text:
The definition of formal Charge
Ok, most people including me know that formal charge is just a book-keeping tool. I know how to calculate it, apply it to Lewis structures etc etc. But I am confused in its very basic definition which says - "Formal charge is the charge on an atom if the electronegativity differences are ignored" . If the electronegativity difference is ignored, then how can an atom have charge ? I know this is not a place to ask such a basic question , but still I am very confused about this miniscule definition.

Take a look at this previous question to see if it helps you out.

@NicolauSakerNeto Thanks but I know all that (how to calculate and use formal charges). I just want an explanation to this definition.

In terms of your definition, formal charge is the charge on an atom when electronegativity differences, regardless of relative electronegativity, are ignored. You must realise that a formal charge is not the same as a real charge. They are used simply as a bookkeeping method only. But, Polarity considers electronegativity rather than formal charge.

A good description appears in the book, "Chemistry A Molecular Approach".
In the molecule of hydrogen fluoride, we know that it has a dipole moment, and fluoride is slightly negative. Ignoring this information (electronegativity difference) and we share the bonding electrons equally, the formal charge can be calculated, and both of them are determined to be 0.
In reality, we know that fluoride has the electron from the hydrogen most of the time (due to electronegativity difference). We say the hydrogen has a partial positive charge and the fluoride has a partial negative charge.
From this, it is apparent that the formal charge is different from the charge on the atoms in a molecule.

Does the formal charge affect the overall charge seperation? I mean in carbon monoxide formal charges are -1 on carbon and +1 on oxygen but considering only electronegativity we should expect the negative charge on oxygen whereas the electrostatic map potential shows some negative charge on carbon. So can we say that the overall charge depends also on formal when and only when the proposed lewis structure have formal charges opposite to these according to electronegativity differences?

@adosar Actually, the formal charge in carbon dioxide is +4 on carbon and -2 on oxygen. Since formal charge is for bookkeeping, it has no influence on what the charge distribution in the molecule is.

I said for the carbon monoxide.

@LDC3 I think the electronegativity difference indicates which of the 2 atoms is attracting the "bonding electron density" more strongly. By assuming equal sharing, you know that the charge on oxygen is +1. The further polarisation of shared electron density towards oxygen would perhaps reduce +1 to a δ+ partial charge (since the oxygen atom can't completely attract the shared electron density; that would make the bond predominantly ionic )

The formal charge is the charge on an "ion" that results when all valence electrons participating in bonds are assumed equally shared between this "ion" and the others it is bonded to. If we consider, for example, nitrogen bonded to three carbon atoms using its 3 bonding electrons then it loses, assuming equal sharing, 1/2 an electron to each carbon but gets back a half share in an electron from each carbon. Net charge on the nitrogen: 0. Now assume that the unshared pair form a second bond with one of the carbons. The nitrogen loses a half share in each of those two electrons but does not get back a half share from the carbon in this case. Thus the nitrogen has a formal charge of +1. But this is a formal charge. A bookkeeping convenience - nothing more. In actuality the electronegativities of carbon and nitrogen will determine where the electrons spend their time and thus the apparent charge distribution. Obviously it would be quite tedious to try to figure out what those charge distributions would be. It's handy sometimes to assume, when considering covalent bonds, that they are perfectly covalent.

It's best to think about why electronegativity is used in this definition of formal charge and perhaps replace it with a concept that is familiar. One definition of electronegativity is the ability of an atom to attract electron density towards itself. So if we think of a simple diatomic series with (lewis pairs added) like: H:H,  H:Br, H:Cl and H:F, with electronegativities H < Br < Cl < F, then those two electrons ":" are closer to F than they are to Cl, which in turn are closer to the Br atom and in H:H they are shared equally. Now the formal charge is the number of electrons ascribed to the atom minus the atomic number. So the question is how do we ascribe an electron to each atom? If we try to include electronegativity arguments then we would ascribe more electrons to F then to Cl in those diatomics and the formal charges in HF and HCl would be different. However, if we ignore the electronegativity and let the atoms share the electrons equally (like in a lewis diagram) then we arrive at the familiar formal charge definition that you have noted.


text:
Do filled Orbitals also hybridize?
Recently I was watching a video on Valence Bond Theory, on the MIT Open-Course youtube channel. The teacher there said that

For hybridization to occur, electron promotion from fully filled orbitals to empty ones is a must.

But here in a book it is specifically mentioned that

Electron promotion, though common, but is not 'a must' for hybridization.

Also, it is clearly written that sometimes even the filled orbitals take part in hybridization.
I tried finding some examples of such a case on the web, but could not. It would be a great help if someone tells me even one example where the fully filled orbitals also take part in hybridization, with an explanation why it is so.
The link for the 41 min long video is here.

I tried to watch that... $\ce{PH3}$ cannot be described in terms of VSEPR or hybridisation. The bond angle is about $93^\circ$. I could not watch any more after that.

@Martin I never said anything about PH3 and she DID NOT explain PH3 by VSEPR theory. If you had watched the video full you would have got my point. I think since you did not watch the video and neither you bothered even reading my question , so you should not comment anything that is not related to it.

I have carefully read and edited your question, and as I just saw I have to do this again since you changed the tags to the topic that they aren't. I really wanted to answer this question, but now I am more than only annoyed by your suggestive tone.

I can understand Martin's first comment; the lecturer does imply VSEPR and hybridization for $\ce{PH3}$ (a molecule relevant to the question) when she says "the lone pair pushes down the bonds below 109.5°". This is not true; the lone pair is in an orbital which is essentially the same as the phosphorous $3s$ atomic orbital, and hence has quasi-spherical symmetry about the phosphorous nucleus. Such high symmetry cannot induce directional repulsion. Martin's view (shared by many others) is that hybridization doesn't exist in the first place. This is fair as a type of answer.

@NicolauSakerNeto She said that just in a casual way, though she was just trying to explain that the bond angle will be decreased due to the replulsion from lone pair. She is teaching to first year students, she need to specify such detail, that too on open course. But still I dont understand why am I commenting this PH3 thing again and again. I want to know an example where filled orbitals hybridize, nothing more.

@Martin I don't care about the tags. I want an answer to my question, which I don't think you can give ( I am not being rude ). You can go on editing if you like. I am sorry if you found me rude, but I still think there was no need of your comment.

@NicolauSakerNeto how do yo know the lone pair which is "essentially" the same as the phosphorous 3s orbital?

@NicolauSakerNeto I am not opposing hybridisation. Just the concept in which it is usually treated, as a rigid theory that is required to form certain bonds. Since the bonding angle is $93^\circ$ it would be wrong to neglect the influence of the s orbital. || Dear Hardik, after studying five years of chemistry, a doctorate and currently a postdoc in computational/ theoretical chemistry, I think I would be able to answer your question. Maybe my comment was a bit impulsive, and I apologize that I have offended you with it, I just wanted to point out an obvious flaw of the lecture.

@Dissenter Look at ron's answer, explaining Coulson's theorem. You can calculate the hybridisation at the phosphorus centre based on the bonding angle and you will see, that the only place the lone pair can go is the (mostly) s orbital.

@Martin I don't doubt your knowledge at all. You have a large reputation, it indicated you know much more chemistry than me. Just the point was I needed an example, which LDC3 has given.

Whatever you were trying to say, I clearly understood it wrong, I apologize for any inconvenience. I would like to add one last thing, your book mentioned electron promotion, this concept is outdated by at least twenty years now. I highly recommend to switch to a different book, e.g. Chemistry of the Elements, by N. N. Greenwood, A. Earnshaw; Inorganic Chemistry: Principles of Structure and Reactivity, by James E. Huheey et. al. ...

@Martin Yes you are right, this concept is outdated. The book although mentioned it as a separate small note. And the teacher in the video too, thats why i got confused. I will sure try to read the book you recommended.

You mean like ammonia? If the filled orbital did not partake in hybridization, then the molecule would be flat.
Water would be linear if the filled orbitals did not partake in hybridization.

Actually, one of the lone pairs is unhybridized (basically pure $\ce{p}$) and the other lone pair is roughly $\ce{sp}$ hybridized.  The two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized.

@ron If that was the case, then the bond angle would be different fro what you measure.

How so?  The O-H bonds are $\ce{sp^3} hybridized so the expected H-O-H angle would be around 109 degrees, close to the experimentally observed value.

@ron I didn't quite see what you wrote, but it is incorrect since you cannot have $sp^3$ hybridized and $sp$ hybridized and a $p$ orbital on the same atom. $sp^3$ hybridization requires an $s$ orbital and 3 $p$ orbitals, which uses all the orbitals so you cannot also have an $sp$ hybridization and a $p$ orbital on the atom. BTW, water has a bond angle of 104.45°.

Sure you can.  2 sp3 orbitals require half of an s orbital and 1.5 p orbitals, leaving 1 p orbital for one lone pair, and half an s orbital and half a p orbital for the sp orbital the other lone pair to exist in.  This hybridization scheme is consistent with the Photoelectron spectrum data for water.

