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Why polymorphs form?
We always hear about the different arrangements of molecules that lead to the formation of polymorphs, my question here is why would a polymorph form in the first place? The material was given enough time to solidify in  a proper arrangement, why did it choose to solidify in a form that isn't the most stable?

This is a very broad and very interesting question, but the basic point is: polymorphs exist either because

their formation is controlled by kinetics (rather than thermodynamics, which would favor the most stable phase), or
the thermodynamic equilibrium between the various polymorphs is influenced by the synthesis conditions (temperature, pressure, presence of templating agents and other species in solutions…), and once the material has crystallized in a given form, it does not spontaneously undergo a transition to the most stable phase (again, because of kinetics)

Thanks for your reply, I actually understand the second part of your answer, but I am unaware of how the kinetics would control the formation of the polymorph crystal. Materials solidify in a way to get to the most stable form (from a thermodynamic point of view). How would kinetics change that rule?

@Error404 “Materials solidify in a way to get to the most stable form” — not always, the most notorious counter-example being glassy materials. Polymorphs are also example where the thermodynamically favored structure is not formed, for example because another phase crystallizes more rapidly

That's absolutely correct. An amorphous forms when the rate of cooling, or solidification is higher than the rate of crystallization. But what I am wondering about here is that polymorphs (not amorphous) are given enough time to re-arrange themselves, why have they chosen to be in a metastable form (high energy state) when they can crystallize into the crystalline form (lower energy state)? lets not assume any special conditions (pressure or other factors) during recrystallization.

@Error404 There's not a simple unique answer. Maybe an example of a similar phenomenon is crystal growth: although it is a crystallization process, not all facets of a crystal grow at the same speed, giving rise to specific crystal shapes depending on the material (and synthesis conditions).


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Why sponge can't absorb mercury
Recently I have watched a video demonstration with a sponge and mercury and the sponge can't absorb mercury like it does with water.
Why is that?

I'm a bit puzzled at the usage of the “everyday chemistry” tag, because mercury is quite toxic. I don't think many households still have mercury in them, nor should they!

@F'x : LOL! then you should remove it from post.

@F'x That explains this persistent twitch I have all of the time!

@jonsca You'd be amazed… Until 1950's, people used mercury-based teething powders for babies, because the mercury made  their gums softer (this book is a very interesting history of poison)

@F'x have you ever heard about thing that measure temperature?

@SenadMeškin yes, and I've also heard that mercury thermometers are being phased out (not recommended for home use), and outright banned in some countries

@F'x same as cigarettes :)

I think the key feature is the surface tension of mercury-air which is extremely high (about 7 times water-air which is already high), combined with the chemical nature of the sponge, which is typically cellulose.
Since cellulose is a big polysaccharide it has a reasonably favorable interaction with both water (due to the OH groups) and alcohols and alkanes (due to the carbon chain). This causes the contact angle of water, alcohol and alkanes with the sponge to be low (at least lower than 90 degrees) thus allowing capillary action to suck the liquid into the sponges pores. The metallic nature of mercury has an unfavorable interaction with the sponge thus resulting in a contact angle above 90 degrees. This will not allow capillary action because the cosine of angles above 90 degrees is negative, therefore the sponge cannot soak up the mercury.
A sponge made out of metal (silver is common I believe) would be able to soak up the mercury because it has a favorable interaction in terms of surface energy.

A simple way to summarise this would be to observe that mercury doesn't wet cellulose but water does.

You are asking wrong way. Sponge does not absorb almost anything. Apples, sand, dust, wires,...
The point is: Why can sponge absorb water?
Because the interaction of water and the sponge are similarly or even more favorable than interaction inside water. The nature of this effect is capilary action.

I don't see how “apples, sand, dust and wires” can be compared with water and mercury, as they are not liquids… A sponge adsorbs water, ethanol, alkanes, acetone, and most other liquids. Mercury does stand out.

You exactly got the point. Molten metal is technically a liquid, but stands out. As will probably Gallium, if somebody could try. Maybe also ionic liquids? All and all, capillary action is the answer. I probably overexaggerated around the important point, which is why it works for water, not why it does not work for mercury.

It's not about capillary action. It is about whether mercury will wet the surface. Only then is capillary action important.

source: Stack Exchange

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How can I formulate a rate equation using a solid catalyst?
When formulating a rate a equation two experiments must take place: one in which the reactant concentrations are varied, and one in which the catalyst concentrations are varied.
If I am given a sample of a solid catalyst, such as $MnO_2$, how can I change its concentration? Because I can't dilute a solid and increase the volume, which is how I'd do it for a catalyst in solution.

You will perform the reaction inside a container or solution of some particular volume. This means that you can change the catalyst concentration inside that volume by adding more or less catalyst in different experiments. So you don't increase the volume of the solid (i.e. change its density), you change the mass of solid inside the reaction volume.

The rate equation is actually defined for the activity of each reactant and product. In most cases, the activity can be approximated to the concentration.In the case of solid catalysts, you'll have to use activity. I'm not sure how the actual value of activity is calculated, but the solid catalyst's activity is increased when you increase its surface area. That can be one way to change its effect on the rate equation.This link has some mention of the activity based rate equation and some examples.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Torsion angle symbol
I'm reading up on molecular modelling and have come across some terms I should probably have learnt a long time ago but always scared me off a bit with the Greek letters. I guess I'm a big boy now so it's probably time to clarify...
The textbook uses the lowercase symbol omega ($\omega$) for torsion (a.k.a. dihedral) angle while Wikipedia gives the uppercase, $\Omega$, plus both upper and lower case phi ($\Phi$, $\phi$) and psi ($\Psi$, $\psi$) in its description which from reading I'm starting to understand designate particular geometrical cases from the article, though it looks like it might be a little difficult to remember at first.
My textbook also gives in the list of symbols and physical constants [lower case] tau, $\tau$, but I can't see any mention of this on the Wiki page.
Firstly I wanted to ask is the use of upper/lower case totally arbitrary, and secondly is tau just a non-specific way of expressing torsion angle compared to the other 3 (which is the only use I can imagine for it?) Or is this use totally confined to molecular modelling?

From personal experience as a computational chemist, we have always use the symbol $\tau$ when referring to a torsional angle.  Also, a torsional angle is not equivalent to a dihedral angle (even though a lot of people interchange the two).  Whatever symbol you use, just be sure to define it explicitly somewhere and you're golden. Secondly, as far as I'm aware, casing is more situational than anything else.

Right, I've not been introduced to a formal definition of either, I had a working understanding of torsion angle from a diagram of ethane in a staggered conformation and just Googled 'torsion angle', the first result for which was Wiki:Dihedral angle... Care to explain the difference as you see it? (or anyone else reading this!)

I've written out about 4 halfway attempts trying to explain this only to realize its really difficult because I have never really given it much thought.  I can give an example however.  Imagine a CH$_4$ molecule with atom labels H1,C2,H3,H4.  The angle between the following four atoms H1-C2-H3-H4 would be a dihedral angle.  Imagine Glycylglycine (Glycine dipeptide) and you wanted to describe the cis- or trans-like nature of the peptide bond.  You would measure the angle C-C-N-C which would be a torsion angle (where the C-N bond is the peptide bond).  I could explain this better with pictures.

For the record, I just went and asked a couple of professors still lingering around here and they were both unable to explain it to me.  Will follow up on this once my boss is back in town next week.

It is a habit to assign greek letters (most often lowercase) to angles. In case of "normal" angles, it's mostly $\alpha, \beta, \gamma$, whereas for dihedrals some higher letters, as you mentioned, $\phi, \psi, \omega$. As noted in comments, whatever you use, clearly state what you mean.
The difference between dihedral and torsional angle is more conceptual, then formal. In torsion you discuss rotation around a bond, dihedral is general angle of two planes, as mentioned in the $\ce{CH_4}$ example. The formal expression for both of them is the same, indeed.
To be noted, there are some well known widely used dihedral angles, where you should stick to the common notation. The most prominent example is the protein backbone arrangement, see Ramachandran plot. In this case you should not invent your own notation.

"In torsion you discuss rotation around a bond" I didn't know this was the actual defining factor for a torsion even though its how I've always implemented it.  A torsion is still the angle of two planes no?

You are right, torsion angle is measured as angle of planes ABC x BCD, where you expect bond B-C to be the one, which rotates. I.e. central bond in butane.

I liked your answer.  I just never realized it was not a 'formal' definition.


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What is an algorithm for making a net neutral molecule from a set of guessed positive charges?
Given a set of partial charges for the atoms in a molecule, which are educated guesses but do not add up to zero, we would like to zero them in an intelligent way.
For example, if the net charge is negative and we see a +0.1 atom bonded to a -0.9 atom, we should probably shift the -0.9 towards zero to make that section of the molecule more neutral.
There is also chemical information which can be used, eg, keep oxygen atoms positive and don't let the charge on hydrogen be greater than 1.0.
Is there a name for this class of problem? What algorithms can be applied?

I'm assuming you are talking about charge equilibration. Google "charge equilibration" and you will find a bunch of papers on the subject.

Although you did not mentioned specifically, it is likely that you want to perform some Molecular mechanics force field calculations. The charges are important quantity, so you need more than educated guess to obtain usable results. Nonetheless, the problem of charge equilibration is solved by "Polarizable force fields".
It should be noted, that they are often more expensive, less reliable and less accurate than standard FFs.


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Competing resonance and inductive effects in a substituted benzene
A molecule of phenol is much more inclined to under go an electrophilic substitution reaction than a molecule of benzene because the $\ce{-OH}$ group is highly reaction favoring. From what I understand, this is because when the $\ce{-OH}$ group is attached, it contributes a resonance structure involving a double bond between the oxygen and the ring, which stabilizes the intermediate arenium ion.
However, from what I also understand, oxygen is also a highly electronegative atom and therefore inductively draws electron density away from the benzene ring. Despite these two competing effects (electron donating via resonance and electron withdrawing via induction), the overall effect of the $\ce{-OH}$ group is electron donating. This also applies to $\ce{-NH2}$ group.
Am I right in concluding that resonance effects are stronger than inductive effects, if not in general, then at least for electrophilic substitution of substituted aryls?

Considering benzoic acid, I would say no because the carbon of $-COOH$ can freely rotate but still be in its plane, but still we say that it is a meta directing group. When I do get the time I will post my answer.

