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Why does increasing the available surface area of a pure solid not effect equilibrium?
Take a reverseable reaction of a gas and a solid, $$\ce{A(g) + B(s) <=> C(g)}$$
The rate of the forward reaction will depend on the surface area of the solid available. If I take that system at equilibrium and add more of the solid, especially if it's in a form with a large surface area this should change the rate of the forward reaction, but this doesn't change the equilibrium. The closest to an answer I've found is that the rate of the reverse reaction increases by an equal amount so that $K$ remains constant, but no justification for that is given.
Increasing the rate of the forward reaction increases the concentration of $C$, which should increase the rate of the reverse reaction, that makes intuitive sense. But I can't see any obvious reason why that should be equal. Presumably my textbook is being a bit reductionist here.
Is it making a simplifying assumption and not bothering to mention that assumption? (the reactions are being considered as independent of surface area? There are unconsidered phenomenon involved where the activity of a pure solid can't be treated as unity and we're only dealing with an approximation?).
Or are there more complex factors at play that result in the rate of the reverse reaction increasing by the same factor as the rate of forward reaction, and the book just doesn't bother to discuss them?

Forget the concentration of C. It doesn't change at all. Just why would the rate of the forward reaction increase? Because of the increased surface area? That's right, but... wait, the reverse reaction occurs on the same surface, so its rate would also increase by the same factor.

Again that argument is reductionist and boils down to "because it does". C could decompose regardless of A being present and can do so anywhere in the volume. Which means the increased surface area from adding more of A shouldn't nessicarily increase the rate of  the reverse reaction by the same factor. Which means something else is going on to make that true.

There is a subtle factor that you're overlooking, and that is how fast does the system come to equilibrium. So with lots of surface area for B the system will come to equilibrium faster, but the value of the equilibrium constant will be the same.

You can change the amount of $\ce{B}$ available and the forward rate, but that doesn't change the rate constant. And the equilibrium constant is directly related to the rate constants of the forward and reverse reactions:
$$K = \frac{k}{k'}, \text{where }k\text{ is forward rate cosntant, }k'\text{ is reverse rate constant}$$
As @MaxW points out, you'll get to equilibrium faster, but it's the same equilibrium.
Edit:
Here's another way to think about it. Consider the reaction:
$$\ce{H2O_{(l)}<=>H2O_{(g)}}$$
Does the equilibrium between liquid water and water vapor change if you increase the surface area of a pool of water? No, because even though evaporation is faster, there's also more area for condensation to take place.
In your specific case, the reverse reaction from $\ce{C}$ needs to interact with some amount of solid to redeposit $\ce{B}$. Otherwise, you're making $\ce{B_{(g)}}$ and that's not the same thing...
Edit 2 (for OP's comments):
Notice that your reaction is actually extremely unrealistic. Gases don't just form solid or liquids on their own; they need some nucleation seed. Otherwise, there would be no point in seeding rain clouds.
As @IvanNeretin points out, invoking microscopic reversibility, you'd need to invoke these nucleation points in your reaction, forward and backward. These points are of course whatever surface is available on $\ce{B}$.
This makes the reaction:
$$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{<=>}} n B_{*} + C_{(g)}}$$
$\ce{B_{*}}$ refers to the surface. Notice that it cancels out so it doesn't affect the equilibrium constant, but having more surface area will increase the forward rate if you increase the surface. It will also increase the reverse rate, which is why the equilibrium constant can stay the same.

None of that describes the physical process of why however, which is what I'm interested in. If the case of the evaporation of water it easy to see what going on physically.
The equilibrium is a result of the vapor pressure which is constant wrt surface area. In a closed system that evaporation will increase the ambient pressure until it matches the vapor pressure, at which point equilibrium is reached.

That's not any different from your example! The two gases maintain an equilibrium.

Just to note, the catalysis reaction:
$$\ce{A_{(g)} + B_{(g)} + n B_{} \overset{k}{\underset{k'}{<=>}} n B_{} + C_{(g)}}$$
has nothing to do with
$$\ce{A(g) + B(s) <=> C(g)}$$
which implies consuming the macroscopic amount of solid.
Such reactions are not "extremely unrealistic". They include routine commercial metal purification, rectification of uranium isotopes and restoration of filament in halogen lamp.

That's not true. If your want to create a solid from a gas, it does not just happen. A solid implies interactions that a simple reaction equation does not capture. This is not a catalytic reaction. The surface on the left is not the surface on the right. "Cancelling" them is even dubious, but I'm trying to express the properties of a solid that the OP's equation cannot convey.

One last comment. If you think the original reaction is fine as written, consider running the reverse reaction on a single molecule of $\ce{C}$. Do you get back solid $\ce{B}$? If you think the answer is yes, please define what a single molecule solid looks like.

Kinetic and equilibrium equations, as well as state of matter definitions, are statistical and can't be applied to single molecule.
I agree with the logic of your answer to the original question: the activity of solid phase appears at both sides of equation.
Your writing with nB(*) at both sides, and denounce of gas-solid reactions were just misleading at first reading.

@sa7 Sorry about any confusion...

If you pulverize B (s) to have more surface area, the system will still reach the same equilibrium. All it'll do is make the system reach that equilibrium faster. But it's still the same equilibrium.


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Combustion of saturated hydrocarbons vs combustion of unsaturated hydrocarbons
I read in my science book that saturated hydrocarbons burn with a blue flame (indicates complete combustion) and unsaturated hydrocarbons burn with a yellow flame (indicates incomplete combustion) due to the highier percentage of carbon in unsaturated hydrocarbons as compared to saturated hydrocarbons. So I think the reason why unsaturated hydrocarbons don't burn completely is the presence of double and triple bonds in unsaturated hydrocarbons which are more stronger than single bonds because of which they are difficult to break.  There was a question that :-
Why is a mixture of ethyne and oxygen used for welding?
And the answer was(as written in the book) :
So that ethyne burns completely without producing soot but according to me the issue is not about availability of oxygen.
As I saw in the reactions of combustion, unsaturated hydrocarbons require less oxygen to burn than saturated ones. For example :-
$$\ce{C2H6 + 7/2 O2 = 2CO2 + 3H2O}$$
And
$$\ce{C2H2 + 5/2 O2 = 2 CO2 + H2O}$$
Here we can see that unsaturated hydrocarbons require less oxygen as compared to saturated hydrocarbons.
So my queation is :-
Is the answer given in the book correct. If yes, why?
$$Thank you$$

very related (possibly a duplicate): http://chemistry.stackexchange.com/questions/46925/why-do-unsaturated-hydrocarbons-burn-with-a-sooty-flame-while-saturated-hydrocar?rq=1

In fact, according to my answer to the related question, your book is wrong unless it specified specific conditions of burning.

You are viewing it from the wrong perspective.
In the case of normal burning, Generally, the oxygen is the limiting reagent (therefore some of the hydrocarbons left behind ).
Let's take 1 mole of Oxygen.
So you have to take  0.285 mol of ethane (0.57 mol carbon )  and 0.4 mol of ethyne (0.8 mol of carbon).
It's quite clear ethyne has more carbon content for a given amount of oxygen. As if some hydrocarbon get left, Ethyne will give more carbon content.
https://www.youtube.com/watch?v=FJi4ODPQSE0
(Note: If concentration of oxygen is very high and enough temperature is given then both will burn with Blue Flame, as it is done in case of welding
https://www.youtube.com/watch?v=hKMDxag8yAo )
Now In the case of welding, welding is done in presence of excess of Oxygen, so there is no question of limiting reagent of oxygen.
What matters is Auto Ignition point, Ethyne has much lower auto ignition temperature that is 305 °C as compared to ethane which has 515 °C.
Of course, you pointed out above that for, given 1 mol of ethyne and ethane, ethyne need less oxygen to be burned, so it can be taken in quite concentrated form.
You can take maximum 80% Ethyne (concentration in air), it will burn.
Whereas ethane can't be burn as its concentration exceeds 12.5% (in air).
More concentrated heat implies Higher can be temperature achieved.

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Obtaining calcium
I have an issue with a way of extracting calcium from a muscle tissue (from a steak, to be exact). I would like to calculate the concentration/amount of calcium stored in a piece of the muscle, yet I did not find an adequate way of doing so.
Reading about calcium and calcium carbonate extractions I figured that maybe passing an electrical current in a solution would release the calcium, though I am not 100% sure it would work. Maybe dissolving the tissue first would work? Is it possible at all?
Any insight will be appreciated.

You can incinerate the sample in an oven. Place the sample in a flat dish made of porcelain or (in particular if the sample releases a liquid with a high salt content) a good stainless steel (e.g. 1.4571). Make sure that fresh air can reach the sample. Incinerate at least for 4 h at a temperature of 400 °C. (If the sample is large or contains large amounts of fat, you may need to heat slowly over a longer period.) If the resulting ash is still black, continue to heat at a temperature of 700 °C until you obtain a white ash.
In order to estimate the required sample size, you may expect that 100 g of fresh beef contains about 75.1 g water and yields 1.23 g ash that contains about 3.5 mg calcium. [Source: German Messanleitungen für die Überwachung radioaktiver Stoffe in der Umwelt und externer Strahlung]
Let the ash cool down and wet it with water. Dissolve the calcium salts with concentrated hydrochloric acid. You have to be careful since the ash usually contains significant amounts of carbonate salts, which will fizz and foam when you add the acid.

This is easy part of the problem. How he's going to deposit or titrate the microamounts of calcium in this dirty solution?

@sa7 the solution is not all that dirty and yes, there are special pigments selectively binding to Ca2+ and allowing for complexonometric titration. The method, of course, needs to be found in special literature, ideally one that contains sample preparation procedure for living tissue in the first place.

btw, dissolution of remains may prove difficult because calcium gives insoluble salts with phosphate present in the living tissue.\

Given the small amount of calcium, complexometry may not be the ideal technique to use. You're probably better off with atomic emission spectrophotometry. If you're rich or very very lucky I'd go for ICP-MS as the plasma should get rid of any organic remains that may interfere.

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Explanation for increasing wave number trends for hydrocarbons
At my university, we had an organic chemistry exam question that stated,"The trend in the values of the ir absorbtions for the c-h bonds of methane ($\ce{CH4}$: $2995\,\mathrm{cm}^{-1}$), (ethylene: $3080\,\mathrm{cm}^{-1}$), and (acetylene: $3305\,\mathrm{cm}^{-1}$) can be explained by...
The answer was " by the increasing bond strength of the carbon-carbon single, double, and triple bonds. I thought the correct answer was  "increasing s character of the carbon atoms". Could I get an explanation as to why the first response is valid?
Thanks!

