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What is the structure of diferric hexaphenolate?
I was studying about the ferric chloride identification of phenol, where one product formed is diferric hexaphenolate. But I am having trouble drawing out the structure of it. With ferric ion having a 3+ charge, it must be attached to three phenolate ions. But how can this ferric triphenolate be combined with the other Fe(III) ion and three more phenolate ions? And dative bonds from the other three phenolate ions to the ferric triphenolate doesn't seem to work.
I searched about it but I only found information about ferric triphenolate. I am now skeptical even about the existence of diferric hexaphenolate.

Related: What is the product of the chemical reaction between phenol and ferric chloride?

Are you looking for the structure of $\ce{Fe[Fe(OPh)6]}$? Look at these papers: http://science.sciencemag.org/content/118/3064/327    https://www.nature.com/articles/1651012b0    https://pubs.acs.org/doi/abs/10.1021/ja00966a011


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Energy levels in an atom
I have been thinking about energy levels of an atom. When we study line spectra of hydrogen atom we say that when electrons jumps back from a higher shell to a lower shell it emits photons of certain frequency, but in that example we say that hydrogen has infinite energy levels and electron can jump back from any level. But when we study bonding we say that some atoms lack in d orbitals and for example say Nitrogen has no d orbitals, but when doing line spectra it has infinite energy level. How these two statements hold up with each other?

The H atom has an infinite number of energy levels spanning a finite energy range. This range is 13.6 eV, the ionisation energy, and is equal to the Rydberg R in energy. In the simplest (basic) theory the energy is $E_n=-R/n^2$, where n is the principal quantum number, $n=1..\infty$) thus as the energy increases the energy levels become closer to one another. The energy also rises (becomes less negative) as n increases.
In addition, for each level n there are other orbitals of nominally the same energy which describe the angular momentum & shape of the orbital. The orbitals are  1s, [2s, 2p], [3s, 3p ,3d], [4s, 4p, 4d, 4f] etc. where the levels in brackets are nominally of the same energy. (However,  interaction between electrons changes these energies but only slightly compared to their total energy ).
The light emission (fluorescence) you refer to comes from transitions between any two of these levels (subject to energy and angular momentum conservation). Other types of atoms behave similarly but because there are multiple electrons the equations describing the energy become far more complex.
Thus H atoms do have d orbitals just as do N atoms, but in their lowest energy state there are not enough electrons to fill any of these. The d orbitals only start to become filled as one reaches Sc in the transition metals.
In H and N atoms higher orbitals can be reached, by for example by absorbing photons or imparting energy from fast moving electrons in a discharge. However, in bonding, d orbitals will not be involved in the ground state bonding orbital of a molecule if there are not enough electrons and not enough energy to initially fill the d atomic orbitals. As soon as there are enough electrons, as in transition metal complexes, d orbitals become essential to understanding bonding.
Note: (I'm ignoring  ideas such as hybridisation between orbitals)

Note in one case it talks about jumping from a shell to another shell and then in another it talks about the types of orbits.
An atom has an infinite amount of shells available to it, but in each shell there are a finite amount of orbits allowed.
There is a nice discussion of that here : Difference between shells, subshells and orbitals
Nitrogen is filled (in the ground state) at the n=2 shell, so it only has l=0 (s) and l=1 (p) orbits, it can't have d orbits in the ground state.
For more information : Quantum Numbers,
Atomic Orbitals, and
Electron Configurations

Note that because of the increase in energy levels of shells, only hydrogen has an infinite amount of shells with negative potential energy.

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What roles do neutrons play in an atom?
An atom typically consists of electrons, protons, and neutrons. Electrons are negatively charged and participate in bonding to stabilize the atom.
Conversely, protons are positively charged and balance the charge of the atom. In addition, their positive charge attracts the negatively charged electrons.
What role do neutrons play in an atom?

You're only considering the electro-magnetic force. Protons and neutrons interact via the strong force, which is far stronger than electro-magnetism - that's how the nucleus can hold together against the electro-magnetic repulsion of the individual protons in the first place. Of course, reality is even more complex - you can't just keep adding neutrons to make atoms more stable :)

"electrons participate in bonding to stabilize the atom" doesn't make very much sense.  The stability of a molecule is determined by its electrons, but each atomic nucleus in that molecule is, AFAIK, no more or less stable than it would be if stripped of all its electrons and floating in the vacuum.

It is a strangely anthropomorphic question. None if the particles has job description.

Sorry I don't have enough reputation to add a comment, but I think you can find you answer from here

Neutrons bind with protons and one another in the nucleus through the strong force, effectively moderating the repulsive forces between the protons and stabilizing the nucleus.$^{[1]}$
$\ce{^2He}$ (2 protons, 0 neutrons) is extremely unstable, though according to theoretical calculations would be much more stable if the strong force were 2% stronger. Its instability is due to spin–spin interactions in the strong force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the $\ce{^2He}$ nucleus a negative binding energy. $\ce{^3He}$ (2 protons, 1 neutron), on the other hand, is stable, and is also the only stable isotope other than $\ce{^1H}$ with more protons than neutrons.$^{[2]}$

$^{[1]}$ Wikipedia, Neutron, Beta Decay and the Stability of the Nucleus
$^{[2]}$ Wikipedia, Isotopes of Helium

Also I's point out that (1) Neutrons confounded the relationship between atomic number and atomic mass as chemistry was being understood as a science. (2) In general the mass of different isotopes has a very small effect on chemical reactivity.

In a few more words, physicists right now are confident in saying that there are four fundamental things that happen:

Protons and neutrons stick together. (The "strong nuclear interaction".)
Neutrons sometimes "fall apart" into a proton, electron, and antineutrino. Sometimes this can happen in reverse, too. (The "weak nuclear interaction", also known as "beta decay" or "radioactivity".)
Positive charges repel other positive charges and attract negative charges. (The "electromagnetic interaction", also known as "chemistry" and "light".)
Things fall down. ("The gravitational interaction".)

(I said that these were in order of "decreasing everyday strength," but that's not very precise given that these things scale differently with distance etc. But this is the rough order that you should be thinking about the problem you're interested in.)
Everything else you are used to is caused by these 4 fundamental interactions. For example when you are sitting in a chair, secretly the force holding you up is a force of electron clouds around nuclei repelling each other, so this is chiefly the "electromagnetic" forces at play opposing the "gravitational" forces pulling you down.
There is also a slight subtlety which these 4 interactions do not completely cover, but every physicist knows about it: it says that "two identical particles cannot remain in an identical state." This usually means that those particles have to occupy higher and higher-energy states. It turns out a lot of the structure of the periodic table comes from this rule! This rule ultimately says that the number of columns you add (when you add columns to the periodic table) must be twice the next odd number: so you see that we start off adding 2 columns, then we add 6 columns, then we add 10 columns, then we add 14 columns; the physics says that the next number of columns to add would be 18 and that the pattern is seen when you divide by 2, you first add one pair, then three pairs, then five pairs, then seven pairs: increasing odd numbers. And this is just because each new electron needs to be (a) orbiting further away and (b) possibly spinning faster.
So as a nucleus gets bigger and bigger, a similar story happens. The neutrons and protons cooperate due to the strong nuclear interaction. They, it turns out, really like to stick to each other! But then the second effect takes hold: if a nucleus has too many protons, they have to be in really spinny high-energy states in the nucleus, because the lower-energy states are already occupied by other protons! But there are lower-energy neutron states which are unoccupied. At some point it becomes energetically favorable for a proton to reverse-beta-decay into a positron plus a neutron plus a neutrino, so that the neutron can fall into that lowest-energy state.
So that's what the neutrons do in the nucleus: they are "about as sticky" as the protons but they are different particles that can occupy the other states.
Now you might also think, "oh, those protons are repelling each other also, due to the electromagnetic interaction." And that is true, but it is a weaker effect than either of these. That effect basically balances out at a certain atomic number, which happens to be Iron. All of the smaller atoms are driven more by the strong nuclear force to want to "fuse" together into larger atoms, trying to be iron. And all of the bigger atoms are driven more by the electromagnetic repulsion to want to "fission" apart into smaller atoms. (But of course until you get to intrinsically-unstable atomic numbers like Uranium, you can still have little stable states of larger-than-iron atoms, where if you fire a neutron into the nucleus it might fall apart but for right now it's jiggling around safely.)
See also: Wikipedia's articles on stable nuclides and nuclear binding energies.

Good answer, but electromagnetism is much stronger than weak interactions.

@JG I mean, that's true in a technical sense (the mass of the W boson makes the weak force between two nucleons much smaller than the electromagnetic force between them), but I struggle with how to better convey this idea of "looking at each nucleon individually, the first thing you should think about is its ability to stick to other nucleons, the second thing is its ability to transmute into other nucleons, and finally you should think about its electromagnetic repulsion of other nucleons." Maybe I'll just delete the aside of "decreasing everyday strength".

I think the issue is that the stability of a nucleus hinges on the competition between strong and electromagnetic forces, and weak interactions occur when they would increase the resulting binding energy. Weak forces aren't taken into account in the semi-empirical mass formula.

In chemistry, the neutrons are important as they determine the spin of a nucleus which determines if and how it is observable by NMR. Like 1H is spin 1/2 and 2H is spin 1. The neutron also adds mass to the atom.
In chemical reactions however, the nucleus is not involved and nuclear reactions are more a topic in physics.

Neutrons play very important role in an atom. They provide stability to the atom and also prevent protons from repelling one another.

The answer would be more complete if you explained at least briefly how the neutron provides stability and prevents protons from repelling each other.

OK Sir, why not. If we drive an empty car, we will feel that the vibrating very much because of unstable, and if we keep some mass in the car then we will feel comfortable to drive and the car will be moving very smooth. Because mass provide the car stability. In physics We can say that mass is inertia and inertia always providing stability. Proton having positive charge and same charges repel each other. Here neutrons play a role of a boundary between the protons to prevent the repulsion.

You have good points on your answer, could you edit it and perhaps expound on it so that others can better understand how neutrons provide stability and prevent protons from repelling one another? Also, big warm welcome to Chemistry.SE!

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Equating emissions to mass loading
I want to estimate aerosol mass loading by dividing emissions of the particular aerosol by the associated area. Assuming the air had no other particles in it to begin with - is this procedurally correct?

Could you describe your system a little more thoroughly?  When you say "associated area" do you mean volume or the area of some surface on which the aerosol is depositing.  There will be suggestions that this is off-topic as not being about chemistry.  What is the composition of the particles and what type of surface is your chamber, container, etc.?  Humidity can lead to increased aerosol mass; what kind of water vapor concentrations are present in your system.  My questions are too much to answer in comments, I'm suggesting you re-phrase your description and question.