Often, but not always, mesomeric displacement leads to a shift in prevailing over the inductive effect.
Pi-electrons are at the peripheral orbitals. Association with the nucleus is less strong than at the sigma-electrons. The ionization potential of pi-electrons is smaller and chemical bond is more polarizable. Therefore, the dipole moment associated with the mesomeric effect can prevail over the dipole moment associated with the inductive effect.
However, there are groups in which the inductive effect prevails, such as halogens.
Also, you can read about hyperconjugation. Hyperconjugation can have influence to the mesomeric and inductive effects too.
Source: Reutov  O.A., Kurts A.L., Organic Chemistry, vol. 1 of 4, ch. 2.2.2, MSU

thank you for the informative answer. However, my prof does not believe in the hyperconjugative effect and only considers the inductive effect - which often seems to have the same effect as the hyperconjugative effect. Also can't the hyperconjugative effect be lumped under resonance/mesomeric effects?

@Dissenter This effect (hyperconjugation or sigma,pi-conjugation) - something between a mesomeric and inductive effects. It doesn't have a strong influence in bond C-H, but bond between carbon and metals we can't describe without it.
Mesomeric effect of halogens is very different from all other substituents

Am I right in concluding that resonance effects are stronger than
  inductive effects, if not in general, then at least for electrophilic
  substitution of substituted aryls?

No, this is not a general phenomenon.
Consider the case of electrophilic aromatic substitution of fluorobenzene.  Fluorobenzene undergoes electrophilic nitration roughly 10 times slower than benzene, yet it strongly directs ortho and para.  Nitration of fluorobenzene produces primarily ortho (13%) and para (86%) product, only a trace of meta product is observed.
The fact that that substitution occurs primarily at the ortho and para positions signals that the fluorine has a +R effect.  The fluorine is donating electrons through resonance into the benzene ring - but only to the ortho and para positions; resonance structures cannot be drawn that donate electron density to the meta position (this applies to both the ground state and transition state resonance structures).
Fluorine is much more electronegative than hydrogen, so inductively fluorine will remove electron density from the ring (-I effect).
Since the reaction rate for electrophilic nitration is decelerated in fluorobenzene compared to benzene, this indicates that there is less stabilization (less electron density to stabilize the positive charge) of the transition state in the fluorobenzene reaction than in the benzene raection. If the fluorine substituent is donating electron density through resonance, but overall the electronic stabilization has decreased, then the -I effect must overwhelm the +R effect in this case.
There is a continuous rebalancing of inductive and resonance effects as we move through the series benzene, toluene, dimethyl aniline, anisole and fluorobenzene.  With dimethyl aniline and anisole, the resonance effects far outweigh the inductive effects, but by the time we reach fluorine the inductive effect outweighs the resonance effect.

what about cases other than halogenated benzenes? I believe that the general trend of resonance > inductive effect holds for these cases (at least according to my prof).

Yes, the general trend holds for the common groups like alkyl, OR and NR2.  But obscure groups like -SR (maybe when R=CF3), pentafluorophenyl, etc. may also violate the "general" trend of resonance effects outweighing inductive effects.

"resonance structures cannot be drawn that donate electron density to the meta position" ... how do resonance structures correlate with reality? Why can't more e- density be delocalized to the meta position? Aren't all the p-orbitals connected anyway?

@Dissenter  1) Resonance structures correlate pretty well with reality, that's why they still find use.  2) All p-orbitals are connected, but some have very small coefficients in a given molecular orbital, effectively disconnecting them.  Perhaps you can run an MO calculation on toluene at school, I suspect some orbital(s) close to the HOMO would have  smaller or oppositely signed coefficients for the meta carbons.

What's the physical reasoning behind small coefficients?

@Dissenter  The wave function is telling us that there is not much charge (or opposite sign charge, depending upon the sign of the coefficient) at that position.

How can we rationalise the relative lack of electron density at the meta position without resonance structures and save functions

@Dissenter  You can't.  Chemists brominated toluene and found that the o- and p-positions were substituted.  They repeated the experiment with nitrobenzene and found that the meta isomer predominated.  They realized that inductive effects alone could not explain their observations, so they looked further and applied\developed resonance theory.

Who downvoted this answer? I gave this answer the bounty because I thought it was very through and well-written; if the downvoter has any comments/suggestions I'd be open to hear them (and I'm sure ron would be too).

(I thought the previous answers were incomplete, mainly because previous answers have talked about halo arenes, particularly about fluorobenzene)
Your question reminds me of a doubt I had once. Consider the two structures shown below:
Can I say these two are geometrical isomers? It sounds a pretty dumb question, but here is the thing. One could argue that whenever we show that the $-COOH$ group is meta directing, we draw resonating structures that have a double bond between the carbons of the $-COOH$ and the carbon of benzene to which the group is joined to. Therefore one could say that the rotation of the carbon is restricted and hence are geometrical isomers.
But the thing is, that the rotation of the carbon in $-COOH$ is actually free(not exactly free, since it has to be in the same plane as the rest of the benzene, since it has $sp^2$ hybridization which confines it in a plane) and the $\pi e^-$ of the $-COOH$ group remains with it and doesn't get localized in the benzene ring. But this leads to a very natural question here- then how can we explain that $-COOH$ is meta directing? By inductive effect. Or at least that is what my teacher said.
-R (or -M) is shown by a group if a more electronegative element is attached by a double bond directly then it decreases electron density at ortho and para positions wrt meta and hence the group is deactivating. Considering the case of benzoic acid, electronegativity of oxygen plays a vital role , mainly because it attracts the $\pi e^-$s of the ring and decreases electron density over the ring. Although we can say that resonance does take place (which doesn't contribute much to explain why $-COOH$ is meta directing, as explained above), but the more contributing reason here is that oxygen pulls the electron density towards itself making the ring deactivated. So therefore in this case inductive effect plays a vital role here, even more important than resonance.
It could also be explained as to why $-COO^-$ is activating by the above. Since a -ve charge over a very electronegative element (oxygen) makes it very stable and kinda "calms it down" from attracting more electrons, therefore oxygen won't attract the electrons of benzene so much therefore relaxing the electron density of benzene a little bit and hence make it ortho and para directing.
Hence, as we can clearly see the electronegativity of oxygen here plays a vital role in making $-COOH$ meta directing and $-COO^-$ ortho and para directing. This is the best example to disprove your conclusion because here the overall effect by the oxygen is attracting electrons demonstrates a case where inductive effect is more dominant than resonance.

I fail to see how these two molecules you draw are different and how they refer to the point you are making, especially when you consider that the carboxylic group to be freely rotatable (It's about 7 kcal/mol, so this is an acceptable assumption). However, in its minimum conformation, the $\pi$ system is of course delocalised over the whole molecule, withdrawing more electron density from the ring.

^ That is my exact point @Martin. The carboxylic group is freely rotatable. When we draw the resonant structures of the system and show the delocalisation of the $\pi e^-$ of the carboxylic group over the ring, there is a $\pi$ bond b/w the carbon and benzene, and one could say that the structures I drew are geometrical isomers. But they aren't. There is no difference between the structures I drew. I explained it in my answer. This was a doubt I had before.

But how does that explain that the carboxylic group is meta directing? If it was only the inductive effect, then clearly the COOH must be drawing a lot of electron density from the $\sigma$ system. Hence the para position would be least affected by this group.

@Martin I have also addressed this in my answer. To some extent, the $\pi$ electron of the carboxylic group does go in the ring by resonance. It is just that the more dominant reason for the $-COOH$ group to be meta directing is the pulling of $e^-$s by the oxygens. I do not deny to the fact that the $\pi$ electron of the carboxylic group doesn't get delocalised.

And who down voted this? What is wrong here? I'm open to your suggestions.

Well, I did not downvote, but It is really hard to follow your train of thought. I still do not understand what your point is. And don't be offended by downvotes, I get a couple of them every now and again.

Did you at least get my point @Martin ? Then you might help and edit the question.

I downvoted for two reasons: (1) The compounds shown do not have ortho, meta, and para positions. They only have four positions. (2) Resonance DOES account for why a carbonyl is a meta director. Both resonance and inductive effect play a role. The question is looking for instances where resonance and inductive effect are competing. The carboxylate analog could be an effective example, but that hasn't been demonstrated in this answer. If that can be explained more clearly, and the irrelevant part about the diacid removed, it could be a useful answer.

I guess you didn't get my answer. I had a doubt in which I learned that the rotation of the carbon in carboxylic group is actually free and not restricted because the pi electron of the carbonyl group DOESNT delocalised in the benzene ring. This is a hard fact to digest and that is why I used my doubt as an example to prove my point. Moreover, when I finished explaining my doubt and why was I wrong, then I was talking about benzoic acid (I was referring to ortho, meta and para positions in benzoic acid ((1))).

Although I agree to the fact that resonance does contribute to carbonyl being meta directing, but the more DOMINANT factor for it being deactivating and $-COO^-$ being o-p-directing is inductive effect and not resonance. Here the effects aren't competing with each other, but the inductive effect is more dominating here. It wasn't necessary to give only an answer which showed inductive and resonance effects competing with each other. I do hope that now I'm clear enough @jerepierre ?

@Rohinb97 The digression about the diacids greatly confuses your answer. I don't argue that the carbonyl pi bond will not donate electrons to the aromatic ring. The relevant resonance structure (to explain the carbonyl as a meta-director) is the ring donating to the carbonyl. Just because there is free rotation doesn't mean that the resonance contribution is zero. In any case, what is the evidence that inductive effect is dominant over resonance? The question poses groups with opposite effects. We can look at the product ratios to determine which effect is dominant.

I'm sorry if I wasn't clear enough. But I have edited my answer and I hope you do get my point. And to answer the above, the statement that the ring donates the electrons is technically wrong, it's the carbonyl group which attracts it, but anyways, my point is that the inductive effect is more dominant than resonance effect because the pi electron of the carbonyl group doesn't get delocalised in the ring (because if it had, then the above structures drawn would've been geometrical isomers) and hence resonance takes place in a small amount. But still we say that $-COOH$ is meta directing. Why?

Because the oxygens attract the electron density of the ring and therefore makes it deactivating. Since resonance does take place(very little), it increases electron density at meta wrt ortho and para and hence carbonyl is meta directing. The above can also be supported by the fact that $-COO^-$ is an activating group. A -ve charge over an electro-ve element makes it more stable and decreases it's electro-vity to some extent. Hence the major effect responsible for carboxylic being meta directing is nullified, and hence is ortho and para directing.

