text:
Why are vinylic and arylic carbocations highly unstable?
I have checked the internet and read quite a few books, but I still am not able to understand why vinylic and arylic carbocations are highly unstable.
What I found while surfing the internet is: For arylic carbocation, the carbon bearing postive charge cannot participate in resonance with adjacent carbons because it's p orbital is perpendicular to the other orbitals. But, I didn't really understand what this means.

Background
S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital.  In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.
This is the basis of the inductive effect.  For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - the methyl group is inductively electron donating to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.
What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.
Answer
Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon.  For a positive center to be attached to a more electronegative group is destabilizing.  Hence the vinylic cation is less stable than a typical alkyl cation.
Things are even worse with the aryl carbocation on the right.  Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry.  A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$.  These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.

I was taught that electrons from an sp3 carbon move to an sp2 carbon due to hyperconjugation, that is: the shift of a bonded electron into the unhybridised p orbital of the sp2 carbon. Is the shift of electrons due to a combination of both this and inductive effect?

Yes, that's how I see it, a mix of resonance and inductive effects.

@ron for hyperconjugation to occur, must the orbitals hyperconjugating be in the same plane?

Yes, for hyperconjugation orbitals have to be in the same plane.

The instability of a vinyl cation cannot be blamed on its sp2 orbital being very electronegative as the positive charge resides in a p orbital, not an sp2 orbital.  The instability derives from the inability of that p orbital to overlap with the the sp2 orbitals of the carbon on the other end of the double bond. The bond angles of that carbon are too large (120*) and their highly electronegative nature prevent stabilization of the cationic center.  The only stabilization available to the vinyl cation comes from the saturated sp3 carbon on the other side and its stability is therefore similar to a primary cation.

We need to know one fact i.e. more the s-character in the hybridisation of a compound more is its electronegativity. This due to the fact that s-orbital is nearer to the nucleus and is thus attracted more towards the nucleus leading to an increased electronegativity.
As the carbon of vinyl carbocation is sp- hybridised (50% s-character, 50% p-character) it is highly electronegative. And positive charge on electronegative atom is highly unstable


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Vibrational Spectrums of HI, HBr, HF, and HCl
Do you all know of any large graphs for the vibrational spectrums of HI, HBr, HF, and HCl? I need graphs that are large enough to see the spacing between the peaks in the spectra of HCl, HBr, and HI.
Alternately, if you know of any raw data sets for any of the above, that would work. Thanks in advance.

Have you tried searching NIST?  Here is a link to the IR spectrum of HBr.  (http://webbook.nist.gov/cgi/cbook.cgi?ID=C10035106&Mask=80#IR-Spec) The others should be easy to find as well as these are simple systems that have more than likely been previously and extensively investigated. (I don't know how large you need the spacings. This first inquiry should make for a good point of reference). Also, do you only care about experimental data or will spectra generated from computational methods be sufficient?

Computational methods would be fine, I'm just trying to use it to show the effects of isotopes in molecules to high school students.

Actually, I spoke too quickly.  These diatomics will only have one fundamental vibrational mode (one peak) so computational methods on a single gas-phase diatomic won't help you there (i.e. there won't be any splitting) .

For $\ce{HCl}$ I would suggest this paper.

The fine structure in the diatomic vibrational spectra is based on various population of the rotational levels.
The best demonstration of isotopic effects in vibrational spectra is to switch hydrogen for deuterium.
The frequency of vibration (from Hooke's Law adapted to a harmonic oscillator) is given by:
$$\nu = \frac{1}{2\pi}\sqrt{\dfrac{k}{\mu}}$$
where $k$ is the force constant of the oscillator and $\mu$ is the reduced mass:
$$\mu=\dfrac{M_1M_2}{M_1+ M_2}$$
Switching $\ce{HCl}$ to $\ce{DCl}$ has a much larger effect than the two chlorine isotopes (35 and 37):

$\ce{H^{35}Cl}, \ \mu\approx\frac{35}{36}=0.972$
$\ce{H^{37}Cl}, \ \mu\approx\frac{37}{38}=0.974$
$\ce{D^{35}Cl}, \ \mu\approx\frac{70}{37}=1.892$
$\ce{D^{37}Cl}, \ \mu\approx\frac{74}{39}=1.898$

The difference between $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$ is in the third decimal place.
Here is a nice Google Image search for the vibrational spectra of HCl and DCl

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Why doesn't the boiling point of an aqueous solution depend on intermolecular attraction?
I understand that boiling point is a colligative property for aqueous solutions, and that it empirically depends on the mole fraction of the solute, rather than the identity of the solute. I understand the van't Hoff factor / ionic strength and why it matters. I also get that an increase in entropy is the primary driving force for boiling point elevation.
My question is, why does the influence of intermolecular attraction go out the window for solutions, as opposed to pure substances? Shouldn't the identity of the specific dissolved species - that is, their charge and their size, affect the intermolecular attractions in the solution? Shouldn't greater intermolecular attraction increase the boiling point by reducing vapor pressure, just as it does for pure substances? For example, neutral molecular solutes as opposed to strong electrolytes, or large ions like iodide vs smaller ones like fluoride.
Is the effect simply too small to matter compared to the entropy factor, or am I missing something theoretically?

I think your suspicion that the non-ideal intermolecular attraction influences the colligative properties is justified. Have a look here for example. Especially for ionic solutes there is a difference between the expected (ideal) and the observed behavior. But usually the entopy factor is dominant.

GOOD QUESTION! That is the nastiest p-chem question I've seen.  Atkins, Physical Chemistry 2nd edition (the readable one), pp 228-232pp.  Boiling point elevation is presented as purely an entropy effect, given two assumptions:
1) The solute is non-volatile.
2) The non-volatile solute does not dissolve in the solid solvent.
Both assumptions are explored in Chapter 10.
A solute increases the entropy of the solution but does not change the entropy of the vapor at a given P and T.  A higher T is then necessary to give the same entropy difference.  I am not satisfied that this wholly addresses a strongly interactive solute versus a non-interactive solute versus a strongly negatively interactive solute,
http://www1.lsbu.ac.uk/water/kosmos.html
Kosmotropes and chaotropes
http://www.amazon.com/Physical-Chemistry-Peter-Atkins/dp/1429218126
Atkins 9th edition is current.
"There is almost no chemical concept that Atkins can't make unnecessarily confusing"
http://www.amazon.com/Physical-Chemistry-A-Molecular-Approach/dp/0935702997/ref=zg_bs_16052591_2
Library - it's pricey.
http://www.amazon.com/Thermodynamics-Engineering-Approach-Yunus-Cengel/dp/0073398179/ref=dp_ob_title_bk
Library - it's very pricey.

Perhaps the enthalpy of interaction between the solute and solvent does not change as the solvent evaporates (at least until the concentration of solute gets high enough that solute-solute interactions become important) so the entropy is the only factor that changes.

You are not wrong to assume that the solute-solvent interactions matters. They indeed do and will influence on vapor pressure of the liquid. The caveat here is that this property, like many others of solutions, is primarily described for ideal solutions.
In an ideal solution, the solvent-solute, solvent-solvent and solute-solute interactions are all considered identical (Chemical Principles, P. Atkins & L. Jones). So even though the interactions can't be ignored (like it's done for ideal gases), they shouldn't influence on the properties since they are all the same.
It's no surprise then that real solutions will have a closer-to-ideal behavior as they get more and more diluted, since, in practice, there are proportionally fewer and fewer solute-solvent interactions happening.