Given the context of your question, I assume you must mean "the associated volume", not "the associated area".
Other than that, you are exactly correct.  For example, given the following parameters:

aerosol emissions rate = 0.010 mg $\mathrm{hr^{-1}}$
  duration of aerosol emissions = 1 $\mathrm{hr}$
  container volume = 1 $\mathrm{m{^3}}$

Then the aerosol mass loading would be given as:

total aerosol mass = 0.010 mg $\mathrm{hr^{-1}}$ $\mathrm{x}$ 1
  $\mathrm{hr}$ = 0.010 mg
aerosol mass loading = 0.010 mg / 1 $\mathrm{m{^3}}$ = 0.010 mg
  $\mathrm{m{^{-3}}}$

If you really meant area and not volume, as in surface area of the aerosols, then your statement would be over-simplified and you would need to include other assumptions made concerning particle size and density etc. in order to determine whether the way your statement is correct as written.


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What does peroxide value exactly measure?
I understand that peroxide value gives a measure of the extent to which an oil has undergone oxidation, however, I looked and couldn't find any source which explains the chemistry behind PV value. More specifically, how is pv value related with oxidation?
Wikipedia states that "the peroxide value is determined by measuring the amount of iodine which is formed by the reaction of peroxides (formed in fat or oil) with iodide ion." But what is the reaction of the peroxides? And how is it related with oxidation?
Thank you

The test measures the amount of hydroperoxide by a redox reaction with iodide:
$\ce{2ROOH + 2H+ +2I- -> 2ROH + I2 + H2O}$
In the course of the reaction, the hydroperoxide is reduced, while iodide is oxidized to iodine. In principle, this is a variation of the iodine clock reaction of hydrogen peroxide.

How does this show oxidation?


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The Compressibility of a substance, dependent on
My Chemistry book states that (talking about compressibility equation):

"A sample with Volume V is subjected to a pressure increase, delta P,
  and the resulting change in volume, change in V, is measured.  The
  ratio is divided by V; thus, the tabulated value depends only on the
  substance being measured and not on the geometry of the sample."

I'm not sure exactly how they are drawing this conclusion.  How does dividing by (-) volume cause the compressibility factor to not be dependent on the "geometry" of the sample?
Thanks.

It doesn't. Generally speaking, this is not true. A solid may have different compressibility values along different directions. Though the book is not talking about solids, I guess.

@IvanNeretin Is it true for gases?

Yes. That's because gases (and liquids, for that matter) are isotropic.

"(of an object or substance) having a physical property that has the same value when measured in different directions."

Can you clarify what that means? What property of a solid is different when measured in "different directions"?

Generally speaking, just about any property which can be measured in different directions. Thermal and electric conductivity, speed of sound, etc.

Oh okay.  I guess I'll learn about that later, Thanks.  I thought it was talking about volume which lead to some confusion, but it makes sense now.

If you apply an external pressure on an object and analyze the deformational response of the object (say using Hooke's law in 3D), you find that the state of stress and strain within the object are completely uniform and independent of direction within the object (if the object has direction-independent properties to begin with), irrespective of the shape of the object.  This is sort of the same kind of thing that happens in thermal expansion, where the object strains uniformly in all directions, and the strains are independent of the object shape.


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How do atomic radius and nuclear charge influence ionisation energies?
This is actually a multiple choice question.

Why is the first ionisation energy of neon higher than that of fluorine?

A: Fluorine is more electronegative than neon
B: Neon has a complete octet, but fluorine does not
C: The atomic radius of fluorine is less than that of neon
D: The nuclear charge in neon is greater than that in fluorine

I understand why A is incorrect and why B is correct. But I am confused about C and D. I did try to search online but my questions just increased.
Firstly, why is the atomic radius of fluorine less than that of neon? Both fluorine and neon are in the same period which means that they have the same amount of shells. Since the nuclear charge of neon is more, shouldn't the nucleus attract the electrons in the outer shells closer? Which would result in neon having a smaller atomic radius than fluorine.
Secondly, according to a source, the largest factor is that the 2p sub shell in neon is full whereas in the fluorine it is not. They said that a full sub shell means that the atom is more stable which is the way neon has a higher first I.E. but since the 2p sub shell is full, wouldn't there be mutual repulsion?
I'm sorry if this is a lot, I'm trying to get a head start for my next education level and I cannot find any textbooks for that level's chemistry. I've been reading a summarized guide book and trying to do assessments whilst trying to find guides online as well.

I don’t fully understand the bit about mutual repulsion, sorry.

When we are considering atomic radius:
The atomic radius of noble gases are considered the highest in a period since we only consider their Van der Waals radius (distance between two similar atoms in their closest approach). Although technically they should have lower atomic radius due to increased $Z_\text{eff}$.
As for ionisation energy:
The ionisation energy of noble gases are way higher than the other elements of the same period due to stable configuration

‘The atomic radius’ is a somewhat wavy concept. In quantum mechanics, there is no such thing as an atomic radius which would imply an ‘end of the atom’. Instead, electrons are said to occupy orbitals and these orbitals technically extend into infinity albeit at very low values (i.e. they converge towards zero pretty rapidly on macroscopic scales). Therefore, as soon as you are talking about quantum mechanics or calculations, you need to define an arbitrary cutoff value as an ‘end point’ or ‘radius’ of your atom. These cutoff values are typically chosen so that the probability of locating an electron inside the border are $95~\%$ or some other similar percentage.
For the same reason, it is nontrivial to determine experimental atomic radii. If you were to just experimentally measure the electron cloud, all you would see is fog getting less dense outwards from the centre. Hence why experimental values rely on measuring compounds and attributing the distance between nuclei (whose positions can be accurately determined) in part to one atom’s radius and in part to the other’s. Neon has, to the best of my knowledge, not been persuaded to form compounds with other elements yet, so there is no way to experimentally measure its radius.
Varius calculation cutoffs do indeed state that neon’s atomic radius is smaller than fluorine’s — at least within the theory employed. Other definitions such as the van der Waals radius give a different picture because they rely on different means of calculation. For a comparison of different definitions and values, check out the Wikipedia data page on atomic radii. Note that especially the van der Waals radii do not follow the periodic trend of decreasing across a period as the other definitions largely do.
With that out of the way we can dismiss C as factually probably incorrect.
The larger nuclear charge means the electrons in the shell experience a greater potential. Ionisation means removing one of those electrons from the atom against the potential that is pulling it back. Just by means of a simple thought experiment, we can see that removing an electron from a nucleus with $10+$ charge is harder than removing one from a $9+$ nucleus. (In addition, the greater nuclear charge also leads to the orbitals being less diffuse i.e. the radii being smaller regardless of the cutoff used. This is indeed as you said.)

C : is incorrect because the atomic radius of fluorine is greater than that of neon, not less, as you correctly identify. Defining/measuring/calculating these radii can be a bit tricky - I find Mills' model most helpful in this case. It calculates $\ce{F}$ to be 20% larger than $\ce{Ne}$ (and thus the first ionisation energy ~20% lower).
D : We all know that first ionisation energies follow a seesaw pattern across the elements, so greater nuclear charge does not by itself imply greater first ionisation energy. It all depends on the number of electrons screening that charge...
References:
Randell L. Mills, The Grand Unified Theory of Classical Physics
(ISBN: 978-0-9635171-5-9)

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Stability by using charge/radius rule
I found a question

Which is more stable among the following?
(a) $\ce{CaCO3}$
(b) $\ce{MgCO3}$

Answer : (a)
Explanation was given as:

$\ce{Mg^2+}$ has smaller radius than $\ce{Ca^2+}$ and thus charge/radius of $\ce{Mg^2+}$ is greater , hence it is less stable.

While when reading about the coordination compound, I found the sentence

Higher the charge of central metal atom ion i.e., greater ionic potential
charge/radius then greater is the stability.

In first case higher the ratio less is the stability.
In second case higher the ratio more is the stability.
Is the difference in the answers in both cases due to a difference in the salts (i.e. simple and  complex salt) or is any one of them wrong?

Please allow me to answer the question in two parts.
Stability of carbonates
We observe this order when we compare the thermal stability of group I-A carbonates: $$\ce{Li2CO3 \lt Na2CO3 \lt K2CO3 \lt Rb2CO3 \lt Cs2CO3}$$
On similar grounds, the thermal stability of group II-A carbonates can be predicted.
This order is in accordance with Fajans' Rules. Here's the table we need to glance at from the Wikipedia page.

$$\begin{array}{|c|c|} \hline
\mathrm{Ionic} & \text{Covalent}\  \\ \hline
\text{Low positive charge} & \text{High positive charge} \\
\text{Large cation} & \text{Small cation} \\
\text{Small anion} & \text{Large anion} \\ \hline \end{array}$$
  Polarization will be increased by high charge and small size of the cation.

Thus, larger the ratio of charge and radius, higher is the polarization power of the cation, more is the covalent character in the compound and lesser is the ionic character; Ionic compounds are more stable when exposed to heat. A higher ratio will therefore mean lesser stability.
Stability of complex compounds
Thermodynamic stability of complexes, on the other hand, is determined by the Irving–Williams series, according to which general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is:
$$\ce{Mn^2+ < Fe^2+ < Co^2+ < Ni^2+ < Cu^2+ > Zn^2+}$$
Owing to this observation, a lot of theories were formulated to explain this phenomenon. One such theory was that since Crystal Field Theory (CFT) was based on the idea that a purely electrostatic interaction existed between the central metal ion and the ligands, the stability of complexes should be somehow related to the size by charge ratio (otherwise known as the ionic potential) of the central metal atom. [Source]
An interesting thing to note here is that Wikipedia has stated in its article that:

None of the explanations can satisfactorily explain the success of the Irving–Williams series in predicting the relative stabilities of transition metal complexes.

Thus from manganese till copper, it can be said that higher the ionic potential, more is the stability.

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Does latex protect against acids?
I heard somewhere that some plastics are pretty good at resisting decently strong acids. To what degree is this true? Does latex protect against corrosives? Are plastics good at "resisting" bases or other strong solvents?
I'm not a chemist, so I apologize if my question seems dumb. I couldn't find much info on the subject of latex vs corrosives via Google.
Also, I don't plan on handling any dangerous chemicals; I'm just curious.