I think that you're correct. The C-C bond strength should only explain the C-C wavenumber. The C-H bond strength is correlated with hybridisation of the carbon in question and as you suggested an increasing s character leads to a stronger bond. Sorry if you saw my earlier answer (now deleted), I entirely misread the question in my haste.

But could the first answer technically explain the reason for the C-H trend? For example, If the C=C bond is stronger than the C-C bond, then that's an indication that the H bond to the C will be stronger if attached to the C=C? I'm just playing devils advocate.

Methane has two vibrational normal modes with C-H stretches, $3026 , 3156 \pu{ cm^{-1}} $ and the other molecules similarly have different modes in which CH bonds stretch. (In a normal mode all atoms move in synchrony but with different phases depending on symmetry). Thus its not as simple as suggesting amount of s character in a CH bond as all such bonds move together in such a way that there is a fixed 'in and out of phase' relationship with one another. The correlation more a sort of 'rule of thumb' than a strict scientific explanation. I feel sure the 's' character is the answer expected.

There is no such thing as increasing or decreasing s character of a carbon atom. See, carbon atom has but one s-orbital; it can't have more, it can't have less. You may, though, speak of a bond having more or less s character.

As orthocresol suggested in his comment, you are correct. First of all, methane does not contain a $\ce{C-C}$ bond, so that answer could not explain the wave number form methane. In addition, you can think of bonds as tiny harmonic oscillators. Specifically, they behave like a mass on a spring, absorbing frequencies that match the compound's frequency of oscillation. Remember that for a mass on a spring, $\omega = \sqrt{\frac{k}{m}}$, where k is the spring's stiffness (related to the strength of the bond) and m is its mass. This is why stronger bonds (such as a $\ce{C=C}$ bond over a $\ce{C-C}$ bond) have a higher frequency, and therefore wavenumber (these two quantities are directly proportional). Likewise, the $sp$ orbitals in acetylene are stiffer than the $sp^2$ orbitals in ethylene, which are in turn stiffer than the $sp^3$ orbitals in methane. This explains why methane has a lower wave number than ethylene, which in turn has a lower wave number than acetylene.


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Atom-condensed softness matrix
In the ACKS2 polarizable force field paper, I found a thing called the atom-condensed softness matrix. In another paper, I found this expression for it:

$$
\chi_{kl} = 2 \sum_{i}^{\text{occ MOs}} \sum_{j}^{\text{unocc MOs}} \frac{\langle\psi_{i}|g_{k}|\psi_{j}\rangle\langle\psi_{j}|g_{l}|\psi_{i}\rangle}{\epsilon_{i} - \epsilon_{j}} \delta_{\sigma_{i}\sigma_{j}},
$$
  where

$\epsilon_{i}$: orbital energy of the $i$th KS orbital
$\chi$: (non-interacting) response matrix
$\psi_{i}$: spatial orbital
$g_{i}$: potential basis function

What is the physical meaning of it, or at least what information we can get from this matrix?
If in KS-DFT we consider the system as non-interacting, why do we consider interaction between two species in this equation? (if I understand it right - between molecules $i$ and $j$)

Your first misconception is what i and j mean. These are indices of molecular orbitals.
A response matrix is something used in respone theory. It stems from Perturbation Theory.
Response Theory is a mathematical formalism to compute time-dependent molecular properties (in theoretical chemistry, that is).
This pdf (no guarantee for anything) describes it pretty well and concise.
http://www.lct.jussieu.fr/pagesperso/toulouse/enseignement/molecular_properties.pdf

Thank you very much. One more question then - in calculations I get matrix size of 10x10 for every atom in molecule (i.e. for ethylene it would be 60x60) - does this depend on basis function used and does every matrix element represent interaction between different orbitals (for s, px, py, pz, d etc.)?

@BrigitaP. How many MO do you have? How big is your basis set for every atom? Over what does the matrix iterate? What integrals are being calculated there? The answers to these questions should answer yours. I think you would help yourself the most if you get a book about Computer Chemistry (Jensen maybe) and read a little bit in there.


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Formulae for energy of photon (E=hf and E=hv)
I have seen the energy of a photon given by the formulas:
$$E = h \cdot f \tag{1}$$
Where $E$ = energy of the photon, $h$ = Planck's constant, $f$ = frequency of radiation (Source: BBC article)
I've also seen it given as
$$E= h \cdot \nu \tag{2}$$
Where $\nu$ stands for frequency (Source: Wikipedia article)
But in this topic I've also seen
$$f=\nu/\lambda \tag{3}$$
Where $\nu$ = speed, $\lambda$ = wavelength (Source: Simple English Wikipedia article)
Is there any difference between equation 1 and equation 2? And is equation 3 unrelated?

If you're talking photon energy with the Planck equation, you should be using $c$ for speed.

I understand your confusion. They all are actually the same formulas. It is because lets start with the first equation,
$$E=h \cdot f$$
But we know that frequency can be written in terms of speed formula,
$$ v = f \cdot \lambda$$
Which can be rearranged into the following,
$$ \frac{v}{\lambda} = f $$
If you put that expression into the above equation, then that becomes
$$ E = f \cdot \frac{v}{\lambda}$$
Hence the same formula can be written in many ways.

Your first two equations are just the same. $f$ usually stands for frequency and in the second equation you have just used $\nu$ for it. That is the frequency of the photon that is coming in or being released.
Whereas your third equation actually comes from the speed formula as I have written above,
$$ v = f \cdot \lambda$$
Rearranging for frequency, you get:
$$ \frac{v}{\lambda} = f $$

Thanks for the answer @MathCurious314 :) So is it just a matter of v standing for different things in E=Hf and v=f*lambda? And the first two formulas are exactly the same?

There is much extra in this answer, but fundamentally, author correctly points out that it is the same equation.  Take heed orthoCresol's comment in question

@K-Feldspar exactly. $v$ is being used for frequency in the second while it is being used for velocity in the third. :)

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Degredation of Anusol and Sunscreen
Hopefully this is not TMI. I mistakenly used some expired Anusol recently, and noticed that the product is 12 years expired. One of the listed ingredients is Zinc Oxide, which is also used in sunscreen. This led me to ask myself "why do consumer drug products expire?", which led to the question, "How does Zinc Oxide, one of the active ingredients of sunscreen and Anusol, degrade over time?"

Existence in an oxidative atmosphere means many commercial products may be, and often are, exposed to oxygen.  Many things, and almost all drugs, react with oxygen.  The products of these reactions (oxidations, normally) are different molecules that may be less effective than the original drug.  As such, the efficacy of a drug that is beyond its expiration date is presumably lower, meaning dosing instructions become invalid.  Similarly, water content in the atmosphere may alter reactivity of drugs.
In general, studies to determine expiration dates take large amounts of time, and are hence expensive.  The standard for such an expiration would normally be a remaining purity level of the drug in question.  For example a company might study their drug for a year and note it is still 99% unchanged.  This means it meets a pretty good standard after one year.  From a marketing perspective they just label the drug as expiring after one year.  This allows them to be out of the legal liability because they are disclosing actual data indicating the efficacy of the drug over a period of time, and if a consumer takes the drug outside the expiration date, they are using the drug counter to instructions.  It also tends to generate extra sales for people who keep old drug stocks around, and must buy new ones to guarantee efficacy.  The actual drug may be still therapeutically useful even if "expired".
Specifically related to zinc oxide, I see no pathway for further degradation.  Wikipedia reports Anusol actually contains zinc sulphate (monohydrate), which I expect to be similarly robust.  However, the other but the qualities of the product ( presumably an ointment or lotion ) may have degraded such that its application may be less effective.  Additionally the Anusol plus contains a drug called Pramocaine Hydrochloride.  This compound is likely to oxidize overtime.


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Why does my voltaic pile produce a higher voltage then predicted?
In my chemistry class we are building our own voltaic cells and trying to produce as much current as possible with materials we are given using redox chemistry.
My lab partner made a battery without thinking about the actual chemistry and it produced about 1.7 V per cell and 5.5 V for 3 cells in series. We tried different voltmeters to confirm that our readings were not a hardware issue.
After consulting our textbook and our teacher we realized that the theoretical voltage produced from our battery should be .71 V per cell and about 2.1 V in series of 3 cells.
Our battery setup was as follows:
Mg(s)|Al(NO3)3(aq) || Mg(NO3)2(aq) | C(s)
with magnesium as an electrode and the strongest reducing agent and ionized aluminium as the strongest oxidizing agent with sodium nitrate in the salt bridge. I understand that the aluminium ion solution should be switched with the magnesium ion solution so that the aluminium doesn't cause the magnesium to oxidize.
My question is why and how does this battery produce such high voltage? We used distilled water to clean all of our beakers but even then I don't think a little bit of copper or other ions dissolved in our electrolyte would be enough to produce that much voltage and sustain it for the hour we spent trying to figure this out. So how did the battery produce a higher than expected voltage?
I have a diagram of what our battery looked like here https://docs.google.com/drawings/d/1u1ZjI9vHCNIeAACrRZmL3CqP_AgshX-HKg-00sS0rvc/edit?usp=sharing

Did you use the nernst equation to account for using non-standard conditions?

No I had no idea it had existed as we had just started learning about this recently. I will look into the to figure out if the nernst equation provides the extra voltage that was recieved.


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Jamming Theory or DLVO Theory for Oobleck
After doing some research I came across two theories that explain the shear thickening effect of Oobleck.  The DLVO theory which involves the suspension of molecules via two opposing forces [attractive (Hamker/Van de Waals) and repulsive (electrostratic)].  When a force is applied the attractive force (which reigns over smaller distances) temporarily overcomes the repulsive electrostatic forces.  This causes the particles to clump together and the mixture takes on properties of a solid. My question is which theory is the correct one in the case of oobleck.  The Jamming theory states that the sudden application of pressure, displaces the water that the starch molecules are suspended and move freely in.  This causes that particle to "jam" together (due to the low pressure zone created) effectively locking them in place.  This temporarily results in the oobleck becoming a solid.
**I know the description of the DLVO theory is quite watered down but I just wanted to speed up the answering process (if that's even possible) More info on DLVO Theory


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IUPAC name of the tert-pentyl group
What is the IUPAC name of tert-pentyl? I know that isopentyl is 2-methylpentane and neohexyl would be 2,2-methylhexane, but what about tert?