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Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not?
Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not? And is there a reason why the products of reaction 2) are written separately, instead of "2OH"?
1) $\ce{OH- + H2O <=> H2O + OH-}$
2) $\ce{O^{2-} + H2O -> OH- + OH-}$

Correct, all reactions are reversible, just to different extents. Think about the combustion reaction in your vehicle. Is that reaction reversible? Sure, in that nothing is 100% irreversible. The reversing of the combustion just might take a while. But it's not reversible to any significant extent in the short and medium-term; i.e. you're not going to be re-capturing your combusted gasoline and be putting that back together as usable fuel for your vehicle.
Also, about the reaction of oxide ion with water - that reaction goes practically 100% because oxide ion is a very strong base. The strongest base supported in high concentration by a medium is the conjugate base of the medium. (Leveling effect). For water, that would be hydroxide ion (remember that conjugate bases lose a proton). Oxide ion is the conjugate base of hydroxide ion. We would expect oxide ion to be an even stronger base than hydroxide ion because of oxide ion's even higher negative charge density - we're talking about a negative two charge here (versus negative one for hydroxide) and we're missing a hydrogen proton (which only further decreases the size of the oxide ion and further increases its negative charge density). If you think about this Coulombically, then of course oxide ion is going to be a stronger base than hydroxide ion - opposite charges attract (recall that water has two partial positive charges/dipoles stabilized by its hydrogen atoms). F=kq1q2/r^2 and one of the q's here is massive.
Also about how the products are written - it doesn't matter how the products are written. Either way you are creating two HO- (hydroxide ions). Write hydroxide with the negative next to the O (so either -OH or HO- are acceptable) because that's what you're doing - heterolytic clevage of the O-H bond in H-O-H, which (in this case) creates a negative formal charge that is mostly stabilized by the oxygen atom (NOT the hydrogen atom). Plus it's the hydroxide ion not the oxyhydride ion (whatever the hell that is).
On second thought, when writing -OH, make sure the "-" is clearly a superscript, lest people think you're talking about a hydroxyl group.

Hydroxide is the strongest base in water (leveling effect).  The first posted reaction is symmetric.  Given reactant and product sides of the equation, how do you tell them apart?
Oxide is a much stronger base than hydroxide.  The second reaction spontaneously proceeds to the right (leveling effect).  If you look up the Brønsted basicities of oxide and hydroxide, you can write a simple thermodynamic ratio for their relative abundances at equilibrium.
There is no reason not to write $\ce{2 OH^-}$ or $\ce{2 (OH^{-})}$.

All reactions are reversible but some are more reversible than others. Some reactions are so little reversible that they are practically irreversible, like that reaction of yours between oxide anion and water.
It all comes down to comparing the stabilities of reactants and products. The actual question is, why is oxide anion so unstable compared to hydroxide ions? One approach would be to think how would you make the oxide anion from a water molecule. Clearly, you can remove one hydrogen from a water molecule with a mild base but taking the second one would require a much stronger base. Take a look at the pkb value of hydroxide ion and visualize the equilibrium populations in your head.
Another possible approach would be to compare the electronic structures of the species. Oxide anion has twice the negative load of the hydroxide ion. Hydroxide ion has one hydrogen to relieve the stress.


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Why are alkali and alkali earth metals indicative of a strong acid?
Why are alkali and alkali earth metals indicative of a strong acid? Does anyone know why it is so? Or tell me what is indicative of strong acids?

Is $HCl$ is a strong acid? I don't see a alkali metal anywhere.

Alkali and alkaline earth metals are highly electropositive, and thus the electron density is shifted over the hydroxyl group. In essence, it is the -OH that is cleaved and not just the proton. You are actually referring to bases.

What protons get cleaved? That is not in my textbook.

Ok, then how do we tell if something is a strong acid? How is hydroxyl groups related to bases, not to be confused with hydroxide anions?

Okay, we're sticking to the Arrhenius theory. Here's the executive summary: Bases and acids often have hydroxyl groups. Whether the entire hydroxyl group is cleaved and leaves as the hydroxide anion or whether a proton is cleaved and leaves as the H+ cation depends (mainly) on the electronegativity of the central atom. Less EN/more electropositive --> probably going to have its hydroxyl group cleaved to form the hydroxide anion. More EN/less electropositive --> probably going to just have its proton cleaved away.

So with this in mind we can rationalize why KOH is a base while phosphoric acid (which has three "HO" groups) is an acid. K: very electropositive. P: more electronegative. The K loses a negative HO- group and is stuck with a positive charge (K+). That's okay; K is EP; it's fine with a positive charge. $\ce{(HO)_3PO}$ just has a H+ cleaved away since the oxygen (and the P) are both EN and are able to stabilize the remaining negative charge. Underlying these considerations is the fact charge is conserved (very useful if you study equilibria in analytical chem!)

Alkali and alkali earth metals are not indicative of a strong acid; if anything, they're more indicative of a strong base, but this is one of those half-truths/Santa Claus/tooth fairy level "truths" that general chemistry textbooks tell students. Just because something has an "-OH" in it doesn't make it a base or an acid automatically. The textbook is trying to placate readers with a simple rule that is sometimes right so that when you go into the lab and see $\ce{NaOH}$ (sodium hydroxide), you recognize that's a base. $\ce{KOH}$, that's a base too. $\ce{Ca(OH)_2}$, that's a base too.
But what about $\ce{(HO)_3PO}$? What about $\ce{(HO)_2SO_2}$? The two molecules I just mentioned are phosphoric acid and sulfuric acid, respectively (not phosphoric and sulfuric base). The formulas are, in addition, written with respect to structure, to illustrate that both have $\ce{OH}$ groups.
And what about phenol? That's an organic molecule with an OH group hanging off. What about acetic acid, $\ce{H_3CCOOH}$? That has an OH group on the end hanging off (no, it has no O to O bonds). None of these are bases (at least to any significant extent).
One must realize that to be a base - an Arrhenius base (as it seems that your textbook is introducing you to - one must have
a) An electropositive element. An electropositive element would be a group one or group two element and is able to stabilize positive charge no problemo. Example: Sodium, Na. We often find sodium as sodium ion: $\ce{Na^+}$. But do we find any of the electronegative elements - say - fluorine - as $\ce{F+}$? Hell no. $\ce{F-}$, maybe. That's because those electronegative elements are better at stabilizing negative charge while electropositive elements are better at doing the opposite. Remember that chemistry is just one big free energy optimization game; in simpler terms - stability is paramount and is the underlying goal of chemical processes.
b) The electropositive element must be attached to an OH group. Through heterolytic bond clevage, the OH group will leave the electropositive group as the hydroxide ion, $\ce{HO^-}$. The heterolytic bond cleavage (in this case) leaves the electropositive element short of one electron but remember that's perfectly okay - electropositive elements are alright with a positive charge.
This happens with an Arrhenius base (water is implied)
$\ce{KOH ->K^+ + HO^-}$
This is what the Arrhenius theory might lead to to believe (which isn't true; and again, water is implied):
$\ce{(HO)_2SO_2 -> 2HO^- + SO_2^2+}$
Rather, this is the proper (first) ionization of sulfuric acid. Remember that sulfur is much more electronegative than one of the group 1 or 2 elements. This, together with the inductive effect of the attached oxygens allows the sulfur to stabilize a negative charge gained through heterolytic cleavage of the O-H bond. Note that I do not mention resonance as a factor (although textbooks might imply this as a factor) because sulfur does not generally access its d-orbitals in bonding, and hence, there is no double bond (and therefore a set of electrons to delocalize) in either sulfuric acid or the hydrogen sulfate ion. Both are best represented as charge-separated species with a +2 formal charge on the sulfur. In other molecules, however, resonance stabilization is a factor in their acidities - e.g. take the acetate ion when considering acetic acid.
$\ce{(HO)_2SO_2 + H2O -> H_3O^+ + HOSO_3^-}$
So in essence the aforementioned two factors (electropositive element and OH group) make an Arrhenius base. Unfortunately the Arrhenius definition of acids and bases is rather limited and if you study chemistry further, you will forget the Arrhenius definition and supplant it with the Bronsted-Lowry definition of acids and bases and the Lewis definition of acids and bases (at the least).
And if you are really looking for what is indicative of a strong base or acid, you just assess conjugate stability. A strong acid has a very stable conjugate base. A strong base has a very stable conjugate acid. A weak acid has a less stable conjugate base. A weak base has a less stable conjugate acid.
Example: $\ce{HCl}$ is the poster child of strong acids. Its conjugate base is $\ce{Cl^-}$, which is basically unreactive in water solution, thereby making $\ce{HCl}$ a strong acid. Remember that IF $\ce{Cl^-}$ were more reactive, then the reverse reaction (un-dissociation, if you will) will be favored, and this isn't going for $\ce{HCl}$'s cause - which is losing its hydrogen proton and keeping it lost.
$\ce{HCl + H_2O -> H_3O^+ + Cl^-}$
This reaction doesn't go backward to any significant extent; i.e.
$\ce{Cl- + H_3O^+ -> not much of a reaction}$
Therefore, $\ce{HCl}$ is a strong acid because it basically ionizes 100%; there is no reverse reaction.
Bottom line for the OP: Take everything your general chemistry textbook has to offer with a grain of salt. Right now your textbook is helping you develop a hammer to take to chemistry problems - crude, but sometimes effective. Later in your studies you will progress to finer tools - the chisel and scalpel, if you will. Here are some common fairy tales offered by general chemistry textbooks:
1) "$\ce{OH^-}$" - the negative formal charge in the hydroxide ion is mainly stabilized by the more electronegative element (as common sense would dictate, and that element would be oxygen). So write $\ce{HO^-}$ and if your teacher asks you what the hell that is, smack him or her down with the science.
2) $\ce{sp^3d^2}$ hybridization and other explanations of hypervalent molecules with d-orbital utilization. This myth started with the misreading of a chemical paper which made use of d-polarization functions (quantum mechanic stuff) as saying that hypervalent molecules access their d-orbitals on a regular basis. D-functions are not d-orbitals. Species such as sulfate ion are best represented with a +2 formal charge on the central atom. Linus Pauling was wrong about Vitamin C and he's wrong about bonding too. So is your textbook (likely, unless it was updated since 1990, the year in which a seminal paper was published which definitely refuted the use of d-orbitals in hypervalent molecules). Note that textbooks may be "updated" yearly but most of the time the only changes are 1) a new photo of a chemical molecule on the cover and 2) a new edition stamped onto the spine.
3) The existence of the $\ce{H^+}$ proton in water. No. This is Coulombically (and comically) unlikely. The high positive charge density of the bare proton means that it is hydrated by water almost immediately in water, creating the hydronium ion. So in writing reactions, write $\ce{H_3O^+}$, not $\ce{H^+}$.
4) The 90 degree bond angle. No, methane ($\ce{CH_4}$ does not have any 90 degree bond angles. 90 degree bond angles are actually relatively rare in chemistry. So no, a methane molecule does not look like a set of cardinal axes with H's stuck on the ends where North, East, South, and West are supposed to be. It's a tetrahedron; draw it like a tripod or draw it with projections.
5) $\ce{NaCl_(aq)}$ and other "aqueous" species which actually exist in their dissociated states in water. There is no unionized sodium chloride in water. Proof - put some table salt in water. Do you see anything in the water? No, all the crystals dissolved. So don't write $\ce{NaCl_(aq)}$. Your teacher may ask you to write the net ionic equation for reactions; in my opinion, this is the only correct equation one can right.