There are all sorts of resources for picking the best gloves for handling any given type of chemical. This resource describes latex as being good for protecting against bases, alcohols, dilute water solutions, and fair against aldehydes, ketones.
Indeed if you check the Glove Type and Chemical Use table towards the bottom of the page, you can see that latex fairs pretty well against most acids. These ratings give you an idea of how long it takes to penetrate the glove when exposed to that chemical. As you can also see from this table, neoprene gloves are slightly better for handling acids, and though it's not shown on the table, PVC gloves are also good.

Why are plastics good for protection against acids? Why don't they dissolve like other materials?

It depends on the acid. I would recommend Neoprene gloves instead of latex gloves. I did this excerpt from this guide that I think is taken from the website that ringo mentioned:
VG= Very Good G= Good F=Fair P=Poor
Acetic acid VG
Chromic Acid (50%) P
Citric acid (10%) VG
Hydrochloric acid G
Hydrofluoric acid (48%) G
Lactic acid (85%) VG
Lauric acid (36%) F
Lineoleic acid P
Maleic acid VG
Nitric acid (limited service) F
Oleic acid F
Oxalic acid VG
Palmitic acid VG
Perchloric acid (60%) F
Phosphoric acid G
Potassium hydroxide VG
Sulfuric acid G
Tannic acid (65%) VG
Sodium hydroxide VG


text:
Molecular orbital diagram for nitrogen monoxide, the nitrosyl cation and the nitrosyl anion
As I understand it, all of them should have same order of energy of molecular orbitals as a model for $\ce{NO}$ should withstand ionization and adding an electron.
JD Lee Pg 109 shows the same diagram for $\ce{NO}$ as the one quoted in here. I couldn't find $\ce{NO+}$ except in here, where $\sigma_{2p_z}$ has a higher energy than $\pi_{2p_x}$ and $\pi_{2p_y}$.
If it is correct, then, there is $s$-$p$ mixing in $\ce{NO+}$ but not in $\ce{NO}$?

There is sp mixing in all of the three compounds, but they are sometimes not showed to have simplicity.

Note that $\ce{NO+}$ is isoelectronic with $\ce{N2}$ and $\ce{NO-}$ with $\ce{O2}$.

"If it is correct, then, there is $s$-$p$ mixing in $\ce{NO+}$ but not in $\ce{NO}$?" what is the "it" referring to?

Sorry, I know that sp mixing is present in essentially all of them, I wanted to see if they have the same order of energy of 2p sigma bmo and 2pi bmo. The "it" was the second link. I think the following example can clear my mind : is HOMO of CO an antibonding or bonding MO when it takes part in reactions and not ionization?

s-p mixing isn't an ON/OFF switch. It's always present but in O2, F2 it just isn't a sufficiently large effect such that it changes the relative ordering of the energy levels. In B2, C2, N2 it is large enough. You can generalise to the heteronuclear diatomics.

As usual with these simple molecules: It's not as simple as it seems. Here it actually doubles down quite a bit more.
The short and very dis-satisfactory answer is: It is very, very complicated. For this reason, I am reluctant to present a molecular orbital diagram. However, I will provide calculated MOs. While it may be tempting to just look at the pictures, the rationalisation behind it is more important. What will follow now are a few attempts at that rationalisation.
In the nitric oxide series, $\ce{NO+}$ is surprisingly the simplest. Why is that surprising? Because it is isoelectronic with $\ce{CO}$; and this molecule is one of the most complicated ones there are. You can read a bit more about it here: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?
I do compare $\ce{NO+}$ more with $\ce{CO}$ rather than $\ce{N2}$ to which it is also isoelectronic, because of the asymmetry introduced by the different nuclei. The molecule is the simplest to explain, because it is a singlet ground state. All electrons are neatly paired, and spacial symmetry can easily be matched with the degeneracy of the orbitals.
The π-orbitals are not the HOMO, instead it is the σ lone pair at nitrogen.
The next more complicated is $\ce{NO-}$, which is isoelectronic to $\ce{O2}$. In this case it is hard to find a less symmetric isoelectronic version. It is well known, that the ground state is a triplet because of the degeneracy of the π-orbitals. While we would expect the orbital ordering to be the same as in $\ce{NO+}$, it is not. This is mainly due to the fact, that occupied orbitals behave differently than unoccupied orbitals.
In some way, because of the paramagnetic character, we have a situation wich cannot be described by a single molecular orbital description. You need at least two: one for α-electrons (or spin up) and one for β-electrons (or spin down). The good thing however is, that both of these descriptions may still be matched with spatial symmetry.
The bonding π orbitals become higher in energy than the σ lone pair because the anti-bonding counterparts are partially filled.
The most complicated molecule in the series is the radical $\ce{NO}$. There is absolutely no simple description. Because of the unpaired electron, you need at least two descriptions again. However, this is not enough because now one of the spin-state description cannot match the spatial symmetry. This is because one electron cannot occupy two orbitals at the same time in an MO description. Because of that (from a technical point of view) the orbital degeneracy has to be broken. Physically this is not possible.
The molecule has to be treated with a multi-reference approach, i.e. one MO description for every possible configuration. It is probably safe to assume that the orbitals of π symmetry don't split that much and are higher in energy than the σ lone pair.
Below you can find the calculated MO. The level of theory used for $\ce{NO+}$ was RMP2/def2-QZVPP, because this is equivalent to the level the other molecules were treated, which is ROMP2//UMP2/def2-QZVPP. One can clearly see how the orbital symmetry for $\ce{NO}$ breaks the degeneracy of the pi orbitals.
For $\ce{NO+}$ MOs 5,6 and 8,9 are degenerated. For $\ce{NO-}$ MOs 6,7 and 8,9 are degenerated.
Colour code: blue/orange, α, β-occupied; purple/yellow-orange, α/β-occupied; red/yellow, unoccupied; contour value, 0.1.

With a little more literature research (prior to answering the question) one would have found the following publication:
Renato P. Orenha and Sérgio E. Galembeck. Molecular Orbitals of $\ce{NO}$, $\ce{NO+}$, and $\ce{NO–}$: A Computational Quantum Chemistry Experiment. J. Chem. Educ. 2014, 91 (7), 1064–1069.

Just a random side-question: What’s the dipole moment’s negative and positive sides in $\ce{NO+}$?

@Jan That's not easy to be answered since the calculated dipole for ions is dependent on the coordinate system. When running the calculations I did not pay attention to where I put the centre of mass. I'd need to rerun the calculations with that in mind, or think of another way to analyse.

@Jan, for your entertainment: What is the dipole moment direction in the nitrosonium ion?


text:
How to calculate the amount of aluminium that has been oxidised?

I have a sample of pure aluminium (100%), 1 kg, stored in "normal condition"
After some time the weight of sample has increased by 1%.
I assume that aluminium has oxidized to $\ce{Al2O3}$

I know the weight before and after, and my question is: how to calculate how much aluminium has been oxidised?

First of all, welcome!  Starting point: from the molecular weights of Al and O, and the chemical formula of the oxidized product as you've given it, you can solve your problem.   Have you tried this?  Can you try it and show what you come up with?  Once you've made that effort others will be more motivated and better able to help answer your question (though you may answer it yourself if you try this!)

Yes, I have tried :), just wanted someone to confirm my calculation: As I understand the weight of 3 particles of O is almost same as 2 particles of Al - so simple calculation is: 1% of weight increase means 1% of oxidation - but I am not sure if this method of thinking is correct?

Absolutely.  Just post your calculation and I (and I'm sure others) will be happy to review and comment on it.

Initially, 1 kg of $\ce{Al}$ is present.
After some time, oxidation increases the weight by 1%, or 10 g (e.g. 10 g of oxygen has been added).
The atomic mass of oxygen is 16.0 g/mol.
The molar ratio of $\ce{Al}$ to $\ce{O}$, from the molecular formula $\ce{Al2O3}$, is $\mathrm{\frac{2~mol~\ce{Al}}{3~mol~\ce{O}}}$.

$$\mathrm{\frac{10~g~\ce{O}}{16.0~g~O/mol~O}=0.625~mol~O}$$
$$\mathrm{\frac{2~mol~\ce{Al}}{3~mol~O}*0.625~mol~O=0.417~mol~\ce{Al}}$$
$$\mathrm{\frac{27.0~g~\ce{Al}}{mol~\ce{Al}}*0.417~mol~\ce{Al}=11~g~\ce{Al}}$$
Note that even though only a single significant figure is given for each of steps 1 and 2 of the problem, I assumed that they should read "1.0 kg" and "1.0%" rather than just "1 kg" and "1%" for the purposes of illustrating the above calculation. Thereby, 0.42 mol or 11 g of aluminum was oxidized to $\ce{Al2O3}$.


text:
Does bonding happen naturally?
For example, how do the bonds in NaCl form?
I understand that the elements need to obtain an octet structure to become more stable, by either accepting or losing electrons.
In NaCl, from my understanding, the sodium atom loses an electron and gives it to the chlorine atom, in turn gaining a +1 charge and then causing chlorine to have a -1 charge.
The new compound formed is NaCl. How does it form ? Do i need to place the respective atoms close to each other? What is needed to them to bond? Also, what is needed for bonding for other types such as covalent and metalic? Does it happen naturally ?

Bonding surely does happen naturally, in that it did exist before we humans came along and started placing atoms this way and that.

Your second and third sentences go a long way toward answering your question.  "Nature" wants to be in it's most stable (low energy) state, and swapping/sharing electrons so that everyone gets a complete octet is one way in which the most stable chemical state is achieved.

The collision theory aims to explain how reactions happen.
In the theory, when the particles of the suitable reactant collide with each other, the particles with the right conditions (orientation, velocity, etc.) will react and form new chemical elements.
It is quite difficult to form $\ce{NaCl}$ from sodium metal and chlorine gas alone. If you have to produce $\ce{NaCl}$ from sodium metal and chlorine gas (which you would not), you would do it in an aqeuous environment:
$$\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g) ^}$$
$$\ce{Cl2(g) + H2O(l) -> HCl(aq) + HOCl(aq)}$$
$$\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O}$$
But in this method $\ce{NaOCl}$ is produced as a side product, so I do not recommend using this method.

It is quite easy to make $\ce{NaCl}$ from sodium and chlorine. It is neither safe nor cost-effective.

I mean directly in an anhydrous environment.

Drop sodium metal into chlorine gas. After the dust settles, the shrapnel will be coated with salt. The reaction is spontaneous and violent. The process is not cost-effective because both elements are manufactured by electrolysis of sodium chloride.

@BenNorris Are you sure that it is not because of the oxide layer?