The retained names “tert-pentyl”, “isopentyl”, and “neopentyl” were still included in the 1993 IUPAC recommendations.
However, according to Subsection P-57.1.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these retained names are no longer recommended. The preferred prefix names are as follows:

“tert-pentyl”: 2-methylbutan-2-yl
“isopentyl”: 3-methylbutyl
“neopentyl”: 2,2-dimethylpropyl

P-57.1.4 Retained prefixes no longer recommended
The retained names phenethyl (2-phenylethyl) for $\ce{C6H5-CH2-CH2-{}}$; benzhydryl (diphenylmethyl), for $\ce{(C6H5)2CH-{}}$; isobutyl (2-methylbutyl) for $\ce{(CH3)2CH-CH2-{}}$; sec-butyl (butan-2-yl, 1-methylpropyl) for $\ce{CH3-CH2-CH(CH3)-{}}$; isopentyl (3-methylbutyl) for $\ce{(CH3)2CH-CH2-CH2-{}}$; tert-pentyl (2-methylbutan-2-yl, 1,1-dimethylpropyl) for $\ce{CH3-CH2-C(CH3)2-{}}$; and neopentyl (2,2-dimethylpropyl) for $\ce{(CH3)3C-CH2-{}}$; are no longer recommended; the first name in parentheses is the preferred prefix name.

There is a really nifty chart for this on the pentyl group page of Wikipedia:

$$ \small
\begin{array}{lcc}
\hline
\text{Name} & \text{Structure} & \text{IUPAC status} & \text{IUPAC recommendation} \\
\hline
n\text{-pentyl} & \ce{-(CH2)4CH3}\ & \text{} & \text{pentyl} \\
tert\text{-pentyl} & \ce{-C(CH3)2CH2CH3} & \text{Allowed} & \text{1,1-dimethylpropyl} \\
\text{neopentyl} & \ce{-CH2C(CH3)3} & \text{Allowed} & \text{2,2-dimethylpropyl} \\
\text{isopentyl} & \ce{-CH2CH2CH(CH3)2} & \text{Allowed} & \text{3-methylbutyl} \\
sec\text{-pentyl} & \ce{-CH(CH3)CH2CH2CH3} & & \text{1-methylbutyl, pentan-2-yl} \\
3\text{-pentyl} & \ce{-CH(C2CH3)2} & & \text{1-ethylpropyl, pentan-3-yl} \\
\hline
\end{array}
$$

What would tert-Pentyl look like using a Skeletal Formula ?

@jimpix like this.

Last one in chart is wrong. The right formula is -CH(CH2CH3)2.

That is now corrected.


text:
Can we create human muscles artificially?
Can chemical processes create sarcomeres? Not the entire cell with mitochondria, just what basically makes the muscle contract, myosin and actin, in the goal of reproducing our muscles. The source of energy and the activation of troponin could then be controlled externally.

Interesting. But I believe we aren’t that far yet.

I don’t think this question is too broad: it is very well-defined and asks about a single point of interest. I also think that it is not opinion-based: either there are papers in the literature that say ‘yes’ or there are no such reputable papers. In any case, I am voting to leave open.


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How fast do it take a gas kinetic energy to react to intense heat?
Let's say I have  argon gas in a small container. And opened the flap of the container and stuck a rode that was around 2000-3000 Degrees Fahrenheit, And was only in there for a millisecond. How long will it take the heat to transfer to the gas raising the kinetic energy, causing extreme pressure? Let's also add the gas was frozen to around -5 F. I need to know how fast it take to change the kinetic energy of a gas. And how to make it happen as fast as possible?
P.S Would the heat spread evenly in that short of time frame?

I think you are mistaken - while the rod is placed in the container, it will heat up the air via a process known as convection. According to wikipedia, the rate of convection is given by $\frac{dQ}{dt}= hA(T_a-T_b)$ where $\frac{dQ}{dt}$ is the rate of heat transfer, h is a constant for the material, A is the exposed surface area, $T_a$ is the surface temperature of the object, and $T_b$ is the temperature of the gas around the material. My guess is that even at those temperatures, one millisecond will not be enough for the increase in pressure to be as extreme as you are looking for, as the gas around the rod will quickly heat up, but this heat will take a relatively long time to diffuse. As the air around the rod heats up, the rate of heat flow from the rod to the environment will slow down, meaning that only a small amount of the gas in your container will have heated up to a significant temperature.
When you remove the rod, no new energy enters the system. In other words, the average kinetic energy of the gas is now at its maximum. All that will happen is that the energy will become more uniformly distributed, following the second law of thermodynamics. In fact, if you were to get some sort of extreme pressure effect, it would most likely occur very shortly after the rod is removed. At this point, there are a lot of individual high energy particles in the air, which could create a pressure wave, even though the average energy of the particles has not substantially increased. It would quickly be dampened by lower energy particles in the gas, but the velocity of such a wave could be approximated with root-mean-square speed for a gas.
In general, you are correct that the fastest way to change the temperature of a gas is to make it really cold (please note that that a frozen gas is a solid, not a gas. I assume that you just meant that the gas was really cold.) and the inserted object really hot; however, in addition to this, you can make the process even faster by increasing the surface area of your hot object.


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Why exactly does atomic radius increase down a group?
I realize this should be a fairly basic question, but I'm still not quite satisfied with what I've been told from numerous sources. The general explanation seems to state that as we go down a group, there are more electron shells present to contribute to a "shielding" phenomenon, where inner electrons cancel out part of the inward force from the nucleus with a repulsive force. Thus with more energy levels present and more shielding, valence electrons begin to reside further and further away from the nucleus.
But at the same time, textbooks make a big deal about the concept of effective nuclear charge, which states that the "effective" force a given valence electron feels is a function of the net charge of the nucleus and non-valence electrons combined—the effective nuclear charge. I find this contradictory, since as we go down a group, the effective nuclear charge remains constant (equal numbers of protons and non-valence electrons are added). So what am I missing here? Why does shielding matter? Is the force on a single valence electron a function of the net nuclear charge or do we have to look at it from a more piecewise level, considering how other electrons are in between it and the nucleus and how it interacts with those individual electrons?

The description of $Z_\mathrm{eff}$ you gave is a bit too simplistic. As you correctly said, that would predict the same value of $Z_\mathrm{eff}$ for all elements in a group, which is not true.
In general we have $Z_\mathrm{eff} = Z - \sigma$ where $Z$ is the nuclear charge (that solely depends on the number of protons) and $\sigma$ is the shielding constant, which reflects the electron-electron repulsions but is not a simple function of the number of core electrons.
One of the earliest models to determine $\sigma$ was described by Slater in 1930.1 These are easily found online as "Slater's rules". These are quite simplistic, and so people tried to find better ways to calculate $\sigma$. Nowadays, one very popular source for values of $Z_\mathrm{eff}$ are the Clementi values.2 That's still quite long ago, so you can imagine that since then, we have come up with even more complicated ways to calculate it.
A peculiarity is that the values of $Z_\mathrm{eff}$ actually increase down the group. At the very least, that should dispel the myth that $Z_\mathrm{eff}$ only depends on the number of core electrons.

(Source: Keeler & Wothers, Chemical Structure & Reactivity: An Integrated Approach, p 264)

"Surely that should mean that the atomic radii decrease down the group!" you say.
I don't fault you for thinking that at all. However, that's not what we see: the atomic radii clearly increase going down the group. The second factor is the principal quantum number, $n$.
In the hydrogen atom, an electron in the 1s orbital has a much smaller radius3 than an electron in the 2s orbital. That's something that comes right out of the quantum mechanical description of the hydrogen atom.4 The same can be said for other atoms: a 2s orbital in lithium is smaller than a 3s orbital in lithium.
Note that I didn't compare a 2s orbital in lithium with a 3s orbital in sodium. Why? That's because sodium has a larger effective nuclear charge, and this serves to contract all the orbitals in sodium relative to those in lithium. Essentially, the orbitals are pulled inwards by the larger $Z_\mathrm{eff}$.

So, as you can see, going from lithium to sodium, there are two competing factors:

Larger $n$ suggests that the outermost electron should be further away from the nucleus.
Larger $Z_\mathrm{eff}$ suggests that the outermost electron should be nearer to the nucleus.

Unsurprisingly, it turns out that the variation of atomic radius is dependent on a combination of the two. We know that the atomic radius increases going down the group, so that must mean that the increase in $n$ (going from 2 to 3) outweighs the increase in $Z_\mathrm{eff}$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula.
Notes and references
1 Phys. Rev. 1930, 36 (1), 57
2 J. Chem. Phys. 1963, 38, 2686
3 I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.
4 I'm avoiding the circular logic that "higher energy means further away". It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period.

Okay, this was actually quite helpful - so I guess all I can really say without getting into quantum mechanics is that the increasing energy of the valence electrons (which necessitates that they be further from the nucleus) outweighs the slightly increasing effective nuclear charge.

Yeah, pretty much. You can see this trend in the first ionisation energy as well.

How does the presence of electrons in the d-orbital affect this? This question popped up in my head when I saw the covalent radius of Group $15$ elements (you can check this: https://ptable.com/). Now, here the covalent radius increases by almost $30pm$ when we go from N to P. But, from As onwards the change in covalent radius isn't that big. Why is it so? The only difference in the electronic configuration is that there are d and/or f electrons from As onwards. If anything, it should only increase it more because now the shielding effect is not only due to $4s$ but also due to $3d$ electrons.

I will try to be simple. There are 3 things which affect the atomic radius of an atom.

Nuclear charge: nuclear charge tends to decrease the atomic radius of the atom. Higher the nuclear charge smaller will be the atomic radius.
Shielding effect of inner shell electrons: the shielding effect of inner shell electrons decreases the effect of nuclear charge on outer shell electrons and hence an increase in shielding effect is associated with an increase in atomic radius.
Number of shells: it is common that the higher the number of shells in an atom, the farther out the outer electron will be from the nucleus and the greater the atomic radius will be.