Overall, this is a pretty solid answer for a beginner to acid-base theory. It's simple and precise while touching on the right topics that may be investigated further by the reader. If I can suggest a slight couple of fixes, it would probably be best to balance the charges in the equation which produces hydroxyl ions out of sulphuric acid (even if the reaction itself is unreal), and I do believe that right after describing the actual ionisation reaction, you mean to say Arrhenius acid instead of base.

Thank you for the feedback! Very excited to hear that my answer was decent! I have balanced the charges in the imaginary ionization of sulfuric acid. About Arrhenius acid - I did mean to say Arrhenius base - in a reference to the two things I listed previously about what made an Arrhenius base. It is true however that it is confusing I mention Arrhenius base right after talking about an acid. Point taken and edit made :). I've tried to be as thorough as possible and as chemically correct and perhaps even a bit "fun" here as possible precisely because I know am speaking to a new chem student :)

Is this college/ AP level chemistry? Man, I have never been taught 30 percent of this stuff.

What haven't you covered? I learned this in my second semester of college.

I'd just like to add a small commentary to Dissenter's nice answer. Not only is it important to analyze the structure of the solute to determine its acidic or basic behavior, but also the structure of the solvent. We are very used to thinking about aqueous solutions of acids and bases, but that's just a particular case, and limiting oneself to water hinders a deeper understanding of acidity and basicity.
A classic example is to consider nitric acid, $\ce{HNO3}$ as a solute. In liquid water as a solvent, nitric acid is a strong acid, and its behavior is represented by the formal reaction:
$$\ce{HNO3 + H2O -> NO_3^- + H3O^+}$$
However, if pure acetic acid $\ce{H3CCOOH}$ is used as a solvent (also known as glacial acetic acid), nitric acid is a weak acid, behaving like so:
$$\ce{HNO3 + H3CCOOH <=> NO_3^- + H3CCOOH_2^+}$$
But the real fun is what happens if pure sulfuric acid $\ce{H2SO4}$ is used as a solvent. Here, nitric acid is a weak base! Instead of the following reaction:
$$\ce{HNO3 + H2SO4 <=> NO_3^- + H3SO4^+}$$
, what happens in reality is:
$$\ce{HNO3 + H2SO4 <=> H_2NO_3^+ + HSO4^-}$$
The take-home is just that acidity and basicity are somewhat more fluid concepts than they might at first appear from Arrhenius theory, so avoid being too strict about classifying and memorizing what substances are acids, bases or neither.

Exactly, and to expand on what Nicolau mentions, we can discuss the leveling effect. A chemist (can't remember who) found that some acids - e.g. hydrochloric acid and nitric acid - have the same strength in water. How can this be? They're clearly different, yet they have the same strength in water. This is because these strong acids are "leveled" in water solvent. Take-home:
The strongest acid that can exist in any solvent is the conjugate acid of the solvent. The strongest base that can exist in any solvent is the conjugate base of the solvent.
So for water, the conjugate acid (remember that conjugate means inverse; acid means proton-losing, so inverse/opposite of proton losing is proton gaining) is the hydronium ion, which is formed precisely by tacking a proton to water (this is a Lewis acid/base or coordination chemistry reaction).
And the conjugate base of water is the hydroxide ion. So the strongest base that can exist in water solution is hydroxide ion - anything stronger will be leveled to hydroxide ion. Take the following leveling reactions as examples:
$\ce{HCl_(l) + H_2O -> H_3O^+ +Cl^-}$
Here pure, liquid hydrochloric acid is leveled to its conjugate base and hydronium ion.
$\ce{S^2- + H_2O ->HS^- + HO^-}$
Here, a strong base, sulfide ion, is being leveled to its conjugate acid and hydroxide ion.
So I agree, don't try to memorize any list of strong acids or bases - such a list is just about useless and often the list only works for situations in which water is a solvent.
Also this discussion reminds me of a good quiz question my teacher once asked me:

If one draws out the Lewis structure for the solvent, one finds that there is a lone pair on the nitrogen. This parallels ammonia, $\ce{H_3N}$, which also has a lone pair on the central nitrogen atom. The point was for us to draw the parallel between ammonia and liquid methylhydroxylamine. Upon realizing that, we can then realize that $\ce{H_2S}$ is a strong acid in this solvent because even in water solvent, $\ce{H_3N}$ is effectively protonated by $\ce{H_2S}$ - the two undergo a large extent reaction. One can ascertain this reaction quantitatively through calculating a complex K for the following reaction:
$\ce{H_2S + H_3N <=>H_4N^+ + HS^-}$
$K_{above~reaction}=\frac{[HS^-][H_4N^+]}{[H_2S][H_3N]}=\frac{K_a(reactant~acid)}{K_a(product~acid)}=\frac{(1.1*10^{-7})}{(5.6*10^{-10})}=196=reaction~progresses~far~to~the~right$
Or simply by comparing the $\ce{K_a}$ values of the reactant acid and the product acid. Off the top my head (I basically got the $\ce{K_a}$ values of all the acids memorized because of how much we studied B/L A/B chemistry), they are (also mentioned above in the calculation of the K for the reaction):
${K_a(H_4N^+) = 5.6 x 10^{-10}}$
${K_a(H_2S) = 1.1 x 10^{-7}}$
So as we can see, we are going from a stronger acid (not strong acid) to a weaker acid. It makes sense that this equilibrium reaction would be a large extent reaction to the right; strong to weak; waterfalls; high to low, etc. Analogize as you see fit.
OP: Let the chemistry talk to you; don't let the rules dictate everything because in chemistry, the "rules" are mostly illusions.
EDIT: And further expanding on the fact that acidity and basicity are rather fluid concepts, consider this myth: neutral water has a pH of 7.0. No. The only requirement for neutrality is $\ce{[H_3O^+]=[HO^-]}$ in a solution. Consider the following reaction:
$\ce{2H_2O + heat ->H_3O^+ +HO^-}$
The reaction is stoichiometric - i.e. hydronium and hydroxide ions are created in a 1:1 ratio.
At 25 degrees Celsius, the concentration of both ions produced in this auto-ionization reaction is ${1.0 *10^{-7}}$ M.
However, if you heat up water, the reaction goes further to the right (Le Chatlier's principle - disturb a system at equilibrium and the system will try to return to equilibrium; in this case the system will try to consume the extra heat through further auto-ionization).
So what does this mean? It means that hot water (i.e. steam) is dangerous.
Hot water is BOTH acidic and basic (i.e. corrosive) - and highly so. So don't come into contact with super-heated water/steam. Contact with skin may require a visit to your dermatologist or quite possibly your undertaker.


text:
Ionizable hydrogen atoms for acids
Given that hydrogen atoms that are attached to a compound must be ionizable for the compound to be an acid, what prevents a hydrogen from being ionized? Is it solubility. If the bond is nonviolent, water could not dissolve the compound, and so hydrogen atoms are not ionizable. Is this correct?