@ron You said (in your answer above) that it is a mathematical concept. What you are doing here is picking and choosing what you like (and using that) instead of following the rules. If you use a $p$ (or $s$) orbital in one hybridization, you cannot use it any other way. It may help explain data using your scheme, but for others following the rules, it will confuse them entirely.

see http://isites.harvard.edu/fs/docs/icb.topic776365.files/lecture%202.pdf  water is discussed on p. 18.  Let me know what you think.

Let us continue this discussion in chat.

Can't now, should I post a comment here when I have a block of time? Never used chat before.

@LDC3 Thats the answer. In ammonia the 2s and all 3 2p orbitals hybridize and 2s is completely filled. No electron promotion occurs. I now think why I could not get such an easy example. Thankyou so much. Your one line answer was all I wanted.

@ron Your comment was not correct. In water also the 's' orbital which gets hybridized to sp3 is fully filled, and that hybrid orbital holds the lone pair.

Water would not be linear, since the s orbital is filled and the p orbitals are perpendicular to each other. Also there are no "rules" for hybridisation, as hybridisation refers to the contribution of different orbitals of the same atom within one molecular orbital. The source given by @ron shows that quite clearly. It also states the most important thing about hybridisation: It is in response to bonding.

First off, hybridization is a concept chemists developed to help explain reality (their observations).  Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction.  Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form "directed" bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce.  Finally, energetically speaking, hybridization is a net neutral process.  When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs.  So if some electrons are "promoted", then some are lowered in energy as well
For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals.  The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon.  In the case of phosphine ($\ce{PH_3}$) the situation is reversed.  The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes.  The central phosphorous atom remains unhybridized.  In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.
Personally, I like the concept of hybridization and find it quite useful.  But you have to be careful where to apply it.  Hybridization is best when it is used to explain reality after the fact.

Not sure what you're getting at, but I thought my answer might help the OP have a better understanding of the subject.  If you feel the answer isn't helpful, you are free to down-vote and\or constructively criticize it.

It does help, I was just pointing out that you could also include an example that the OP requests. It's easy to veer from the requested answer when there is so much to explain.

I agree with almost everything you say, but not the part where the new resultant AOs have an average energy. It is just not possible to assign energy values to hybrid orbitals, as these are no eigenfunction of the Hamilton operator (They do not produce the correct quantum numbers).

@Martin 2 questions, 1) if I theoretically mix 50% carbon 2s AO and 50% carbon 2p AO to create an sp orbital, why wouldn't its theoretical or estimated energy be 0.5E(s) + 0.5E(p); 2) if I mix two carbon sp AOs to form an sp-sp MO I can experimentally determine its energy (PES spectrum of acetylene) and see if it is halfway between the 2s and 2p energy levels for atomic carbon, right?

Theoretically you could do that, but the value you obtain would be a number without a physical meaning. One has to be very careful in using this number even as an estimate, since it is not an observable state. Hybrid orbitals are directionalized (localised) orbitals, that would violate the spherical symmetry of an atom and also the quantum numbers. The concept of hybridisation is very useful, in describing a general bonding pattern. It will give you wrong results for (excitation) energies, since if you distort the molecule (in any way) bonding and therefore also hybridisation changes.

You will always measure the MO, which would be the linear combination of two sp AO, or in MO theory terms 22s + 22p. I am pretty certain that the contributions of these orbitals are not equal, since the hydrogen cannot polarise the orbital as well as the carbon. And then there is this nasty thing called electron-electron repulsion. This affects filled molecular orbitals, resulting in bonding orbitals less bonding than virtual orbitals are antibonding. So I am pretty sure, that the energy you measure for this MO is nowhere near the energy halfway between 2s and 2p.

@Martin I should have been more clear, I was referring to the sp-sp C-C sigma bond in acetylene, not the C-H bond

Your point was very clear. But both bonds will not involve an idealised sp orbital. The second carbon has a much larger effect on the polarisation on the carbon atomic orbital, than the hydrogen on the other side. In a very idealised theory, the hybrid AO for the CH and CC Bond would be the same. Since these are subject to a different external field (of the nuclei) they cannot be the same any more. Therefore the hybridisation will also change, because as you stated, this is only a response to bonding.

@Martin Yes, of course!  Thank you for reminding me.

With hybridisation, you only hybridise the number of orbitals you need i.e. that are filled. So in methane, with four bonds, you have four electron pairs which need to fit somewhere. So you need four orbitals. Therefore we say that three 2p orbitals hybridise with the 2s orbital, making four sp3 hybrids (the superscripts denoting the number of each pure orbitals involved) each with 1 electron in them. Note that we had to invoke the idea of hybridisation here, since the electron configuration of carbon is 1s22s22p2, and this would only give us two bonds. We know experimentally that this isn't true (in other words, methane exists!) and so have to conjure a trick to work around it.
Alternatively, let us look at the CH3+ cation. This has three bonds, and so three electron pairs. This time we need to invoke the idea of hybridisation to give us three orbitals. So we say that two 2p orbitals hybridise with the 2s orbital, to give three sp2 orbitals. This leaves us with one p-orbital that we didn't hybridise, which would be a non-bonding orbital (in fact, empty), and so this lies perpendicular to the plane of the molecule.
For another example, to clarify further, consider ammonia. Now ammonia has three bonds, but it also has a lone pair. We count the lone pairs in our hybridisation scheme. Therefore we also need to hybridise to obtain four orbitals, like in methane, despite only having three bonds. Thus ammonia is also sp3, mixing three 2p and one 2s orbital. Note however that in contrast to methane, three of these hybrid orbitals are singly occupied, and one is completely filled (the one which contains the lone pair).
Now, we can look at cases by which orbitals do not hybridise.
For heavy atoms such as lead and bismuth (the same groups as carbon and nitrogen respectively) hybridisation does not occur. This is referred to as the inert pair effect. In this case inert pair refers to the unwillingness of the 6s2 orbital to undergo hybridisation. This is because heavy atom self-self bonds i.e. Pb-Pb are incredibly weak. We can imagine that some energy is expended in the hybridisation process because we are mixing s orbitals with p-orbitals of the same principal quantum number, of which the p are always higher in energy (said to be non-degenerate). So the hybrid orbitals are higher in energy than the pure s, and ever so slightly lower than the pure p (but not by much, as the p-character of an sp3 hybrid is 75%). So, the energy gained in forming a heavy atom bond such as Pb-Pb isn't enough to warrant hybridisation, since hybridisation requires energy.
Notice that I've used anthropogenic words to explain this i.e. (we hybridise as opposed to the atoms hybridise). This is deliberate. It's unclear whether hybridisation really occurs or not; indeed more sophisticated theories can be used to explain the phenomena. Hybridisation is more of a caveat used to explain something we couldn't readily explain with the previous level of theory. It's perhaps telling that we only invoke hybridisation when we need to. It's a characteristic of a bad theory when that happens. The idea is invoked readily in chemistry however, so you would do well to familiarize yourself with the terminology.If you're interested more in this topic, look up molecular orbital theory for a more nuanced explanation.

Molecular orbital theory is not a competing interpretation. In MO terms, hybridisation is the mixing of different kinds of atomic orbitals of the same atom into a molecular orbital. It is still invoked as a mathematical concept and only treated as such. There can also by hybridisation in virtual (unoccupied) orbitals. The theory that is employed (by organic chemists mostly) is a very crude approximation. Hybridisation is always a result of bonding and therefore hybrid orbitals cannot be assigned an energy value.

@Martin, I'm not sure I completely agree. In inorganic chemistry, for example, you use the idea of a symmetry adapted linear combination of orbitals (SALCS) to rationalise the bonding in, say, octahedral complexes, instead of saying that they are sp3d2. Nowhere in the MO diagram of such complexes does it mention hybrid orbitals of this kind. Unless of course the process of taking SALCS is considered hybridising, which is something I'm not familiar with?

The symmetry adapted linear combination is somewhat closer to the truth - meaning a less crude approximation - but it is still an approximation. Even everything we know so far about orbitals is an approximation, since knowing the exact wave function is not possible, we can still consider different atomic orbitals contributing to a molecular orbital. Any linear combination of ao towards an mo of the same atom is therefore a hybridisation. You cannot neglect the influence of this when working within the framework of linear combinations.

I see, thanks Martin, I understand your point. I'll edit my contribution accordingly to reflect this.

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Finding orbit radius using the Bohr model and Rydberg equation
To start with its a homework problem, quite lengthy.

A particle of mass equal to 208 times the mass of an electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr model of the atom is applicable to this system,

Derive an expression for radius of $n$th Bohr orbit.
Find the value of $n$ for which the radius is equal to the radii of first orbit of hydrogen.
Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.