Certain. The oxide layer would slow this reaction down.


text:
Are there any stable complexes where an (N-unsubstituted) imine is bound eta^2 to a (late) transition metal?
I've been looking for an example of a stable complex where a $\ce{C=N}$ is coordinated $\eta^2$ to a (late) transition metal, but could not find any. Do they exist at all?
For $\ce{C=C}$ there are countless examples, Zeiss' salt, $\ce{Ni(COD)_2}$, etc.
If no or only very few $\ce{C=N}$ analogs exist, what's the reason?

http://pubs.acs.org/doi/abs/10.1021/om050240t  Not exactly what you are looking for, but an example of $η^2$ imine on a rhodium complex. However their use of $η^2$ seems to refer to the entire connected imine moiety rather than through the double bond.


text:
Mechanism for palladium-catalysed spirocyclisation of conjugated diene
I'm trying to work out the catalytic cycle for the following process:

I gather that the palladium will be reduced from Pd(II) to Pd(0) in the process and that the 1,4-benzoquinone is a redoxidant. However, I am not sure where the palladium would be initially inserted (possibly one of the double bonds?) and how the chlorine ends up on the final product.

Notice also the full equivalent of benzoquinone. This is used to oxidize the eliminated Pd(0) basck to Pd(II).

You don't have to wonder about which alkene coordinates to the Pd(II) centre. In this case, they are conjugated and therefore can coordinate together as a diene, which is a 4-electron donor. Many pi complexes are known ranging from alkenes (2e donor), to allyl (3e), to diene (4e), to cyclopentadienyl (5e), ... so this is not something really surprising.

Coordination of the diene to Pd(II) makes it more electrophilic. The pendant hydroxyl group simply does a 5-exo-trig cyclisation to form the C-O bond, generating a Pd(II)-allyl species. In terms of stereochemistry, the OH group has to attack anti to the metal. Deprotonation can happen with acetate, or the hydroquinone dianion that benzoquinone is later reduced to.

This Pd(II) allyl species is an intermediate in the well-known Tsuji–Trost allylation. In this case, the Pd-allyl species is being generated in a different way, but goes on to react in the same way. Chloride being a relatively soft nucleophile attacks the allyl moiety, again anti to the Pd, leading to a Pd(0)-alkene complex. The stereochemistry of the chlorine isn't indicated in your product but this is my guess at what's going on.

Dissociation of the alkene directly gives the desired organic product and reoxidation of Pd(0) to Pd(II) with benzoquinone regenerates the catalyst.

Looking this reaction up on Reaxys, it seems that it is taken from a 1991 article by Bäckvall and Andersson.1 The mechanism posited in the paper is identical to what I have written above, and on top of that my guess at the stereochemistry was correct: the authors indicate that the product obtained was 99% cis (referring to the chloride and ether oxygen in the product).
Interestingly, with $\ce{Li2CO3}$ instead of $\ce{LiCl}$, the product obtained has a trans acetate. Carbonate isn't a great nucleophile so it doesn't act as one, but it can deprotonate acetic acid. The acetate ion thus produced then forms a bond to Pd(II) before it migrates to the allyl system.

Reference

Bäckvall, J. E.; Andersson, P. G. Stereocontrolled oxaspirocyclization of conjugated dienes via palladium catalysis. J. Org. Chem. 1991, 56 (7), 2274–2276. DOI: 10.1021/jo00007a002.


text:
What exactly happens while washing clothes
I am reading about soaps , micelles and all that. I understand how soap behaves in water contaminated by oil. Soap helps in emulsifying the oil into water.
Let us suppose now we have dirt(i.e oil, grease etc) not alone in water but present on a some cloth. Why can't we put the cloth just in water to get it clean ? What precisely is the need for agitation in a washing machine or by any other method ?
I searched and found that agitation produces foam but according to this post foam does not alter the washing process anyhow.

Agitation or mechanical scrubbing[either by hand or machine] helps in the emulsification of the grease or oil patch

When you wash a piece of cloth, even without soap, it's better to agitate. The vibration, friction, etc. will help speed up the process of mixing the dirt/grease with water.
Even the very definition of "agitation" uses the example of washing. From Vocabulary.com

Agitation is the act of stirring things up, like the agitation of a washing machine that moves the water, detergent, and clothes around and gets the dirt out

And as you rightly pointed out from another question, "the froth has little or no effect on the detergent action".

The modern definition of washing is "washing can be defined as both the removal by water or aqueous surfactant solution of poorly soluble matter and the dissolution of water-soluble impurities from textile surfaces". This is in the book "Laundry Detergents" written by E. Smulders. There is a whole chapter on the physical chemistry of the washing process. You can read that.
What is the purpose of the washing machine? Washing machine provides the mechanical agitation and the modern ones also heat the water (esp. the European style front loaders).
This is akin to using a stirrer in a chemical reaction or using a stirrer to dissolve difficult to dissolve solids in a given solvent and using heat to accelerate this process. The purpose of agitation is to constantly bring the clothes and fibers in contact with fresh and (ideally concentrated) detergent solution. This is exactly why the front loaders are far better than these vertical type washers commonly seen in the US and Asian countries. Front loaders uses a concentrated solution (a smaller volume) vs. gallons of water in vertical machines. With apologies, to other manufacturers, not all washers were created equal (just like cars).
Now there is not only oily dirt but there is physical dirt as well. You need mechanical action to dislodge it from the twisted and woven fibers of the clothes. German washers especially Miele, Siemens, Bosch outperform in cleaning tests on the so-called washing grades of A, B, C, D, E etc, so they way you agitate also matters and how long you agitate.
If you let the clothes soak in a detergent solution, you are simply relying on diffusion process to help emulsify or dissolve the oily dirt. So imagine the difference, an oily spot soaked in detergent vs. an oily spot constantly being exposed to fresh detergent solution. Consider the analogy of copper sulfate crystals sitting at the bottom of a beaker vs. a stirrer continuously shaking the crystals. It takes a couple of months to make the entire solution blue (diffusion alone). A stirred solution takes a few minutes to dissolve the crystals. This is why soaking alone is an inefficient process. So much so for advocacy of washing machines or the need of a proper mechanical agitation. Some manufacturers also tried ultrasonication but apparently this no better than an ordinary washer.
The rest of magic of cleaning clothes comes from the detergent chemistry-which is far too complex to discuss in a single post. You have surfactants, bleaching agents, bleach catalysts, enzymes, water softeners, builders, foam regulators, EDTA, optical brighteners, anti-sudsing agents, soil anti-redeposition agents, perfume and God knows what not. Finally you have to protect your washing machine from corrosion so detergents also have anti-corroding agents.
Last but not the least, they have to ensure that none of these chemicals are carcinogens, mutagens, skin irritants  and they don't damage the sewer system and these days the environment. Salute and hats off to detergent chemists who make our clothes clean everyday.
Kirk Othmer Encyclopedia and Ulmann's Encyclopedia of Chemical Technology will be great resources for detergent chemistry. They have chapters upon chapters on these topics.

I was just about to post "Doesn't anybody here know how soap works?" in meta, but you've come through with an answer, thank you! I'm curious about "...emulsify or dissolve the oily dirt." Are there some aspects of removing "oily dirt" when washing clothes are are indeed more like dissolution than emulsification?

May be I used the words loosely. See how soap "dissolves" oil. I am using dissolve in a loose sense. If you have fruit stains, you are then actually dissolving those stains and also bleaching them. Enzymes may destroy protein type stains.

In older times, when clothes were boiled in lye, you would actually soapnify the oil stain by coverting it into soap :-) Many washermen in South Asia still practised this for large sheets. The German machines have a Kochwasche program- which is almost the same as cooking the laundry. Literally it washes clothes in boiling hot water.

sure, for example I think that the membranes of dead skin cells are certainly broken down by the detergent releasing the contents of the cells, which themselves are broken down. Some of that organic material could be said to be emulsified while certainly some proteins and nucleic acids will just go into solution.

Right, it has a fascinating chemistry. I had fascination with clean clothes and with detergents and read a lot about them since I was in school from those Encyclopedia :-)

@uhoh, Are you still not satisfied with any answer?

Since I'm not the OP I can't accept answers. Leaving the bounty to run the full period before accepting maximizes the number of  views of the page and possibly up votes for you. In some ways SE is a little bit like a game. I don't care much about my own reputation but anything I can do to increase reads and votes for others I try to do. Also, even after the bounty expires there's still a silent 24 hour "grace period" to help forgetful people like me :-)

And if I forget again I just add a second "replacement bounty" which costs me double! 1, 2

A grease/oil spot of about a milligram would make a droplet about 1 mm in diameter. This is too large to stay suspended in water - it would float to the surface, where it would combine with any other tiny droplets, even if these droplets were coated with surfactant. When you withdraw the clothing from the dirty water, or drain the water thru the fabric, the oil/grease would be trapped again on the fabric.
However, if you agitate all the 1 mm droplets, each covered with a monolayer of surfactant, in the presence of more surfactant in the water, the droplets will deform, elongate, and split, generating more surface area (the extra surfactant molecules in the water will rush to coat the oil/grease). As agitation continues, the tiny droplets become smaller and smaller, always covered with a monolayer of surfactant, because you put enough into the water at the beginning of the washing cycle.
Finally, the droplets become small enough that their buoyancy is negligible compared to Brownian motion (maybe 0.005 mm or less) and they act like large dissolved molecules. At this size, the particles are very difficult to filter out of the water using ordinary fabric, which is good, because you don't want to trap them, you want to drain them away.
One way to make an emulsion of small particle size is to add an organic acid (e.g., oleic) to an oil (e.g., canola), then pour this into a sodium hydroxide solution. The acid saponifies in intimate contact with the oil and the oil disperses into small droplets coated with oleate anions. Sodium ions are also in solution. This happens spontaneously thru a chemical reaction.
However, if you drop canola oil into a sodium oleate (a surfactant) solution, it floats without significant emulsion forming. This is typical of ordinary oil/grease - you don't have a readily available chemical reaction to form small droplets. But if you shake it you break the oil into tiny droplets, and you will get an emulsion that is stable (i.e., doesn't separate into two layers) for a while. The emulsion will be stable longer if the droplets are smaller. You have substituted mechanical energy for the chemical energy, and the surfactant acts to maintain the stability.
If you have insufficient surfactant, the droplets won't get small enough to become stable in water. If you use excess surfactant, you will get smaller particles and a more stable emulsion, and lots of foam.

Soap can mix with both water and with oil. Why? The soap molecule has
two different ends, one that is hydrophilic (polar head) that binds
with water and the other that is hydrophobic (non-polar hydrocarbon
tail) that binds with grease and oil.