Not all hydrogen atoms in a compound are equally acidic. The acidity of a hydrogen at a certain position in a molecule depends on its chemical environment, i.e. the solvent (if the compound is dissolved), which atoms and functional groups it is attached to, and how well the negative charge of the corresponding conjugated base is stabilized. A quantitative measure for the acidity of a proton is its $pK_a$ value which is defined as follows (reference):
$$pK_a=-\mathrm{log}_{10}K_a$$
with $K_a$ being the equilibrium constant of the dissociation reaction in water $\ce{HA + H2O \rightleftharpoons H3O+ + A-}$, in which the hydrogen is abstracted from the compound by heterolytic bond cleavage. $[\ce{H2O}]$ can be assumed to be constant (approx. 55 mol/l) and is therefore included in $K_a$:
$$K_{a}=\frac{[A^{-}][H^{+}]}{[HA]}$$
Let's take formic acid ($\ce{HCOOH}$) as an example. As the name of the compound suggests, it has at least one acidic proton, and if all hydrogens would be equally acidic and dissociate from the molecule, the reaction would look like this:
$$\ce{HCOOH + 2H2O \rightleftharpoons 2H3O+ + COO^{2-}}$$
However, this does not happen in aqueous solution, only one of the protons is acidic:
$$\ce{HCOOH + H2O \rightleftharpoons H3O+ + HCOO-}$$
The reason for this is that both hydrogens are situated at different positions in the structure:

With a $pK_a$ of 3.77, the hydrogen of the carboxyl group is significantly more acidic than the hydrogen bonded to the carbon atom. The typical $pK_a$ of an aldehyde hydrogen (H bonded to the carbon of a carbonyl group) is in the range of 17 (source), so such a position cannot be deprotonated in water.

And the reason the hydrogen of the carboxyl group is more acidic than the hydrogen directly bonded to the central carbon atom has to do with the electronegativity of carbon (or lack thereof).

Your argument would be much stronger if you consider aqueous solution in your equations, i.e. $\ce{HA + H2O <=> H3+O + A^{-}}$.

@Martin Thank you for your suggestion. I have edited my equations accordingly.

Ionization here (in terms of acids) refers to heterolytic bond clevage, not homolytic bond clevage (which would create a hydrogen atom, complete with its one electron). Unfortunately in the Brønsted-Lowry definition of acids, we are dealing with hydrogen protons - i.e. hydrogen atoms stripped of their lone electrons.
Of course we could take a whole hydrogen atom from a molecule, i.e. we could do this with water, and what we'd end up with is a hydrogen atom and a hydroxyl group. Unfortunately the lone hydrogen atom is rather unstable by itself - that's why at standard state conditions you find not the hydrogen radical but instead diatomic hydrogen. But then the bare hydrogen proton is rather unstable too in its own right; it's solvated in water actually (as the hydronium ion; it may even be solved by multiple waters).


text:
Preparation of phenylamine/aniline
This is the procedure I carried out in class:

Reduction of nitrobenzene by using tin and $\ce{HCl}$ acid
After the completion of the reaction, $\ce{NaOH}$ added to neutralize excess acid and dissolve any $\ce{Sn(OH)2}$
Steam distillation is carried out to obtain suspension of phenylamine
Salting carried out where $\ce{NaCl}$ salt is added. Aqueous layer discarded
Solvent extraction: dry ether added and mixture shaken vigorously. Aqueous layer discarded
Distillation under reduced pressure carried out and solid phenylamine left in flask

My questions:

Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?
Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

I'm not familiar with the procedure, so I have some questions. It doesn't sound like a distillation in step 6, it sounds like you just evaporated the solvent. Did you actually collect phenylamine in a receiving flask? Also, in step 3, what was collected in the receiving flask and what remained in the boiling flask?

@LDC3: In step 3, I collected a cloudy suspension of phenylamine and in the boiling flask the leftover tin and aqueous layer containing some NaOH etc. In step 6, obtained solid phenylamine (This was the procedure our teacher told us to follow but I want to know the reason behind each step... the last step seems a bit unnecessary to me... Shouldn't simple distillation be enough? )

Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?

You could have performed any procedure to remove the solvent:

Simple distillation  - This would be slower than what you did and could overheat the undistilled aniline causing decomposition.
Vacuum distillation - This is what you did, perhaps even on a special apparatus called a rotary evaporator. This method is still a distillation, even if your product was not distilled. Something was distilled.
Evaporation on a hot plate (or steam bath)-

Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

Steam distillation is suitable because the aniline is already in an aqueous suspension. If do vacuum distillation, you would need to

Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure distillation)
Vacuum distil the aniline

If you look at what you did, it is the same steps in a different order:

Distil the aniline
Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure
distillation)

Either is probably suitable. In the end, the steam distillation is easier and less hazardous than the full-blown vacuum distillation you would need.

This was extremely helpful. Just one thing: Is steam distillation used in scenarios where the two liquids are immiscible and reduced pressure distillation when there are miscible liquids but one of them will decompose if too much heat is supplied (like phenylamine)?

Steam distillation requires immiscible liquids to work well.

and distillation under reduced pressure works with miscible liquids (as in this example, ether and phenylamine)?


text:
Which process is used to separate Hydrogen and Oxygen from Water?
What's the process called where you can use electricity to separate the oxygen and hydrogen from  water? Is it electrolysis? Can it be applied to other liquids?

Problem in your question: should be be O and H, because $\ce{H2O}$ (water) is made of H (Hydrogen) and O (Oxygen). $\ce{CO2}$ is carbon dioxide, for reference.
And yes, electrolysis can be performed on many compounds, not just water.
An electrolytic reaction is where an electric current is used to drive an otherwise non-spontaneous chemical reaction. Splitting water is just one of such reactions.
You can read more here.

When I was younger I went to the Science Center in Toronto  (about 1975 I think) and there was an exhibit about Electolysis  showing the splitting water phenomenon (Ithink) BUT they also had a part of the exhibit that could take some oxygen and hydrogen and somehow re-integrate them into H2O (without any minerals or additives in it). A kind of REVERSE ELECTROLYSIS, is there such a process? Did the Science Centre in Toronto have such an exhibit in the mid-70's?


text:
How does carbonic acid cause acid rain when Kb of bicarbonate is greater than Ka?
How does carbonic acid cause acid rain when $K_b$ of bicarbonate is greater than $K_a$? Should it not create an alkaline solution?
$K_a = 4.8 \times 10^{-11}\ (mol/L)$. $K_b = 2.3 \times 10^{-8}\  (mol/L)$. These are the values for $\ce{HCO3-}$.
These numbers are from a school book that I read, but it's not in English. It's called "Kjemi 1" by Harald Brandt

What is the ${K_a}$  of carbonic acid? Note that sources differ in their ${K_a}$ values, and especially for carbonic acid, since there are two kinds - a pseudo-carbonic acid/hydrated carbon dioxide and the real thing (which exists in equilibrium with hydrated carbon dioxide but in a small concentration - about 4% of what what appears to be carbonic acid is true carbonic acid, with the rest simply being $\ce{H2O*CO_2}$.
Either way, I find that the ${K_a}$ of the mixed carbonic acid is about $4.2 \times 10^{-7}$, which is greater than $1.0  \times 10^{-7}$, and this implies that a solution of carbonic acid alone should be acidic no matter what. This suggests to me that your numbers are wrong; would you mind sharing your numbers and their source if possible?
EDIT: I see that you have updated your numbers. Nonetheless, I believe that your ${K_a}$ for carbonic acid is wrong; that number looks suspiciously like the ${K_a}$ instead for hydrogen carbonate ion (or the bicarbonate ion). According to Wikipedia, the ${pKa}$ of carbonic acid, is 6.3 (and this is taking into account any aqueous carbon dioxide). Convert this to a ${K_a}$ value and we get about $5.0 \times 10^{-7}$. This is in-line with the value I obtained from a copy of Daniel C. Harris' Qualitative Chemical Analysis.
EDIT 2: I think you've realized your mistake; as you say, the values are for $\ce{HCO_3^-}$, which is the hydrogen carbonate ion.

