi_{\ce{H_2^+}}\rangle = N(|1\rangle + |2\rangle).$$ Thus, quantum superposition is used when there is more than one way an event can happen; in this case, our event was finding the electron at $x'$ ; we superpose the alternative amplitudes to get the total amplitude of the event of finding the electron at that coordinate.

Let's consider $\ce{H_2}$; we describe first the VB description; again starting with Feynman's words:

As our next two-state system we will look at the neutral hydrogen molecule $\ce{H_2}.$ It is, naturally, more complicated to understand because it has two electrons. Again, we start by thinking of what happens when the two protons are well separated. Only now we have two electrons to add. To keep track of them, we’ll call one of them “electron a” and the other “electron b.” We can again imagine two possible states. One possibility is that “electron a” is around the first proton and “electron b” is around the second, as shown in the top half of Fig. We have simply two hydrogen atoms. We will call this state $|1⟩.$ There is also another possibility: that “electron b” is around the first proton and that “electron a” is around the second. We call this state $|2⟩.$

So, we can write the general state $|\psi_{\ce{H-H}}\rangle$ as the superposition of $|1\rangle$ & $|2\rangle.$ We can again prove the superposition from point no.(2) but this time we've to use $|x_1,x_2\rangle$ instead of $|x\rangle$ that is, $$\begin{equation}\langle x'_1,x'_2|\psi_{\ce{H-H}}\rangle\\=\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|\psi_{\ce{H-H}}\rangle dx\\= N'\left(\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|1\rangle dx +\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|2\rangle dx\right)\end{equation} ;$$ where $N'=\text{normalisation
constant}.$ This evidently proves the superposition $$|\psi_{\ce{H-H}}\rangle= \frac{1}{\sqrt 2}(|1\rangle + |2\rangle)$$  where $N'= \frac{1}{\sqrt 2}.$
Come to the MO description of $\ce{H_2}$(only the bonding orbital is concerned here to shorten the query). This is where my problem begins...
This is quoted from J.D.Lee's Concise Inorganic Chemistry:

Consider two atoms $A$ and $B$ which have atomic orbitals described by the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}.$ If the electron clouds of these two atoms overlap when the atoms approach, then the wavefunction for the molecule(molecular orbital $\psi_{(AB)}$) can be obtained by a linear combination of the atomic orbitals $$\psi_{(AB)}= N(c_1 \psi_{(A)} + c_2\psi_{(B)})$$
[...]Suppose the atoms $A$ and $B$ are hydrogen atoms ; then the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}$ describe the $1s$ atomic orbitals on the two atoms. [...]

Now, wavefunctions are nothing but probability amplitudes; since _molecular orbital is a quantum superposition of the atomic orbitals; it must satisfy the point no.(2) like all other superposition described above.
But I'm not understanding what base states are used in this case; in VB theory, $|1\rangle$ represented the state of having one electron around one $\ce{H}$ atom & the other electron around the other $\ce H.$ So, $\langle x_1,x_2|1\rangle$ meant the amplitude to find electron $a$ at $x_1$ & electron $b$ at $x_2.$
But what should I use in this case? How to represent $\langle x_1,x_2|\psi_{\ce{H-H}}\rangle$ using point no.(2) in MO theory?
I'm not getting how to apply the point no.(2) here; $\psi_{(A)}$ represents the amplitude of only one electron; same with $\psi_{(B)}.$
For applying point no.(2), there must be two ways for the event to find the electrons at $x'_1,x'_2$_ so that point no.(2) yields the result $\psi_{(\ce{H-H})_{(g)}}= N( \psi_{(H)_{1s}} + \psi_{(H)_{1s}}).$ But I'm not getting how $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes of the same final event of finding the electrons at $x'_1,x'_2$; if they were not representing the amplitudes of getting to the same final event, how could they be superposed by point no.(2)? In VB theory, the base states are defined so that the alternative amplitudes define the same final event & so they were superposed; however how should I interpret the base-states in MO theory, so that $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes to the same final event? How should I apply point no.(2) in case of molecular orbital to prove the superposition?

Same story as with the previous question: quantum superposition is always expressed mathematically as a linear combination, but the converse is not necessarily true. Not each and every linear combination expression has something to do with real physical quantum superposition. In many cases it is just a mathematical trick.
In the question about resonance we've already faced a situation when we could only formally think about a linear combination as a superposition of some imaginary states. Same story here: molecular orbitals are just linear combination of atomic orbitals. As for the case of resonance, the states you think are superimposed (atomic orbitals this time) are not even states of electrons in a molecule. The only "real" (though approximate) states of electrons in a molecule are molecular orbitals, not the atomic one, thus, strictly speaking, there is no superposition out there.
Besides, you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule. To construct (approximate) many-electron wave function you need to make (antisymmetric) product of $n$ molecular orbitals, where $n$ is the number of electrons. And since each and every molecular orbital describes just one electron, all your analysis of molecular orbitals for two-electron case of a hydrogen molecule is plain wrong.