Now, I did the first part and got the answer correct. Here's what I did.
Suppose the mass of the particle revolving is $M$, its speed is $v$, and $M = 208 m_{e}$. Electrostatic force is the centripetal force. Therefore
$$
\begin{align}
\frac{Mv^2}{r} &= \frac{(ke)(3e)}{r^2} \\
v^2 &= \frac{3ke^2}{208m_{e}r}
\end{align}
$$
From the Bohr model,
$$
m_{e}vr=\frac{nh}{2\pi}
$$
where $h$ is Planck's constant. Therefore,
$$
v = \frac{nh}{2\pi m_{e}r}
$$
Squaring it,
$$
v^2 = \frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}
$$
Equating the two equations which have $v^2$ in them,
$$
\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2} = \frac{3ke^2}{208m_{e}r}
$$
After solving for $r$, we get something like this,
$$
r = \frac{n^2h^2}{4(\pi)^{2}3ke^{2}208m_e}
$$
All of the above is correct. The problem is in the second and third parts; when I put $r =\pu{0.53*10^{-10} m}$ I do NOT get the required answer. To approach the third part, I started with the standard Rydberg equation,
$$
\frac{1}{\lambda} = \mathcal{R}Z^2 \left( \frac{1}{n_f^2}-\frac{1}{n_i^2} \right)
$$
I plugged in each value, $n_i = 3, n_f = 1, Z = 3$; but again didn't get the answer correct.

The answer to the second part is 25 $(n=25)$; and to the third is 55.2 picometers.

To answer the second part:
We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$.
Part one has a mistake, as it is
$$
\begin{align}
&&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\
&&Mvr&=n\hbar\\
\implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2}
\end{align}
$$
We also know the Bohr radius:
$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$
Therefore we can write and cancel:
$$
\begin{align}
&& r&=a_0\\
&& \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\
\therefore&& Z\frac{M}{m_e}&=n^2\\
\therefore&& n&=\sqrt{Z\cdot208}\approx25
\end{align}
$$
The third part:
The Rydberg Formula is given as
$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
with the Rydberg $\mathcal{R}$ constant defined for a photon emitted by an electron. We'll assume that the mass of the nucleus is 7 atomic units (three protons + four neutrons). Taking into account that $m_p\approx 1836m_e$, we arrive at
$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{7\cdot1836m_e}}$$
Now the Rydberg constant has to be modified to include the mass of the particle:
$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$
With $\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$ (wikipedia), I got to $\lambda_\mathrm{vac}=55.6~\mathrm{pm}$.
Without taking the reduced mass into account, i.e. $\mathcal{R}\approx\mathcal{R}_\infty$ I arrived at $\lambda_\mathrm{vac}=54.8~\mathrm{pm}$.
Both values are reasonably close to the solution given.
(If the question was really about the muon, the more accurate weight ratio is 206.77 and the corresponding wavelengths 55.1 pm and 56.0 pm.)

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text:
Is the photoelectric effect possible for semiconductors?
Is the photoelectric effect possible for semiconductors?
I imagine that this might be, if possible, a two-photon process: excitation of an electron from a filled to conductance band, and then on to ejection. Is anyone familiar with such materials, or even whether this is possible?
I can't readily think of a theoretical reason against it. Two-photon events are often less likely than one-photon, however I could imagine irradiation by a laser of bandwith covering both the band-gap wavelength and that necessary for ejection would do it.
If it does indeed happen, I'd be glad to hear a short summary of how it differs from the photoelectric effect of metals.

Given that Jaques Pankove's excellent book on 'Optical Processes in Semiconductors' devotes chapter 13 to 'Photoelectric Emission', you can rest assured that the photoelectric effect is alive and well in semiconductors.  While somewhat more complex than emission from simple metals, there is no need for two-photon processes. The 'tricky' part is that you have to look at where the electron states are, and then how much energy is needed to take it out of the semiconductor.
The work function is the energy difference from the Fermi energy to the vacuum level.  There may not be any states at the Fermi energy though.
The electron affinity is the energy difference from the bottom of the conduction band to the vacuum level.
One has to take the band structure into account, since direct and indirect transitions will likely have different energies required.
Also, surface band bending may make interpretation of experimental results more difficult. Again, perusing Pankove may prove useful.

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text:
Why does charcoal absorb smells from the air, and where do those smells 'go' once absorbed?
I've read that using blocks of charcoal can remove some odors from the air by placing them in the vicinity of the odors. (Maybe not as efficiently as other forms of carbon, as pointed out in the comments of that link, but something nonetheless.)
What is the chemical process by which charcoal can do that, and once the odors are 'gone', where do they 'go'? In the case of 'toxins in gases', what typically is done to them upon 'deodorizing' by the charcoal block - are they collected, neutralised / transformed upon contact, or something else?

The scent molecules are adsorbed by the charcoal.   The higher the surface area of the charcoal the more smell it can absorb

I know, but I'm wondering what chemical process this absorption is :). I've even read that they can be re-used after some 'airing out' so, if that's true, it's interesting to know what the process is! 'Stuff' (including odors and toxins in gases) can't just disappear, something happens to them! I'm wondering what it is in this case.

It is a form of carbon that has very large surface. The odors and impurities are held to the surface with Van der Waals forces. The capability of removing notable amounts of these impurities or odors is thanks to the extraordinarily large surface.

Its simple adsorption. The smells/ chemicals are stuck at the surface, and once you expose the activated charcoal to sunlight for a few hours, it release the smells and is again activated and charged, and good to use.
A good quality, with high adsorbent surface area activated charcoal kit can work good for a period of about 1 year, however you need to keep it in sunlight at-least once a month.


text:
How to Experimentally Quantify the Response of a Receptor to a Ligand?
In medicinal chemistry, there are multiple reasons why one would want to experimentally determine the effect of a particular molecule on a receptor e.g. measuring how 'strongly' that molecule brings about a chemical response (an agonist) or how strongly it prevents a response (an antagonist).
If we were looking at an insulin receptor antagonist (for whatever reason), then it would be easy to quantify the strength of the antagonist by a simple blood test of the test subject (human or animal), to see how much insulin was present.
What happens when we're faced with a scenario inherently difficult to quantify? How do we measure the action of a drug then?
Say we've found a natural neurotransmitter receptor agonist and want to compare this to a synthetic drug of a similar nature. I'd struggle to see how we'd experimentally be able to analyse the amount of neurotransmitter present in the synapses in each of the cases to compare.
Do we then look at less direct, perhaps qualitative methods of analysis? Such as 'state of agitation' or 'how many times the subject's arm twitched in a minute'?

What you're asking about is a wide topic.
You have the basic ideas of agonist and antagonist, but there are also molecules called inverse agonists. Every receptor has some base level of activity, even without an agonist to activate them, or in the presence of an antagonist. Inverse agonists push this activity even lower. There are even "partial agonists" which are molecules that activate receptors, but not as well as the endogenous ligand, so act like competitive inhibitors because they occupy some of the receptors the endogenous ligand would have. In addition, the binding site is important. The endogenous ligand binds to a location called the orthosteric binding site, but other molecules may bind to other locations on the receptor. These are allosteric binding sites. Allosteric binders may activate or inhibit a receptor, but they may also be modulators. Modulators don't directly activate a receptor, but allow a stronger response when another ligand binds at the orthosteric site. Here's a schematic to help show how different types of ligand affect receptor activity.

Anyway, ligands have two important properties, affinity and efficacy. Affinity is how strongly a ligand binds to a receptor, and efficacy is how the ligand affects the receptor's activity.
Before we can dig too deep into agonists or antagonists, we should think about receptor theory. Receptors, like all proteins, have complex 3 dimensional structure, or conformation. This conformation is affected by solvent, pH, temperature, and other factors, but always has some fluctuation. Classical receptor theory used a lock and key model, where an inactive receptor bound a ligand and became active. However, this did not explain the base activity and activity of inverse agonists, so an equilibrium based model was developed. Here, the population of receptors has an ensemble of conformation states, some of these are active. When an agonist binds, it drives the equilibrium of the population toward the active state by stabilizing the active conformation. This diagram shows a simplified example of how drugs affect the equilibrium. Ri is inactive receptor, Ra is active. D is drug. Ra leads to some activity, while Ra-D leads to strong activity.

Affinity is relatively straightforward to measure becuase the same idea can be applied to almost any receptor-ligand system. Usually, a radiolabled version of the drug is added to a known amount of receptors in a binding assay. The goal is to determine a dissociation constant, Kd. Kd is used to quantify how well a ligand binds to a receptor, and smaller numbers are tighter binders.
Look at this reaction: D + R <--> DR with forward rate constant k1, reverse rate constant k2. ( how to get the right arrows here? )
At equilibrium, k1[D][R] = k2[DR]
and $K_d =\frac{[D][R]}{[DR]}$ = $\frac{k_2}{k_1}$
So incubate the radioligand and receptor until some DR complex is formed, then seperate the bound and unbound ligand so you can measure D and DR. The seperation method is important because if it's too slow the DR complex will revert to D and R, creating inaccurate data. But this is commonly done using dialysis, centrifugation, and vacuum filtration. Dialysis works well when dissociation rate is slow. You put receptor in one side and drug in both sides. It's slow, but it's theoritically most accurate becuase you can measure the true steady state by taking an equal volume of solution from each side of the membrane without affecting equilibrium. But you also need a large supply of receptor and drug.
Centrifugation works when you can immobilize the receptor on a large particle, usually you do this by using whole cells. You add an excess of of drug, then spin it down and take off the supernatent. Timing is still an issue and free drug can get caught in the pellet.
Vacuum filtration is usually considered the best because it can seperate the components very quickly, by pulling the solution through a filter and trapping the receptor on the surface. However this requires specialized and expensive equipment.
Once you measure bound and free drug at several concentrations, you can do a double reciprocal plot
Measuring efficacy is different because you need to choose an activity to measure. Idealy it will be objective and easy to quantify. If the receptor is an enzyme, you can measure its enzyme activity. If it activates a gene transcription factor, you can use cells modified to express green fluorescent protien or luciferase when that transcription factor is activated. If it's an ion channel you can measure a cell's membrane potential. If you cannot measure anything else, you can use behavioral assays. There are several assays used to measure mouse activity and try to get at certain aspects of their brain and are often used to study depression and other mental illnesses. These are subjective. Efficacy is quantified by EC50, the concentration at which response is half of maximum response. Smaller EC50s are more efficacious.
This field is much much larger than my answer. For more you should read this article : http://www.ncbi.nlm.nih.gov/pubmed/15063082


text:
Normal and heavy water mixing
This question seems silly, but this just came in my mind while i was learning about hydrogen.
Means will there be any reaction when $\ce{H2O}$ will be mixed with $\ce{D2O}$?