What do you mean? I asked specifically for HCO3-: "Kb of bicarbonate is greater than Ka?".

Okay, I think we need to revisit your original question about how carbonic acid can make a solution acidic. We need to consider what's in a solution of carbonic acid. In a solution of carbonic acid, we have 1) water and 2) carbonic acid in the main. We do not have bicarbonate ion in a sizeable amount in the solution; therefore, what controls solution pH is carbonic acid. There is no need to consider the various K values for bicarbonate ion because it's simply not in solution at a high concentration relative to carbonic acid - i.e. carbonic acid is a weak acid and does not ionize much.

Okay, but is it H2CO3 or HCO3- that causes acidic rain? My problem is that according to my book, HCO3- + H2O produces an acidic solution, thus giving acidic rain. But at the same time it states that HCO3- will react as a base, because it's Kb >> Ka

True, $HCO_3^-$ will react as both an acid and a base. So bicarb ion is amphoteric. And you are right that bicarb ion is a stronger base than it is acid. I believe that acid rain is caused by carbon dioxide being solvated by water to make "carbonic acid." In this solution of carbon dioxide, there are multiple equilibria, and the reaction of bicarb as an acid is one, but I wouldn't say it's the principle, pH driving reaction. That would be the ionization of "carbonic acid."


text:
The solubility of haloalkanes in water
Haloalkanes aren't very soluble in water because they can't form hydrogen bonds, and the energy required to break hydrogen bonds in water etc is higher than the energy released when new bonds between the haloalkane and water are formed.
But as fluorine forms hydrogen bonds why couldn't a fluoroalkane form hydrogen bonds with water thus become more soluble in water?

Related: Why is dichloromethane immiscible in water?

I think the answer lies with the fact that alkanes are basically non-polar, even if they have halogen substituents. Water, on the other hand, is polar, so the solvent and solute are incompatible.

What is it about alkanes that makes them non-polar? Wouldn't the asymmetric distribution of electronegativity, especially with fluorine, create a permanent dipole?

Alkanes are saturated carbon chains. The geometry about each carbon is tetrahedral and tetraheral shaped molecules can be non-polar if all their attachments are the same. Yes, adding a fluorine will create a dipole, but there's more than there just being a dipole; we must ask how significant is the dipole. The answer is usually not that significant.

I'm sorry to nag, but why is it not usually significant?

I like to think of fatty acids - lipids - which make up the phospholipid bilayer membranes. These, yes, have a permanent dipole; they have hydroxyl end groups (which makes them acidic). However, they are very very long carbon and hydrogen chains. So even though they have a permanent dipole, this dipole is weak, and they are essentially non-polar, rendering them not very permeable to substances such as water. As a result, there is a need for transmembrane water transport proteins - aquaporins.

Yes I initially thought that for long alkanes the dipole across the whole molecule is insignificant, but then I considered something like methanol and couldn't really justify why this is important.  Methanol has the electronegative oxygen, sure, but this is less electronegative than fluorine, and hence I can't see why methanol is polar when fluoromethane isn't. Also, the strength of the dipole across the C-F bond itself won't change as the alkane increases in length, and this is where the hypothetical hydrogen bonding would occur.

@Joe Fluoromethane is polar and actually quite soluble in water (2.295 g/L, from Wikipedia). With increasing length of the carbon chain, this dipole is not affected to a great degree. However, when the hydrophobicity of the alkyl chain overcompensates the effect of hydrogen bonding to the fluorine, the solubility of the organofluoride will quickly decrease with increasing chain length, as a comparatively small part of the molecule is readily solvated by the water molecules.

Yep. Some of the smaller organofluoride and halogen substituted alkanes are indeed soluble but as you stated, halogenalkanes are generally insoluble, and that's because most halogenalkanes are long enough as to render the permanent dipole insignificant.

@Joe, in the case of methanol, the hydroxyl group is both an H-bond donor and an H-bond acceptor. Haloalkanes can only function as acceptors. Furthermore, in the case of fluoroalkanes specifically, the small atomic radius and high electronegativity of fluorine makes it very weakly polarizable. Also, the short C-F bond length lowers the dipole moment (by comparison to, e.g., C-Cl bonds), since dipole moment is a function not just of charge but also separation distance.

Come to think of it, this is not quite obvious. Organics with oxygen bound to carbon seem to be able to form hydrogen bonds; acetone, for example, is infinitely miscible with water. One might think that fluorine in similar compounds (being more electronegative and thus bearing higher negative charge) would be even better at accepting hydrogen bonds. This is not the case, though.

Yes, it's true that fluorine forms Hydrogen bonds with water (and related solvents). However, fluoroalkanes are not appreciably soluble in water because the Hydrogen bonds are simply not strong enough. Fluorine forms only one Hydrogen bond (compare it to 2 of oxygen (in an alcohol) and 3 of nitrogen). Plus, the electronegative nature of fluorine is further suppressed by +I effect of the alkyl groups (specific to fluoroalkanes)
For instance,
the solubility of Fluoromethane:(2.295 g/L)
the solubility of Methanol: completely miscible
Hope this helps.

The simple truth is that the only molecule where F forms hydrogen bonds is H-F. In C-F bonds, the F does not form hydrogen bonds. We may argue on the explanations, but those are the facts. Just remember what happens when a drop of water falls on a clean pan coated with Teflon.

Perhaps you could elaborate on why this is. This sounds like an interesting answer but at the moment it is lacking in detail.

Anything that dissolves in water must overcome the inter-molecular forces (H-bonding here) between water molecules first.

The insolubility of fluoroalkanes can be explained on the basis that energy required to break the bonds between water molecules is higher than the energy released when new bonds are formed between fluoroalkanes and water.

The haloalkanes are only very slightly soluble in water. In order for a
haloalkane to dissolve in water, energy is required to overcome the
attractions between the haloalkane molecules and break the hydrogen
bonds between water molecules. Less energy is released when new
attractions are set up between the haloalkane and the water molecules
as these are not as strong as the original hydrogen bonds in water. As a
result, the solubility of haloalkanes in water is low. However, haloalkanes
tend to dissolve in organic solvents because the new intermolecular
attractions between haloalkanes and solvent molecules have much the
same strength as the ones being broken in the separate haloalkane and
solvent molecules.

You are not answering the question... as to why a flouroalkane doesnot dissolve (appreciably) in water.

I felt that fluorine would be able to form hydrogen bond only when it is initially bonded to hydrogen atom so that it gets significant electron cloud, in our case alkyl group is attached to fluorine so it becomes difficult for fluorine to hogg electrons now when fluoroalkane is dissolved in $\ce{H2O}$ attractive interactions between fluoroalkane and $\ce{H2O}$ is not strong enough to break hydrogen bond which binds $\ce{H2O}$ molecules. Thus fluoroalkene is not much soluble in water as it won't be able to form bonds with $\ce{H2O}$ molecules.

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text:
What chemical test can differentiate secondary and tertiary alcohols?
I have to perform a chemical test to determine whether the solution is a secondary or a tertiary alcohol. How can I do this? Please explain the procedure and the apparatus.

Or you can try the very simple Lucas' test using zinc chloride and concentrated hydrochloric acid:
$$\ce{ROH + HCl -> RCl + H2O}$$
It works best with tertiary alkohols, slow with secondary alcohols and not at all with primary alcohols.

Add sulfuric acid to a chilled aqueous solution of the alcohol and sodium nitrite.
Record the UV spectrum.

Tertiary alkyl nitrites show an absorption band at $\lambda$ = 400 nm, secondary and primary alkyl nitrites only below.

A secondary alcohol can be further oxidized to a ketone (reference), while a tertiary alcohol cannot. For example, when you use dichromate as oxidation agent, you will observe a color change of the solution from orange to green. A tertiary alcohol will not react unless your oxidizer is too concentrated.

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text:
Equivalent Cyclohexane Structures?
Are these structures equivalent?
Image 1:
http://i.minus.com/jnojtCx6wYvqD.png
Image 2:

Yes, the two structures you've drawn are the same; both substituents are in axial positions.  The two structures shown in the link you provided are not the same; they are conformational isomers.  The one on the left is the same as the one you've drawn with both substituents in the axial position.  The one on the right has both substituents in the equatorial position.  These two conformational isomers exist in equilibrium with the (e,e) isomer favored.  The two conformers interconvert by a cyclohexane ring flip.