When greasy dirt or oil is mixed with soapy water, the soap molecules
arrange themselves into tiny clusters called micelles.

The water-loving (hydrophilic) part of the soap molecules sticks to
the water and points outwards, forming the outer surface of the
micelle.

The oil-loving (hydrophobic) parts stick to the oil and trap oil in
the center where it can't come into contact with the water.
With the oil tucked safely in the center, the micelle is soluble in
water. As the soapy water is rinsed away the greasy dirt goes along
with it. Micelle Picture from Wikipedia
So basically, the soap is attracted to the fat/oil/grease because of
its fat-loving side but then tears up the grease by pulling it into
the water using its water-loving side. Sounds kind of like a football
play--surround the oil particles and move them away from one another.

Ever wonder why it is easier to clean dirty, greasy hands (and other
things) in hot or warm water rather than cold water? It is because the
fats and oils soften or melt in hot water, which allows them to attach
more readily to the hydrophobic end of the soap molecule. In turn,
that makes it easier to rinse away.

Credits: entire answer is copied from https://www.chagrinvalleysoapandsalve.com/blog/posts/how-does-soap-work/  Without any edits


text:
Selectivity of olefin cross-metathesis reaction
I'm looking at the reaction shown below which uses a ruthenium catalyst:

I am confused because the ruthenium catalyst which is used in the reaction should be Z selective due to the features on the catalyst because the adamantyl group favours the side bound pathway i.e. the Z isomer and the di-iso-propyl phenyl group blocks the path to the E alkene.

Did you perform this reaction and get that result? Are you looking at some publication that publishes this?

Is there an error in the reaction?  The reactant has 5 carbons, and the product has 8.  If this is supposed to be a dimerization, what happens to the two missing carbons?

@CurtF. There doesn't appear to be a mistake. This is standard olefin metathesis. Ethylene is the assumed by-product.

I see.  Olefin metathesis is cool.

@CurtF. They did give out 3 Nobel Prizes for it in 2005...

@Jan just looking at a publication

Please add/cite your source then =)


text:
Tin (II) chloride is white while tin(II) iodide is red. Why?

Tin (II) chloride is white while tin(II) iodide is red. Why?

The answer given in my book says:
Bigger anion has higher polarizability; more polarization greater is the intensity of colour(valence shell electrons are loosely bound to the nucleus).
This answer doesn't seem to be very explanatory. Even google doesn't offer anything. It would be great if someone could explain it in a better way, focussing on the following two points:
Firstly, how does more polarization lead to greater intensity of colour?
Secondly, how is intensity anyways related to frequency of radiation?

(2)In molecules the intensity of electronic transitions is not primarily determined by frequency but by overlap of wavefunctions via Franck-Condon factors. (1) The polarisation does not give greater intensity of colour but shifts energy levels into visible spectrum. ('White' means solid is scattering light and probably absorbing in UV).  Possibly (I'm not sure here) but the reason for this could be due to some Charge Transfer Complex formation between iodine and tin complex being stronger and hence of lower energy that that due to Cl. But it could be that the whole structure is different.

If I had to guess, I would say ligand to metal charge transfer.


text:
How to make graphite from wood/charcoal
Would be very grateful if someone had the answer or can point me in a direction. I've searched a lot but didn't find it yet. It could benefit others as well.
I do not have a high tech lab, but I can invest, would prefer methods that don't require above \$100 or \$300, but am open to it. I can build a furnace and weld.
Edit: As I don't have 50 rep points to reply, I will answer here to comments the first comment as to what use I need it for;
It is for water Nanofiltration of 1 micron or lower ,and secondly batteries making.
--- Second comment; thanx alot for the document, I will check it and tell you about it.

What do you want to use the graphite for? the use you intend for it can really change the method used for the synthesis

graphitization of carbon materials occurs well above above 2500 celsium, so I suspect there is no pragmatic means to do so. Graphite is available online, though.

@permeakra never heard of microwave ovens? lasers maybe? this technology can reach far above 2500 "celsium", those who buy online can't call themselves physicists

There was a method patented in 1896 whose claims would seem to meet your requirements.  Even with your restricted budget, used laboratory equipment companies can probably provide you what you need.
An excerpt from the patent states:

"My invention relates to the production of graphite from any of the
  cheap and common forms of carbonaceous materials, such as mineral
  coal, coke, charcoal, gas-carbon, and carbids and the like; and it has
  for its object to provide a process or method of cheaply and
  practically producing pure graphite..."

The reagents used are common, cheap and readily available.  The patent states that:

"[the mixture is] composed of powdered coke, sand, salt, and sawdust,
  the coke forming about fifty per cent., by weight, of the mixture, and
  these materials are thoroughly mixed and introduced into the electric
  furnace..."

The key issue for you will be designing the high-temperature inductively heated furnace, though the patent describes this thoroughly, including diagrams.
The availability of used equipment to fashion the device as stated is of course uncertain, and you will have to be creative with what is available.  I would start my search with LabX, with whom I have no affiliation whatsoever.  I hope this does "point you in the right direction", and Good luck!

so, what about the professor Roop Mahajan who creates graphene from coal by grinding coal to coarse powder which is later ball-milled to nano-sized particles which later reacted with nitric acid to convert it into graphene oxide ,which are later reduced to pure graphene ? you just need to press it in order to make graphite (ACS Appl. Nano Mater. 2023, 6, 15, 14594–14608)


text:
Emergency shower for nitric acid spills
Disclaimer: I'm asking this because I am curious.  Not because I'm actually handling nitric acid.
In another answer I was reading, the poster mentioned that if you spill nitric acid, you shouldn't pour water on anything but a small spill, because it will generate heat (maybe enough to flash-boil the water?), but if you spill nitric acid on yourself, you should have an emergency shower ready.
Except, wouldn't the emergency shower be using water and cause the exact same problem?  Is this just a case of deciding that possible burns from a hot liquid, while under a cold shower are much better than definite acid burns?

Anyway, with nitric acid (unlike sulfuric) heat is not much of a problem.

Note that the standardized flow rate of safety showers for laboratories is 60 l/min if no requirements are stipulated by national legislation (e.g. 30 l/min in Germany). We could use such values to make a rough estimate of heat and temperature for a given amount of nitric acid.

Concentrated sulfuric acid has another danger. It is slippery.

If you spill nitric acid onto the table, you yourself are unharmed and you can use a cool head to decide what to do next. If the spill is small, pour water on it to both dilute it and dilute the heat — remember water has a high specific heat capacity. If the spill is larger, I would probably try adding sodium hydrogencarbonate to react away the acid before wiping it up. Adding larger amounts of water will work, too, but you need to get them from somewhere and then need to get rid of the resulting puddle.
If the spill is on you you want to get the acid off your clothes and body immediately. The quickest and most effective solution is to pour water over your head and lots of it. This is what the emergency showers are for. Lots, here, has two functions: it not only quickly dilutes the acid to a less harmful concentration but also (because it is washing down) serves well to actually remove the heat — remember again that water has a high specific heat capacity. Emergency showers have a high throughput for exactly this reason.

So more  caveats about an emergency shower. (1) Using an emergency shower isn't the time for modesty. You strip bare ass naked under the flow of water.  (2) Contaminated clothing should never be removed over your head to avoid eye exposure. Bandage scissors  should be available at the shower. There isn't time to go looking for scissors if an accident occurs.

True that; I decided to only concentrate on the effects of water + acid here ^^

Adding water to acid may result in some of the acid going somewhere else in relatively uncontrolled fashion.  If the acid is on the table,  and the "somewhere else" is on you, that's bad.  If the acid is on you, however, and the "somewhere else" is pretty much anywhere, however, that's good.

This all comes down to a matter of relative amounts of acid and water. A large spill on the floor (say, 2L) would require water to be poured onto it, and the initial volume would be small compared to the acid volume. If you used a volume of water that would fill a typical garden bucket(9L or 30 seconds of flow from a tap), this would not mix spontaneously, and there would be parts of the spill flow where acid concentration is much higher, causing problems.
In the case where a spill occurs on a person, the volumes of acid resident on the individual are much smaller, and the relative amounts of water much greater. There may only be a matter of tens of mLs of acid remaining on clothing, and much less on the skin. The volume of water applied from a safety shower is much greater than this, and can dilute the acid before adverse reactions occur.
So, in the case of an acid spill in the lab:

Remove the individual, remove clothing, immerse in safety shower
Evacuate local vicinity as requried
Contain spill, and use chemical absorbent to decontaminate hazard.


text:
Is there a difference between Close Cubic Packing and Face Centered Packing?
I see some places use close cubic packing (ccp) and face centered packing (fcp) interchangeably. Are they the same thing? Does one provide more information than the other?

The Wikipedia entry for "Cubic crystal systems" says:

...Face-centered cubic (abbreviated cF1 or fcc, and alternatively
called cubic close-packed or ccp)

In another entry it is stated:

...face-centered cubic (fcc) (also called cubic close packed)...

The text of the above references make no statement about one term providing any more or less information than the other.  This does not preclude the possibility that some organization or another may have a preference for one term or the other, but the meaning seems to be the same.

No need for the seems to and other decertainisers; they are ;)

Typically close-cubic packing refers to the high packing factor of FCC systems. It can be used interchangeably with FCC but usually used with metallic materials only, where the atoms are all the same.
Face-centered cubic is used for metals and other materials as it describes the Barvis lattice of a material, which an FCC packing may not yield the highest packing factor in a non-metallic material.


text:
Difference between intercalated lithium and metallic lithium (Li-ion batteries)
I have a question about Li-ion batteries. I sometimes see the two terms - intercalated lithium and metallic lithium - used when talking about this type of batteries and I was wondering what is the difference between the two?
I understand that the intercalation means that the Li cations are transferred from one electrode to another and that is why it is called intercalated lithium (or am I wrong?). But what about metallic lithium? What does that mean?
I found this on Wikipedia:

Li-ion batteries use an intercalated lithium compound as one electrode material, compared to the metallic lithium used in a non-rechargeable lithium battery

But what makes them different? In other words, what are the differences between intercalated lithium and metallic lithium? Can anyone help me to understand this?