I've written my whole answer again in these comments now @jerepierre. I do hope you get the point I've been trying to make.

@Rohinb97 As stated in the question, the hydroxyl group is inductively withdrawing and resonance donating. Based on the product distribution of an EAS reaction of phenol, we can see that resonance dominates over the inductive effect. You assert that the directing effect of -COOH is primarily due to inductive effect because of free rotation which disrupts resonance to the ring. However, the C-O bond in phenol is freely rotating as well, yet the resonance effect still dominates. This tells me that free rotation alone is not enough to conclude that inductive effect can dominate over resonance.

@jerepierre I guess you are right. Didn't think this. Thanks. But it is a fact that resonance doesn't take place (so much) in benzoic acid. The pi electron is tightly held up by the two oxygens and is in resonance with the lone pairs of the next oxygen. The pi electron of the carboxylic group doesn't delocalise in the ring. But still it is meta directing. Inductive effect is more dominant in this case.


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Free automatic organic synthesis Linux command line program
I'm trying to find a free automatic organic synthesis Linux command line program which take mol or smile files as input. I have random molecules and I'd like to know how difficult above a baseline they will be to synthesize. An implementation of synthetic accessibility and not a full step wise synthesis program would also be acceptable. Thanks for the help.

I would be very surprised if such a program exists for anything but a small subset of similarly structured molecules.  Additionally, refer to the help section with respect to questions that are purely of a simulation nature.

A general solution to this problem, by the way, would be literally worth zillions of dollars.

And probably a Nobel Prize (with its additional 8 million Swedish Kronor).

@bobthechemist awh well then you can successfully be surprised as I found a solution.

The SAscore is now implemented in RDkit and available for download.

If you're wondering what "SAscore" is, see the following open-access article: "Estimation of synthetic accessibility score of drug-like molecules based on molecular complexity and fragment contributions" http://www.jcheminf.com/content/1/1/8


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Why is there a difference between O2 and B2 sigma 2p molecular orbitals in diagrams?
The molecular orbital diagram for $\ce{O2}$ says that the  sigma 2p bonding molecular orbital is lower in energy than the pi 2p bonding molecular orbital. Why is this not the case in the $\ce{B2}$ MO diagram?

It is, though, isn't it?

This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same symmetry and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.
However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the energy difference of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is less, facilitating intermixing.
You can find this topic in any higher level chem books. Hope you got my point; thank you.

Could you provide an example or link to a book that contains this information?

https://www.google.co.in/?gws_rd=cr&ei=7H7XUt6XNsKUrgftrIGADg#q=mixing+of+molecular+orbitals

Boron radius - 80 pm, Oxygen - 65 pm. Funny-funny-funny

exactly oxygen is not bigger in size....


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Are there any tricks I can use to pull SO2F2 out of the air?
A massive apartment complex near where I live is being fumigated with Vikane gas (sulphuryl flouride, $\ce{SO2F2}$).  The process will take a long time (weeks) because there are a lot of buildings, and their tents appear to have rips in them, clearly visible: several neighbors have complained of symptoms of exposure to the gas, and I'm worried about it in my home.  Obviously we should try to get them to stop / etc., but that's a matter of law and politics not meant for a chemistry site.
In the meantime, what I would like to know is if there is any safe way I can use to pull the gas out of the air if it should waft its way into my home.  My understanding is that activated charcoal will not work for this molecule.  I've heard that either strong bases (usually photo-developing chemicals are mentioned) or heavy water vapor reacts with the gas; but I'm not sure what that reaction is, or if it would result in something better or worse than the gas itself.  I've tried Googling but with just a high school chemistry exposure some decades ago, I can't make much headway.  Please note that I do not have access to any chemicals or machines I can't readily buy in a nearby store.  My goal is to turn it into some liquid or solid, or bind it to such, so that I can't breath it in.

Pumping all the air in your house through a system is going to take a fairly serious set of fans, much less continuously running a system to try and remove the gas from all the surrounding air as quickly as it diffuses in.

I guess I should have clarified; I realize it's unfeasible to remove all trace of the stuff; I just want to mitigate damage as much as possible.  For instance, they're also spraying Termidor (fipronil), which has carbon chains.  I put out activated charcoal and set a few fans blowing, just to try to pull out of the air what it can.

Without some sort of high-performance fan/filter system, it's going to be seriously difficult to mitigate things to any significant extent. Here's a kicker for you: the gas you refer to is significantly denser than air, and will tend to sink. Unless you spend a lot of time near the ground (or have small pets), putting lots of fans without a dedicated filter system will probably kick the gas up and increase your exposure.

Sulphuryl Fluoride is chemically fairly stable, considering it's a pesticide - toxic effects are caused by chemical reactions which means they can't be inert.
However, according to the Vikane documentation, it's not UV stable. UV lighting is fairly widely available, but TBH a sunny day will do far more.

So is the chemical stable or unstable? You contradict yourself.

@Dissenter: Rad my answer again. Chemically stable, not UV stable. There's a reason I qualified both uses of "stable".


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Calculation of Atomic weight percentage in AAS
I am currently doing doping studies in a system similar to this ($\ce{A_{x}B_{1-x}}$) and I have to find atomic weight percentage of $\ce{B}$ element using atomic absorption spectroscopy(AAS). I have prepared different PPM of solutions according to the weight percentage of $\ce{B}$ element.
Example to prepare 10 atomic weight percentage of $\ce{B}$ in $5~\mathrm{ppm}$ solution, I have used this equation to calculate molecular weight:
$$\text{Actual mol weight}=\text{mol.wt}(A)\cdot x + \text{mol.wt}(B) \cdot (1-x).$$
In this case $x = 0.9$ and $1-x = 0.1$.
From this, I have prepared $5~\mathrm{ppm}$ of a solution of the doped material assuming that $10\%$ of $\ce{B}$ is doped completely (but in reality only very low percentage is doped).
I got the below data from AAS:
B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)
    10          5                 1.5
    20          5                 0.9

From the above data, I have to calculate actual atomic weight percentage of $\ce{B}$.
My colleague has suggested this method in order to find $\text{Actual atomic weight percentage}$
$$\text{act.At.Wt.%} =\frac{\text{Obs.AAS.Conc(µg/ml)}}{\text{Sol.Conc(ppm)}}\cdot \text{At.Wt.percentage.of.}\ce{B}$$
Using this I got the results below.
B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)   Actual B (At wt %)
10                5                  1.25            2.5
20                5                  0.9             3.6

But I have doubt in this method as you can see in the table the observed AAS concentration for 10 and 20 At wt % $\ce{B}$ is 1.25 and 0.9 (µg/ml) respectively. You can see that the AAS concentration of 20 At.wt % is lower than 10 A.wt %. But when we calculate actual $\ce{B}$ At.wt.% we can see that the 10 A.wt %  is lower comparing 20 A.wt %. Can you tell me whether the equation suggested by my colleague is correct or not? If not can you suggest any other method in order to find the actual atomic weight percentage?

Relating mass fraction (ppm) with concentration (μg/ml) without knowing the density of the solution is impossible. You have to create your solutions with given concentrations (μg/ml). You have to weight a given mass of your substance and fill with water to a certain volume. You have to use volumetric flasks therefore. Consider the rules for applying volumetric flasks for getting correct results.

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text:
Oxygen or carbon with a positive charge? (comparison of canonical structures)

Which of the contributing structures of the resonance is more stable?

I'm watching a video lecture by a professor of my college where he puts this question to the class. The class unanimously says B. At first I thought they were wrong, but then the professor agreed with them.
Is this true? The reason he gave was that oxygen's octet would be complete in B, hence it contributes more. But I don't understand this.
1) Isn't the octet complete in A too? It has 2 covalent bonds - 4 electrons (2 from sharing) and 2 lone pairs. Adding up to 8.
2) Oxygen being more electronegative than carbon, would pull the electrons towards itself, leading me to think that it would be more comfortable with a lone pair rather than another bond. Also, won't it prefer to lose out that '+' charge which the bonding's burdened onto it?
Am I wrong?
Another thing which made me believe that Prof. may have made a mistake while talking about the more relevant contributor, is that when he added Br$^-$ to the compound, he added it to the canonical structure A. And don't you explain reactions with the most relevant contributor, i.e. B?

Answer
The presence of an extra bond in the canonical structure B, along with the completion of the carbon valency makes it more stable than A. The reaction with Br$^-$ is best explained looking at A, though, because the incoming nucleophile ends up bound to carbon, not oxygen.
Thanks, everyone!

I guess the octet is complete in both the cases. But for the stability of these resonance structures, points to considered in decreasing order of importance are higher number of bonds; complete octet; charge separation; +charge on electropositive and -charge on electronegative elements.

So you say that option (B) is more stable?

While adding Br$^-$ to the compound, why do we take the reaction further by adding it to compound (A), then? :/

Between A and B, look at how the octet rule is satisfied for the carbon atom (and not the oxygen)…

@mikhailcazi That is because, relatively, as compared to other carbons, the one with a positive charge in one of the resonance structure is slightly positive and hence more prone to attack of nucleophile $\ce{Br-}$. It doesn't add on the oxygen because the lone pairs of oxygen in both the resonance structures repel the incoming nucleophile, which cannot form a bond with oxygen.

@mikhailcazi in general, oxygen does pull electrons towards itself in most resonating structures but can also donate an electron pair (acquire positive charge) if there is a positive charge on an adjacent carbon.

@SatwikPasani Just for the sake of thoroughness, an oxygen pulls more electron density toward itself, not actual electrons themselves.

@SatwikPasani okay, so option (B) is more stable because 1) extra bond 2) carbon octet complete. Thanks! This helped.

Octet is not complete for carbon in (A). Option (B) is more stable with more octet satisfying atoms,

Which of the contributing structures of the resonance below is more stable?

Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities.
What the question is asking is "Which contributor is more important in describing the structure and behavior of the hybrid?" Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).

The difference between the two structures is just the amount of space available to the lone pair. If you think about it in terms of the particle in a box approximation, the path length for that lone pair is much longer when in the double bond state, so it will be a lower energy state.

Structure A does NOT have on octet on the carbon that has the plus charge. Structure B has an octet on every atom. Thus, structure be is more stable.