Yes, a reaction known as "hydrogen exchange" will take place.  When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture.  Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.

Statistically, you would get that ratio, but unfortunately the $\Delta H_f$ for $D_2O$ is about 9 kJ/mol different from $H_2O$ so that the equilibrium is not that easy to determine. http://www1.lsbu.ac.uk/water/data.html

@ LDC3 Excellent point LDC3, I'll edit my answer.

@LDC3 the relevant quantity is not the difference vetween D2O and H2O, but the difference between their average and HDO. We would expect HDO to fall nearly halfway between H2O and D2O in enthalpy and free energybif formation, so the statistical model would likely be accurate in any fluid sample.


text:
How hazardous are peroxide-able ethers if stored properly?
This question is directed towards laboratory scientists and technicians who actually deal with ethers that can develop peroxides upon concentration (e.g. diethyl ether, THF, dioxane). I am already very familiar with the literature, the normal precautions (dispose of after 1 year, test for peroxides, etc.), how to remove peroxides (shake with ferrous salt or metabisulfate).
My question is how hazardous are peroxides in reality if stored properly. If a peroxide-able ether upon concentration is stored in a cool dark cabinet, is rarely opened, has an inhibitor (BHT or ethanol), and argon is used above the headspace, it is likely that the ether can be used for longer than a year?
Again, I need the input of an actual practitioner. Simply quoting written material will not help me, as I have consulted LOTS of written material and I believe I understand the theoretical issues. I am looking for practical experience.  I have heard informally that some ethers are available in European countries as OTC solvents, and so it leads me to think that the standard warnings may be too severe. Thanks!

I believe, that when I was still a student in university, we kept and used DEE and THF for more than a year, especially when dried and kept in inert atmosphere. I think most safety concerns apply to quantity. There are probably other reasons why you should not use solvents after a year or so...

One of may previous labs used some diethyl ester and THF, and no accidents occurred. As long, as they are properly stored (no direct sunlight, tightly sealed bottle, a tiny bit of $\ce{KOH}$ on the bottom of the bottle) they are reasonably safe. Of course, I would be a bit ... concerned in case of work with diisopropyl ester, but not so much.

@permeakra: Thanks! Can you answer these two questions I have: 1) do I need to pre-dry (I would assume yes with something not so water reactive), and 2) how much to use (I would assume you don't need much)? You seem to agree with my answer to 2). And I have no diisopropyl ether and agree that it is a higher order of storage danger!

DEE and THF dissolve quite a bit of water, so boiling and then distillation of marketed solvent  over excess of sodium or potassium hydroxide is common. In case of extra large quantities, another pre-drying stage may be economically justified, but in common labs it is usually dried with alkali hydroxide directly. The amount of hydroxide required for drying is very dependent on quality of the raw solvent, so no suggestion here. I would use roughly 1-2 table spoons per liter unless it is extremely wet.

As far as I remember, it was a small amount. Again, 1 table spoon per liter should do.

@permeakra: Thanks again. But my concern is that using such a water reactive drying agent first may produce a fire hazard. Might it not make sense to pre-dry with something less water reactive? I have a list of such agents.

@trb456 DEE and THF are fire hazards on their own.  Both DEE and THF have flash points below zero, bp at 35 and 66 Celsium respectively and autoignition temperature at roughly 160 and 320 C respectively. Given that, room temperature is already above flash point for both of them, but since they boil well below autoignition temperature, they are not likely to ignite as far as you heat liquid solvents. Even if heating source (i.e. energetic reaction) will have high temperature, the temperature of the vapors will be equal to bp, which is not enough to autoignite. Granted, THF is safer.

@trb456 after some thoughts I wouldn't use sodium metal on large quantities without pre-drying or testing, though.

Peroxides are no fun.  There have been explosions when ether solutions were concentrated to dryness.  What you really want to do is store your ether over potassium hydroxide - hydroperoxides are slightly acidic, and the base forms hydroperoxide salts and pulls them out of solution.
Also, if you keep solvents for more than a year you clearly aren't turning them over fast enough.  Keeping flammable material that you don't need just increases the danger in case of fire.

Thanks for this. I've tried looking into this but could not find hard procedures. Two questions I have: 1) do I need to pre-dry (I would assume yes with something not so water reactive), and 2) how much to use (I would assume you don't need much)?

Having accepted Abel Friedman's very sound answer, I wanted to append a coda of sorts. After posting this question, I tested all my opened peroxide-able ethers (diethyl, THF), and there were no measurable peroxides despite being well past the 1 year mark. I double-checked that the test was valid with hydrogen peroxide, which produced a strong positive for peroxides.
I conclude that the standard recommendations for peroxides are similar to the warnings on MSDS sheets, in that they err on the side of extreme caution. As Abel Friedman says, peroxides are no fun! So testing and careful storage (full airtight containers, argon in the headspace, dark cool location, inhibitors) are essential. But if you do these things, the "store-by" dates do seem to be excessive.
As to the valid point that I might not be turning over my solvents fast enough, I realized that the converse could be true in a working lab. That is, solvents are used frequently, and thus are left out in the light and opened frequently. If these are not used quickly enough, there is a real risk of peroxides. But it would seem that dark airtight storage (and inhibitors!) does retard peroxide formation well beyond the recommended limits.


text:
What is the wavelength of an electron?
The kinetic energy of an electron is $1.67 \times 10^{-17}\mathrm{J}$. Calculate the wavelength ($\lambda$) of the electron.
I know the formula $\lambda= \frac{h}{mv}$; where h is Planck's constant.
How do I find $\lambda$ without $v$ (velocity) and $m$ (mass)?

The mass of electron is never given in question you should remember it. Mass of electron is $9.109 \times 10^{-31} \mathrm{kg}$.
Now read the question Kinetic Energy is given, so from that you can easily find velocity of electron.
$$K.E.=\frac{1}{2}mv^2$$
so You will get $v= 6 \times10^6 \mathrm{m/s}$
Now you can substitute value of m and v in formula $\lambda = \frac{h}{mv}$
You will get wave length of electron as $0.12 \times 10^{-9} \mathrm{m}= 0.12 \mathrm{nm}$

You do not even need to laboriously calculate the velocity of the electron. All you require is the mass (which should be in your knowledge), and the kinetic energy - already given.
See, momentum P = mv;
And kinetic energy = 1/2mv*v
Now, if you multiply and divide the KE expression by mass m, you get -
(1/2)(mv)(mv)*(1/m) = (1/2)P^2(1/m)
So, all from here, you can cross-multiply and find momentum directly, and hence substitute in de-Broglie's formula to find the wavelength of the electron.


text:
Biological Consequences of Asteroid Mining—Death by Isotope?
It's been documented that NASA hope to capture an asteroid in 2025, and have subsequent aims to mine that asteroid. If if this is successful, we would expect other asteroids to be mined in the future.
A consequence of this is that relative atomic masses of elements mined—those with two or more stable isotopes—will no longer be faithful to our current periodic table. Ruthenium alone, a high-value rare earth, has five stable isotopes alone, ranging from $^{98}\ce{Ru}$ to $^{102}\ce{Ru}$. The relative abundances of these isotopes are bound to differ in other places other than Earth, and so would the weighted average.
Say after mining our ruthenium, we use it in an electronic device and consequently that device is thrown away. For argument's sake, also say that this ruthenium—of an isotopic distribution never before seen—leaches into the environment and later comes into contact with biological systems. Could it be possible that compounds hitherto considered non-toxic on Earth become toxic to biological systems by virtue of a newly realised isotopic discrimination?
As an example, $^{13}\ce{CO2}$  is effectively discriminated against in uptake by plants in comparison to $^{12}\ce{CO2}$  (due to it being a diffusion limited reaction). Ruthenium is purported to be carcinogenic, but the most abundant isotope on Earth of Ruthenium is $^{102}\ce{Ru}$ which could be effectively non-toxic by virtue of the fact that  compounds containing these atoms takes so long to diffuse, they don't diffuse across biological membranes at all.
If a sample of Ruthenium from the moon found its most abundant isotope in $^{98}\ce{Ru}$, then this would be expected to diffuse much faster than its heavier counterparts. Therefore could it be possible that a compound previously thought of as non-toxic could become toxic via this effect. Does this sound plausible?
NB: I've deliberately missed out other routes of exposure such as ingestion to only consider diffusion controlled reactions.