Now, if we combine all these three effects, we will arrive at the following conclusion:
On coming down the group,

Number of protons increases → nuclear charge increases → atomic size decreases.
Number of shells increases →  atomic radius increases.
The shielding effect increases (since now electrons from previous shells also contribute to the shielding effect) → atomic radius increases.

As you can see that there are two effects which favour the increase in atomic radius and one which favour in its decrease. Often (in other words) we say that the combined effect of shielding effect and increase in number if shells outweigh the effect of increased nuclear charge and hence the atomic radius increases.

Atomic radius increases down the group and decreases from left to right in a period.
We have to keep in mind 3 main points while considering this trend:

Elements of the same period have the same amount of shells and differ only in the number of electrons in the outer shell.
The shielding effect of the outermost electrons on the electron in the same shell is almost negligible.
The nuclear charge increases from left to right in a period and hence the effective nuclear charge increases on going left in a period.

Therefore the electrons get more attracted to the nucleus, hence the atomic radius contracts when we go from left to right in a period.

Down a group means that the valence electrons are on a higher energy level (they are further away from the nucleus). You would also expect that going down a group, nuclear charge increases (more protons), however, the electrostatic strength down a group is relatively the same. Why? well because there are more shielding due to more energy levels so overall nuclear attraction is similar.
Z eff = Z(no. of p) - S(shielding e) <-- finds effective/overall nuclear charge. Let's take oxygen and sulfur for example. Z eff for oxygen = 8-2=6 and Z eff for sulfur = 16 - 10 =6. This means that the overall electrostatic attraction between the nucleus and the valence electrons are similar, so the major factor must be that the electrons are on a higher energy level, so they are further away from the nucleus, that results in larger radius. So going down a group always mean larger radius.

But is saying that radius increases because the valence electrons are on a higher energy level and thus further away sort of answering the question with itself?

This is factually incorrect.


text:
IS DMSO-o a σ-donor, a π-donor, or both?
DMSO is an ambidentate ligand that can bind in a κO fashion via the oxygen lone pair and in a κS fashion via the sulfur lone pair.
I know that DMSO can behave as a π-acceptor when it bonds a metal at the sulfur, but what is the symmetry of the $\ce{O\bond{->}M}$ MO when DMSO binds at the oxygen? I have some diagrams of DMSO's HOMO and LUMO MOs, and it seems clear that DMSO-κS is a π-acceptor due to the LUMO's potential to overlap with a $\mathrm{t_{2g}}$-type d orbital, but the HOMO looks like it would do a π interaction. This confuses me because I know from IR that the $\ce{S-O}$ bond is weakened when DMSO binds at the oxygen, and there are some lower energy occupied MOs that would have σ symmetry overlapping with $\mathrm{d}_{z^2}$, for instance.

Is DMSO-κO a σ-donor, a π-donor, or both?
(keep in mind that there are some more σ-donor-looking MOs at energy levels below the HOMO)
Also; The reference below (1) says that DMSO's tendency to bind as DMSO-κS is dependent on the metal having a high enough "electron charge density" to have a sufficient back donation contribution to the $\ce{M\bond{->}S}$ π bond. What does this mean? The paper also says that ruthenium and rhodium prefers to bind as DMSO-κS. Based on Pearson's HSAB concept, I would expect that ruthenium would prefer DMSO-κS because it is a soft acid as a result of its larger second-row ionic radius giving it lower charge density (and S is softer than O), but the paper seems to be saying the opposite. What do they mean by "electron charge density"?
(1) Panina, N. S; Calligaris, M. Inorg. Chim. Acta 2002, 334, 165-171.

I did some more research and I've noticed from some crystal structures that DMSO-o binds in a nonlinear fashion, so probably only one phase of the HOMO lobe is interacting with a d orbital, and this would make it a σ-donor. If it was a π-donor, it would bind linearly (like with dxz). This is confusing, though, because depopulating this antibonding HOMO should increase S-O bond order, and DMSO-o has decreased IR freq., so there must be some involvement of the lower energy occupied MOs (which are more bonding with respect to the S-O bond).

The first thing to remember is that a σ-symmetric bond is always better than a π-symmetric one. Orbitals that can bond with each other are typically oriented in one direction or another (except for s-orbitals, but those can only participate in σ bonds, anyway). Approaching these orbitals from the direction ‘their lobes point towards’ and in the direction ‘your own lobe is pointing towards’ results in a much better overlap than if you attempted to create a π type interaction. Thus, typically any σ interaction will be preferable to any π interaction between a given set of two bonding partners.
To be perfectly honest, I am not aware of any ligand which does not initially bond as a σ donor.
The question remains what will happen if we add π effects into the picture. π effects can either be of the π-Lewis acidic type (e.g. $\ce{CO}$) or of the π-Lewis basic type (e.g. $\ce{F-}$). In the case of DMSO, the LUMO has much greater contributions on sulfur than on oxygen — in line with the general expectation that the LUMO is centred on electropositive partners while HOMOs centre more around the electronegative partners.
Also in line with what oxygen does in other ligands, I would expect DMSO-κO to be a π base, due to the probability of another populated orbital having significant contributions on oxygen. Contrarily, the picture of the LUMO shows how well sulfur in DMSO-κS can act as a Lewis acid.
For Lewis acidic bonding, i.e. π backbonding (also written as M$\leftrightarrows$L), it is beneficial if there is significant charge density on the central metal, i.e. if the metal has a somewhat Lewis basic character. Often, this will be enhanced by the presence of electrons in d-orbitals of the $\mathrm{t_{2g}}$ type.
For Lewis basic ligand bonding, i.e. π forward bonding (also written as L$\rightrightarrows$M), it is beneficial for the metal to be in a Lewis acidic, electron-poor state — often with empty d-orbitals.
In the DMSO-κS context, remember that sulfur carries a partial positive charge and is thus already electron deficient somewhat ($\pm 0$ oxidation state, but it almost has the same electronegativity as carbon, so it is not far away from $\mathrm{+II}$). Therefore, the σ forward bond will be weak per se. Only if sufficient stabilisation through backbonding can be supplied will a DMSO-κS complex be sufficiently stable.

so you think that DMSO-o is π-basic, then? What about my observation that DMSO-o binds with a M-S-O angle of <180˚ while DMSO-s binds with M-S-O=180˚? I would think that the π-donating interaction from my HOMO would have to bind to a t2g d orbital in a linear fashion, so the nonlinear bonding seems to indicate that it might be more σ-basic.

@gannex Note that it does not have to be the HOMO doing the sigma donation. I’m also having a hard time imagining the kappa-S version having a $180^\circ \angle(\ce{M-S-O})$ — that would imply that the methyl groups be dangerously close to the metal!

Oop, you're right. The angle is indeed less than 180˚ for both. I was thinking of the wrong MO. That makes sense given the LUMO I showed... what wouldn't make sense would be a nonlinear bond with DMSO-o given the HOMO I showed IF it was a π-donor interaction.
If it indeed does not have to be the HOMO doing the sigma donation, then things make a lot more sense... but I would've thought it would have to be the HOMO. I have a few lower E OMO's that would fit with the observed IR frequencies and bond angles, and I will upload a picture later on...

@gannex Yes, it doesn’t have to be the HOMO that’s doing the dative bonding but it often is. It just has to be an orbital that is reasonably centred on the donating atom and oriented towards the accepting one.

OK well if it doesn't have to be the HOMO then theory is looking a lot more in line with observations :-)


text:
Charge to mass ratio of an electron
The charge to mass ratio of an electron is -1.7588×10^8 coulomb/g, and is fixed. My textbook, however says that as the speed of an electron increases, the value of charge to mass ratio increases. How is this possible? Shouldn't the mass and charge of an electron always be the same? How is the ratio changing?

Ever heard of the relativistic effects (time dilation, length contraction, etc)? Those are pretty counterintuitive.

That should be the other way around. The charge is not a function of velocity, but the mass is. Faster electrons are heavier, think "dynamic mass" or "relativistic mass". As a result, the mass to charge ratio increases with the velocity.

The electric charge on an electron is fixed, regardless of its speed. The apparent mass of an electron INCREASES with its speed (relative to the observer). So charge/mass will decrease as mass → ∞. The major problem with these statements is the use of "mass" to mean "apparent mass" or "observed mass". More and more (and probably the majority now days) of the Physics community reserve the term "mass" to mean "rest mass" (which is the mass observed at zero velocity relative to the observer). This is a bit more complicated than it seems: consider a gas. The center-of-mass velocity may be zero (relative) but obviously each molecule has its own velocity not dependent on center-of-mass. Fortunately, in the range of temperatures chemists are concerned with, relativistic velocities aren't seen, so no correction is necessary between a molecule's rest mass and its apparent mass. Apparent mass, observed mass, relativistic mass, dynamic mass, are (usually) all different ways of expressing the fact that the energy of an object affects its mass; E= mc² and that energy is the sum of potential and kinetic terms, where the kinetic term is ½mv². Hence the mass you will observe depends on how fast the object is traveling relative to you (or your observational equipment). The Special Relativity equation for mass is m = m0 ÷ √( 1 - (v/c)²) where v is the particle's speed (magnitude of its velocity), c is the speed of light, and m0 is its mass when v=0, i.e. its rest mass. As you can see, as v approaches c, the denominator approaches 0 and so the observed mass gets larger without limit.
I should also note that the generalization I made, that chemists need not "worry" about relativistic velocities, isn't entirely correct. For instance, the color of gold is due to relativistic effects (otherwise it would be silver just like the other more common metals). In general, atoms do not have relativistic velocities, but larger atoms have electrons with relativistic momentum, which can be thought of as relativistic speed. There's no reason why atoms can't have relativistic speeds, other than the obvious problem of how we'd contain and observe them. If there wasn't any "chemistry" at relativistic speeds, we'd have fusion power already - confinement of plasma is "hard" because it interacts with the container.


text:
What constituents in soft drinks are the cause their strong pungent smell?
Wikipedia and PubChem tells us that $\ce{CO2}$ is odorless in lower concentrations but can have a mildly pungent acidic smell at higher concentrations.
However, there is no pungent smell from chemical reactions such as the combustion of cooking gas, effervescence from carbonate upon the addition of an acid, and the sublimation of a piece of dry ice.
But many soft drinks and soda waters (including Sprite, 7 up, Pepsi, Coca-Cola, and Thums Up) emit, open opening, a strong burning smell that may cause severe irritation to eye and nose. The same burning is experienced during the burping (or belching) that comes out after consuming a glass of such drink
So, I want to know, is this due to $\ce{CO2}$? Maybe the manufacturers pump too much $\ce{CO2}$ into their drinks that it reacts in a different way than $\ce{CO2}$ from LPG-oven flames or carbonate/acid reactions? Or maybe that 'burning effect' is due to some mysterious additive?