RULE 1: resonating structure with more number of bonds wins and (B) has 1 more bond than (A) hence it wins

Structure A is more stable. Reasons are:
1) The O has a complete octet with 2 electrons per bond around it(based off of octet rule, not formal charge or oxidation state) + 2 lone pairs.
2) The O is not positive. O prefers to be negative if charged at all because of its electronegativity being so high.
3) Carbon is not so determinant on charge. It is fine being positive, negative, or neutral.
4) Each carbon that isn't positively charged has its octet from C-C-C or O-C-C + 2 H's.

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text:
Stoichiometry: efficiency
My book gives me the following problem:
"A mixture of $125.0 g$ $\ce{N2}$ and $32.0 g$ $\ce{H2}$ reacts to $36.5 g$ $\ce{NH3}$. Calculate the efficiency."
My method to do this is - one I learned from my teacher, to put every given molecule's amount of substance (moles) in a table with 3 steps: start (S), gone (G)/formed (F) and rest (R).
$125g\:N_2 \leftrightarrow n=4.462$ moles
$32g\:\left(3\right)H_2\leftrightarrow n=16.13$ moles
\begin{matrix} &N_2&+&3H_2&\longrightarrow &2NH_3\\S&4.462&&16.13&&\\G&4.462&&13.386&F&8.924\\R&0&&2.74&&8.924\end{matrix}
($\rightarrow$ abundance of $H_2$)
Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?

I don't know what is wrong with your calculator, but when I divide 32.0 by 2.00, I get 16.0, not 16.13.

@LDC3 true, while $M_{\ce{H2}}\approx2.02$ g/mol and therefore it should rather be $15.84$ mol.

Yes, that is completely correct. Let me summarize the approach.
If you look at the chemical reaction:
$$\ce{N2 + 3H2 \rightarrow 2 NH3}$$
You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$

The formula weight of $NH_3$ is 17, not 16.

@LDC3 Oops ... corrected it


text:
Why does equilibrium constant change when the reaction is multiplied by a number and nothing in the real has been changed?
As far as I know, a large equilibrium constant shows that the a reaction like the one below  wants to complete:
$$\ce{B <=> A}$$
If A and B are gases, then K is:
$$K_1=\frac{[A]}{[B]}$$
but when I multiply the reaction by 20 , K is
 $$K_{20}=\frac{[A]^{20}}{[B]^{20}}$$ while the same number of moles of B are consumed. If $\small K_1 = 10$ then $\small K_{20} = $$10^{20}$ so the reaction must be approximately complete. Why does this happen while the reaction materials didn't change?

First of all, you switched the numerator and denominator in the definition of $K$. If $\ce{A <=> B}$ , then $K=\frac{[B]}{[A]}$. Products over reactants.
Notice that, if $K$ is large, then mathematically it is necessary for either the numerator to be large and/or the denominator to be small. Therefore, qualitatively, if $K$ is large, then you either have a lot of products, or little left of reagents, both of which imply a reaction which occurs close to completion.
Now here's the problem - we were speaking qualitatively. Your first sentence ("a big equilibrium constant shows that the reaction wants to get complete") is only a rule of thumb. It happens to work well because in reactions where equilibria descriptions are quantitatively reliable/exist at all, the stoichiometric coefficients are relatively small when written with the lowest integers (using the lowest integer coefficients is almost always done). The rule of thumb is borne out of repeated calculations under these implied conditions. They are not the case for your second equilibrium example, so such a rule should not be expected to still be valid.
Indeed, though $10^{20}$ is an impressive number at first, when you actually go do the calculations to find the relative amounts of reactants and products, you will at some point have to take its twentieth root, promptly turning it back into a much smaller number. Another way to think about it is that with such large stoichiometric coefficients, even a small change in concentrations is amplified enormously by the large exponents and has a huge impact on the value of the reaction quotient; a small increase in products will go a long way to reaching the equilibrium condition.

somebody said this example : A+B--->C+D then if k = 2 when we solve $$\frac{x^{2}}{(10 - x)^{2}}$$ then 5.85 moles of the reactants are consumed . for 100A+100B--->100C+100D , k = 2^100 and if you solve $$2^{100} = \frac{x^{200}}{(10 - x)^{200}}$$ , the answer will be the same . so the rule is false!

forgot to say that moles of A and B are before the beginning 10 , I think it is wrong because they had multiply first moles by 100 so we would have 1000 moles of each reactant , Am I wrong?

@user2561 When you multiply the global reaction by any number, you don't multiply the mole amounts of any species you're trying to solve the problem with; that's an entirely different piece of information related to the problem itself (a boundary condition, as they say), not the reaction . If you did everything correctly, the answer must be the same in either $A→B$ or $100A→100B$ if you use the same values of $[A]$ and $[B]$ in both situations. If the answer is not the same, then that's an indication you did not properly change the value of the equilibrium constant.

You're a bit muddled up.
Let's say I have the following reaction:
$$\ce{A <=> B}$$
This signifies that one molecule of $A$ reacts to give one molecule of $B$. The molecularity of this reaction is 1.
The equilibrium constant (say $K_{eq}$) is defined as the ratio of concentration of products to that of reactants.
i.e. $K_{eq} = \frac{[B]}{[A]}$
If a reaction is written as:
$$\ce{2A <=> 2B}$$
This signifies that two molecules of A react with each other to produce two molecules of B. It is not the same as our first reaction!
The equilibrium constant (say $K'_{eq}$) will be:
$$K'_{eq} = \frac{[B]\cdot[B]}{[A]\cdot[A]} = \frac{[B]^2}{[A]^2}$$
The equilibrium constant is used to get an idea of how much the reaction is favoured. If $K = 13$, then the concentrations of the products are thirteen times more than the concentrations of reactants. Therefore, any value of $K$ greater than one can signify that the reaction wants to proceed forward.

There's an issue with your last sentence. The equilibrium ratio between products and reagents is unchanged after multiplying a reaction by any number. Also, technically, the reaction is exactly the same after multiplying the equation by a number. To say otherwise is to suggest that something changes in the kinetics/mechanism. Changing the relevant equilibrium constant is just a trick that ensures proper bookkeeping, and does not alter any parameter of the reactants/products/reaction.

I was told by my professor that saying 2A -> 2B would imply that 2 molecules of A react to form products, although multiplying a number to the equation can be used to compare it to other equations for adding two equations, or comparing stuff like $\Delta H$. Because it quantitatively doesn't make any difference :/ Is that wrong?

The implying part is a problem. The total chemical equation for a reaction actually does not in general convey any chemical information, in its purest form. All it does is count the atoms before and after reaction, making sure they're in the same amount and type. But we often try to add extra chemical meaning, and that can result in tricky situations. The reaction mechanism steps are what should tell you how a reaction works, and in general the mechanism has more than one step and does not involve all species interacting at once, as may be implied by the total chemical equation.

For a reaction $A→B$, we can imagine figuring out that it proceeds by a single step where two molecules of A react to produce two molecules of B. Thus, while $A→B$ is mathematically correct, it does not represent the true nature of the reaction. If you multiply the equation by two to get $2A→2B$, it is still mathematically valid, and you could say it now carries the actual mechanism. But the point is that this is not necessary, as the total equation usually isn't meant to represent the mechanism in the first place. The total equation is fundamentally just a mathematical equality, LHS = RHS.

@NicolauSakerNeto Oh, okay, I get what you mean. Does that mean that my answer is incorrect?

Not really, it should be fine. It's a subtlety, and I think any viewer on this site usually reads both the answer and the comments for it, so you're good!


text:
Why are more easily oxidized elements more reactive than less easily oxidized elements?
Why do the more easily oxidized elements displace the less easily oxidized elements?
Question inspired by Activity Series of Metals

The title is asking "why are more reactive things more reactive?" In fact the whole question seems a little like that.

Ease of oxidation is how easily it loses electrons. Electrons are the basis of chemical reactions. If chemical A is more easily oxidized than B, then B is a relative oxidizer to A, so in terms of the activity series, the elements that are more easily oxidized can react with more chemicals, since they will be able to act as reducing agents to more chemicals.


text:
How do you draw (E)-1-phenyl-1,3-butadiene?
I'm having trouble drawing this molecule because I don't understand how the functional group Phenyl (by -ene suffix) can also be a substituent (by 1-phenyl). Also, what does the 1,3-butadiene mean?

ManishEarth put together the pieces to understand the connection of the atoms to form the molecule.  Yet since the molecular structure requested in the original post is about (E)-1-phenyl-1,3-butadiene, the answer actually is opposite configuration:

By convention, alkenes posses a reference plane orthogonal to the plane of drawing.  In case the alkene is substituted twice by groups attached at the termini of the alkene -- in the upper example a phenyl and an alkyl chain -- these may be either at the same or the opposite side of this reference plane.
For the ease of convenience, two drawings simplifying the substitutents by "R" demonstrate this. In the first example, they are opposing each other, so their relationship is described either by (E) or (trans). The second example shows them at the same side of the reference plane, therefore, this alkene has a (Z)- or (cis)-configuration.

You might ask about the origin of (E) and (Z).  They are derived from German entgegen and zusammen; similar like the Latin (trans) for across, and (cis) for nearby.  Both ways to describe these configurational isomers are found in the literature for one and the same molecule, yet in an article or report, you should stick to either the (E)/(Z) or (cis)/(trans). (And yes, in italics.)

Firstly, the phenyl group is just a benzene with a hydrogen removed:

In other words, $\ce{C6H5 -}$.

the functional group Phenyl (by -ene suffix) can also be a substituent (by 1-phenyl).

The -ene suffix is about the double bonds in the system, in this case it applies to the butane backbone. It has nothing to do with the phenyl.

Also, what does the 1,3-butadiene mean?

1,3-butadiene is a butane backbone with two double bonds at carbon 1 and 3:

The final molecule will look lie this:


text:
Which is most stable conformation of optically inactive butane-2,3-diol?
The most stable conformation of optically inactive butane-2,3-diol is:
(A)
(B)
(C)
(D)
I chose C as the correct answer but, in the answer key it is given to be B. C is the staggered form and the most stable one. It can also have hydrogen bonding between -OH and -H groups. B can also have hydrogen bonding but there would be repulsion between the two Me groups. So, which one is correct?

Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on hydrogen boning, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value.
However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct.
The correct answer cannot be A or D since they are representing a chiral diastereromer.