Thank you. I was trying to make a comparison between image 1 and image 2; are image 1 and image 2 the same? Because in image 1, the picture has the chlorine with dashes and the ethyl group with a wedge. I however have it switched. So would both be valid representations of 1-chloro-2-ethylcyclohexane?

Image 1 and Image 2 are not the same.  Image 1 shows two conformers (the a,a and the e,e conformers) of 1-chloro-2-ethylcyclohexane that interconvert with one another via a cyclohexane ring flip.  Image 2 shows just the a,a conformer.  For completeness I'll mention that there are more isomers of 1-chloro-2-ethylcyclohexane, namely the a,e and e,a conformers which again interconvert via a cyclohexane ring flip.

Okay! I just wanted to make sure that both images were images of 1-chloro-2-ethylcyclohexane.

Also when you say (e,e) isomer is that referring to the conformational isomer with the equatorial bonds because you say "favored" in equilibrium - and as I understand, placing the substituents in the equatorial positions helps minimize the diaxial interactions/steric strain?
Also I see what you mean by how they're non-equivalent; one is favored at equilibrium.

Yes, (e,e) means both substituents are in the equatorial position; (a,a) would have both in the axial position.  In equilibria, conformers with substituents in the equatorial position are generally favored over conformers with axial substituents.  If you look at a model of cyclohexane you'll see that an axial substituent at position 1 has an unfavorable steric interaction with hydrogens (or substituents) at the axial 3 and 5 positions.

Thank you. I'll have to get myself a model kit. I understand that the chair flips but I still want to see the mechanism/how it flips.

@ron This link to the image shows the enantiomer of molecule that is pictured above. The left figure has the ethyl group up and the chloride down, which is the same as axial, axial in one of the conformers of the boat (the other being equatorial, equatorial).


text:
Natural Bond Orbital analysis: Significance of stabilization energy determined by 2nd order perturbation
PREFACE: I am no expert on this topic.  My questions at the bottom may be off base.  I have some experience with symmetry-adapted perturbation theory (SAPT) when it comes to analyzing intermolecular interactions.  The total interaction energy of a system can be quantified in four different energy components including exchange, induction, electrostatics, and dispersion.
Natural Bond Orbital (NBO) analyses translate the complex quantum-mechanical wavefunction into a more tangible Lewis-dot-like formalism (natural Lewis structure).  The introduction of the following series of equations has been largely influenced by the presentation of Weinhold and Landis (2012).
Important Terms:

Lewis-type NBOs - filled 'donor' orbitals (e.g. $\sigma_{AB}, \sigma_{CD}, \cdots$)
Non-Lewis-type NBOs - vacant 'acceptor' orbitals (e.g. $\sigma_{AB}^{\ast}, \sigma_{CD}^{\ast}, \cdots$)

Lewis and non-Lewis NBOs mix:
Consider a wavefunction, $\Psi$, comprised of two components, a known natural Lewis structure wavefunction, $\Psi^{(L)}$, and a non-Lewis correction wavefunction, $\Psi^{(NL)}$, such that
$$\Psi = \Psi^{(L)} + \Psi^{(NL)}$$
The natural Lewis structure wavefunction is simply a fictitious representation of an exact physical system where resonance effects are non-existent.
Now imagine a corresponding Hamiltonian, $\hat{\mathcal{H}}$, which can also be treated as an additive scheme with corresponding components to each piece of the wavefunction:
$$\hat{\mathcal{H}} = \hat{\mathcal{H}}^{(L)} + \hat{\mathcal{H}}^{(NL)}$$
Given these two equations and the form of the Schrödinger Equation, $\hat{\mathcal{H}}\Psi=E\Psi$, there must also exist a similar construction of the energy of the system, $E$, such that
$$E = E^{(L)} + E^{(NL)}$$
These three equations provide the foundation for a "systematic 'perturbation theory' analysis", where $\hat{\mathcal{H}}^{(L)}$ is the unperturbed Hamiltonian with a corresponding (known) eigenfunction, $\Psi^{(L)}$, and energy eigenvalue $E^{(L)}$.  Note that $\Psi^{(L)}$ is not only known, but it is related to an 'idealized single reference picture' in which each Lewis-type NBO has exact double occupancy for closed shell systems.
As with Hartree-Fock, the limit of a single-reference method can be reached by an infinite expansion of a basis ($\Omega)$ in the one-electron eigenvalue equation.  For the natural Lewis structure, this can be written as
$$h^{(0)} \Omega_i^{(L)} = \varepsilon_i^{(L)}\Omega_i^{(L)} \qquad \mathrm{where~~} i = 1,2,\cdots,n$$
where $\varepsilon_i^{(L)}$ are orbital energies of Lewis-type NBOs.
Because $h^{(0)}$ also contains the non-Lewis contributions such that
$$h^{(0)} \Omega_j^{(NL)} = \varepsilon_j^{(NL)}\Omega_j^{(NL)} \qquad \mathrm{where~~} j = n+1,\cdots$$
Because natural Lewis structures contain zero resonance, the donor (Lewis-type NBOs) and acceptor (non-Lewis-type NBOs) orbitals do not interact due to their orthogonal nature
$$\int \Omega_i^{(L)\ast}h^{(0)}\Omega_j^{(NL)}d\tau=0 \qquad \mathrm{for~all~~} i,j$$
However, a real-world $1e^-$ Hamiltonian operator, $F$ (i.e. Fock operator, Kohn-Sham operator, etc.), will have donor-acceptor interactions such that
$$F_{ij} = \int \Omega_i^{(L)\ast}F\Omega_j^{(NL)}d\tau\neq0$$
leading to mixing (delocalization) which connects the non-Lewis type NBOs with Lewis-type NBOs.
Second-Order Perturbative Treatment of Delocalizations
This mixing of the donor and acceptor orbitals can be treated with second-order perturbation theory.  Even though $\Psi^{(L)}$ ignores interactions with non-Lewis acceptors, it contains more than 99% of the total electron density, $\rho_L$, offering a good starting point for an unperturbed wavefunction.
The mixing of donors and acceptors lead to an overall energy lowering ("stabilization"), a quantum mechanical phenomenon.  Consider two interacting orbitals, as schematized below.  One is doubly-occupied with a lone electron pair, $n$, and the other is an unoccupied antibonding orbital (here a $\pi^{\ast}$ orbital).
$\hskip1in$
The overlap of the lone electron pair, $n$, with the vacant antibonding orbital, $\pi^{\ast}$ causes an change in energy of the lower occupied orbital, $x$. This change in energy between the non-mixing orbital with $n$ and the mixing orbital $x$ (denoted as $\Delta E$) is referred to as 'stabilization energy' of the electron delocalization.  This energy (in kcal mol$^{-1}$) is determined via the following equation:
$$\Delta E_{ij}^{(2)} = \frac{-q_i \left| F_{ij} \right|^2}{\left(\varepsilon_j^{(NL)}-\varepsilon_i^{(L)}\right)}$$
where $\varepsilon_j^{(NL)}$ is the energy of the non-Lewis NBO (i.e. $\pi^{\ast}$),  $\varepsilon_i^{(L)}$ is the energy of the orbital occupied by $n$, and $q_i$ is the occupancy of the orbital ($q=2$ in the above figure).  The 'stabilization energy' $\Delta E_{ij}^{(2)}$ as determined by second-order perturbation treatments is commonly abbreviated as $E(2)$.
Recap:
So we have introduced (albeit very poorly) the concept of a separable wavefunction into natural Lewis and non-Lewis type formalisms.  These two states mix, allowing for delocalization of electrons, a phenomenon which leads to lower energy states.  This energy lowering is commonly called 'stabilization energy'.
Physical Significance?
Unfortunately, after such an introduction, I am still at a loss as to what all of this really means.  As a theoretical chemist (I'm a chemist, theoretically speaking...), I am interested in the physical significance of this type of quantity [$E(2)$].  However, I am unsure how significant this type of value truly is.  Do these delocalizations (energy stabilizations) lead to an overall lower energy of the system?  Would a system with more of these delocalizations be expected to lie lower in energy than a comparable system without delocalizations?  Can that type of correlation be made?
I guess I just do not see any usefulness in computing these $E(2)$ values for any reason because, while it may tell you that particular donor-acceptor interactions lead to this energy lowering, it doesn't offer much more than that.  What am I missing here?
TL;DR - What can I use $\mathbf{E(2)}$ values for?