Yes, it is $\ce{CO2}$ and not some additive. It is just that we can't feel it in low concentrations, exactly as Wikipedia says.

http://chat.stackexchange.com/transcript/message/33376929#33376929 there is phosphoric acid. http://chat.stackexchange.com/transcript/message/33377067#33377067  links here

Yes there is; so what? It is not like you pour Coca-cola down into your nose to smell the solution. Phosphoric acid per se is not very volatile. Sure, some of it might get to your nose in the form of tiny droplets, but those are responsible for a minor component of the smell at best. I can attest to that. I did smell pure $\ce{CO2}$ out of curiosity.

and that pure CO2 was just same as cold-drinks? or more? or less (approx)?

Yes, pretty much so.

@IvanNeretin you could write an answer.

CO2 doesn't irritate. I believe burning is caused by acidic droplets. You can prove/disprove this by arranging experiment with some membrane to retain the droplets.

@sa7 Have you ever tried breathing nearly pure CO2?

Nope, I think I would loose consciousness very soon.

It is definitely $\ce{CO2}$, nothing else.

Yes, that strong smell is mostly $\ce{CO2}$, possibly with a faint tint of volatile aromatizers from the beverage.
Our sense of smell works in surprisingly non-straightforward ways. Indeed, the smells of the same compound in high and low concentrations are sometimes perceived as quite different, to the point of being totally unrecognizable. A classic example is skatole, which may produce either a pleasant flowery aroma or a fecal stench, depending on the concentration alone. Looks like $\ce{CO2}$ is just a somewhat less extreme example of the same effect.
Air with 5% of $\ce{CO2}$ doesn't smell much of anything (not that it is OK to breathe it for long), which leads most people to conclude that $\ce{CO2}$ is odorless. This is quite wrong. If you try to inhale some pure $\ce{CO2}$ (which I do not recommend or endorse), that's when the smell kicks in. Your breathing just stops instinctively, as if you were trying to inhale water. By that moment, though, some $\ce{CO2}$ reaches your olfactory nerves. I am at a loss of words as to what it feels like. Being slapped across your face with a shovel seems like a moderately adequate approximation. Besides, you feel that familiar refreshing sharp odor of an open Coca-cola bottle.
So no, this is not due to some mysterious additive.

+1'd for "Our sense of smell works in surprisingly non-straightforward ways".  It is surprising fact that scatole, which is one of the bad-smelling component of stool; produce a pleasant aroma when the concentration is low!. Same has been told about indole, too!. However It is tough to accept because fecal-odor smells bad even-when it comes from distance. And after purging; when the bathroom-is cleaned, the remaining trace of smell is never felt pleasant.


text:
Am I limited by the pKa value of an weak acid when preparing a buffer solution of it (and its salt)?
If I was to make an acidic buffer system with the ph of, for instance, 4, would I be limited with my choice of weak acids? That is, how big does the difference between pH and pKa have to be in order for the weak acid not to be usable? Or is that not the case? I am very well aware that the optimal weak acid for has to have a pKa value quite close to the pH of the wanted buffer because they offer the most buffering capacity.
Also - Can someone hook me up with some equations applicable for making/dealing with buffers other than the Henderson Hasselbach equation? Are there any equations that can help me predict the buffer capacity?

There is a pretty good treatment of this topic at http://www.chembuddy.com/?left=pH-calculation&right=pH-buffer-capacity

The gist is that the further the pH is from the pKa then the less buffer capacity the solution will have. If the absolute difference between the pH and the pKa is more than 2 pH units then there is essentially no buffer capacity.

If I was to make an acidic buffer system with the ph of, for instance, 4, would I be limited with my choice of weak acids?

Yes you are limited to weak acids that have a pKa of +/-1 of your desired buffers pH value. This is a rule of thumb used in many labs.Some theroetical background is quite nicely explained on wikipedia wikipedia#buffering  capcacity.
This reference also includes some useful equations related to buffering capacity.


text:
Hammett plot shape: dependent on substituent or mechanism?
So, we were shown a Hammett plot that, on the left was linear, but on the right was concave up.  How would I explain the shape of the Hammett plot?  Does it depend on whether the substituent is an EDG or an EWG?  Is it based on the mechanism of the reaction/rds?
Also, a hypothetical question was proposed to us regarding the generation of Hammett plots: how would you generate a Hammet plot for a ring with multiple substituents? The sigma value, sigma+ value, etc. are based on a single substituent, right? So, if you were to have multiple substituents, then what would you do? Would the plot be 3D with two axes for sigma or would you just compare the groups and see which one has the larger effect?


text:
Why is the reduction by sugars more efficient in basic solutions than in acidic ones?
Why is it said that reduction by sugars is much more efficient in basic media than in acidic media as in the following quote taken from Satyanaryana’s Biochemistry, 4th edition page 16?

In the laboratory, many tests are employed to identify the reducing action of sugars. These include Benedict’s test, Fehling’s test, Barfoed’s test etc. The reduction is much more efficient in the alkaline medium than in the acidic medium. (sic emphasis)

Is it something to do with the reagents used because many of them are just basic solutions, but if the reagent happens to be an acidic solution then will be that statement valid?

Fun fact about the quote: All three of those tests are the same reaction with mildly different conditions. But note that Tollens’ reagent requires the same conditions.

On a side note, Barfoed actually mentions slightly acidic conditions in his original article DOI.

This is based in the underlying redox rection. If we take e.g. mannose and attempt to oxidise that, the (unbalanced) half-reaction we need is the following:
$$\ce{C6H12O6 -> C6H10O6 + 2e-}\tag{Ox1}$$
To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about alkaline solutions, let’s do the latter:
$$\ce{C6H12O6 + 2 OH- -> C6H10O6 + 2 e- + 2 H2O}\tag{Ox2}$$
As you can see, the oxidation of sugars consumes hydroxide ions. If you increase the concentration of hydroxide, that will typically increase the reaction rate.
Of course, one should always check what the reduced compound is. In your case, it is always a complexated copper(II) ion and is reduced by the following equation:
$$\ce{2Cu^2+ + 2 e- + 2 OH- -> Cu2O v + H2O}\tag{Red}$$
Even here, if we set up the half-reaction, we notice that we are consuming hydroxide. Thus, the overall redox reaction is:
$$\ce{C6H12O6 + 2 Cu^2+ + 4 OH- -> C6H10O6 + Cu2O v + 3 H2O}\tag{Redox}$$
If a reaction consumes hydroxide ions, it is enhanced by basic conditions. Likewise, if a redox reaction consumes protons, it is enhanced by acidic conditions.

so implicit in this is that if I increase acidity then the reaction will proceed backwards, since to balance electrons we need to add protons on product hand side? Also in what form does the hydrogen is removed in the unbalanced reaction?

@JM97 I learnt to set up redox half-reactions in the following way: 1) identify the redox pair and the electron difference; add them. 2) balance the charge using $\ce{H3O+}$ ($\ce{H+}$) or $\ce{OH-}$, depending on conditions. 3) balance the mass, usually be checking the number of oxygens or hydrogens. The hydrogens in $(\mathrm{Ox1})$ do not participate in redox chemistry, hence they are ignored until we arrive at the third step which is after charge balancing. Thus, they are initially missing.

Also, the reaction won’t necessarily go backwards just because the pH value is reversed to acidic. For some redox reactions (i.e. bromine dis- or comproportionation) it does, but for others the Gibbs free energy difference is too large and simply nothing will happen given suboptimal conditions.

@vapid I decided to go from the aldehyde form to a linear mannonic acid form, please do not assume I wanted to go to any other form. Adding something like $\ce{+ [O]}$ was not appreciated and not desired for the first equation. Finally, I ask you please never to change British spellings to American spellings, especially in my posts.

What is this $\ce{C6H10O6}$ compound then? If you wanted a linear mannonic acid as you wrote, then the formula is $\ce{C6H12O7}$.

@vapid Good thing to point out; make it mannonolactone instead. Which would mean starting from the furanose or pyranose; doesn’t matter.

Mannolactone is OK, but I guess the mannonic acid (not mannuronic, my mistake) is a main product.


text:
Structure of IO2F5(2-) dianion
This question came in my exam to draw the Structure of $\ce{[IO_2F_5]^2-}$
I am confused. Here's what I reasoned:
Iodine will have hybridisation $\ce{sp^3d^3}$ so it will be pentagonal bi pyramidical in shape. I can't reason where will be the position of oxygen atoms... will they be in the axial position or in the pentagon. According to Bent's rule, more electronegative atom occupies the position which have more bond length , smaller bond angle so by this reasoning the fluorine should be on axial position and the oxygen atoms in the pentagon.
Is my reasoning correct or there is something more to it?

Aren't the oxygen atoms singly bonded with excess negative charge (that's why the 2- above the bracket)? The more electronegative atoms should be axial. There is a good MO explanation why, and I think that someone mentioned it in an answer. I will search for it.

Don’t use hybridisation involving d-orbitals for main group elements; it is wrong.

@NeeleshVij See this question and the ones linked therein.

See, before answering the question, let's look at the orbital splitting that takes place. $$sp^3d^3\equiv sp + p^2d^3$$
Since double repulsions are much more than single bond. Thus, they occupy axial, more $s$ containing orbitals, the $sp$ orbitals.
Also note what Bent's rule says:

The more electronegative atom prefers to stay in the orbital having less $s$ character, while the lone pair and double bond prefers to stay in the orbital having more $s$ character.
The orbital occupying more space around that central atom will have more $s$ character. Thus justifying the answer.

Hope it's useful!

See my comments on the question on the topic of involving d orbitals in main group hybridisation.

Jan, I don't clearly see the relation between your comment and the answer to question that you mentioned. Please elaborate your opinion. Also note that I've taken these concepts from a standard book, which I believe should be right. Kindly enlighten me.