Is the gauche effect applicable here? Hyperconjugation stabilises 2,3-difuoroethane. So does it increase the stability of butane-2,3-diol as well?

well what if the condition of optical inactivity is removed? then wouldn't it be  A

how is A chiral by the way

If the condition of optical activity were removed, it would be D (fewest gauche). If you are having trouble determining that A is chiral, try drawing it in a different style (perhaps line-bond with perspective wedges). Or, rotate it into an eclipsed conformation and note that while one pair of substituents line up in mirror-symmetry, the other two do not. A is the R,R enantiomer, B and C are the meso compound, and D is the S,S enantiomer.

@PrakharSankrityayan - It sounds as if you are on the verge of asking another question, and the comments are not the place for it.

I disagree with the notion of a net dipole moment of zero being worth it to give up hydrogen bonding. As I learnt, hydrogen bonds are so good on the energy scale that if the solvent cannot hydrogen bond, the molecule will do everything to do so in itself — hence why 1,3-diketo compounds should always be present as the intramolecularly hydrogen-bonded enol tautomer in unpolar solvents.

@BenNorris could you please check once again .A is coming out to be S,R and not R,R.     Please clarify if I am wrong.thanks.

@aniline - actually A) is S,S. Both chirality centers are numbered: 1) OH, 2) other chirality center C, 3) methyl, 4) H. For the front center, these are arranged clockwise, but the H is forward (all three are forward actually). For the rear center, these are arranged counterclockwise with the H back. Convert it to a perspective drawing if you need to.

What do you mean be perspective drawing.. I can draw saw horse structure from it but how do I know which group is above and which is below?

The argument made by Ben can be supported by computational chemistry. I calculated three conformers of the compound.
Conformation B1 includes an intramolecular hydrogen bond, while in conformation B2 the proton was rotated away from the second hydroxyl group. Conformation C is the staggered one. The displayed structures were optimised in the gas phase.

Computations were performed with Gaussian09 rev. D at the DF-BP86/def2-SVP level of theory. Solvent effects were estimated with the polarised continuum model, taking the dielectric constants as $\epsilon=78.3553$ for water and $\epsilon=2.0165$ for cyclohexane. Energies in $\mathrm{kJ\,mol^{-1}}$. The energies of each row are relative to conformer C.
\begin{array}{lrrr}\hline
\phantom{\hspace{3cm}}
                 & \hspace{2cm}\mathbf{B1}
                 & \hspace{2cm}\mathbf{B2}
                 & \hspace{2cm}\mathbf{C}\\\hline
\text{gas phase} & -7.2      & 10.2      & 0.0      \\
\ce{H2O}         & -7.0      &  5.3      & 0.0      \\
\ce{C6H12}       & -7.3      &  8.8      & 0.0      \\\hline
\end{array}
As you can see, the intramolecular hydrogen bond accounts for much more than the repulsion between hydroxyl and methyl moieties. While the dihedral angle in B2 is $\angle(\ce{OCCO},\mathbf{B2})=-75.2^\circ$ - larger than expected -, it is smaller than expected in B1 with $\angle(\ce{OCCO},\mathbf{B1})=-57.0^\circ$.
We can assume that the hydrogen bond supplies about $-12$ to $-16~\mathrm{kJ\,mol^{-1}}$ depending on the solvent, which is in quite good agreement of what is known in the literature.
It is remarkable, that the energy difference between B1 and C in the various solvation states is negligible, but in the open form it has quite a significant impact, i.e. a polar solvent stabilises the open form more than an non-polar solvent. This is somewhat a little oversimplification because the used solvent model is quite crude. To obtain a concise picture one would, of course, have to increase the level of theory and treat the solvent explicitly.
Take home message: If you have two hydroxyl groups in proximity, don't forget that there might be an internal hydrogen bond stabilisation.

Huge "Thank yous" to pH13 for creating this awesome picture!

Didn't quite get the home message. I am just a newbie in opticak isometism so I wanted to confirm. Are you saying that hydtogen bonding internally is decreasing energy of the molecule more than the staggered configuration

Stabilisation due to H-bonding(intramolecular) is far more stabilising factor in this case than the destabilisation between interactions of the adjacent methyl groups. Again this depends on solvent (in aprotic solvents 'C' could possibly be more accurate). If this was from a competitive exam book , say one for JEE or AIPMT (seems very familiar, hence the reference) , then the solvent is assumed to be water, in which case 'B' is correct.


text:
Temperature & time necessary to carbonize all organics
Suppose that I have a cast-iron pot which my awful roommate has vigorously scrubbed with soap and left to soak in soapy water over the weekend. The seasoning has been destroyed and replaced with awful-tasting soapy contaminants.
It seems to me that the simplest and best way to deal with this is to heat the pot over a campfire until only black carbon remains, then clean and reseason.
Not knowing exactly what is in my awful roommate's dish soap, and being a very fussy foodie, I want to be conservative here... so how hot would I have to heat the pot, and for how long, to be quite sure that virtually nothing but carbon remains?
Edit: Compare using the self-cleaning cycle of an oven for similar purpose, reported to work well...

There's a lot of exciting, and not well defined, chemistry that happens with seasoning cast-iron cookware.  Essentially, you are creating a cross-linked polymer film that is hydrophobic, providing a "stick-free" surface.  There are many many websites claiming the "best way" to season (or re-season) cast-iron cookware.  Few of these sites (and I'll let the OP and visitors search for those on their own) are based on science.  (Although I am not going to question tried and true family secrets, passed down from generation to generation.)  One site in particular makes a fairly convincing argument for seasoning that is science-based.  I suggest this approach can help restore your pot.
I suspect what has happened is that the scrubbing did remove the seasoning and exposed base metal to water and oxidation - your pot rusted.  The "soapy contaminants" is probably iron oxide rather than something from the dishwater, although there are such products as metal soaps which are metals chelated to long-chain fatty acids.
As for heating the pot to burn off all the residue - I doubt this is the most effective use of your time.  Scrubbing with acidified water to remove the iron oxide and a long seasoning session with an easily cross-linkable oil should get your pot back to (or better than) original condition.

It actually is just a hypothetical - came up in a Reddit discussion - do you happen to know the answer to the question as posed literally, even though I'll take your advice if I ever need to reseason a pot for real?


text:
Best reagent for reducing a ketone in the presence of an alkene

The following reduction is best accomplished with:
(a) Zn/Hg, conc. HCl
  (b) N2H4, C2H5ONa
  (c) H2, Pd
  (d) All of the above

My thoughts: By looking at the question it is evident that only the C=O group should be affected. H2/Pd is used for hydrogenation which might affect the C=C double bond as well. Option (a) gives the reagents which are used for the Clemmensen reduction which is used to convert carbonyl groups to alkanes. So, I think (a) is the correct answer. Is that right?

The Clemmensen reduction is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction.
Choice (b) is the Wolff–Kishner reaction, which also converts ketones into alkenes. It requires strong base (not acid) and works on almost all ketones.
Choice (c) will react with the alkene also, and so cannot be correct.

The double bond will not retain its position in the WK nor in the Clemmensen. The product will not survive the acidic conditions of the Clemmensen. I have reduced polycyclic, aliphatic ketones by the Clemmensen method. The answer is e) none of the above. If the OP is interested in what can go wrong by these two methods, I will provide suggestions.


text:
Will bridged compounds the undergo SN1 reaction?
$\mathrm{S}_{\mathrm{N}}1$ reaction involves only one molecule in the rate determining step. So, the molecule which undergoes $\mathrm{S}_{\mathrm{N}}1$ reaction should be stable when it forms a positive ion. $\ce{(CH3)3C^+}$ is considered stable due to inductive effect from the three carbon atoms and also hyper-conjugation. But, what about bridged compounds like 1-bromobicyclo[2.2.1]heptane:

When $\ce{Br^-}$ is abstracted the positive ion formed will be stable and therefore this will undergo $\mathrm{S}_{\mathrm{N}}1$ reaction. Am I thinking right? Please help me correct myself.

angle strain prevents this molecule from undergoing $\mathrm{S_N1}$ reaction

Very interesting question… I would give two answers, at different levels:

At undergrad level, I think the carbenium ion formed at the bridge head is less stabilized that “regular” ternary carbenium ions, because it cannot adopt the planar trigonal geometry which would be ideal for it.
At higher level, this is much more complicated: the norboryl cation is an example of non-classical ion…

Bridgeheads carbocations are a violation of Bredt's Rule, which, for the reasons mentioned, are not common.

As already pointed out by previous contributors, 1-bromobicyclo[2.2.1]heptane will not undergo substitution via $S_{N}1$. A fully solvated and relaxed bridgehead cation is unlike to be formed. Whether a short-lived ion pair in a common solvent cage is possible... who knows.
But fortunately, the title compound isn't as dead as it seems: I'm sure I've seen it converted to the corresponding Grignard reagent and then quenched with $\ce{CO2}$.

If this compound can undergo nucleophilic substitution, it can only be via SN1 fashion, as the reverse side attack in the anti-bonding $\ce{C-Br}$ orbital is blocked by the other bridgehead.
The reason that is often given why bridgehead carbon cations are not stable is strain. Strain itself is a concept that is difficult to grasp and it more often depends on what you choose as a reference system. For the sole qualitative purpose the cation is more strained than a non-constricted tertiary cation.
Another reason that is often given is that the carbocation cannot adopt a trigonal planar structure. While this is technically true, the deviation from an ideal trigonal surrounding is quite small; it's just about 25°. (DF-BP86/def2-SVP)

I believe the reason for this instability is a bit different. In $\ce{C+(CH3)3}$ the hydrogen carbon bonds can neatly align with the empty p orbital of the central carbon. Hence electron density can be transferred into this orbital or from another point of view, the positive charge can be delocalised into the neighbouring bonds.

The above is just one example configuration calculated at DF-BP86/def2-SVP. It is not the lowest structure, but one chosen for the purpose of showing the overlap in the hyperconjugation. Since the molecule is not rigid at all and the methyl groups rotate, the stabilising effect will be there most of the time, in this conformation it is easiest to see.
That stabilising effect is not present for the carbon hydrogen bonds in proximity to the bridgehead. Those hydrogens actually point away from the cationic centre, hence overlap is not at all possible. Therefore the charge cannot be stabilised by those bonds.

However, there is a small stabilising configuration between the carbon-carbon bonds and the unoccupied orbital and this is the reason, why bridgehead carbocations become more stabilized with larger ring structure.