Intercalation is the addition of lithium ions into a host material without significantly changing the host's structure. In my research, I will often prepare test cells where the cathode is an intercalation material, like $\ce{CoO2}$, which can accomodate (or "intercalate") layers of lithium ions between the oxygen layers in the crystal lattice to form $\ce{LiCoO2}$. It's not that there are layers of cathode material surrounding lithium, but rather there's only 1 material, but lithium ions can move into and out of it.
$$ \ce{CoO2 + Li+ + e^- -> LiCoO2} $$
I mostly use lithium metal for the anode because it's easy to work with and has a stable potential, making it a good reference electrode:
$$ \ce{Li^+ + e^- -> Li_{(s)}} $$
In this setup we have an intercalation cathode ($\ce{LiCoO2}$), and a lithium metal anode. The major downside is that when you cycle lithium metal, it forms dendrites which can puncture the separator and short-circuit the cell, leading to catastrophic failure. To solve this problem, commercial cells use graphite anodes, which can intercalate lithium ions:
$$ \ce{C6 + Li+ + e- -> LiC6} $$
This solves the dendrite problem, but makes the battery less efficient. Since dendrite formation is only a problem with repeated cycling, non-rechargeable ("primary") batteries just use lithium metal instead. Lots of researchers have tried to solve "the dendrite problem" for decades, with little success.
The diagram below shows a lithium ion battery with an intercalation cathode on the right: the green parts are lithium, the blue parts are transition metals (Ni, Co, etc.) and the red spheres are oxygen. The anode on the left is the graphite intercalation anode, but you could replace it with a metallic lithium anode.

There is a lot of confusion about lithium as it exists in a lithium ion cell. Most articles are written by non-chemists who do not understand the fundamental difference between lithium existing as an ion (a positive ion or cation particle) and as a metallic atom (an electrically neutral particle). The plain fact is that one electrode of a lithium ion cell contains lithium oxide (or other ionic compound of lithium, such as a phosphate). In that electrode, the lithium exists as lithium ions as part of a lithium salt.
The other electrode can be just metallic lithium, but it has been found to be more practical, safe and effective for the life of the cell for the lithium to be layered with thin sheets of graphitic carbon, called graphene which is a bit like pencil lead.
A suitable electrolyte separates the two electrodes. Lithium hexafluorophosphate dissolved in an organic solvent, such as the ester-like solvent, ethylene carbonate, is usually used.

Intercalated in this context basically means inserted, or interposed.  So an intercalated lithium-ion battery has sheets or films of material separating layers of lithium.  According to this Wikipedia entry:

Separator materials in lithium ion batteries must not block the
  transport of lithium ions while preventing the physical contact of the
  anode and cathode materials, e.g. short-circuiting. In a liquid cell,
  this separator would be a porous glass or polymer mesh that allows ion
  transport via the liquid electrolyte through the pores, but keeps the
  electrodes from contacting and shorting. However, in a thin film
  battery the electrolyte is a solid, which conveniently satisfies both
  the ion transportation and the physical separation requirements
  without the need for a dedicated separator.

A lithium-metal battery is simply a conventional, non-rechargeable battery in which metallic lithium serves as the anode.  There is a table of the common cathode materials used in lithium-metal batteries here.

Thank you for the explanation. But I can find Lithium Metal even in the context of rechargeable Li-ion batteries (see: http://www.oalib.com/references/13559284). So when they are speaking about intercalated Lithium is it the atoms and when they write Lithium-metal is it the structure (as in the second link you added: Lithium batteries are batteries that have lithium as an anode?)? And can one say that an ion is a metal or it has to be in a metallic bond to characterize it as a metal?

The separator in a lithium-ion battery is independent from the question of whether the electrode are intercalation-based or not. Even with lithium metal and a conversion material, you still need a separator.


text:
How do I identify resonating structures for an Organic compound
Basically, my concepts regarding Resonance are not totally clear. I just need some help, on how to find the no. of resonating structures, given a molecule, and how do I draw those structures. Some examples would be pretty helpful. Thanks

Essentially, resonance structures are explained by Resonance Theory. Resonance theory looks to answer the question: How do we reconcile the differences between Lewis Structures of species and their observed characteristics? There are a few rules that must be followed when determining resonance of compound.

Resonance occurs in species for which there are two or more valid Lewis structures.

For a species, each valid Lewis structure is called a Resonance Structure or Resonance Contributor. In these cases, you are simply moving an electron or pair of electrons to another location in a molecule. An example is HCO2-.There are two possible resonance contributors in this molecule. As shown in the example, resonance structures differ only in the placement of their valence electrons, not their atoms.

Resonance structures are imaginary; the one, true species is represented by the resonance hybrid.

The Resonance Hyrbid is the weighted average of all resonance structures. An example of the resonance hybrid occurs in either of these molecules.

The resonance hybrid looks most like the lowest energy (most stable) resonance structure. Essentially, the least formal charges on the molecule, or most evenly distributed charges.

A few tips I have for drawing resonance structures based on Lewis structures for different molecules include:

Move away from a negative charge. If presented with a negative charge on a specific atom, it means that there is an excess of electrons on that atom. Either take electrons away from the atom, or move a bond that it is involved in.
Move towards a positive charge. A positive charge represents a lack of electrons, this means that you should move electrons to the atom with the charge, or move a double/triple bond to allow it to become involved in another bond.
Move away from a radical. A radical does not represent a charge, as it is only a single electron, but you should (in most cases) move electrons away from the radical.

Also, always remember that electrons are the only things that move in resonance structures. They can also not be added or subtracted, but only moved. Make sure not to move/add/remove any atoms, and you cannot move electrons to an atom that is already saturated with bonds for example, an sp3 hybridized Carbon as it is already saturated with Hydrogens.
Best of luck!

Wow. Thanks. That should hopefully clear some facts for me

Yeah for sure, let me know if there are any other questions that I can help with! :D

There are more than two resonance contributors in $\ce{HCO2-}$. 2. There is no most stable resonance structure! 3. "Essentially, the least formal charges on the molecule, or most evenly distributed charges." This might be purely based on experience and is not at all true. The prime counter example is CO.


text:
What chemical equation proves that KHCO3 can extinguish a fire?
In a dry powder fire extinguisher, there is a chemical known as $\ce{KHCO3}$ and I was wondering how $\ce{KHCO3}$ works in a chemical equation to extinguish the fire.

$$\ce{KHCO3 -> KOH + CO2}$$

Please also see this answer which contains a subsection and a reference for how $\ce{NaHCO3}$ works. It is more that just the decomposition yielding $\ce{CO2}$; surpressing chain reactions tying up radicals is also key here (and heat of vaporisation and heat capacity).

Related: http://chemistry.stackexchange.com/questions/42130/potassium-vs-sodium-bicarbonate-in-fire-fighting

According to this Wikipedia entry:

Decomposition of the bicarbonate occurs between 100 and 120 °C (212
  and 248 °F):
$$\ce{2 KHCO3 → K2CO3 + CO2 + H2O}, \Delta H>0$$

So, you produce a nonflammable powder and two nonflammable gases, and you absorb heat from the fire in an endothermic reaction.
The picture here is more complex than just a chemical equation however. The $\ce{KHCO3}$ also acts as an energy-absorbing material and provides a solid surface on which free radicals can be destroyed.

Tried \DeltaH and it came out bad.  What was wrong?

Thx @OscarLanzi .

Thx @NilayGhosh.


text:
Processing Chitosan from Crab Shell
I am trying to extract chitosan from crab shell (Japanese snow crab). I want to dissolve chitosan in water and mix in gelatin to obtain some type of gel, that I would then test for my Master 2 Architecture thesis.
However, because I am from an architecture background I am having a hard time to understand the process of chitosan extraction.
I got 1M $\ce{HCl}$ (500 mL) and 1N $\ce{NaOH}$ (500 mL). We gathered around 3 kg of crab shell. I am following the guide of two researchers (here and here), but I can't really understand what they mean in terms of percentage, and dosage of the acid.
Could you please help me in very simple terms?
What does the percentage relate to (solute and solvent)?
Do I need to add water with the acid I have and then mix it with the shell?

Are you sure that your first link (http://here) is correct?


text:
What element is/used to be called Fb?
What element used to be called Fb? It was supposedly found in a meteorite in Antarctica and I can't find any references of it.

To my knowledge, there is no element with the symbol Fb. There are however, some with b as the second letter of the element symbol, such as Rb, Pb, Sb, Tb, and Yb.

I'm guessing a P looked like an F somewhere and that it contains lead...wild guess.  Any idea at all of where you saw the reference? On-line, text book, sci-fi channel ;)

https://boards.4chan.org/sci/thread/8615928 ?

@Secret that's very reassuring. Thanks for digging it up.

Its a text-question in a book. Everything is made up... Its an exercise

I'm voting to close this question as off-topic because it is based on a hypothetical textbook question posted on 4chan.

Given that Facebookium ($\ce{Fb}$) does not exist, it is most likely a typo/transfer error for the first character of the element symbol.
The possible choices are $\ce{Rb}$, $\ce{Pb}$, $\ce{Sb}$, $\ce{Tb}$, and $\ce{Yb}$.
Among these, lead ($\ce{Pb}$) is very likely to be the right one.
In fact, there is a publication in Earth and Planetary Science Letters, 1973, 20, 391-394 where the authors note in the abstract

In the four meteorites Yamato (a), (b), (c) and (d) recently found in Antarctica, the trace elements W, Re, Os, Ir, Pt, Au, Hg, Tl, Pb, Bi, Th and U were determined by spark source mass spectrometry. The concentrations of these elements were also measured in the meteorites Orgueil (C1), Murray (C2), Allende (C3), Abee (E4), Allegan (H5), Mocs (L6) and Johnstown (Ah).


text:
What indicators can be used when substituting iodine with other halogens in the Winkler method?
I'm a chemistry student who is doing an investigation with respect to the Winkler method. The Winkler method is one which is used to find the oxygen content in water. Primarily, the halide which is used is iodine. However, I want to make use of other halogens such as fluorine, chlorine and bromine, together with their respective halides. However, I am stuck on how I will actually do the titration, since the indicator used for the case of iodine is a starch indicator. I don't exactly know which indicator I should use for the other halides.
My question is this: What indicators should I use in the titrations involving the other halides?

Note that the iodine is formed in solution from the iodide reagent.  I'm pretty sure you don't want to go there with the other halides.  The acidification step liberates iodine, and if other halogens acted analogously, you would then form the more volatile bromine, chlorine or fluorine gases.  No thanks. I doubt any of those would work except for possibly bromide anyway.  What's wrong with iodide?

Related: http://chemistry.stackexchange.com/questions/27450

There's nothing much wrong with it, it's just that I'm just doing an investigation for other halides within the method. Like, just trying to see how other halides might affect the results obtained.