P.S. And to be honest, I don't understand what you did for one-electron $\ce{H2+}$ case as well. To me it looks like you have too many misconceptions in your head currently. I suspect, you're jumping the gun by trying to learn the applications of quantum mechanics to chemistry having no really solid background in theory itself.

So, let's look on what you're saying on $\ce{H2+}$ case. Following Feynman, we write the general state of the system $\lvert \psi \rangle$ as a superposition of the two basis states $\lvert \psi_1 \rangle$ and $\lvert \psi_2 \rangle$,
$$
  \lvert \psi \rangle = c_1 \lvert \psi_1 \rangle + c_2 \lvert \psi_2 \rangle \, .
$$
Fine. And then you decided to "reverify the superposition of the states as". What do you want to "reverify" out there? Superposition is asserted, you don't need to "reverify" it whatever it means.
Yes, you can switch from the state vector formalism to the wave function one by taking the inner product of both sides with position eigenvectors $\lvert x \rangle$,
$$
  \langle x \vert \psi \rangle = c_1 \langle x \vert \psi_1 \rangle + c_2 \langle x \vert \psi_2 \rangle \, , \\
\psi(x) = c_1 \psi_1(x) + c_2 \psi_2(x) \, ,
$$
where the wave function is defined as usual, i.e. $\psi(x) = \langle x \vert \psi \rangle$. That's fine, though trivial: you indeed get the (weighted) sum of the corresponding wave functions which are the probability amplitudes. But that is because it just an alternative form of a superposition expression: it tells the same thing as before, just in different words. We didn't "reverify" anything, merely restated the superposition one more time.
Anyway, that would be the thing I at least understand. But look what you did! A complete mess! First, you forget the coefficients in the linear combination. Secondly, you used the resolution of identity for absolutely no reason. Thirdly, normalisation constant just appeared out of the blue. Finally, what you think you get at the end is in fact the starting point of a journey: it is an assertion Feynman did.
So, for me all you algebra appears as random manipulations of symbols you've seen somewhere but don't really understand.

+1; I am not bothered at you critisizing me; you've every right  but what I'm bothered is why you didn't understand the $\ce{H_2^+}$ case; I've just used the point no.(2) & the continuum of base states the same way Feynman explains in chapter 16. It deduces the correct result. So, what is the wrong there?

@user36790, no, I do not criticise you. In fact, I like your questions. You ask a questions on things which many do not bother about but just take for granted. The problem is that your questions are not well suited for Q&A sites, such as Chem.SE: they are too long, not sharp focused, and usually consists of few questions rather then one.

Now, let me explain what in particular I do not understand about $\ce{H2+}$ case right in my answer.

You are right, sir that  you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule.: in Atkins' Physical Chemistry book, he wrote that MO orbitals are one-electron wavefunction . But I didn't pay heed as every book  says that each molecular orbital is occupied by two electrons; so if it were a one-electron wavefunction, how could it be occupied by two electrons?

@user36790, it is a long story. In short there are two kinds of orbitals out there: spin orbitals and spatial orbitals. For now, just think of each spatial orbital as consisting of two spin orbitals. So, when it is said that a particular spatial orbital is doubly-occupied, what is actually meant is that both its consisting spin orbitals are singly-occupied.

As I said, you don't quite understand what you're talking about. When you use the resolution of identity by introducing $\sum_{i=1}^{2} |i⟩⟨i|$ in the middle of the bracket $⟨x|S⟩$ you clearly assert that ${ |i⟩ }{i=1}^{2}$ is a basis. And since, it is a basis, I could write $|S⟩ = \sum{i=1}^{2} c_i |i⟩$ without working out any algebra out there. You already asserted superposition by introducing the resolution of identity, you don't need to prove anything out there.

I suggest the following: stop reading few book concurrently. Read just one of them, then, once you finished, start the next one. First, you should build a solid foundation of quantum theory in your head. I didn't read Feynman myself, but many recommend it, so I think you could continue to read it. Or, even better, since you have so many misconceptions, start from the very beginning.

Then you are right, sir; but see this; there the author introduced superposition in the same way as I've written. Nevertheless, my main point is  I can write these stuffs for $\ce{H_2}$ in VB description as the base states are defined so. However, I couldn't write those in case of MO description as I considered $\langle x_1,x_2|$ to be the base-state of position. But molecular orbitals, as you said, are one-electron wavefunction.