Like always there are a couple of different points at play, that help stabilise certain positions, or cease to do so.
So with the right motivation, a SN1 reaction might be possible.

Colourcode
Occupied orbitals are displayed in orange and blue, while virtual orbitals are displayed in red and yellow. Phases are chosen randomly.

Well, t-butyl cation isn't really stabilized by inductive effects only. Interaction of empty $p$ orbital of the central carbon with the $\ce{C-H}$ sigma bonds are really important in that manner. And they are not present in norbornyl cation due to strain.
So: no, it is very unlikely for that compound to undergo an $S_N1$ reaction.

The bridgehead carbon in the pictured compound is $sp^3$ hybridized. Upon solvolysis of the bromide ion, it should rehybridize to $sp^2$. However, this does not occur because the bridgehead carbon is too strained. Therefore, a bridged compound will not undergo an $S_N1$ mechanism st the bridgehead carbon.

Few bridge head carbons (1-bromobicyclo[2.2.1]heptane) are observed to give the corresponding alcohols with aqueous $\ce{AgNO3}$ at $150~\mathrm{^\circ C}$, i.e. With $\mathrm{S_N1}$ slowly. It has been suggested that the extra bonds in the larger ring help to relieve the strain in the formation of carbocation which tries to assume a planar configuration. (source: Cengage organic chemistry, by K.S. Verma)


text:
Electrolysis of aqueous copper (II) nitrate
There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.
Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.
However, looking at the half reaction for copper:
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  (E^\circ &= \pu{+0.34 V})
\end{align}
Compared to nitrate:
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) + 3e- &-> NO (g) + 2H2O (l)}&
  (E^\circ &= \pu{+0.96 V})
\end{align}
Since it produces more energy, I believe the second half reaction will occur instead.

I begin by finding all the species/ions I have:

$\ce{Cu^2+}$
$\ce{NO3^-}$
$\ce{H2O}$

Taking a look at the Standard Reduction Potentials at 25°C:
For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$):
\begin{align}
  \ce{NO3^- + 4H+ + 3e- &-> NO + 2H2O}&
  (E^\circ = \pu{+0.96 V})
\end{align}
As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$):
\begin{align}
  \ce{2H2O &-> O2 + 4H+ + 4e-}&
  (E^\circ &= \pu{-1.23 V})
\end{align}
For the balanced reaction I have:
\begin{align}
  \ce{4NO3- + 4H+ &-> 4NO + 2H2O + 3O2}&
  (E^\circ &= \pu{-0.27 V})
\end{align}
Some observations that can be made are that:

Nitric gas is forming at the anode
Oxygen gas is forming at the cathode
At least $\pu{0.27 V}$ must be put in

My question is:

Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.
What would happen if you placed a necklace at the cathode?

Are the electrodes inert or made of any reactive metal, (say for example copper)?

@SatwikPasani The electrodes are inert, they are two solid slabs of copper just sitting in solution.

Then the copper electrodes might also participate in the reaction at the anode. $$\ce{Cu_{(s)}->Cu^2+_{(aq)} + 2e^-}$$

@SatwikPasani $\ce{Cu_{(s)}->Cu^2+{(aq)} + 2e^-}$ has a potential of +0.34V while $\ce{NO3^{-}{(aq)} + 4H^{+}{(aq)} + 3e^{-}->NO{(g)} + 2H2O}$ has +0.96V. Since it produces more energy, wouldn't the stronger oxidizing agent be used instead, i.e, $\ce{NO3^{-}}$?

Have you considered the overpotential of nitrates on copper surface. Although I am not sure, considering the kinetics (overpotential), cooper is oxidised in preference to Nitrate at the anode.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since
$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$
Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  E^\circ_\mathrm{ox} &= \pu{-0.34 V}
\end{align}
Your second half-reaction is a reduction (as written)
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) +3e- &-> NO (g) + 2H2O (l)}&
  E^\circ_\mathrm{red} &= \pu{+0.96 V}
\end{align}
The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:
$$
\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -> 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\
E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}
$$
How does this change your analysis?

This seems to be a thought experiment, and many factors have been considered, but not enough to correspond to reality.
One item to consider is that copper is electrolytically purified by psssing current between a thin cathode of pure copper and a thick cathode of impure copper. An electrical power supply drives the copper (in solution) from the anode (+) to the cathode (-), and less reactive metals precipitate while more reactive metals remain in solution. Sulfate is the most usual anion.
But nitrate could be used as well. Oh, but nitrate is an oxidizer, so it complicates the issue. Perhaps. At high concentrations (say, >25%), nitric acid is such an oxidizer that it will passivate iron, and not dissolve it. Reduce the acid concentration to 10% or less, and iron will dissolve in it, because the acidity function outweighs the oxidizing function. So let's consider our copper nitrate to be at a moderate concentration (~10%), and without added acid ($H^+$), which is an important reactant in all the oxidation reactions.
With pH near neutral (pH = 7), the potential of the nitrate reduction half cell drops from +0.96 volts to +0.408, so then the combined cell voltage is only +0.068. What this means is that a copper block in a solution of copper sulfate will corrode slightly, consuming hydrogen ion. The reaction will not proceed as far as developing pH = 8 before stopping. The copper blocks would show a bit of staining or etching.
When an external potential is applied, one reaction is that $Cu^{+2}$ is forced off the anode into solution and some other $Cu^{+2}$ in the solution is driven onto the cathode to maintain neutrality everywhere. But nitrate ion is a a bit more active than a mere spectator ion and some will be reduced at the anode to $NO$. The pH will increase because $H^+$ is consumed.
At the cathode, copper will plate out, not oxygen. I can imagine copper hydroxide being precipitated at the anode and $NO$ gas being liberated there as well, but it could get more complicated, as $NO$ reacts with $OH^-$.


text:
What is the charge on the copper ion in antlerite?
Antlerite is $\ce{Cu3(SO4)(OH)4}$.
I'm trying to get the charge on the copper ion in this substance.  So far I have the mass of $1$ mole of antlerite, its percent oxygen content, and the number of individual hydroxide ions in a $\pu{955 mg}$ bead of the substance, if that information helps at all.
I need help to solve this question.

Ok, here is a short answer. Since you already know the chemical formula it's actually quite easy provided you are familiar with the process of determining the oxidation numbers in chemical compound. From the Wikipedia article (and the way the chemical formula is written indicates that too) you can get the information that the $\ce{SO4}$ part of the formula is sulfate and sulfate ions carry a charge of $-2$, i.e. $\ce{SO4^{2-}}$. Furthermore, in most cases $\ce{H}$ gets an oxidation number of $+1$ and $\ce{O}$ gets an oxidation number of $-2$ so that the 4 $\ce{OH}$ units in the compound each have a charge of $1 - 2 = -1$, so you have $\ce{OH^{-}}$, i.e. hydroxyl groups - this is supported by the way those $\ce{O}$'s and $\ce{H}$'s are grouped, if you see "$\ce{OH}$" in a compound it usually means "hydroxyl group". So, you have one sulfate group with a charge of $-2$ and four hydroxyl groups with a charge of $-1$ each, giving you a charge of $-6$ in total. But your compound $\ce{Cu3(SO4)(OH)4}$ is neutral, i.e. has a charge of $0$, so the three $\ce{Cu}$ have to counteract the negative charge exerted by the anions, in other words you have to distribute a charge of $+6$ between the three copper atoms. If you assume that the charge is evenly distributed, i.e. that all the $\ce{Cu}$ are the same (this need not necessarily be the case, you can also have mixed oxides like $\ce{Fe3O4}$ where you have $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ within the same mineral), you get to a charge of $+2$ for each copper ion, i.e. $\ce{Cu^{2+}}$.


text:
How to detect H₂SO₄ (aq) and NH₃
We have two bottle named $\ce{H2SO4~(aq)}$ and solution of $\ce{NH3}$ but we don't know their name actually. What I have to do to detect these solution  by only using :

$\ce{CuCl2}$

or

$\ce{NaCO3_{(s)}}$

Take two test tubes, add moderate amount of one unknown solution to one of the test tube and the other unknown to another. Add little $NaCO_3$ salt to both of the test tubes. The test tube which gives brisk effervesence was initially filled with $H_2S0_4$, and the other test tube will be having $NH_3$.

On mixing 2 parts of $\ce{NH3}$ with 1 part of $\ce{CuCl2}$, following reaction will occur.
$\ce{CuCl2 + 2NH3 -> Cu(NH3)2^{2+} + 2Cl-}$
This will give you soluble deep-blue copper-ammonia complex: $\ce{Cu(NH3)2^{2+}}$
and also, $\ce{NH3}$ does not react with $\ce{Na2SO4}$.
And $\ce{H2SO4}$ with $\ce{Na2CO3}$ will follow the reaction below :
$\ce{H2SO4_{(aq)} + Na2CO3_{(s)} -> H2O + CO2_{(g)} + Na2SO4_{(aq)}}$
As $\ce{Na2SO4}$ solution is colorless, so you can perform any appropriate test to identify the evolved gas i.e. $\ce{CO2}$.
A simple method used to test $\ce{CO2}$ is :

Bubble the gas through limewater (calcium hydroxide). If carbon dioxide is present, the limewater will turn cloudy and a milky white precipitate will form.


text:
Is there a mnemonic to study the spectrochemical series?
Mnemonics help to by-heart things which are difficult to do so and which are important. I found mnemonics for studying the periodic table, electromagnetic series etc which were of great help. So, is there a mnemonic to study the spectrochemical series since knowing it will be of great help to find the geometry of a given co-ordination compound?
The spectrochemical series is a list of common ligands for transition metals in order of increasing ligand-field splitting parameter $\Delta$, which is the energy difference between the sets of $\ce{d}$ orbitals:
$$\ce{I-} < \ce{Br-} < \ce{S^2-} < \ce{SCN-} < \ce{Cl-} < \ce{NO3-} < \ce{N3-} < \ce{F-} < \ce{OH-} < \ce{H2O} \approx \ce{C2O4^2-} \text{ (oxalate)} < \ce{NCS-} < \ce{CH3CN} < \ce{py}\text{ (pyridine)} < \ce{NH3} < \ce{en}\text{ (ethylenediamine)} < \ce{bpy}\text{ (2,2'-bipydiridyl)} < \ce{phen}\text{ (1,10-phenanthroline)} < \ce{NO2-} < \ce{PPh3} < \ce{CN-} \approx \ce{CO}$$

Is there a mnemonic device for remembering this series?