I think you are missing that the number behind $E(2)$ can tell you a lot about the energetic cost when breaking resonance, which in turn affects reactions in which the molecule analysed takes part.

@tschoppi So say a model dimer system has an E(2) of -10 kcal/mol for a particular intermolecular interaction and has an overall electronic binding energy of -20 kcal/mol.  Does that mean that half of the binding energy is due to this E(2) value?  Can that type of correlation even be made?

I think that's exactly the kind of information that you'll get out of this energy. However, this is just based on gut feeling and as such isn't an answer backed up by references.

This value gives you a hint on how "accurate" your Lewis structure is. The smaller it is, the better it will be described by Lewis. As being theoretically a chemist myself, this is the only thing I could cook up in an instant. Please always keep in mind, that NBO is an Analysis only and it is strongly dependent on what you want to find. ({default} NBO for $\ce{H2O}$ gives a sp- and p-lone-pair orbital...) But I like your question very much, if I would have time...

@Martin I hate NBO analyses because its qualitative by nature.  You can't converge results, unlike something like SAPT.  I suppose that is why I've never used it in the past.  But now that I'm using it, I'm having a hard time accepting it (because it doesn't do things it isn't supposed to do that I want it to do).  NBO seems 'nice' but 'not really useful' in my opinion but then again my opinion is really based on inexperience.

I guess this is rather something for a chat than discussing it here (we already broke that record). From my experience NBO is a fantastic tool for bonding analysis. It will give you the structure that every chemist will understand. The Lewis concept itself is a bit of a stretch for most molecules, so is any tool that tries to force it onto them. However, NBO gives you more robust and reliable charges (NPA) and they might present you with a nice set of local orbitals that impresses every experimentalist.

@Martin I don't question its usefulness as a way to qualitatively interpret the wavefunction into something more tangible that most people will understand.  I simply find its non-ability to be a quantitative and convergent approximation to be a complete turn off for me.  I've seen it used in the literature plenty of times to quantitatively describe something but it cannot really be used for that.  I'd rather rely on something that gives you can indication as to how 'correct' something is (i.e. convergent methods).  *sigh

If you are looking for physical significance I suggest you consider whether it is measurable,  I.e., does a Hermitian operator exist which will give you E (2).

@Deathbreath The associated operator is indeed Hermetian.  However, this is not really the 'problem'.  The problem lies with approximating the wavefunction as an additive scheme involving 'Lewis' and 'Non-Lewis' components.  This approximation is largely unphysical if I recall correctly.

@LordStryker That is actually exactly the crux. Non-Hermitian operator = unphysical. Hermitian operator= find experiment that measures said operator('s expectation value). The HF energy is also "unphysical" in that it does not describe any real system (so far). It does not mean that its  contribution cannot be measured.

LordStryker, I took some small(?) editorial liberties in my edits to the question -- please feel free to change (back) anything you don't like.

The bounty was very kind, LordStryker - thank you!

TL;DR: Lewis $\to$ Non-Lewis $\mathbf{E(2)}$ values have no direct physical significance, are intrinsically un-measurable, and serve only to quantify the extent to which the "real" wavefunction for a system deviates from the fictional idealized Lewis-structure wavefunction.
$E(2)$ values do, however, correlate with a variety of trends in chemical bonding and reactivity, and thus can be helpful in interpreting experimental or computational data or in highlighting potentially interesting lines of inquiry.

From a comment:

LordStryker: This approximation is largely unphysical if I recall correctly.

You do recall correctly. I'm reading Weinhold's Valency and Bonding (1st ed.) currently, and the first chapter is peppered with definitions of artificial Hamiltonians and Fock operators. For example, the first example he gives is of a Hamiltonian operator for independent (not field-averaged!) electrons:
$$
\hat h = \hat t\!_\mathrm e + \hat v_\mathrm{ne},
$$
where $\hat t\!_\mathrm e$ and $\hat v_\mathrm{ne}$ are the kinetic energy and nucleus-electron interaction operators, and the electron-electron interaction operator $\hat v_\mathrm{ee}$ is completely absent!  If I'm reading the text correctly, the proper Hartree-Fock Hamiltonian is then defined in terms of $\hat h$ and a "perturbation" operator that is essentially just the missing electron-electron interaction term:
$$
\hat H^\mathrm{\small HF} = \hat h + \hat H^\mathrm{\small (pert)} \equiv \hat h + \hat v_\mathrm{ee}
$$
I don't think he actually ever uses these definitions in the course of developing the NBO methodology, but it's instructive that this is the first example he chose in defining the perturbation approach that underlies it.  To the best of my ability to determine:
The core of NBO analysis is founded upon selecting non-physical, but chemically intuitive, reference states, and quantifying the extent of the departure from these fictional references that is required in order to reach the "real" wavefunction of interest.
In other words, Martin is exactly right:

This value gives you a hint on how "accurate" your Lewis structure is. The smaller it is, the better it will be described by Lewis.

Responding to a couple of the specific questions at the end of the post:

Q: Do these delocalizations (energy stabilizations) lead to an overall lower energy of the system?

Absolutely, when compared to the fictitious idealized-Lewis reference.  In his 2012 book, Discovering Chemistry with Natural Bond Orbitals (the one cited at the top of the original post), Weinhold illustrates quite explicitly how coercing the wavefunction not to exploit these delocalizations leads to appreciably more-positive energies.  Again, though, the calculations carried out with these delocalizations prohibited are entirely unphysical, so it's not as though a system "knows it should delocalize as much as it can to get more stable"—any real system will already intrinsically exhibit all available delocalization that provides increased energetic stability.

Q: Would a system with more of these delocalizations be expected to lie lower in energy than a comparable system without delocalizations?

Yes. This is how NBO explains the spatial patterns of, e.g., hyperconjugation. In the Wikipedia article on the topic, for example, the role of hyperconjugation in establishing the energetic favorability of the staggered conformation of ethane is discussed.  Per the below figure (public domain), the staggered conformation allows electrons in a given $\ce{C-H}$ $\sigma$-bonding orbital to delocalize into the $\sigma^*$-antibonding orbital of a parallel $\ce{C-H}$ bond on the vicinal carbon:

From Valency and Bonding, p228, this favorable hyperconjugative delocalization is reflected in $E(2)$ values of greater magnitude:

Second-order perturbative estimates [$E(2)$ values] indicate that each trans-like donor-acceptor [vicinal $\sigma \to \sigma^*$] interaction stabilizes the [staggered-conformation] molecule by $\pu{2.58kcal mol^-1}$, compared with only $\pu{0.89kcal mol^-1}$ for the cis-like interactions [in the eclipsed geometry].  The smaller gauche-like stabilizations ($\pu{0.20kcal mol^-1}$ at $60^\circ$ in the staggered conformer, $\pu{0.70kcal mol^-1}$ at $120^\circ$ in the eclipsed conformer) diminish the difference somewhat, but still preserve a significant hyperconjugative advantage for the staggered conformer.

So, while it's experimentally impossible to quantitatively "measure the energy" of an ethane molecule in which hyperconjugative delocalization is forbidden to occur, the $E(2)$ values provide support for the qualitative argument of hyperconjugation as a significant element of the preference of ethane for the staggered conformation.  The relative (in)stability of various chemical features found across a wide range of systems can be examined by calculating judiciously selected $E(2)$ values.