Can you provide a source for plant's not using C13? Because I know they use at least some C14, that's how radiocarbon dating works. And by your argument, C14 should diffuse even more slowly than C13.

I didn't say that they don't use it, but that there is an active discrimination. And yes C14 will diffuse even more slowly. Radiocarbon dating relies on absolutely minute quantities of C14. As for evidence, the δ13C isotope signature is well known in biogeochemistry for example.

Also, to allude to C3 and C4 plants, carbon dioxide uptake is different in the different species. In C4 it occurs via the Hatch-Slack pathway which is less diffusion controlled than C3's Calvin-Benson.

I like the question, but my guess is that every answer would boil down to "we don't know, here are some proposed (or: performed) tests about it [refs 1, 2, 15 and 42]" ...

@tschoppi, I realised it's more of a discussion than a question with a definite answer, although those references might be what I'm looking for!

The largest biological consequences of isotopes are found, of course, with hydrogen.  Both deuterium and tritium are toxic.  Since many biological processes rely on resonance effects, too much deuterium messes up metabolism.  Tritium, while having the same effect as deuterium, adds the beta decay radiation dose.

This is an interesting question and you raise a number of points, let's step through them.

A consequence of this is that relative atomic masses of elements
  mined—those with two or more stable isotopes—will no longer be
  faithful to our current periodic table.

But this is already happening.  $\ce{^235U}$ constitutes 0.72% of uranium found on earth and decays to the stable isotope $\ce{^207Pb}$, which is found in a natural abundance of 22.1%.  Before we get to the asteroid the abundance of various isotopes and their weighted average mass is already shifting.

98 Ru , then this would be expected to diffuse much faster than its
  heavier counterparts

Diffuse much faster?  First off, we're talking about diffusion, not chemical reaction.  Elements or compounds usually enter the body by ingestion of one sort or another (eating, breathing) rather than diffusion, and these ingestion pathways wouldn't involve isotopic discrimination.  That is, if we breath air that contains isotopes in a certain ratio, then that's the ratio that will initially appear in our lungs.  Nonetheless if diffusion were found to be an issue, my guess is that isotopic discrimination by diffusion would be a very small effect.  For a gas the maximum separation of two isotopes is given by
$$\mathrm{\sqrt{\frac{[MW~of~compound ~with ~isotope~1]}{[MW~of~compound ~with ~isotope~2]}}}$$
In the case of separating $\ce{^235UF6/^238UF6}$ this amounts to a fractionation ratio of 1.0043 after 1 pass.  Ruthenium is lighter, so the effect would be larger, about 1.02 ($\ce{^102Ru/^98Ru}$) after 1 pass, even less if it is in a molecule.
Chemical reaction within our body of ingested isotopic compounds would also show  discrimination due to primary kinetic isotope effects.  The maximum primary kinetic isotope effect is proportional to the reduced mass as follows:
$$k~ \thicksim ~\sqrt{\frac{m_1 + m_2}{m_1 m_2}}$$
Applying this to $\ce{^102Ru}$ and $\ce{^98Ru}$ yields a primary kinetic isotope effect of 1.020, again, even less if the element is incorporated into a compound.
To me, the effects look small and since shifts in natural abundance have been occurring for a long time here on earth, with no one raising a flag, my guess (and that's all it is, a guess) is that it's not something to worry about.  Still, I'd feel better if a NASA toxicologist looked at the question and said "no problem"

For the second point I realised that talking about diffusion relating to human health is probably a bit daft, although there's no reason why it couldn't happen in plants such that the effect is passed down a food chain. If we apply your formula for gaseous separation to carbon dioxide as mentioned, then this amounts to a fractionation ratio of 1.011, which is less than atomic ruthenium!

If gas centrifuges can sort out U238 and U235, which have a tiny percent difference, could they be used to remove CO2 from air? CO2 is about 40g/mol, and O2 seems like the next heaviest at 32g/mol, so the percent difference is larger and separation should be more efficient. But how much energy would it take to run the centrifuges?

@user137 Interesting thought.  Sounds like it should work, I know that mean free path and membrane porosity are also design factors, and as you point out, energy considerations.

According to https://www.ncbi.nlm.nih.gov/pmc/articles/PMC396263/, the discrimination against uptake of $\ce{^{14}CO_2(g)}$ relative to $\ce{^{12}CO_2(g)}$  by plants during photosynthesis is due to diffusion.  They conclude this because  the strength of the effect is 2%, which corresponds nicely to the Graham's diffusion equation (which you showed in your post:  $\sqrt{\frac{44}{46}}=0.98$).

Whatever the isotopes are for asteroidal material (and they are mostly close to those seen down here on Earth), they are contained in the 5 to 100 tons of meteoritic material that falls onto the Earth's atmosphere (and thus filters down to us on the surface) every day. It will be a long time before the cumulative pollution from asteroid (or lunar) mining can match the meteoritic rate.

Also, if you follow what @ron said, for large isotope effect you generally need lighter elements. So the isotope effect for deuterium and hydrogen can be large. But even in these cases biological effects can be mostly observed only when one use high concentrations of D2O.
I don't think that diffusion control would play any role in the biology of metal ions. If we are talking about transition metals with biological role, nature is generally relies on transport enzymes and similar to carry around them. Isotope effects are most pronounced in chemical reactions with relatively high activation barrier and strong vibrational coupling with the given atom. In practice it means enzymatic reactions with active centers where there is a significant role of the given metal.
Do we even know any enzymes in which's active center Ru plays important role? You can ask this for any elements that we expect to mine, In practice, isotope effects hardly can seen even for iron or cobalt atoms, which are pretty common active centers compared to most other metals. If you study natural iron enzymes, the isotopic ratio of 56/57 iron is not anomalous. So I wouldn't expect much biological effect to anything that heavy or heavier.
If we are going beyond biological function, and talking about metals and toxicity, most of the time the complex formation between e.g. proteins and the given metal ions is the important question to study. I don't think is it much influenced by isotope effects, and I definitely would not expect that a toxic heavy metal became harmless or similar drastic change.


text:
Which d orbitals of sulfur take part in the pi bonds of SO3?

In $\ce{SO3}$, 2 $p\pi-d\pi$ bonds are present. But which $d$ orbitals of
  sulfur take part in these $\pi$ bonds?

The answer says $d_{xy}$ and $d_{yz}$. Someone also told me that crystal field theory supports it as while splitting $d_{x^2-y^2}$ and $d_{x^2}$ are raised in energy compared to the other 3, but that theory has nothing to do here.

and the question is?

It is a long-standing myth that $d$ orbitals partake in bonding of most hypervalent compounds in any appreciable amount. There is little to no $d$-orbital participation in the bonding in $\ce{SO3}$ (see here). There are almost only $\mathrm{s}$ and $\mathrm{p}$ orbitals of sulphur involved.

@permeakra Its specified. I just need one reason why only $d_{xy}$ and $d_{yz}$ participate instead of other 3.

@Philipp Maybe you are right, but I read that in "J.D. Lee Concise Inorganic Chemistry" which I think is more trustworthy than the link you provided.

@Hardik I've seen this issue come up several times here on Chemistry.SE. A few other users mentioned that the supposed d-orbital participation in hypervalent molecules is a relic from the past which is contradicted by actual quantum chemical calculations. Maybe people like Martin can chime in an comment on that.

@Hardik Also, have a look here.

If your textbook still needs to use d orbitals to explain the hypercoordinate molecules, then you should get a new one. For the complete molecular orbitals of $\ce{SO3}$ have a look at this question and answer. Apart from this the question is incomplete. In order to determine which orbitals take part in bonding, you have to know the position of the symmetry defining elements. Otherwise the naming of the orbitals is completely interchangeable. (@Philipp thanks for raising awareness.)

@Martin Thanks for giving some support. And once again, nice answer of yours in the link you provided. Really like those orbital pictures of yours.

@Philipp Yes, its specified ".....these models still appear in some beginning-level college texts" . And the book I am following is a also for a beginner in college. I am still learning chemistry, just passed high school. Sorry if the question was not upto the level of CHEM-SE.

@Hardik There is nothing to feel sorry about. It's still a common thing to be told in universities even though it is an outdated view. I've been told it that way too before diving a little deeper into the more theoretical aspects of chemistry. And some of the historical concepts in chemistry still have their uses today, because they give you some easy to grasp guidelines and help you to develop a good intuition, although there might be some limitations and exceptions. But it is always good to be aware of the pitfalls and to know that there are more advanced theories out there.

@Philipp  I just realised that the first link you gave was a bout sulfite, when I answered, I was like... there is something missing there... where is the pi bond... and then i noticed that there are too many electrons ;) Unfortunately, no one published an article about the electronic structure at the basis set limit.

Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools.
As Philip already pointed out in the comments: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare Wikipedia). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule.
There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:

Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):
 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-88.394 is S1-s=1.00
 Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17
 Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50
 Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33
 Alpha occ 5 OE=-7.922 is S1-s=1.00
 Alpha occ 6 OE=-5.945 is S1-p=1.00
 Alpha occ 7 OE=-5.941 is S1-p=1.00
 Alpha occ 8 OE=-5.941 is S1-p=1.00
 Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17
 Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18
 Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12
 Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11
 Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14
 Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12
 Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20
 Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13
 Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21
 Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15
 Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45
 Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32
 Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18

Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the O2-p orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, 0.15; 0.15; 0.10; 0.10. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible.
If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.

authors: عبدالوهاب دفع الله


text:
How does the vacuum in atoms persist?
From very elementary chemistry I understand that atoms are mostly empty space, with a few tiny little objects whizzing around.
There must then be a vacuum in between these particles. I always thought that nature abhorred a vacuum, and rushed to fill it.
Why doesn't everything clump together in one little ball and get rid of the vacuum? Or are my premises wrong?

There is no vacuum. The space inside atom is by no means empty; it is occupied by delocalized electrons.

nature abhorred a vacuum, and rushed to fill it Thats not nature, what you're saying is air filling up the vacuum due to pressure differences.

I think what you are asking is why electron do not fall in nucleus.
For this you should have general idea of quantum mechanics. Because in the microscopic world nature follows quantum mechanics equations and not classical mechanics equations. Quantum mechanics equations include electromagnetic fields, and their solutions are stable and allow for the existence of atoms, which is what we experimentally observed to start with.
This is general idea of answer given to question physics stack exchange. You can see the detailed answer over in Physics S.E..
This Link will  explain diffrent theories which failed to explain why electron do not fall in nucleus.

Well, the glib answer is that nature doesn't abhor a vacuum or anything else. Abhorrence is an emotional state and the universe, as far as we can tell, is not run by emotions. And the space inside the atom has electrons moving through it so one could argue that it's not a vacuum, at which point, we have no abhorrence, no vacuum and no question.
But that doesn't help. Freddy's answer explains why the electrons don't fall into the atomic nucleus, which is basically because of physics that we've only come to understand in the last hundred years.  Gases can't get inside an atom because molecules are surrounded by electrons, which are repelled by the atom's electrons; a gas can only fill a vacuum if it can move there.

Without quantum mechanics, it becomes a bit sloppy, but the general principle is easy to grasp:
The nucleus is positive, and surrounded by a number of negative electrons matching the charge of the nucleus.
Once the electrons get pulled too close to the core, they start repelling each other strongly, so there is some equillibrium distance.
Why doesn't the core/nucleus pull in the electrons one by one? Well, they fly "in vacuum", so they get ever faster when they're pulled close to the nucleus. The nuclues can't catch them either, because he has not way to store or discard the huge kinetic energy the electron has. So the whole system is stable, in rather the same way the planets and comets surround the sun.
With a bit of QM, there is the additional point that pairs of electrons fly in a formation ("orbital") that keeps them at an extra distance, and all pairs of electrons have a different formation. That makes sure the whole system stays symmetric and thus stable. Electrons are individualistic, they hate to be indistinguishable from the others. Which of course they are. ;-)

There's also the fact that electrons aren't little round balls.

@bon  I thought they were yellow. What's you point?

pairs of electrons fly in a formation is perhaps slightly misleading. They don't fly at all they just exist in space with a probability of being at any one point.

The non-quantum mechanics explanation does not work. Planets have no problem crashing into the sun (what we see is the result of most matter falling into the sun, and some matter, planets, having just the right orbit to stay put). Also, the electron repulsion idea fails for one electron systems (e.g. hydrogen atom, which is common in the universe, https://iopscience.iop.org/article/10.1088/0004-637X/698/2/1467/pdf).


text:
Why is formic acid considered an acid if it is a reducing agent?
My understanding was that since the acceptance of Lavoisier's revised Easter Memoir of 1778, all acids were understood to be oxidising agents.*
However, formic acid is often listed as a reducing agent (i.e. the opposite of an oxidising agent), despite being an acid.
Please can you reconcile this apparent contradiction for me?
* What is meant by "oxidising" or "oxidising agent" has, of course, changed since then, now referring to electron transfer rather than necessarily involving the element oxygen.

might be easier to describe iron as both reducing agent in Fenton reaction and oxidizer in H-W reaction, just depends on who wants the electrons more.

I realize this is an old post, but since it's risen to the top again, I figured I would mention that the issue isn't with formic acid but, rather, with (modern interpretation of) Lavoisier's memoir. Simply put, formic acid is an acid because of its pKa. Additionally, it is a reducing agent because it oxidizes to $\ce{CO2}$ in the course of reducing other compounds (through hydrogenation).

Formic acid is a carboxylic acid with the formula of $\ce{CH_2O_2}$ and is easily oxidized to carbon dioxide. Other carboxylic acids are not as easy to oxidized.
Added:
If a compound gets oxidized (as formic acid can), then another compound has to be reduced. The compound that gets oxidized is called a reducing agent. The compound that gets reduced is called an oxidizing agent. Most carboxylic acids cannot oxidize other compounds (the acid is not strong enough).
Lavoisier was referring to mineral acids ($\ce{HNO_3, H_2SO_4, }$...) in the text. These acids can oxidized other compounds (such as changing iron into an iron ion).

Thanks, but please could you expand on your answer to address the more fundamental question of how an acid can be a reducing agent (or, if you prefer: why this substance, that is a reducing agent, is still called an acid)?

OK, I'm learning! From Wikipedia: "In organic chemistry, reduction more specifically refers to the addition of hydrogen to a molecule... In organic chemistry, good reducing agents are reagents that deliver H2." This is deeply weird to someone like me who's fairly ignorant of modern organic chemistry, but it explains the first part of your answer; and that in turn makes the next part of it comprehensible. It still, however, does not explain why formic acid is considered an acid.

Formic acid is classified as a carboxylic acid. A carboxylic acid is an organic compound that can release a proton when mixed with water. They are different from minearal acids since not all the protons are released when the compound is mixed in water (only a fraction of the compound is disassociated). Formic acid is easily oxidized to carbon dioxide, whereas most other carboxylic acids require stronger oxidyzing agents. http://en.wikipedia.org/wiki/Carboxylic_acid

Thanks for the explanation of why carboxylic acids are acids and why, therefore, formic acid is an acid. I've accepted your answer :)

A molecule might be an oxidising agent under the electron transfer definition, and simultaneously be a reducing agent under the alternative definition: i.e. that reduction is a loss of oxygen or a gain in hydrogen.
For your specific question, formic acid can be deemed to be an oxidisation agent since it loses a hydrogen during protonation of another molecule. The really silly way to think about this is following deprotonation, formic acid has a higher percentage of oxygen than it did before, i.e. it possesses more oxygen than before.
The case of formic acid being a reducing agent is due to the more modern definition of electron transfer. To see the mechanism of this for clarity, see here. It's clear that the formic acid is donating an electron pair (i.e. itself is being oxidised) and is making the other molecule accept those electrons (i.e. that molecule is reduced). The formic acid is therefore a reducing agent.
So your original issue is due to two different definitions of oxidisation and reduction which still exist to this day, and which are by no means mutually exclusive!
EDIT: Of course, the case of formic acid is more simple because it isn't the acid doing the reducing, it's its conjugate base i.e. formate. So formic acid is an oxidising agent under the proton definition, yet formate is an reducing agent under the electron transfer definition.

Thanks. I followed your link, but it didn't clarify anything for me, not least because it seems to be talking about deprotonated formic acid rather than simply formic acid. I'm still finding this all a bit baffling, so further clarity would be welcomed!

Yes of course! You've got it right (I mentioned it in my edit). I meant in general that it's possible for a molecule to be an oxidising agent and a reducing agent under the two different definitions that exist, but formic acid is more simple: formic acid is the oxidising agent however it's formate that is the reducing agent!


text:
Which is the correct way to name pentan-3-one?
I have seen this organic compound named in two different ways.

According to some sources it is written pentan-3-one.. Other sources call it 3-pentanone.
Which one is correct?

Whether you write pentan-3-one or 3-pentanone both are correct.
But according to IUPAC rule pentan-3-one this is correct format of writing the name of organic compound. So I prefer you to write pentan-3-one rather than 3-pentanone.
You should also read this.

Adding to Freddy's answer I would say that, pentan-3-one is preferred overall due to the fact that as molecules start to carry more and more functional groups, it becomes ambiguous of which functional group the number in the front is referring to without knowing the order of priority of naming functional groups.

Some people call it diethylketone.


text:
What element or compound could be used to make the ocean more basic?
The ocean is becoming acidic. This is not a natural evolutionary change but happens through human negative effect on the environment. What element or compound could be used to create a more base ocean? How much of it would be needed? How could it be distributed?

Calcium carbonate has been used to remediate acid rain damage in lakes and ponds.  The scale needed in the ocean might make this unviable.