It will likely not work with other halides. As far as I understood the winkle method, the oxygen is first fixated with $\ce{Mn^2+}$ in basic conditions:
$$\begin{align}
\ce{Mn^2+ (aq) + 2 OH- (aq)&-> Mn(OH)2 (aq)}\tag{1}\label{alkalinemanganese}\\
\ce{2Mn(OH)2 (aq) + O2 (aq) &-> 2 MnO(OH)2 (s) v}\tag{2}\label{bindingoxygen}\\
\end{align}$$
The sample is acidified and iodide is added. This is oxidised to $\ce{I2}$, which forms a complex with remaining iodide.
$$\begin{align}
\ce{MnO(OH)2 (s) + H+ (aq) &-> Mn(OH)3 (aq)}\tag{3}\label{acidicmanganese}\\
\ce{2 Mn(OH)3 (aq) + 2 I- (aq) + 6 H+ (aq)
&-> 2 Mn^2+ (aq) + I2 (aq) + 6 H2O}\tag4\label{iodide-ox}\\
\ce{I2 (aq) + I- (aq) &<=> I3- (aq)}\tag5\label{triiodide}\\
\end{align}$$
The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions.
$$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$
The reason why this works is because you have the equilibrium $\eqref{triiodide}$ which produces the  triiodide ion, which is relatively stable in aqueous solution. Iod by itself does not dissolve in water (much). This complex again forms a very colourful complex with starch, the switch can easily be detected by the human eye.
Polyatomic ions of other halogens are not unheard of (ref. Trichloride ion (c.f. triiodide ion) and therein), but they are a lot less stable. I am not aware that they form a complex with starch so they won't be that easily detected.
Another aspect, I don't want to go much deeper here, is that the redox reactions necessary $(\ref{iodide-ox},\ref{triiodide-reduction})$ for this procedure to work might not work at all with other halides.
The original method was proposed in the 19th century and is still applied today. It is a quite accurate and robust method. There is not really a reason to tamper with it.

But suppose that I did want to try doing this experiment anyways, what kind of indicator should i use for each halogen? The thing here is that there may be no point in tampering with it but I'm still quite curious how this may work with other halogens and really want to do this investigation.

@AbhijeetVats If the species necessary to detect does not form, there is no way you can use any other molecule to detect it. It simply is not possible.


text:
Tartaric acid chirality

In the image above, four variations of tartaric acid are shown:

The variant in the top left is the "basic" skeletal formula of tartaric acid.
The one on the top left is D-(-)-tartaric acid.
The one on the bottom right is L-(+)-tartaric acid.

How would you name the one on the bottom right using the dextro- and levo- method instead of the S and R method?

https://en.wikipedia.org/wiki/Meso_compound

You can’t really name it according to D/L.
This nomenclature requires you to draw a Fischer structure wherein the skeletal atom labelled ‘1’ by nomenclature rules is at the top; then, the orientation of the bottom-most functional group determines whether the compound is dextro, D (pointing to the right) or levo, L (pointing to the left. Below you can see Fischer depictions of all three tartaric acids.

Figure 1: D-tartaric acid, L-tartaric acid and meso-tartaric acid from left to right.
Now do one of the following: either rotate your PC screen/mobile device by $180^\circ$ or rotate your head by $180^\circ$.* You should notice that both D- and L-tartaric acid still have the bottommost hydroxy group pointing to the right or left, respectively. Both carboxylic acids are identical, so we would have to mark one carbon atom by using $\ce{^13C}$ to distinguish the top and bottom. The orientations lead to the same deduced stereochemistry.
This is not the case in the third molecule. Before rotation, both hydroxy groups are pointing to the right-hand side, afterwards they are pointing to the left-hand side. But, as I already stated, the two depictions are identical in the absence of isotope marking. Therefore, this compound eludes the D/L nomenclature and cannot be named unambiguously in accordance with it.
Thankfully, though, we have another descriptor that does not directly stem from D/L: it is the meso descriptor. It is used in case a molecule with asymmetric centres also features a plane of symmetry causing it to be identical with its mirror image and therefore achiral. While it is not part of the D/L terminology, it is the commonly accepted name for meso-tartaric acid.

*: Please only perform gymnastics if you can handle them. I decline all responsibility for injuries that may occur. Stack Exchange cannot be held responsible.


text:
Theoretically, Which d orbital participates in sp3d and sp3d2?
This question is regarding the old way (inaccurate) $\ce{PCl5}$ and $\ce{SF6}$ are taught in elementary chemistry.
In the crystal field splitting diagram for triagonal bypyramidal geometry, $\mathrm{d}_{z^2}$ has the highest energy. Now, since $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$ of same shell hybridise, then the $\mathrm{d}$ orbital closest in energy to $\mathrm{s}$ and $\mathrm{p}$ should take part, which should be $\mathrm{d}_{xz}$ or $\mathrm{d}_{yz}$. Instead, the hybridisation is said to be $\mathrm{sp^3d}_{z^2}$.
A similar problem arises for molecules with an octahedral shape. It has $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ which are higher in energy than the rest and have higher energy difference with $\mathrm{s}$ and $\mathrm{p}$ orbitals of the same shell. There is no such problem for $\mathrm{d^2sp^3}$, $\mathrm{d^3s}$, $\mathrm{dsp^2}$.
EDIT: Let me change the question to ask the same thing. In outer shell octahedral complexes ($\mathrm{s,p,d}$ of same principal quantum numbers take part) why are the $\mathrm{e_g}$ orbitals taking part in sigma bond formation when they are furthest away from $n\mathrm{s}$ and $n\mathrm{p}$ orbitals?

There is no such problem for inner shell complexes.

Since the assignment of a coordinate system is completely arbitrary, this question cannot be answered.

Let's assume in sp3d, the terminal atoms/ligands approach from ±z, +x and y=x√3,z=0 line and y=-x√3,z=0 line. My main question was, was I correct in assuming that the orbitals closest in energy hybridise or there is some symmetry issue too? Secondly, how the people came up with the fact that dz2 supposedly takes part, not the rest?

I just want to point out that this concept of hybridization is ancient and if someone has an answer, it doesn't really has any meaning. It's a helpful tool to teach chemistry to students for sure, but that's about it.

None, because neither the $\mathrm{sp^3d}$ of $\ce{PCl5}$ nor the $\mathrm{sp^3d^2}$ depiction of $\ce{SF6}$ are close to the truth.

Edited the question to ask the same thing using metal complexes.

The new version of the question is in no way the same thing. It is very different. And the answer is symmetry.

The answer to this lies in the symmetry of the orbitals involved.
Naturally, an octahedral complex will have the point group $O_\mathrm{h}$ — the octahedral point group. In molecular orbital theory, we will first need to work out how the ligand group orbitals transform; then we can find out which metal orbitals transform identically, i.e. belong to the same irreducible representation; only these will form bonds with each other. To arrive at the group orbitals’ irreducible representations, we need to see how they transform under the given symmetry operations of the group; thankfully, a copy of the character table can be found on meta (Attention: MathJax heavy link! Give the page about 30 seconds time to load!) Since your diagram does not include π interactions, I will ignore them, too.
\begin{array}{lcccccccccc}\hline
\text{Irrep} & E & 8 C_3 & 6 C_2 & 6 C_4 & 3 C_4^2 & i & 6 S_4 & 8 S_6 & 3 \sigma_\mathrm{h} & 6 \sigma_\mathrm{d} \\ \hline
6 \sigma_{\ce{L\bond{->}M}} & 6 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 4 & 2 \\ \hline\end{array}
We now need to use the reduction formula to determine how many times each irreducible representation is part of this reducible representation.
$$n = \frac 1h \sum_{\chi} N \chi_\mathrm{r} \chi_\mathrm{i}\tag{1}$$
Wherein $n$ is the number of times an irreducible representation is included, $N$ is the coefficient of a symmetry operation group, $h$ is the sum of all coefficients ($48$ in $O_\mathrm{h}$) and $\chi$ is the character of the irreducible or reducible representation, respectively. Thankfully, a large number of reducible group characters end up as $0$, easing our calculations. This should be the result:
$$\begin{align}n(\mathrm{a_{1g}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{a_{2g}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{e_g}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times 2 + 2 \times 6 \times 0) &&= 1\\
n(\mathrm{t_{1g}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{t_{2g}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{a_{1u}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{a_{2u}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{e_u}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times (-2) + 2 \times 6 \times 0) &&= 0\\
n(\mathrm{t_{1u}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{t_{2u}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\end{align}$$
We see, that the six σ-symmetric ligand orbitals transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ — this is also shown in your picture. Now that we have this settled, we need to ask ourselves which metal orbitals these irreducibles correspond to. $\mathrm{a_{1g}}$, the totally symmetric irreducible is easy; it must be the $\mathrm{s}$ orbital. Furthermore, the character table tells us that $(x,y,z)$ transforms as $\mathrm{t_{1u}}$ which means that the $\mathrm{p}$ orbitals must transform as that. (You can verify this by performing the same exercise on the three central $\mathrm{p}$ orbitals.)
This leaves the $\mathrm{d}$ orbitals which — for symmetry reasons as there is no pentadegenerate irreducible — have to split up into at least a group of three and two ($\mathrm{t}_x$ and $\mathrm{e}_x$). Again, glancing at the character table tells us that $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ transform as $\mathrm{t_{2g}}$ and $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ as $\mathrm{e_g}$. The first can easily be verified by the same exercise as for the $\mathrm{p}$ orbitals; the second not so easily due to the ‘unnatural’ shape of the $\mathrm{d}_{z^2}$ orbital. However, do trust the calculations that they are equivalent.
Therefore, it is only $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ that can interact with the ligand group orbitals and which therefore form the σ bonds as your scheme shows. And this also makes sense from a purely qualitative point of view: The $\mathrm{d}_{z^2}$ orbital is pointing towards the $z$-axis ligands (and the ring is somewhat interacting with the $x,y$-axis ligands) while each of the four lobes of $\mathrm{d}_{x^2-y^2}$ is pointing towards a ligand on the $x$ or $y$ axis. The $\mathrm{t_{2g}}$ orbitals are pointing in between the coordinate axis directions and therefore between the ligands and are thus nonbonding with respect to σ bonds.


text:
What do you get when you melt and solidify sand?
Would you get something brittle like glass or something like granite?

More like glass, I guess.

@IvanNeretin Why?

Just my gut feeling.

There are many different types and definitions of both sand and glass.  See the entries in the respective Wikipedia pages (https://en.wikipedia.org/wiki/Sand and https://en.wikipedia.org/wiki/Glass).  This should either help you more clearly state what you are asking or maybe even answer your question.