I editing this question to make it clearer what the you were asking about. The spectrochemical series may not be familiar to everyone.

I don't think you're required to memorize this series, and I don't think there's an easy mnemonic…

I would not recommend a mnemonic for the spectrochemical series for a number of reasons:

Often, it depends on both metal and ligand and coordination geometry how big a certain splitting is.
The splitting of some ligands ($\ce{NH3, H2O}$) is even dependent on the pH — even though that could sort-of be overcome by including their corresponding anions in the list.
Any mnemonic is bound to be incomplete. N-heterocyclic carbenes (NHCs) are very important ligands especially in organometallic chemistry, but they aren’t on your list.
Some ligands (e.g. $\ce{NO}$) can display very complex redox chemistry that defies them being ordered in a simple list.
Rather than learning some list off by heart and rote memorising it later, it is much better to know why a ligand is strong- or weak-field and to deduce that property from the ligand’s structure.

The last point is most important. Rather than learning or attempting to remember that chloride is a weak-field ligand, ask yourself why. It is a π donor ligand that will split the $\mathrm{t_{2g}}$ orbitals of the central metal in a bonding–antibonding way, lifting the metal’s $\mathrm{t_{2g}}$ orbitals and lowering the energy difference. That way, I immediately know what to expect of (say) selenide $\ce{Se^2-}$ even if it is not present in the list. The strength of a certain ligand in this discussion is equivalent to the usual trends in the periodic table. Quickly memorised, no long mnemonic and applicable even to new ligands.

The following mnemonic could be used to remember the spectrochemical series:

I brought some colourful sweets from office containing water. Nisha
ate nine in canteen corridor

$\ce{I- < Br- < SCN-  < Cl- < S^{2-} < F- < OH- < C2O4^{2-} < H2O < NCS- < edta^{4-} < NH3 < en < CN- < CO}$

While this mnemonic is slightly better than the other answer it still is not a full English sentence that would make sense to me.

I use the following to remember the spectrochemical series;

I brought some cloth, nylon net from office, collect water near cyanomethane place, among each big pants, place no proper concentration or concern.

($\ce{I, Br, S, Cl, NO_3, N_2, F, OH, CrO_2, H_2O, CH_3CN, py, NH_3, en, bpy, phen, NO_2, PPh3, CN, CO}$ respectively.)

Wait wait wait. How do you distinguish between CN and CO?

This mnemonic doesn’t make sence to me because most of the word lists aren’t English sentences …

This mnemonic is actually just a collection of words that make next to no sense when put together. It doesn't help much either. How is it possible to say that Collect corresponds to CrO2?

Is there any mnemonic to remember this mnemonic?


text:
The violation of a certain rule or principle in the following electronic configuration
I'm to find the violation of a certain rule or principle in the following electronic configuration:
$$\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$$

$n+l$ rule is not violated because: for $\ce{1s^2}$, $n+l=1+0=1$ so it is filled first, for $\ce{2s^2}$, $n+l=2+0=2$ so it is filled second, and so on.
Aufbau Principle is not violated because: the orbitals with minimum energy are filled first with electrons and $\ce{1s < 2s < 2p}$ which is the sequence of increasing energy orbital judged by the $n+l$ rule.

I'm quoting here a paragraph from my chemistry textbook of the chapter Atomic Structure:

The orbitals given by a particular value of $l$ if $n$ is same, have the same energy and such orbitals are called degenerate. Like $l=1$ gives three $p$ orbitals specified as $\ce{p_x, p_y, p_z}$ and the p orbitals are said to be three fold degenerate. These orbitals are filled according to Hund's Rule.

The numerical value of the principle quantum number $n$ for $\ce{2s}$ and $\ce{2p}$ is the same; however, they don't have a particular value of the azimuthal quantum number $l$. For $\ce{2s}$, $l=0$, and for $\ce{2p}$, $l=1$, so $\ce{2s}$ and $\ce{2p}$ are not degenerate. So none of the three principles or rules whether Aufbau's, $n+l$ or Hund's is violated in this electronic configuration.
However, my friend says that there is a violation of the Hund's Rule in this electronic configuration, and according to him the right configuration is: $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$, which in turn violates the Aufbau Principle, because the orbitals of lower energy level have to filled first completely and $\ce{2s < 2p}$.

So, which one of the electronic configurations is right, $\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$ or $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$?

Note: The required electronic configuration is for the electrons in their ground state.

When you write electronic configurations, you should distinguish your orbitals by $n$ and $l$ only, but not by $m_l$ (or $m_s$), for two reasons. Firstly $m_l$ is not well defined in an atom unless you have an external field; secondly because the electronic configuration is a rough approximation that does not distinguish the energies of different states depending on how you distribute the electrons for different values of both $m_s$ and $m_l$. Indeed the distribution of the $m_s$ has a stronger effect than that of $m_l$ in general.

Both are correct although of different energy as has been pointed out. However, you cannot add a label to decide if the electron is in $p_x$ or in $p_y$ or $p_z$ as they are equivalent unless the symmetry is broken by an external magnetic or electric field.

If we talk only about electronic configurations and rules violation then you are correct. The 2nd one (excited state) is referred for hybridization that is sp2. An excited state  configuration is meant only when the particular atom gets ready or excited for forming a bond. Like boron does it with 3 fluorine atoms to form $\ce{BF3}$. So, if we talk about electronic config and rule violations we must go for ground states exceptional cases are $\ce{Cu}$ , $\ce{Cr}$ and $\ce{Mo}$.

No that's wronge.
the first one is right which is called ground state for an atom.
$\mathrm{(2s)^2}$ must be filled first becouse from hybridization concept we know that S-sub shell is more closer to the nucleus then thw P-sub shell  and thus will have lower energy, so $\mathrm{(1s)^2 (2s)^2 (2p)^1}$ is the right configuration for Boron, in this case neither Aufbau rules are voilated  nor hund's rule becouse of lower energy of S-sub shell.

It is violating Hunds Rule..
 " If degenerated orbitals are available then electrons want to live separately with same spin rather than mutually in same orbital with opposite spin."

The upper one is correct because it is according to the rules of electronic configuration and it does not violate any rule, while the lower one is not in line with the Afbau principle which states that the orbitals are filled with electrons in increasing order of the orbital energies.


text:
Is the electron configuration redundant?
For example the electron configuration for oxygen is 1s2 2s2 2p4. But if an element can never have an element in the p orbital unless it has 2 full s orbitals*, then why can't we write 2p4 as the electron configuration for oxygen? Is it not implied that it has 1s2 2s2? I would think so because the noble gas configuration works on a similar principle.
*My teacher told me this.

The configuration you describe is the ground state configuration, the configuration of oxygen in which all electrons are in their lowest allowed energy states.
Substances can also be their excited states, where electrons are not in the lowest allowed energy states. Excited states happen when atoms or molecules absorb energy, usually electromagnetic radiation (but not always).
An excited state for oxygen may involve one of the $2s$ electrons being promoted to one of the remaining $2p$ spots:
$$\begin{aligned}
\ce{O} + E &\ce{-> O}^* \\
1s^2 2s^2 2p^4 &\ce{-> }1s^2 2s^1 2p^5
\end{aligned}$$
This configuration still describes an oxygen atom, but now the $2s$ subshell is partially empty.
The other reason you can't always assume that the lower subshells are always filled is that there are some exceptions to the aufbau principle. Some elements do not follow it, especially the lanthanides and actinides.

Chromium: $\ce{Cr} - \ce{[Ar]}4s^1 3d^5$ ($s$ is not filled so that $d$ can be half-full)
Copper: $\ce{Cu} - \ce{[Ar]}4s^1 3d^{10}$ ($s$ is not filled so that $d$ can be full)
Palladium: $\ce{Pd} - \ce{[Kr]}4d^{10}$ ($s$ is empty so that $d$ can be full)
Cerium: $\ce{Ce} - \ce{[Xe]} 6s^2 5d^1 4f^1$ ($f$-block elements are weird)


text:
How to calculate mole ratio of two volatile compounds in distillation?
I have two volatile compounds (such as ethyl acetate and ethyl propionate). I am going to separate these compounds by simple and fractional distillation.
How do I calculate mole ratios of the two compounds and determine the boiling point temperatures of the first drop based on the liquid and vapor phase diagram?

If you have the vapor-liquid phase diagram, then it is easy to determine the vapor composition above a boiling liquid mixture (which should approximate the composition of your first drop in a simple distillation).
I'll assume you diagram looks like the one below, which is from the Wikipedia article on vapor-liquid equilibrium(and is mostly at the physical chemistry level):

Let's say you have a mixture of components 1 and 2 that is 40 mole percent (mole fraction = 0.4) component one (approximately the blue line labeled liquid on the diagram). Draw a line vertically from that composition to the bubble point (or liquid) line. Then draw a horizontal isotherm at that point (red line on the diagram above). Where that isotherm crosses the dew point (or vapor) line, you can read down to get the composition of the vapor above that boiling liquid (blue line labeled vapor), about 15% Component 1 (and 85% Component 2). This represents a single vaporization-condensation cycle, which is what happens in a simple distillation.
In a fractional distillation, you can have multiple vaporization-condensation cycles (called theoretical plates). If you know how many theoretical plates your distillation column will have, then you repeat the process for the simple distillation several times. For example, if there are two theoretical plates, you have two vaporization-condensation cycles. After the first one, start over with a new liquid composition equal to the vapor composition after the first cycle. Draw a line up to the liquid line, draw the isotherm to the vapor line, and draw a line back down to composition. If you use the example I have been working with on the diagram I have shown, you should get about 5% Component 1. Repeat as necessary.
If you do not know how many theoretical plates you have in your fractionating column, then you can only make an educated guess.


text:
Finding the number of orbitals on a central atom
In $\ce{BeCl2}$ the number of orbitals on central atom, i.e. on beryllium, are 2. In $\ce{BF3}$, the number of orbitals on central atom , i.e. on boron, are 3.
Similarly in $\ce{NH3}$ there are 4, on $\ce{PCl5}$, $\ce{SF4}$, and $\ce{ClF3}$ there are 5. But how?

Consider $BeCl_2$ with E.C $[He]2s^2$. E.C in excited state will be $[He]2s^12p^1$. These two orbitals ($2s$ and $2p$) get hybridised to form $sp$ hybridized orbitals. Thus, in your language number of orbitals on central atom is $2$. Similarly you can try for other compounds.