It is possible to get a coarse estimate for the amount of material needed to directly neutralize all the acid coming from anthropogenic $\ce{CO_2}$ absorbed by the ocean. It turns out that using the material with the highest ability to neutralize acid per mass (lithium metal), we'd need to throw ~100 billion metric tons of the pure substance into the ocean. Compare this with the entirety of the iron mining industry, which currently produces about 4 billion metric tons of iron/steel per year. There is no way we can access the amount of matter required for a simple neutralization reaction.

The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out.
So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$,  with this carbonic acid in a conventional neutralisation. In symbols:
$$\ce{H2CO3 + CaCO3 -> Ca(HCO_3)2}$$
This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$.
But we won't do that!
Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this.
So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it 'permanently'?
An interesting paper came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.
It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments.
Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved.

I corrected your equation and improved the coding. You can enclose the whole reaction equation in $$\ce{...}$$. This way the spacing is better and you safe a lot of typing...

Cheers Martin, still pretty new to the formatting. Will need to write my thesis in this so I'm bound to improve.

Have a look at the documentation of the mhchem package on CTAN.


text:
What is the cause for thermodynamic reversible and irreversible process?
My book defines

Thermodynamic irreversible process is defined as the process in which the changes take place rapidly, instead of occurring in successive stages with infinitesimal changes in quantities so that the system does not get a chance to attain equilibrium and there is an appreciable difference between the driving and the opposing force.

I have some problems in understanding it.

Which equilibrium are they speaking of?
Why does rapidity of the change cannot help it to attain equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?
What does they want to mean by driving and opposing force?

Your question is a good one. But I think the definition of irreversibility in your book is very poor. It is better to define an irreversible process as a process where the overall entropy (of system + environment) increases. An irreversible process cannot be brought back to its initial state, because that would
require an entropy reduction which would go against the second law of thermodynamics. Nevertheless the entropy of a subsystem can decrease at the expense of
an entropy increase of the environment, which at least offsets the entropy reduction of the subsystem.

But your second question (What does the rapidity of a process have to do with reversibility?) is much more difficult to answer in a satisfactory manner that doesn't simply lead to another question.

Let me start by reiterating what @Phillip said in his comment: this is not a good definition for an irreversible processes - a better definition is:

An irreversible process is one in which the net entropy generation term is greater than zero.

All real processes are irreversible. The "process speed" comes into play when we want to make approximations for idealized systems. It turns out that one way to convert a process into a theoretically reversible one is to set the condition that it moves very slowly between thermodynamic states, so that at any given step, it is in equilibrium. The reason that this isn't a good definition for a reversible process is that there are cases where a rapid change is reversible, and cases where a very slow change is irreversible. It's just that in general use, the "very slow process" approximation is commonly used to obtain a reversible process.

Which equilibrium are they speaking of?

Typically they mean equilibrium between the system and surroundings, but it can also mean within the system itself. Equilibrium is another way of saying "maximum entropy" - the matter and energy are as "spread out" as they are going to get. If the entropy is maximized, no more can be generated. In "real life" this means the system is well-mixed and that matter and thermal energy has dissipated throughout the system - there are no temperature or concentration/density gradients. This is easier to visualize using particles instead of energy - imagine an ideal gas in one side of a partitioned chamber, where the partition has a hole in it to allow the molecules to cross into the other section:

This is an example of an internal equilibrium - the particles cannot cross the system boundary, but they can reach an equilibrium within the system. An example of a system in equilibrium with its surroundings would be a container filled with water at room temperature - there are no thermal gradients across the container boundary.

Why does rapidity of the change prevent it from attaining equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?

Whenever a system is perturbed, meaning moved away from an equilibrium state, it takes some amount of time to reach a new equilibrium. This amount of time is called the relaxation time. Different systems have different relaxation times, and if you make changes quickly enough, the system won't have time to relax, and therefore won't be in equilibrium. Using a piston in a cylinder as an example, if you compress the piston by a very small amount, then the pressure inside will be in equilibrium with the pressure outside almost immediately.

What do they mean by driving and opposing force?

Using the piston as an example again, in this case the driving force would be the external pressure on the piston. The opposing force would be the pressure of the gas inside the cylinder. For a very slow process, at any given time the outside pressure and inside pressure would be the same - the volume would change to make sure this is true. For a very rapid change, the gas inside the cylinder would take longer to reach equilibrium, and so the pressure outside would not equal the pressure inside - the driving and opposing forces would not be equal.
It is important to remember that these are approximations that are intended to allow us to convert really difficult (usually unsolvable) differential equations that describe non-equilibrium or dynamic systems into something that is solvable. The trade-off is that we go from a realistic, fully dynamic system to one with constraints - we treat it like a series of tiny jumps between systems that are in equilibrium. In many cases, these approximations don't affect the accuracy of the solution. However, whether or not it will is determined by the time scale of the change relative to the relaxation time of the system.

I'll try and somehow answer this backwards.
3) Driving and opposing forces: Each process is driven by some kind of 'force' (please do not understand 'force' in the traditional physical sense; it's more of 'reasoning, why something would want to do something'). For example, when having a saturated sodium chloride solution, the sodium and chloride ions that are still bound in their ion structure in the precipitate would like to dissolve, the driving force being the increase of global entropy by a greater disorder of the entire system.
On the other hand, every process also has an opposing force, that attempts to force the process back into its original state. In the case of the saturated sodium chloride solution, the opposing force would probably be something along the lines of releasing water molecules out of a solvation shell, and thereby increasing local entropy.
Finally, for an (essentially) irreversible process, like the reaction of hydrogen and chlorine to form hydrogen chloride gas: This is a radical chain reaction. For each reaction step (formation of chlorine radicals, homolytic cleavage of hydrogen by a chlorine radical to form hydrogen chloride and a hydrogen radical, homolytic cleavage of chlorine by a hydrogen radical to form hydrogen chloride and a chlorine radical, and finally radical recombination in whichever combination) there are of course driving and opposing forces. However, the driving force is usually much stronger than the opposing force. For example, while it is energetically favourable to cleave a chlorine molecule with a hydrogen radical to form hydrogen chloride and a chlorine radical, a chlorine radical would never attack hydrogen chloride to liberate a hydrogen radical as HCl is far more stable than either hydrogen or chlorine.
2) The smaller and less rapid a change happens—in our case, dissolving one pair of sodium and chloride is a very small change—the easier it is to just go the other way. If a change is 'rapid' (I don't really like the word here, but I'm not good enough to find a better one), then obviously the driving force is stronger than the opposing force, creating a bias towards the product side.
But I would agree that this part of the definition is especially weak and hard to understand …
1) Any process can be understood as a dynamic one, with the change happening in both forward and backward direction. So even if you were to burn methane—clearly an irreversible process—you can still consider very minute quantities of carbon dioxide and water recombining to form a methane molecule or two (and oxygen). You won't, however, observe a proper equilibrium between methane/oxygen and carbon dioxide/water.
So while reversible processes are always at equilibrium between one side and the other side, irreversible processes are shifted so far to one side, that it doesn't really make sense to define an equilibrium for them, i.e. they 'do not get a chance to attain equilibrium'.
I hope I could help in some way, and I think you should get a better book ;)


text:
How can we measure the atomic radius of an atom?
The definition of atomic radius is half the average internuclear distance of an atom.
But many times I've been told that it is not possible to handle a single atom. Thus how can we measure the internuclear distance of an atom?

As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element.

This internuclear distance can be determined by two methods:

X-rays method
Spectroscopy method

Note: Atomic radius is not a set value for a given atom. Eg. $\ce{Fe^{+2}}$ ion have different atomic radius than  $\ce{Fe+}$. Moreover,  $\ce{Fe^{+2}}$<  $\ce{Fe+}$

I think it is important to note also that atomic radius is not a set value for a given atom.  For example, Fe2+ ions have a different atomic radius than Fe3+ or Fe in metal.  When we talk about atomic radii in minerals (which are crystalline) then there can be many different radii possible.  You can find tables of these radii in mineralogy texts where they've been determined as the spherical size of the atom that fits into its site in the crystal.

I think OP what to know the practice method of determining atomic radius, but we can add this as note. :-)

Also, I'd rather stress that atoms do not "look" clear-cut as in the picture, but rather blurry, so even the specific atomic radius of an atom in a particular state is somewhat less definite than it might seem.

@ZSG - It is more basic than that. The $\ce{Fe^{3+}}$ ion really has a different sizes in $\ce{FeCl_3}$, $\ce{FeBr_3}$, and $\ce{Fe_2S_3}$ for example.


text:
What to do about a un-dissolved substance?
To exterminate a coackroach infestation, I bought solid boric acid (in powder form). Rather than awaiting the cockroaches' encounter with the boric acid, I thought about spraying boric acid onto affected areas or cockroaches. I only took high school chemistry, but remembered solubility so checked its solubility in water. I then compared it with the solubility of other compounds, to see if this were an effective idea:

Boric acid: At 100 Celsius, 27.53 g/100 g of water.
Sodium chloride: At 100 Celsius, 38.99 g/100 g of water.