Yes, standard books did, for quite some time, involve d-orbitals to explain hypercoordination. However, that concept has been proven wrong. Main group elements do not use their d-orbitals in bonding. Your answer suggests they do. Remember that iodine’s 5d orbitals are even higher in energy than its 6s orbitals, yet nobody suggests 6s should hybridise alongside with 5s and 5p. Think about it.


text:
ZnO layer as a passivity layer
I have bought a zinc sheet and with SEM I can see a ZnO layer, which is a passivity layer. But when I connect this sheet to a conductive wire, le zn sheet is conductive too. Is that normal ?  the ZnO layer should prevent the Zn/wire connection from, isn't it ?

Depends how thick the ZnO is, how the wire is attached, and how closely you look at the behavior of the junction.

Bear in mind that ZnO is a semiconductor. It is a wide band gap semiconductor. Nevertheless, it is a poor choice as an insulating layer. More so if it the layer is thin.


text:
Alternative mechanism for carbocation rearrangements
Is there no alternative mechanism for carbocation rearrangements rather than hydride shifts? I've been killing myself over this thing for over two days, and I do believe that hydride/alkyl shifts are not only unfavourable, but unnecessary as well.
What follows might be a little weird to comprehend at the first try, but please, stay with me here...
Let us deal with one of the simplest situations we can think of, say, 1-butanol in acidic medium.
The H+ ion will interact with the lone pair in oxygen and H2O goes out. Till here we are all in the same boat. Now comes the interesting part.

We're all familiar with the classical theory, so I shall not waste time discussing that mechanism (kindly refer to the enclosed series -1 if in doubt).
I shall only point out what seems, to me, a major flaw.
When the primary carbocation is formed, it causes a number of things to happen:
(1) The electronegativity of the carbon holding the positive charge increases drastically.
(2) This increase in electronegativity causes a dissatisfaction in the adjacent carbon, increasing, in turn, the second carbon's electronegativity significantly.
As it is, carbon has a higher electronegativity than hydrogen. now, you've gone and increased it even further. Now, when the carbon(2)-hydrogen bond is cleaved, what would one expect? One would expect that the electron pair is retained by the carbon and hydrogen goes out as H+. The classical theory suggests that the opposite of this is true. How, then is that kind of bond cleavage feasible?
I say that it is not. It is my strong conviction that nature, in fact, doesn't work that way. Let us look at an alternative mechanism to this kind of a reaction.
(I request all of you to kindly assist me in finding the flaws in what I shall put forth now, for if I can think of it, so can every other person on the planet, and it is impossible that some organic chemist didn't think of it before me. Since it has not been accepted, there HAS to be a flaw. What I think has been expressed in the enclosed Series-2.)

According to the alternate theory, the carbon(2)-hydrogen bond is cleaved as one would expect it to. A comparatively stable intermediate, namely 1-butene, is formed.
There is a comparatively higher electron density in the region of the pi bond. Hence, in absence of any other option, the proton stays in the vicinity of this pi bond, due to electrostatic force of attraction.
This system, when given enough energy(via collisions with other molecules), cleaves the pi bond and the electron pair, now retained by one of the carbons   (most of the time with carbon(1) since a primary carbanion is much more stable than a secondary carbanion), is given to H+, thus giving you your desired secondary carbocation.
After the formation of the secondary carbocation, the formation of 1-butene/ 2-butene is but a trivial question, which happens in the same way as what the classical theory suggests.
So... Is there any other theory as well? Is what I propose feasible? We can never know the absolute truth, of course, but how close are we to it?
PostScript: I am very much aware that what I state is not something new. It must have crossed many other minds before mine.

Please do not propose primary carbocations as intermediates without extreme reaction conditions. On top of that, two cations in a single species is even worse.

In defence of the classical mechanism: obviously the carbon with the full positive charge is hungrier for electrons. So it tugs on the bond pair of an adjacent carbon-hydrogen bond. Now both the carbons in a tug of war, the carbon more desparate for the electrons is secondary carbon (or what your example wants). The hydrogen, with its electron pair, tags along onto the other carbon, and the rearrangement is complete.

For the protonation you are proposing with butan-1-ol, I would assume $\mathrm{E2}$ elimination to generate but-1-ene, which can later be protonated to give the but-2-yl cation (and which can then undergo semi-$\mathrm{E1}$ to give but-2-ene). That would be a better choice than proposing primary carbocations.

@FreezingFire, that's JUST what I'm saying. The roguish carbon(1) would take away the bond pair of carbon(2)-H and make a C-C pi bond. Why is the hydride justified?

@Shekhar Well I wanted to imply a concerted rearrangement, i.e. no hydride ion, but I couldn't think of the term"concerted orbital interaction". Jan answers this question very well! (Most of my orbital understanding is from ChemSE answers anyways!)

@FreezingFire Ok... I get what you mean now. Yes, I agree that Jan has given an almost convincing answer, but I still have a few objections. If you could care to look at the comments posted beneath his/her answer(no offense intended, Jan, I don't know your gender), you could see them for yourself and perhaps help in resolving the issue.

Series 1 has an intermediate with two +'s...

Understood... All the more reason for series 2 to be favoured, isn't it?

Scheme 1: Orbital model of a Wagner-Meerwein rearrangement.
You may think you are proposing something new, but you aren’t.
While these Wagner-Meerwein rearrangements are typically referred to as ‘hydride shifts’, that name should not be taken too literally (as in your first scheme). Instead, it is a concerted orbital interaction. There is never any hydrogen cation or hydride anion formed, hydrogen remains in the vicinity. The entire process is concerted and thus proceeds suprafacially.
Writing the central structure as a double bond with proton in the vicinity, as you did, or as ‘two cations coordinating a hydride anion’ is really only a formality. It is not an intermediate in any way, it is a transition state.
Being a transition state, by the way, means that it is a directed process. If the hydrogen went half way, it is going to proceed on (law of inertia, if you wish) and not go back. However, given a sufficiently stabilised cationic system, it may reverse at a later moment in time.

I must admit that I'm in the dark as regards the semi E1 mechanism... And I'm not able to comprehend the inertia thingy. An Alkene is definitely much more stable than a carbocation, and a primary one at that. So, unless otherwise agitated, why would it want to do anything on its own? And about the actual mechanism for the Wagner Meerwein rearrangement, I admit I didn't know that. I'm just a fresher, who had come into this field of study around 3 months ago. It hadn't been taught to us as yet. Thanks for your help. By the way, there wasn't a double carbocation anywhere in what I thought.

I'm really sorry if I gave an impression that I wanted to propose something new. I know it must have crossed many people's minds, and there had to be an error, otherwise it would've been the generally accepted theory.

About the Wagner Meerwein rearrangement: Is an alkene not much more stable than a carbocation? When the system has an opportunity to steal the C2-H bond pair and utilise it as a C-C pi bond, why would it prefer to get into a comparatively higher potential energy state by making another carbocation?

@Shekhar An alkene is much more stable than a carbocation. But here, we also have a "naked" proton, and we have to consider the stability of the molecule along with the proton. It is presumably more stable for the molecule to have the proton bonded covalently than remaining loosely hanging.

@FreezingFire Do you not have naked protons hanging around in an aqueous solution of HCl? There, it's stable. Here, it's not... How?

@Shekhar The interesting thing is, no. There are no naked protons hanging out in aqueous solution of HCl. In fact, generally in chemistry you will not find a naked proton. In aqueous solution of acids, the proton remains attached to a water molecule, in the form of the hydronium ion ($\ce{H3O^+}$).

@FreezingFire you mean a solvated proton, right? H3O+, H502+, and so on? Most commonly accepted one is H3O+. Agreed. But then, you can solvate the whole (Alkene+proton) system. No wait.... Not that.

@Shekhar If you are studying for engineering entrance exams in India (like me), then you must have found teachers telling you about protons falling off of organic molecules. That's just nonsense. You need a base to extract it off the substrate. I would suggest you to visit chemguide.co.uk and explore the site to actually understand what we are learning. Also, ChemSE is a very good resource too.

@Shekhar You do not always have a solvent that can solvate a proton (for example reactions in non polar solvents). And anyways, mechanism-wise, the proton will favour breaking and making bonds concertedly instead of depending on solvation. It is much easier and favourable.

@FreezingFire instead of salvation, let the system get into a carbocation configuration if and when it gains enough energy to do so. This energy would come from the energy given to the system earlier in the process. So the molecules collide (which is a continuous process almost) and it gets into a configure  depending on the amount of energy given to it. And majority of the configurations so attained will be secondary carbocations since this form needs lesser energy to be formed. How's that?

@Shekhar What you say seems correct to me, but probably incomplete. You are definitely thinking in the right direction! But I cannot discuss it further for want of time. And I am not all knowing either! :) I would suggest you to start with reading the organic chemistry sections of chemguide, and keep questioning here on ChemSE (you can edit your question and include your recent doubt). Happy learning!

@FreezingFire I know its incomplete... And I'm trying to get to an end point... But anyways, thanks for the advice, and best of luck to you!

@ShekharUpadhyay Of course, an alkene is more stable than a carbocation. But an alkene + a naked proton together is less stable than a carbocation so the proton will add to the alkene. More stable then an alkene and a naked proton and also more stable than a carbocation is an alkene and a base’s conjugate acid (e.g. if $\ce{NEt3}$ were present anywhere, $\ce{Et3NH+}$ would be formed).

About the question whether elimination or migration: migration is typically faster than elimination, especially if there is no compelling reason for the proton to leave, e.g. if your solvent is $\ce{HCl}$ in $\ce{Et2O}$.


text:
How to pronounce "potassium manganate(VII)"?
The formal names for such inorganic compounds having metals showing of variable oxidation states involve use of Roman numerals, for example "copper(II) sulphate" or "potassium manganate(VII)" or "hexaaquacopper(II)". Of course, common names like "copper sulphate" or "potassium permanganate" do exist, but if I wanted to use the formal ones, how do I correctly pronounce them?

Simple: Just read the roman numerals as a number.

Copper-two sulphate
Potassium manganate-seven
Hexaaquacopper-two.

For negative oxidation numbers use a minus:

Tetracarbonylferrate-minus-two ($\ce{[Fe(CO)4]^2-}$)


text:
Best chair configuration for trimethylcyclohexane matching a specific Haworth projection?

Which option best depicts trimethylcyclohexane shown in the Haworth projection below?

I thought B because

All the methyl groups are equatorial and equatorial groups are more stable.
All the methyl groups look as spread out as much as possible, reducing steric repulsion.