According to the question, it is totally not clear that whether you want to know filled orbitals or total orbital (filled + vacant). Well, but after seeing the examples you gave, it can be concluded that you are talking only about filled orbitals. As I'm not clear and assuming that you are only talking about filled orbitals, so if you are talking about total, leave a comment.
As each orbital contains 2 electrons, therefore it is easy to find number of filled orbitals.
The number of filled orbitals in central atom will be equal to the sum of number of bonds and number of lone pairs present on central atom.
Example
As in case of $\ce{BeCl2}$, $\ce{Be}$ makes 2 bonds (both with $\ce{Cl}$) and have no lone pair, therefore,
$\text{No. of filled orbitals} = 2$
Let us take another example of $\ce{NH3}$.
In case of $\ce{NH3}$, $\ce{N}$ makes 3 bonds (each with $\ce{H}$) and have 1 lone pair, therefore,
$\text{No. of filled orbitals} = 4$
Take another example of $\ce{ClF3}$
In this case, the central atom $\ce{Cl}$ makes 3 bonds with $\ce{F}$ by sharing its one electron in each bond and have 2 lone pairs on it. (As $\ce{Cl}$ have 7 electrons, 3 shared in bonding and 4 left = 2 lone pairs.) Therefore,
$\text{No. of filled orbitals} = 5$
Hope, now everything is clear to you!


text:
Difference between inorganic and organic semiconductors: electronic structure or configuration, or?
Organic semiconductors differ from inorganic semiconductors. In organic semiconductors the molecules are held together by weak van der Waals interactions and in inorganic semiconductors by covalent bonds. So the bonds are different. How do you express the main difference between the twos? Is it the electronic structure or rather the electronic configuration, or something else?

The conjugated $\ce{\pi}$ electron in organic polymers is responsible for the conduction of electricity.


text:
Which has more metallic character: aluminium or magnesium?
Which element has more metallic character - aluminium or magnesium?
I know that metallic character decreases along a period (from left to right) and increases down a group.
Aluminium comes after Magnesium in the same period (period 3) in the periodic table. So, I thought Magnesium would have greater metallic character.
However, I read in a book that Aluminium has greater metallic character. Is this true? If this is true, why?

+1 Nice question,  it depends on how do you define the metallic character...

Yes, you are right that $\ce{Al}$ is more metallic than $\ce{Mg}$.
It is because, $\ce{Mg}$ have fully filled $3\text{s}$ subshell. Due to this, the first ionisation energy$^{[1]}$ of $\ce{Al}$ (578 kJ/mol) is less than that of $\ce{Mg}$ (738 kJ/mol).

Filled and half-filled subshells show a small increase in stability in the same way that filled shells show increased stability. So, when trying to remove an electron from one of these filled or half-filled subshells, a slightly higher ionization energy is found.$^{[2]}$

You can also say that higher levels of subshells are further away from the nucleus which is why it is easier to remove an electron from $\ce{Al}$ than from $\ce{Mg}$.
References:
 1. http://en.wikipedia.org/wiki/Ionization_energy
 2. http://www.grandinetti.org/ionization-energy-trends
Edit: Updated reference(s)

Well I'd quibble a bit about ionization energies being the defining characteristic of metallic character. I think electrical resistivity would be more characteristic. A metal isn't much of a metal in a gas. Ni has a lower ionization potential than copper in a gaseous state (737.1 vs 745.5) but a higher resistivity (69.3 vs 16.78).

I don't think that using Yahoo Answers as reference add value to your assertion... eheeh...

@GM : edited! And now Stack Exchange wants me to write 2 more characters in this comment so I wrote a lot more than just 2. :p

Two types of metallic character
In fact, there are two type of metallic character if you look at the metal from the chemical point of view or if you look at the metal from the physical point of view:

So it really depends on how do you define metallic character

Chemical metallic character
Since metallic character in chemistry is defined as:

the tendency of an atom to lose electrons.

We can understand that it is directly related to electronegativity.

chemical property that describes the tendency of an atom to attract
electrons (or electron density) towards itself

Basically, it is the opposite (that would be Electropositivity...)! An so it increases from right to left. The picture below shows the electronegativity but can be used for the purpose.

I think the reasoning that the book wants you to do is not remembering all the subshell but just the basic chemistry principles.
So from the chemical point of view $\ce{Mg}$ has a greater metallic character compared to $\ce{Al}$.
Physical metallic character
The physical metallic character is defined taking in account the physical properties of the metal. A metal, from the physical point of view, has a high electrical conductivity, high thermal conductivity, and high density.
These properties are harder to predict:

In this case, however, $\ce{Al}$ has a more metallic character compared to $\ce{Mg}$.

Mg is more metallic than Al due to its tendency to lose electrons. As a result, the electropositivity decreases as we go across periods. Electropositivity is related to metalic character and vice versa. As a result, Mg has a greater electropositivity then Al as well as being more metallic.

The metallic character is given by free electrons in the solid metal. Aluminium has, after silver, copper and gold, the fourth highest electrical and thermal conductivity of all metals. Therefore aluminium is considered more metallic than magnesium.

@aventurin it depends if you look at the metal from the chemical or the physical point of view

Ionization is inversely proportional to metallic character, III-A and VI-A Show abnormalities in ionization energy and Al is in III-A and its ionization energy is less than Mg that's it is more metallic than Mg.

As metals can easily remove an electron .Aluminium is more metallic than magnesium because the last electron in Mg enters in s orbital  and in Al it enters in p orbital and s orbital is more penetrating towards nucleus so it is difficult to remove  an electron in Mg.


text:
Energy required to remove an electron from He

The energy required to remove both electrons from the helium atom in its ground state is $79.0\;\mathrm{eV}$. The energy required to ionize helium (i.e. to remove one electron) is -
(A)   $24.6\;\mathrm{eV}$
(B)   $39.5\;\mathrm{eV}$
(C)   $51.8\;\mathrm{eV}$
(D)   $54.4\;\mathrm{eV}$

My thoughts: What I actually did was to take half of the given energy value since energy required to remove electrons depends only on the principal quantum number(and also on atomic number). But, I found my answer to be wrong.
On the other hand just calculating the energy required to remove one electron from $n=1$ state gives the correct answer. Why is it so?

because Helium is more stable in its ground state so energy required to remove first electron will be higher than second.

We can use Bohr's atomic model to calculate the ionization energy of $\ce{He+}$.
The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives $IE_2=E_{\infty}-E_1$ where $E_1$ is the energy in the first bohr-orbit (principal quantum number=$1$). Conventionally taking $E_{\infty}$ as $0$ we use the Bohr's equation to find the energy $E_1$:-
$$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$
 where $Z$= atomic number, $e$= charge on electron, $n$= principal quantum number, $h$= Planck's constant, $m$= mass of electron, $\epsilon_0$= permittivity of free space.
Using $n=1$, $Z=2$ and the appropriate values for other constants we get $IE_2=-E_n=\pu{54.4eV}$.
Now the energy for removing both electron $IE_{net}=IE_1+IE_2$. Using $IE_{net}=\pu{79eV}$and $IE_2=\pu{54.4eV}$ we get, $$IE_1=\pu{24.6eV}$$
which corresponds to option (A), which I assume is typed wrong by you in the question.

Taking half the total value of both ionization energies isn't a bad initial approach. You were just one step shy of finding the answer (without calculations). Each additional ionization energy will be greater than the last because the potential will be greater; the fewer the electrons, the less electron-electron repulsive force. Thus, the first ionization energy is less than half the total. Only answer A satisfies that requirement, so that must be the answer.


text:
What can you use to remove hair gel?
The hair gel is called "lets Jam Gel". It says it uses micro emulsion technology.
What chemicals can you use to break it down or undo the bonds?
I'm trying to get it out of my head. I washed my hair with T sal and cut my hair off but it still grows back patchy and easy to style like it has gel in it. I think it may be in the epidermis or even dermis by now.
ingredients: 42692 SS11 - Aqua / Water Isocteth-20 Paraffinum Liquidum / Mineral oil Glycerin Oleth-2 Propylene Glycol Parfum / fragrance Sorbic Acid Methylparaben DMDM Hydantoin Ethylparaben Limonene Hydroxycitronellal Geraniol Amyl cinnamal Linalool Simethicone CI 19140 / Yellow 5 CI 14700 / Red 4, CI 172200 / Red 3. F.I.L D45949/1

It would help if you added what you're trying to accomplish, and, if possible, an ingredients list off the container. Are you trying to wash it out of your hair, or out of your carpet, for example? And is it this stuff?

ingredients: 42692 SS11 - Aqua / Water Isocteth-20 Paraffinum Liquidum / Mineral oil Glycerin Oleth-2 Propylene Glycol Parfum / fragrance Sorbic Acid Methylparaben DMDM Hydantoin Ethylparaben Limonene Hydroxycitronellal Geraniol Amyl cinnamal Linalool Simethicone CI 19140 / Yellow 5 CI 14700 / Red 4, CI 172200 / Red 3. F.I.L D45949/1

This question appears to be off-topic because it is about unintended side effects of a product that are better addressed by a physician.

More likely by consulting a hairdressing professional.

Yikes.
As far as I'm aware, a thorough application of ordinary soap or washing-up liquid should be able to wash this off using the normal surfactant processes (binding to organic components with lipophilic regions of the molecule). If you're experiencing long-lived after-effects (say, more than a week maybe), you might want to see a doctor -- it's not impossible you could have had some sort of allergic reaction. Oh, and if you're going to wash hair with soap or washing-up liquid, I'd suggest using conditioner afterwards, to reduce the long-term damage incurred by stripping the natural fatty acids and oils off.
I'm not a medical professional or a tricologist, though, so please don't take this as Official Medical Advice.

Thanks, I have tried this but I don't think it worked. The soap really dried out my hair. I think I got it out of my hair but I have a feeling the gel is still stuck under my scalp.


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An alternative to dimethylsulfoxide (DMSO) for cryoprotection?
Is there an appropriate alternative to DMSO? I mean for embryo or oocyte freezing?
I am a PhD student at present. I am going to study this substance and I'm very interested in this field.

Hi, I took the liberty of modifying your question title to be more specific about what you're interested in DMSO for. I hope this is aligned with your intent.