This looks fine, so I poured boiling water into a bottle with the boric acid powder. I shook it and saw the boric acid dissolve. Yet two weeks later, the boric acid seems to have reversed dissolution. Now, it's separated from the water and sits at the bottom of the water.
Can substances do this? What should I do now? I'd need a pot to boil the mixture over the stove. But then, in the interest of health and safety, I couldn't use this pot anymore, so don't want to sacrifice one.
Update: With thanks to and in view of the instructive answers, would it help to refill the bottle, to benefit from the low solubility of boric acid at room temperature? There's only around 10% of water left, with the boric acid sitting at the bottom.

Never pour water into acid, either solid or liquid. Heat is released when acids dissolve in water. This could lead to local boiling and an acid splashing into your face.

You could put the bottle (of boric acid and water) into a pot of water to heat it up. But why bother. You can use the spray as is and when the liquid is mostly gone, you can add more water, mix it every five minutes for about an hour and you can use that to spray the cockroaches.

I think what you forgot is that the solubility of boric acid (or sodium chloride.. or most solids) is lower at room temperature.
So when you dissolved the boric acid in boiling water, there's no problem at first. As it cooled, you will have a supersaturated solution until eventually the boric acid will recrystallize or crash out.
I'd create a new mixture using room temperature water and add only a small amount of boric acid (as mentioned by @K_P, about 5g / 100 mL of water). Let it dissolve slowly while you mix.
Then if you spray, you shouldn't leave large amounts of boric acid around.
Updated Answer
Yes, if there's only ~10% water, you will want to add more (room temperature) water. This will dissolve more boric acid. Stir/shake to dissolve as much as possible, wait a bit and then you can spray again.
As mentioned, you don't want to add water to a hot acid. In this case, adding more room temperature water to a room temperature acid will probably produce heat, but add slowly and carefully.

Usually, when you want to create saturated solution the approach is following:

You heat the water (does not need to boil though)
You dissolve as much solid substance as possible
Let the water cool down
Pour the water through filter to filter out any recrystallised solid compound

The amount you can dissolve is the amount that can be ever dissolved. If the temperature decreases, some of the dissolved compound will recrystallize.

Looking at your link, the solubility of boric acid at room temperature is around 5g / 100g of water. You can either heat it to boiling point to redissolve or you can filter it and this way you will get a 5g / 100g solution of boric acid that you can still use and also recover solid boric acid. You may need to experiment with different DIY filtering devices depending on the particle size of your precipitated boric acid but I guess you made some nice crystals that shouldn't be very hard to filter.


text:
"Carbon snakes" - what physico-chemical process causes the exothermic response?
Making a "carbon snake" requires the mixing of strong sulphuric acid with sugar, as explained in this about.chemistry webpage - the reaction causes rapid dehydration of the sugar and produces steam in the process, which expands and forms the 'carbon snake', following the incomplete chemical reaction:
$\ce{C12H22O11 + H2SO4 → 12C + 11H2O}$  + mixture of excess unreacted water, sulfur dioxide and acid (and presumably a bundle of other chemicals)
My question is, what physical/chemical processes cause this rapid dehydration and exothermic reaction?

Too busy to leave an answer but check this document out: http://ncsu.edu/project/chemistrydemos/Thermochem/DehydrationSugar.pdf

Your equation is not balanced; you did not account for where the sulfur in sulfuric acid went. Some of the sulfur is oxidized into sulfur dioxide if I remember correctly.

@Dissenter yes, I saw that too (that's what I get for copying that formula from the website - clarified it as an incomplete generalised formula)

does any one knows the ingridients of the commercial snakes

$\ce{C_mH_{2n}O_n -> m C + n H2O}$
The hydration reaction of sulfuric acid acts as the driving force because it is highly exothermic (95.33 kJ/mol H2SO4).


text:
Thermodynamically Reversible Processes: How does an infinitesimal step make a process reversible and how is maximum work derived from the process?
My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause?
Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure $P_i$ to $P_f$. Suppose,the process is done with innumerable infinitesimal steps. Then, say at each step, $dp$ is changed in the pressure.
Before the step, gas and the piston were exerting same pressure $P_i$ . Now, when the piston is raised, gas does expand to equalize the pressure,which is $P_i - dp$ . So, at the end of the step, the gas & the piston will exert same pressure $P_i - dp$ and hence will be in mechanical equilibrium.
During the step,the gas and the piston was not in equilibrium;it was at the end of the step  the equilibrium is achieved. So, during each step of the process the gas would be at disequilibrium . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by $dp$ . So, why does the definition tell that at each step,there is equilibrium?
So, I have three questions:

Why is maximum work only obtained from reversible processes?
According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step?
How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions.

Please help.

Check Carnot cycle.

I haven't seen too many texts deal with the mathematical nuances of reversibility in the Carnot Cycle, in fact none!I know it's a really popular pedagogical tool in classical thermodynamics, but for the life of me I never found it at all helpful!

@user5764 Basically all irreversible paths by definition have to "generate" some new (previously non-existent) amount of entropy (i.e, always make $\Delta S_{univ} > 0$) hence will always have some more inefficiency relative to the reversible processes in terms of doing work (e.g, such as needing more input heat, and for the same amount of work, etc). The rest of the discussion on "infinitesimals/quasi-static/etc" just boils down to (listing) conditions for real processes under which said entropy "generation" can be minimized, well, ideally down to zero for the fully "reversible" case.....

Reversibility is a statement that means, the system and the surroundings are perfectly balanced (in equilibrium) at ANY AND ALL instants in time. This allows us to define $T_{\mathrm{sys}} = T_{\mathrm{surr}}$ and $p_{\mathrm{sys}} = p_{\mathrm{ext}}$ ALWAYS. The latter statement on pressure allows then that we can calculate work in terms of system pressure rather than external pressure. The proof from this point is the mathematics to show, irreversible systems always do less work than reversible systems.
Otherwise, the approach you take with your analysis of pressure in steps is the same one that allowed the tortoise to beat Achilles in a race. It is called Zeno's paradox.
In specific answers to your numbered questions ...
1) Accepting the above definitions allows the mathematics to prove that reversible work is always greater than irreversible work. As a physical answer, we can define reversible work as $W_{\mathrm{rev}}$. Any irreversible process must have a point in time where the system and surroundings are not at equilibrium with each other. Let's use this to get work $W_{\mathrm{extra}}$, and call $W_{\mathrm{irr}} = W_{\mathrm{rev}} + W_{\mathrm{extra}}$. The final proof that $|W_{\mathrm{irr}}| < |W_{\mathrm{rev}}|$ is that $W_{\mathrm{extra}}$ is always work done ON the system over a full cycle (where the system returns to its starting point.
2) You mistake that a process taking (very small but) finite steps as a valid replacement for a process that takes infinitesimal steps. We would never go anywhere in life (literally!) when your approach would be valid. See Zeno's paradox.
3) Infinitesimal steps are not part of the definition of reversibility. They are part of the way to solve a reversible process properly. Reversibility is a definition about the state of system and surroundings during any point in time.

I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.
My mathematical reasoning:
A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use.  Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.
Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being reversible. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails.
My analogy:
Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side.
To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.
To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has still ended up in the same place as the irreversible case.
EDIT: How to Achieve Equilibrium and still have Movement
I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the oscillation of position. Even if you've physically pushed your piston a distance ds, it can still be in equilibrium by the very fact that it can still move forwards and backwards in its next spot.
So in fact the periodicity of position is maintained at each step, which is what we mean by the equilibrium is maintained.
If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur.

Sir, what will you say about the disequilibrium during each step? How should I deal with it? Doesn't it contradict the definition?

I've edited my answer to your response, however the main point to note is: "If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur."

If we consider a chamber with a membrane in the middle separating air and vaccum on either sides,what is my system?If i take the whole chamber as my system,there is no work transfer or heat transfer(chamber is insulated) if i remove the membrane.Therefore this justifies that no work is done by gas in free expansion.But the air and vaccum vary in composition,that is they are not in equilibrium,then taking the whole chamber as my system incorrect.Then what should be my system?Just the air half chamber or just the vaccum half chamber?


text:
How does steric hindrance cause sulfur hexafluoride to have almost no reaction chemistry?
Sulfur hexafluoride ($\ce{SF6}$) is, according to the Wikipedia page, a hypervalent substance.  However, according to the page in regards to $\ce{SF6}$ reactions:

There is virtually no reaction chemistry for $\ce{SF6}$. The main contribution to the inertness of $\ce{SF6}$ is the steric hindrance of the sulfur atom, whereas its heavier group 16 counterparts, such as $\ce{SeF6}$ are more reactive than $\ce{SF6}$ as a result of less steric hindrance

The linked steric hindrance provides many examples of how that chemical process works, but specifically, what steric hindrance (or any other) mechanism results in $\ce{SF6}$ possessing virtually no reaction chemistry?

As you can see:

 Sulfur (16) is significantly smaller than Se (34). That size difference
 means that there is more room between the fluorine atoms on Se than on
Sulfur. That is what steric hindrance means. Think of steric hindrance
 as limited access. The smaller the atom the less access there is.
Oxygen is so small that less than 6 fluorine atoms can crowd around it
because of steric hindrance... probably more like four.
So because there is more room between the fluorine atoms on Se there is
more opportunity for other atoms to attack the central atom Se, but
sulfur is completely shielded by inert fluorine.