But the provided solution states that-

Although A,B and D could all be said to represent the chair form, the best depiction will show the methyl groups repelling one another as far as possible. The arrangement in A with two equatorial and one axial methyl groups makes A the best depiction.

Also, when converting Haworth to Chair, do we have to match the upward facing groups on the Haworth to upward facing positions on the chair?
i.e. the Haworth has 2 upward methyl groups and 1 downward facing methyl. But in A, the left equatorial one is facing upward (as axial in that position is downward), the middle axial methyl is upward, and the right equatorial methyl  is downward (as axial in that position is upward). This doesn't match the Haworth.

While you are correct that B should be the most stable conformer of trimethylcylcohexane, the question asked which one best represents the one shown in the Haworth projection. Answer choice B has R stereochemistry on carbon B (see below), while carbon B has S stereochemistry in the original projection. As such, the initially drawn compound is different from the answer choice B. On the other hand, answer choice A has the correct stereochemistry, and also has the maximum number of possible methyl groups in equatorial positions, causing it to be the best depiction of the Haworth projection.
Note: Instead of finding the absolute stereochemistry by using R/S notation, you could also notice that the Haworth projection and answer choice B have different relative stereo chemistry. In the original molecule, carbons B and C are cis, while in answer choice B, these carbons are trans.

As far as this question is concerned, one doesn't even need to look at the absolute stereochemistry; B and C have the methyl groups on entirely different carbons, and between A and D, one only needs to consider the relative stereochemistry (cis or trans) between the two vicinal methyl groups. You're absolutely right, though. Although you might want to clarify which carbons you're talking about with a numbered diagram.

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


text:
Does the weird 4s and 3d rule apply for much of the p-block from Ga-Rn?
In the transition metals, $4s$ has a higher energy level than $3d$, because in ionization, the electrons from the $4s$ energy level are lost first.
I was thinking that this might be the same for much of the p-block from elements Ga-Rn because:
Ga has a electron configuration of: $1s^22s^22p^63s^22p^63s^23p^63d^{10}4s^24p^1$
Ga$^{3+}$ has an electron configuration of: $1s^22s^22p^63s^23p^63d^{10}$

Well the only exception in the orbitals is for $s$ and $d$ orbitals. Meaning you will first remove electrons from $(n+1)s$ and then onto $nd$ orbital. It is only because the $(n+1)s$ sublevel is only lower in energy if there are no electrons in the d sublevel. As per the Aufbau principle, the the s orbital is therefore filled before. Then as the $d$ becomes filled with electrons, then the energy starts to flunctuate relative to one another and the s orbital ends up higher in energy. That is why when electrons are removed, they are first removed from s and then $d$. Luckily this only applies to $s$ and $d$ orbitals. When are you working with $p$ orbitals, well then the $np$ orbital will always be higher than the $(n-1)d$ orbital. Therefore for making ions, you will have to first removed electrons from $p$ then $s$ and then $d$ and so on.


text:
Determination of (R) and (S) configurations from a Haworth projection
Short version of question : How do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?

Details:
I am comfortable enough finding R and S configuration of Fischer projections, but am unsure which groups are coming in or going out of the plane of the page in a Haworth display.
For example:

Figure 1
Taking carbon $3$ for example, the priorities are as shown below in blue:

Figure 2
By definition, horizontal groups are out of the page (i.e. towards us) and vertical groups are into the page, the lowest priority group (hydrogen) is out of the page. If hydrogen was into the page, carbon $3$ would have a clockwise R configuration, but as hydrogen is drawn out of the page, carbon 3 has the opposite, hence an  anticlockwise S configuration.
If I now took carbon 3 in the Haworth for the structure above, I am assuming it is still an S configuration as the groups are still in the same position around the carbon 3 atom, relative to the other groups.
Update: After seeing the following image (annotated with numbered carbons from this source) I am not so sure about that anymore as carbon 4 has different configurations in the Fischer vs the Chair and Haworth.

Figure 3
But how would I determine the R and S configuration of carbon 3 without going back to the Fischer projection?
Mainly, how do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?
I would think carbon 4 is "into the page" as it is further away from our point of view (which matches the Fischer projection, as it is drawn vertically there).
Based off this, carbon 2 would also be into the page (as its the same as Carbon 4 in the Fischer projection). Hence, the $\ce{OH}$ and $\ce{H}$ would be out of the page.
The priorities would be as follow:

Figure 4
If $\ce{H}$ was into the page, as draw it would be R, but as $\ce{H}$ is into the page, its the opposite and hence carbon 3 is an S
So I arrive at the same answer, but mainly due to comparing the Haworth off of the Fischer projection. How would I judge what was in and out of the page without the Fischer projection and only the Haworth?

After reading this I'm not quite clear on what you don't understand. It seems  you've done this correctly. It strikes me you may be asking how to imagine this Haworth diagram in three-dimensions. If this is the case it is time to get a model kit. Or at least use a 3d one online: http://molview.org/?cid=5793

Thanks, that site is very useful. Well my issue was how to determine if the lowest priority groups in a Haworth were into or out of the page (as when writing R or S, we've always had to put the lowest priority group into the page).  But I think  I've figured it out and what I have to do for Haworth is treat groups pointing upwards as out of the page and groups pointing downward as into the page.

In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.

However, I notice that the question uses (R) and (S) a lot and attempts to argue a lot with it, so a note must be dropped on those stereodescriptors, too. It is important to realise that the (R) and (S) stereodescriptors can change even if the conformation of the carbon in question did not invert or otherwise change conformation. This is due to the way priorities are determined in the CIP system. Let me exemplify this.
First, remember that α-D-glucopyranose, β-D-glucopyranose and the aldehyde form of D-glucose are three different compounds with different physical properties. The α and β forms are diastereomers, more specifically epimers, more specifically anomers of each other and e.g. display different optical rotation. The chain-form of glucose is different altogether. I will construct all further discussion of this answer in a way that it does not matter whether we are dealing with α or β-D-glucopyranose, so you may always assume a mixture.
Now let’s consider the absolute configuration of carbon 3 in both the chain and pyranose forms.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbon atoms of the (b) and (c) priorities; initially undistinguishable. Thus, we must go down one level.
Both carbons are attached to the atoms (in order of priority) $\ce{O, C, H}$. Again, this is a tie; we must go down one level.
Both oxygens are attached to a hydrogen; this is identical and the end of the sub-chain. We must go back to the next, lower priority of the former atom.
The two carbon atoms differ: one is the acetal group which expands to $\ce{O, O, H}$. The other is connected to $\ce{O, C, H}$ (carbon 5). Thus, we have resolved the issue.
Going towards lower carbon numbers is the higher priority.

Therefore, carbon $\ce{C{3}}$ has an (S) configuration.
Chain form

Again, oxygen obviously has highest priority.
Equally again, hydrogen obviously has lowest priority.
The two carbons are initially equal.
Both are again attached to $\ce{O(H), C, H}$.
The two subsequent carbons differ again. This time, one is the aldehyde group which expands to $\ce{O, (O), H}$, the other is $\ce{C{5}}$ which expands to $\ce{O, C, H}$. Since a ghost oxygen atom has a higher priority, going up the Fisher projection has the higher priority.

We arrive at the same (S) conclusion as before.

Now, we will construct the same discussion for carbon $\ce{C{4}}$.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbons, initially indistinguishable again. We go down a level.
Both carbons ($\ce{C{3}}$ and $\ce{C{5}}$) expand to $\ce{O, C, H}$. We must go down a level.
One of the two oxygens is connected to a carbon atom. Thus, $\ce{O(C), C, H}$ wins over $\ce{O(H), C, H}$.
Going towards $\ce{C{5}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (S) configured.
Chain form

Bla, oxygen.
Bla, hydrogen.
Blabla, two carbons. Again, we descend a level to $\ce{C{3}}$ and $\ce{C{5}}$.
Both carbons expand to $\ce{O, C, H}$. We descend a level.
Both oxygens are attached to a hydrogen and nothing else. They are indistinguishable.
The two carbons we are now examining are $\ce{C{2}}$ and $\ce{C{6}}$. The former expands to $\ce{O, C, H}$, the latter to $\ce{O, H, H}$. Therefore, Going towards $\ce{C{3}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (R) configured.

Note that while the configuration of $\ce{C{3}}$ did not change, $\ce{C{4}}$’s absolute configuration changed from (R) to (S) upon formation of the pyranose ring. Thus, you cannot rely on absolute configurations to verify a correct Fischer to Haworth transformation for carbohydrates!

Hi Jan, Could you please clarify this comment "In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.@
Which two perpendicular bonds are you referring to?  Thank you.

@K-Feldspar There is a six-membered ring which is drawn lying down in perspective as if going into and coming out of the plane. The bonds of this ring are assumed to be in the direction of the ring, so if you are looking at $\ce{C{2}}$ from outside the hexagon, the bonds to $\ce{C{1}}$ and $\ce{C{3}}$ are pointing behind you. Thus, the bonds that are drawn perpendicular to the plane of the hexagon (i.e. up and down) are pointing towards you. For a carbon atom on the back of the hexagon (behind the paper plane; e.g. $\ce{C{5}}$), the situation is reversed.

Ah! Terrific. I assumed the vertical bonds where just straight vertical based on the way it was drawn. But this makes sense, it is just the opposite of how it is for a Fischer. I.e. horizontal = in to the page rather than out of the page for Fischer, and vertical = toward you (for a front carbon) rather than away from you for a Fischer. Thank you for explaining that Jan.

@Jan "the bonds to C1 and C3 are pointing behind you" Did you mean pointing away from you instead?

@GaurangTandon A long time ago™ and bad memory™, but I think yes, I used you instead of the paper plane for some reason …


text:
What can I use to make evaporating fake snow?
I want to create the effect that it's snowing in my living room. Ideally, there would be a chemical or substance that would be able to be sprayed out of a snow machine or pressurized canister etc. This substance would start off solid and would have a chemical reaction with the air that would cause it to evaporate into thin air, a là Harry Potter. Or even if the reaction took a couple hours it would be ok.