Now you see the main issue with your questions: they are not focused. As it said on the Tour page, "This site is all about getting answers. It's not a discussion forum. There's no chit-chat." You better break you questions apart. For instance, you could easily make a better question based exclusively on your analysis of $\ce{H2+}$ case. We already discuss it in the comments, but they are not suited for it. You should ask your question peace by peace: one question - one particular concept.

The problem is that you think your analysis of $\ce{H2+}$ is perfectly valid, so that you move forward to discuss few other cases, but already at that point I doubt you really understand what's going on.

We talked last in the chat that every quantum superposition is LC but every LC is not quantum superposition; I just want to know how to determine whether a particular LC is quantum superposition or not. How to determine this? It is conclusive that MO is not quantum superposition . If I'm given a LC, how to say that it is not quantum superposition?


text:
What would be the state of water in interstellar space or vaccum?
Consider the interstellar space: Far from radiations, heat, etc. You can imagine conditions equal to vacuum. At these conditions, what will be the state of water; Ice, liquid, vapours, or something else?
What I know is that in lab conditions in a vacuum flask if we do the same experiment with water:
First it starts to boil, as the pressure in vacuum is almost zero. As it takes no time for the water to reach pressure equal to surrounding pressure, vapour pressure is formed easily. Water starts to boil without any rise in temperature.
Second, since energy is spent in boiling all the kinetic energy is used up and the water molecules start to condense forming ice.
The question now is: What are the differences between lab conditions and interstellar space? Shouldn't the results be all the same?

Well, if you suddenly release some liquid water into the outer space, part of it will instantly evaporate and the rest will freeze.

Why so @IvanNeretin ?

Because such is the nature of things. Look at the phase diagram of water, to begin with. Now, the interstellar conditions imply extremely low pressure, far off the bottom edge of that plot. See, there is no liquid there, only vapours and ice.

From Wikipedia: "In cool, dense regions of the ISM [interstellar matter], matter is primarily in molecular form, and reaches number densities of 10⁶ molecules per cm³."

Any ice that forms will evaporate also.

*Interstellar *- significantly far away from any hot celestial bodies and the associated radiation - so ice! Any heat initially contained within the body of water you release will be radiated outward away from the water towards the background temperature of interstellar space - ~ 2 to 3 deg K.

Nice answer @michael  but can u explain why In lab vaccum conditions , why the water first boils and than condenses ? Whats the difference in a Adiabatic closed vaccum flask here on earth and interstellar space , Should'nt the results be all the same ??

@user21891 Firstly, try to edit in this follow-up question into the question post itself. Secondly, please also show us your thoughts on the problem. This lets us know that we don't do your homework for you :)

@chestermiller There are chunks of ice flying around the outer solar system that are fairly small (i.e. minimal gravitational attraction) that have been there for at least 4.5 billion years.  At very low temperatures sublimation just about grinds to a halt in spite of the exceedingly low pressure.

tl;dr: ice
Let's have a look at the phase diagram of water:

In space, both the temperature and pressure are very low. That would put beyond the bottom left corner of the diagram, well within the ice field. So simply speaking, water would be ice.
However, there is the concept of vapour pressure. Any condensed material (just a fancy word for both solids and liquids) has that same material in the vapour around it. That's why you can smell alcohol, even though it's liquid as one example. That's why a puddle of water will eventually dry out even though it's below the boiling point of water.
The way vapour pressure works is that if you look on the phase diagram, for a certain temperature, the liquid-vapour and the solid-vapour lines have certain pressures associated with them. For example, if you look on 50 C, then the pressure on the curve is 10 kPa. Roughly speaking, this means that in atmospheric pressure, one tenth of the air (which is at 100 kPa) has to be water vapour. If it's any less, water will keep evaporating until it reaches that amount. On Earth, wind and diffusion will disperse the water so it will basically evaporate until there are no more liquid water left.
Now let's think what would happen in space. The temperature is very low, so the vapour pressure is very low. Thus, it should evaporate also in space. However, because the vapour pressures at such low temperatures are extremely low, you might end up with a situation where the gravitation of that ice blob is enough to overcome evaporation, so that will be an agonizingly slow process.

Compounding the already low vapor pressure, any water molecule that sublimes carries with it the associated $\Delta H_\mathrm{sub}$, which cools the solid even more, thus lowering the vapor pressure even further until it absorbs enough background radiation to get back to equilibrium

If you are talking about pure (i.e. distilled) water, then the majority of it will most likely evaporate due to the low pressure /see e.g. this video about how water behaves at microscopic level and how raindrops form - about raindrops/. However, if the water is not pure and has some dust particles within it, then these impurifications may "be able to" act as bases for formation of droplets. Thus, I guess there would be some visible snowflakes / ice-balls formed due to the low temperature (despite the low pressure)... In any case, whatever forms, it will be more or less spherically symmetric due to the lack of gravity.