Where I live the sand is eroded granite. The key issue is how quickly you cool the melt. Slow it can crystallize, fast it can be glass.

Well, it depends on the sand. At White Sands National Monument the sand is gypsum. The black sand beaches on Oahu are basalt particles. // Typical sand is mostly quartz particles. Lighting can fuse the sand into a silica glass called fulgurite. Trinitite is the glassy residue left on the desert floor after the plutonium-based Trinity nuclear bomb test.

Look at fulgurites for some examples. https://en.wikipedia.org/wiki/Fulgurite


text:
What are range and domain of Hamiltonian?
What are range and domain of Hamiltonian? Hamiltonian is function so it should have ones. What are they?

Hamiltonian is not a function, but an operator (a function in a broader sense, if you'd like):
$${\hat {H}}=-{\frac {\hbar ^2}{2m}}\nabla ^2+V(\bar{r},t)$$
It is applied to some function (our supposed $\psi$) and yields some other function. So its domain is the set of all doubly differentiable functions on our underlying space (the space itself can be 1D, 3D, or whatever else, depending on the nature of the problem), and the range is even broader than that, because the derivatives (or $V$, for that matter) do not have to be differentiable.

Many more requirements for the domain may be needed in some problems.
Almost all problems in molecular quantum chemistry would ask for square integrability of the wave function, for instance.

That's another story altogether. Sure, a $\psi$ function requires that to be physically meaningful, but who said we can't apply our operator to any doubly differentiable function, even if it is not good as $\psi$?

Thanks. I am asking because it is often refered as a function of system state which represents system full energy which is scalar value. Or in this case it is just constant function defined on underlying space?

As you feed various functions into $\hat H$, you notice that some of them have a peculiar property: the resulting functions are proportional to the input functions, as in $\hat H\psi=E\psi$ (that's Schrodinger's equation, BTW). These are called eigenfunctions of the system, and E denotes energy. See, it is a scalar value indeed, and it doesn't belong to the range of $\hat H$.


text:
Relative strengths of hydrogen bonds

Order of hydrogen bond strength:
$\ce{O-H...N}$ > $\ce{O-H...O}$ > $\ce{N-H...N}$ > $\ce{N-H...O}$. Why is that so?

According to me $\ce{O-H...O}$ should have the maximum strength while $\ce{N-H...N}$ should have the weakest hydrogen bonds based on the fact that oxygen is more electronegative than nitrogen. It would develop a higher $\delta^-$ charge and the hydrogen attached to it will have a higher $\delta^+$ charge which would make the hydrogen bonding stronger.
I can't reason out the order given in my textbook. After observing it, all I can guess is that oxygen being highly electronegative leads to higher $\delta^+$ charge on hydrogen while nitrogen being less electronegative easily shares its electron cloud with hydrogen, making the hydrogen bond stronger.
I am not sure about this reason as according to this it is more stable for hydrogen to make a hydrogen bond with a less electronegative element and possibly even an electropositive element, (unless and until the hydrogen is attached to a highly electronegative element), which does not occur in normal circumstances.
It would be great if anybody could help me figure out the reason for the above trend.

Whoever asked this question in the first place did not understand a single thing about hydrogen bonding. There are obviously too many parameters to consider when comparing "strengths". Not even if we were to keep all distances the same the comparison would work. Any reasoning trying to reproduce this "trend" is similarly bound to fail.

I don't understand why you say that any reasoning to understand this trend is bound to fail, given that there are several ways to measure hydrogen bond strenght. For example, one can measure dissociation enthalpies in gas phase or dissociation constants in a reference solvent. On the computational side, one could use electronic interaction energies of fully optimized dimers. What is the problem with trying to rationalize hydrogen bond strength?

@Martin-マーチン This question now has ~3k views. Whichever textbook is popularising this concept, does seem to be very popular...

@GaurangTondon this concept is used to solve a few questions in jee, and India is full of jee aspirants which is why this question is becoming popular.

https://chemistry.stackexchange.com/questions/133718/do-amines-or-alcohols-have-stronger-intermolecular-hydrogen-bonds

As a short answer, Hydrogen bonds (for elementary chemistry) are formed with donation of lone pair of a lewis base into LUMO of H attached to a highly electronegative atom. A better donor makes a better bond and more electronegative atom makes a better bond. From my experience, the second factor overrides the first one in most cases.
N is a better donor than O and H in $\ce{O-H}$ is a better acceptor than H in $\ce{N-H}$.

"From my experience" simply is not a good enough reason. Also this kind of generalisation is dangerous, and if not completely wrong then at least not right for the right reasons. This backwards fitting to some supposedly found experimental data is wrong and leads to a completely distorted understanding of chemistry.


text:
Do the units of the law of mass actions convey any meaning?
Since the number that you end up with, when calculating the value of the law of mass action for a given reaction often also has a unit, I was wondering if there is any physical meaning to this given unit or if it is basically "useless"?
Example:
$$\ce{CH3COOH <=> CH3COO- + H+}$$
$$K_c = \frac{c(\ce{CH3COO-})\cdot c(\ce{H+})}{c(\ce{CH3COOH})}\qquad [K_c] = \pu{mol L-1}$$

Well, you can calculate difference in number of moles of gases and active liquid species on reactant and product side species using units of $K_c$ and $K_p$, respectively, as these are mostly used for homogeneous reactions.
For example, the unit of $K_c$ is $\text{(concentration)}^n$ if $n > 0$, more liquid or aqueous (ionic) species on product side.
Then, you can predict whether the product is formed or reactant when you increase or decrease external pressure or volume of a container containing a mixture in equilibrium. Temperature's effect would be only heat of reaction dependent.
Hint: write $K_c$ in terms of moles and volume to check effect of volume, write $K_p$ in terms of mole fractions and total pressure to check effect of external pressure.


text:
What is the dipole moment direction in the nitrosonium ion?
In the recent question about the molecular orbitals of $\ce{NO^{$x$}};\ x\in\{-1,0,+1\}$ the follow-up title question arose:

Just a random side-question: What’s the dipole moment’s negative and positive sides in $\ce{NO+}$?

At the time I was unable to answer that question sufficiently, therefore I share the addendum Q&A style. The question further deconstruct the molecular orbital components to the dipole moment.

In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With $r=107.8~\pu{pm}$ on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are attached at the bottom of the post.
I am loosely following the same procedure/ conclusions as in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.) since $\ce{NO+}$ is isoelectronic with $\ce{CO}$.
According to the definition of the dipole moment
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
positive values in the following table indicate that the negative end is at the nitrogen atom.
\begin{array}{lr}\hline
\text{Orbital} & \text{Bond Moment}           \\\hline
\text{MO 1 } (\sigma)            &   -4.8336  \\
\text{MO 2 } (\sigma)            &    5.5232  \\
\text{MO 3 } (\sigma)            &   -0.7891  \\
\text{MO 4 } (\sigma)            &   -3.0061  \\
\text{MO 5 } (\pi   )            &   -1.1642  \\
\text{MO 6 } (\pi   )            &   -1.1642  \\
\text{MO 7 } (\sigma)            &    4.7383  \\\hline
\text{MO 8 } (\pi, \text{virt.}) &   (2.3831) \\
\text{MO 9 } (\pi, \text{virt.}) &   (2.3831) \\\hline
\text{Total } \sum_i^7 \text{MO }i               &   -0.6956  \\\hline
\end{array}
While we see the same polarisation of the molecular orbitals as in carbon monoxide, the magnitude differs and the HOMO of $\ce{NO+}$ cannot compensate for the polarisation of the electron density towards the oxygen.

Appendix
Base calculation for $\ce{NO+}$ after previous optimisation, E(RHF)=-128.972090453 input file:

%chk=mp2qzvpp.chk
#p MP2/def2QZVPP
scf(verytight)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

base

+1 1
N   0.0 0.0 -0.575312
O   0.0 0.0  0.503398

Analysis calculation input file for MO 1

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

analysis

+13 1

(Both empty lines are necessary.)
Analysis calculation input file for MO 2, etc

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)

analysis

+13 1

2,1

The last line needs to contain the complete reordering procedure. The MO to be analysed needs to be in first place. Therefore for MO 3 it needs to read 3,1,2, for 4 4,1,2,3, etc.


text:
Does Propanone trimerise to give 2,6-dimethylhepta-2,5-dien-4-one?
Does propanone trimerise to give 2,6-dimethylhepta-2,5-dien-4-one?
How and under what conditions?

https://en.wikipedia.org/wiki/Aldol_condensation

That compound is more commonly called phorone and it is made by an aldol condensation. However, the cyclic isomer is a little more common: isophorone.

As the comment pointed out, this is an aldol condensation.
I shall demonstrated the acid-catalyzed version.
Tautomer

Propanone exists in tautomer with propen-2-ol. This is an example of keto-enol tautomerism, with "keto" referring to the ketone which is propanone, and "enol" referring to propen-2-ol.
Dimer

One molecule of propanone $(1)$ is converted to its tautomer, propen-2-ol $(2\mathrm a)$, with the help of hydrogen ions.
Another molecule of propanone $(1)$ is protonated to form a cation $(2\mathrm b)$.
The central carbon of the protonated propanone is an electrophile, so it attacks the double-bond in the propen-2-ol. A carbon-carbon bond is formed $(3)$.
After deprotonation, the intermediate 4-hydroxy-4-methylpentan-2-one is formed $(4)$.
It is converted to its keto-enol tautomer and the tertiary hydroxyl group is protonated $(5)$.
Then, after dehydration $(6)$ and deprotonation, our dimer 4-methylpent-3-en-2-one is formed $(7)$.
This is the first aldol condensation step.
Trimer

The mechanism of the formation of the trimer is akin to that of the dimer just above.
Similarly, one molecule of propanone $(1\mathrm a)$ is converted to its tautomer, propen-2-ol $(2\mathrm a)$, with the help of hydrogen ions.
A molecule of the dimer, 4-methylpent-3-en-2-one $(1\mathrm b)$ is protonated to form a cation $(2\mathrm b)$.
The central carbon of the protonated propanone is an electrophile, so it attacks the double-bond in the propen-2-ol. A carbon-carbon bond is formed $(3)$.
After deprotonation, the intermediate 6-hydroxy-2-methylhept-2-en-4-one is formed $(4)$.
It is converted to its keto-enol tautomer and the tertiary hydroxyl group is protonated $(5)$.
Then, after dehydration $(6)$ and deprotonation, our trimer 2,6-dimethylhepta-2,5-dien-4-one is formed $(7)$.
This is the second aldol condensation step.