The phase transition you are talking about (solid to gas) is called sublimation. There are a few compounds that sublime at room temperature and atmospheric pressure, but the one I think would best suit your purpose is dry ice. If spread into the air as a fine powder, it could sublime before it even touches the ground, especially if the air is warm.
It's important to consider the safety aspects of this, though. $\ce{CO2}$ sublimes at -$78º\mathrm{C}$, so if too much of this powder lands on people, you could end up with some cases of frostbite. Additionally, you would need to keep this space well-ventilated since you will be significantly increasing the concentration of $\ce{CO2}$. I wouldn't imagine it would be too hard to meet these standards, though.


text:
Is it possible create crystalline solvate of electrons?
Okay, I know that this reads ridiculous insane, but one can obtain solvated electrons by adding sodium metal to liquid ammonia solution.
The chemical reaction as written in my inorganic chemistry textbook is this:
$$\ce{Na + x NH3 -> Na+ + e(NH3)_x-}$$
A metallic sheen is observed, and it supposed to have an absorption band at about $1500\ \mathrm{nm}$ that expands into the range of visible light corresponding to the solvated electrons.
These electrons can be react with alkynes as very powerful reducing agents to to remove an H from a C joined by a triple bond, also liberating hydrogen gas as such. This occurs in the solution made by an alkali metal, not from pure extracted electrons
$$\ce{RC#CH + e- -> RC#C- + 1/2 H2}$$
So I was wondering if pure electrons can be solvated in ammonia, if anyone could think of any means by which these electrons could be extracted from the layer, still solvated by ammonia, and some reaction chemistry could be used to trap electrons in a carbon cage in the form of a solid? It is already known that potassium can be electrostatically trapped in valinomycin between the carbonyl groups of valinomycin.
While it may be different small molecules have been trapped in fullerenes. If electrons could be solidified via extracting them from the metallic layer in the proposed reaction, and interacting them with a "host molecule" in host-guest chemistry, do you think that these materials would be highly conductive. It seems to me that captured electrons would have a much different chemistry than simply metals like copper which shed electrons when a current runs through them. Perhaps if these electrons could be absorbed into carbon nanotubes, then the conductivity of nanotubes would increase significantly and these electron complexes may be significantly more powerful computationally than silicon chips.
Here is an experiment by the University of Nottingham scientists showing the solvation of electrons: https://www.youtube.com/watch?v=tYjQXjUUvwY
What do you think about this? Any ideas?

Coulombic charge repulsion should make captured, solid electrons impossible.

Electrostatic repulsion is the main problem. For the same reason it would be impossible to obtain in bulk quantities some highly radioactive elements that undergo beta decay.

It seems you're asking about solid electrides. These are very real and very interesting materials.

Here's another very useful resource for electrides http://www.ch.imperial.ac.uk/rzepa/blog/?p=14272 from Prof Rzepa ... his blog is well worth studying.

@NicolauSakerNeto I think that would make for an interesting answer.

At first I would like to advise you reading about electrides, which were mentioned in the comments. Secondly I won't write about electrides because the links given in the comments are much better at explaining them. So I am going to be a little bit abstract with this question. The first problem which you will come through is Coulomb's law:
$$\mathbf F=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{|\mathbf r|^{2}}\frac{\mathbf r}{|\mathbf r|}$$
If you could somehow prepare even $0.01$ moles of electrons, they would be attracted to the ions which they were separated from very strongly even from kilometers away. The second problem is that electrons can't be simply trapped into any substance because of their wavelike nature and they would tend to "stay" in a localised or in a delocalised state in the molecoules. And here comes the two main things which are preventing such things to exist:
1. Charge distribution
If a conducting body is charged, the charge tends to build up on the edges, and in your situation the repulsive force between the electrons would be so huge, they would "kick" each other off on the edges of the pieces of the substance so it would loose it's electrons.An interesting phenomena such substance would have: Let's suppose the substance containing the electrons is a liquid. All the electrons would be located on the surface of the liquid because of the faraday effect.When the surface of the liquid is disturbed waves appear on it's surface. Let's suppose the electrons somehow can't be kicked off from the surface.
Then there would always be electric current in the liquid because the elecrons would always try to get to the most sharp edge on the surface and the liquid would generate extremely complex magnetic fields because of the currents flowing on the surface of the liquid (I am so excited about the idea of such a liquid, I am so sad I have no idea how it would be possible to make it)
2. Quantum mechanics
The band gap determines if a substance is an insulator or a semiconductor or a conductor. And adding more electrons to something doesn't modify the band gaps. Like if I charge a plastic rod with electrons it won't become a conductor. But there is an alternative solution. If you dope the substance it can modify the band gap energy and it can be more conductive but you can't simply trap electrons in something (F-centers are a completely different story). Imagine irradiating carbon nanotubes with electrons. I wouldn't suspect it becoming more conducting .
Anyway the best source for pure electrons is an electron gun
If you want to deepen your knowledge about the nature of electrons and conductivity I advise you learning some solid-state physics, quantum mechanics and electromagnetism.(None of them is a light weekend program)


text:
Does fluorine in FNO3 have +1 oxidation number?
According to my textbook and also Wikipedia, fluorine nitrate is a compound which can be created (though it is unstable). Also my textbook says that fluorine has a $+1$ oxidation state. But how is $+1$ state of fluorine possible?
I read in another textbook that fluorine can only show $0$ and $-1$ oxidation states (even Wikipedia says so).
Can someone explain as to how fluorine nitrate exists with a $+1$ oxidation state of fluorine?

The oxidation state of fluorine is $-1$ like in all other fluorine - oxygen compounds.

Oxidation states are merely human convention, and as such, are quite arbitrary in borderline cases. Fluorine does not even know what is its oxidation state, nor does it care.

Oxidation states are there merely for convenience, they don't reflect real charge distribution in molecules.

"even wikipedia says so" is not one of the best arguments. :)

Wikipedia is actually pretty careful with oxidation state designations.  Their designations for a given element come from a template which you are supposed to edit only if you cite a source that proves your claim, and any such edits are reviewed.  I have actually seen an oxidation state of $+1$ rejected for fluorine because the claim, based (falsely) on the compound $\ce{F5SOF}$, was rejected. The accepted entry for oxidation state $0$ is based on $\ce{AuF7}$, which does give evidence for that oxidation state.

tl;dr The ambiguity is due to an unfortunate incompleteness of the previously (prior to 2016) existing rules. The oxidation state of fluorine in $\ce{FNO3}$ is $-1$ according to the present rules.
As a general reminder, it is important to understand, that oxidation states are a bookkeeping tool only, and that they hardly represent the general bonding in molecules. With that being said, it is a very simple and powerful tool in many situations. As it is very simple, one has to especially pay attention to its shortcomings or ambiguities.
Prior to 2016 the IUPAC recommended an incomplete set of rules, which I discussed in more detail in Electronegativity Considerations in Assigning Oxidation States.
The special case you have cited is very close to the case of $\ce{HOF}$, which has been previously discussed by me on two occasions already: Oxidation state of fluorine in HOF, What should be the oxidation state of oxygen in HOF (hypofluorous acid)?
According to the new rules, which are based on Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail), and of which the sort version is below adapted from the gold book, there is no more ambiguity about fluorine, which will be assigned $\text{OS}(\ce{F}) = -1$.

oxidation state
gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

Applied to $\ce{FNO3}$, you would assign fluoride an oxidation state of $\text{OS}(\ce{F})=-1$, the terminal oxygens $\text{OS}(\ce{O})=-2$, the nitrogen will have its highest possible state $\text{OS}(\ce{N})=+5$, and the bridging oxygen will have $\text{OS}(\ce{O})=0$.

Contrary to that are the pre-2016 rules by the IUPAC according to the gold book (via the Internet Archive):

A measure of the degree of oxidation of an atom in a substance. It is
defined as the charge an atom might be imagined to have when electrons
are counted according to an agreed-upon set of rules:

the oxidation state of a free element (uncombined element) is zero;
for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to
this are that hydrogen has an oxidation state of -1 in hydrides of
active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of
-1 in peroxides, e.g. $\ce{H2O2}$;
the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the
oxidation states of the constituent atoms must be equal to the charge
on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$,
$\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and
$\ce{H2SO4}$ are, respectively: -2, 0, +4, +6 and +6. The higher the
oxidation state of a given atom, the greater is its degree of
oxidation; the lower the oxidation state, the greater is its degree of
reduction.

You can already see that there is a problem in these rules since they are not definitive. The keyword here is most in bullet 3. They are at least incomplete since there is no mentioning of other elements than oxygen and hydrogen.
This will obviously lead to contradictory results. According to the deprecated rules we need to give oxygen an oxidation state of $\text{OS}(\ce{O})=-2$ in $\ce{NO3F}$. This will give us a total of $6$ electrons that need to come from the other elements. Since nitrogen only has five valence electrons, anything larger than $\text{OS}(\ce{N})=+5$ does not make any sense. Unfortunately that leaves one electron, which can only come from fluoride, i.e. $\text{OS}(\ce{F})=\color{red}{+1}$. It is obviously ridiculous that the most electronegative element has a higher oxidation state than one with a lower electronegativity. This is also in no way an accurate representation of the charge distribution in the molecule.

An edited image in which the charge on fluorine is more readable against the dark green background of the fluorine atom:

The above image shows the (NPA) charges of the molecule calculated on the DF-BP86/def2-SVP level of theory. It clearly shows, that fluoride is not positively charged.
One really has to understand that an oxidation state of $+1$ for fuorine does not make any physical sense, and that this set of rules is insufficient. Thankfully this has been addressed in the new definition, which is not perfect, but at least complete.

I find the black font on the dark green fluorine atom so difficult to read that I took the liberty of adding an edited image. The fluorine charge is calculated from balancing against the more readable charges on the other atoms. // Gold heptafluoride has a difluorine ligand in which the fluorine atoms have oxidation state zero and slight calculated positive charge.

I'm completely unsure, why I've chosen this style. @Oscar you could have just replaced the image, instead of adding the text. Anyway: thanks.

Thanks@Martin. The issue I had with replacing the image is you incorporated source information into the image description and I did not want to lose it. (I prefer to put that information into a caption or the accompanying text to make things more "modular".)

Oxidation state and electronegetivity are quite related.
While studying about Oxidation state I have always seen that in a molecule most electronegetive atom gets the negetive O.S and less electronegetive atom gets the positive O.S.
The fluorine is the most electronegetive element in the periodic table that's why it is given a permanent oxidation state of -1 (0 when it is in elemental form).Since I consider my books and resources trustable I will say that in FNO3 fluorine still have an O.S of -1 (remember the electronegetivity concept).

For a nonelemental example of the zero oxidation state, see here.