Q: So say a model dimer system has an $E(2)$ of $\pu{-10kcal mol^-1}$ for a particular intermolecular interaction and has an overall electronic binding energy of $\pu{-20kcal mol^-1}$. Does that mean that half of the binding energy is due to this $E(2)$ value? Can that type of correlation even be made? (from this comment)

I agree with tschoppi: Based on my reading, yes, I think Weinhold would make exactly this kind of argument. I am ill-equipped to discuss in detail the validity of such an argument, however—though I think that there is at least qualitative, maybe semi-quantitative, value to it.


text:
Calculating Commutator of Differential Angular Momentum
I have tried for hours to calculate the commutator of angular momentum in the differential form, but I cannot get the correct answer. This is my first experience with actually checking if two operators commutes, so there may be some beginner's misunderstandings that causes the problem.
I know from my textbook the correct answer is:
$$
[\hat{L}_x,\hat{L}_y] = \frac{\hbar}{\mathrm{i}}\hat{L}_z = \frac{\hbar}{\mathrm{i}} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
$$
The angular momenta are defined as
$$
\hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\
\hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\
\hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}
$$
Sets up the commutator expression, and writes it out:
$$
[\hat{L}_x,\hat{L}_y] = \left[ y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} , z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right]
$$
From what I assume is a well-known algebraic rule for handling commutators, I get the following:
$$
\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right] - \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right]
- \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right]
$$
It was straight forward to find that the two middle terms commute, i.e. they equal zero. Hence, we are left with:
$$
\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right] \tag{1}
$$
Upon evaluating these terms, I let them act on a function $ \psi $. Looking at the first term:
First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on $ \psi $:
$$
\frac{\hbar}{\mathrm{i}} \left(y \frac{\partial \psi z}{\partial z} \frac{\partial \psi}{\partial x} - yz  \frac{\partial \psi }{\partial x}  \frac{\partial \psi}{\partial z}     \right)
$$
The y in the second term above can be factored out from the differentials, because the differential is not done with respect to y. In the first term, however, the z must be differentiated along with psi by using the product rule. Doing this, and cancelling equal terms of opposite sign, yields:
$$
\frac{\hbar}{\mathrm{i}} y \psi \frac{\partial \psi}{\partial x}
$$
Repeating this calculation for the second term in equation $(1)$ yields
$$
-\frac{\hbar}{\mathrm{i}} x \psi \frac{\partial \psi}{\partial y}
$$
Substituting into $(1)$ yields
$$
[\hat{L}_x,\hat{L}_y] \psi = \frac{\hbar}{\mathrm{i}} \left(  y \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial y}  \right) \psi
$$
Now, comparing with the correct answer:
i) My signs are opposite of what they should be
ii) More importantly, I have a $ \psi $ in the differentials which should not be there. My question is, "why, and how do I rid my answer of it?".
I hope I have sufficiently showed what I have done. I have checked my notes several times, and I cannot get rid of those extra psi's in my answer.

Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see here) is
\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z
\end{equation}
Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1].
So, substituting the differential operator expression for $\hat{L}_z$
\begin{equation}
 \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}
into the commutator equation you should get:
\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}
As for your problem with the additional $\psi$:
Everything is fine until you let equation [1] act on a wavefunction $\psi$.
\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \right) \psi
\end{equation}
You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations:
When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following
\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)}
\end{equation}
which is incorrect. The correct way to do this is
\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)}
\end{equation}
where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result.
So, in the case at hand this leads to
\begin{align}
  \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right] \psi &= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\
  &= -\hbar^{2} y \frac{\partial \psi}{\partial x}
\end{align}
and
\begin{align}
  \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \psi &= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\
  &= \hbar^{2} x \frac{\partial \psi}{\partial y}
\end{align}
If you substitute these equations into the commutator equation you get
\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)
\end{equation}
which is exactly the result you should expect.


text:
Line structure of formaldehyde
Would the line structure of formaldehyde, H2COH, simply be a straight line with a OH on the end of the line?

Formaldehyde, $\ce{H_2CO}$, has the following structure

There is no $\ce{OH}$ group, just a central carbon atom with a single bond to each of two hydrogens and a double bond to oxygen.

Thanks. I don't know why I thought there was a hydroxyl group all of a sudden! Guess I was thinking of formic acid!

When you omit the hydrogens, it would be a straight double line with an O at the end.


text:
What is this red substance that has accumulated at the bottom of a plastic bottle full of petrol (see picture)?
About two weeks ago I put some ordinary unleaded petrol into a plastic water bottle, as a temporary measure (long story, don't ask!). It has now formed a reddish liquid at the bottom of the bottle.
The bottle was emptied prior to adding the petrol, but probably still contained a small amount of water when the fuel was poured in. The bottle was stored in ordinary room temperature.
I presume that the fuel has corroded/reacted with the plastic and left this substance, but I have looked to Stack Exchange in the hope of a more scientific answer...
So, what is the red substance and cause of it?

Please note that the brownish 'slug-shape' is merely an artifact of the light refraction...

I assume that the petrol you put in was poured out the same day and what little remained is what turned brown. Is that correct?

No, the bottle was 'clean', then filled to the brim with fuel, sealed, and the red/brown is a liquid that settled on the bottom whilst the bottle is still sealed and full of fuel. The picture above has a yellowish tint: it is still sealed and full of fuel, it is not empty in the picture. Very strange....!

If there is still petrol in the bottle, then the lower layer must be mostly water. Is the fuel an ethanol blend?

It's super unleaded (so 95 octane) from a UK petrol station.... Sorry, I'm no Chemist so don't know!

Any idea what plastic it is?

It might say on the bottom in symbols, or on the neck.

95 unleaded for sure contains good amount of oxygenates and/or alkenes. My bet, traces of water leaded to separation of oxygenates (alcohols) into separate phase and then formed polar phase extracted some colored components from the mixture. However, it is impossible to give exact details without complicated chemical analysis or full listing of fuel composition

Could be a chemical reaction between the fuel and the bottle. Some plastics e.g. LDPE are not very resistant to just the sort of hydrocarbons of which petrol is composed.

Maybe could be some iron oxide or hydroxide from previous storage or ( if you have a iron tap) from it ore maybe directly from the plastic.
Some fuel has red dye in it but I don't think this is the case, because I don't see how it can precipitate.
Maybe it could be a residue of petroleum resin.

It came from the pump, into a jerry can then into the plastic bottle...the fuel in the jerry can is clean and doesn't have this red substance which makes me think it must be reacting with the plastic.

I wouldn't think it would be iron (III) oxide or iron hydroxide in that kind of quantity. I think it could be petroleum resin though.

Do you have any sources to support 'petroleum resin' being present in petrol (gasoline)? It seems unlikely to me.

@nekomatic no, is more an hypothesis...


text:
Equivalent Line Angle Structures?
Are the following equivalent structures for 2,2-dimethylpropane?

I think so, because my rendering simply rotates the cross 45 degrees!
Also, regarding 3,3-diethylpentane, are these both equally valid structures? My rendering seems to look way different but filling in the carbons and hydrogens, both appear correct:

Ah, the swastika molecule …

a) yes, those are equivalent structures of 2,2-dimethylpropane
b) yes, those are equivalent structures of 3,3-diethylpentane


text:
Identifying nucleophiles and electrophiles in an acid-base reaction

Is ammonia a nucleophile and hydrogen the electrophile in the Brønsted-Lowry acid-base reaction of acetic acid with ammonia?
$$\ce{CH3COOH + NH3 <=> CH3COO- + NH4+}$$

I say that ammonia is the nucleophile because I drew a curved arrow from the lone pair on the ammonia to the ionizable hydrogen on acetic acid, and I drew a curved arrow from the bonding pair in the hydroxyl group to the oxygen in the hydroxyl group of acetic acid. It seems that ammonia is the nucleophile because it is attacking the hydrogen proton (a bare proton) and that hydrogen is the electrophile because it likes the lone pair on ammonia.
So if the above is right, this would make ammonia a Lewis base and the hydrogen proton a Lewis acid? Or do we say the entire acetic acid molecule is a Lewis acid?

In my opinion, your statements are both right, this is the IUPAC definition [1]:

Lewis acid: A molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:
$$\ce{\underset{Lewis acid}{Me3B} + \underset{Lewis base}{:\!NH_\phantom{3}} -> \underset{Lewis adduct}{Me3\overset{-}{B}-\overset{+}{N}H3}}$$

You can see that the author says molecular entity (atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity [1]) and the corresponding chemical species (a boarded concept, that in this case I interpret as the combination of different molecular entity), so I think that the exegesis of this definition is that both the acetic acid and the hydron can be called a Lewis acid. And in fact is what usually all the chemists use.
Reference

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


text:
Electron Pushing
I have a problem with this graphic, specifically the right hand side.

Why a "sextet" atom (an atom with 6 valence electrons)? I thought only a select few elements (C, N, and O) actually follow the octet rule.

Are you referring to BF3?

That would be an example, yes.

Where you find this picture? Very misleading. When sextet is mentioned, the better way is to say is lone pair acceptor or Lewis Acid. You simply need one empty molecular orbital on the atom to be able to do all the steps in the picture. In the case for common B, C, N and O they happen to be preferring sixtet. However, for heavier element, they are wrong. If you change the $BF_{3}$ to $FeCl_{3}$ in the picture, the reaction will be also true. But the iron center is not sextet, it is $3s^{2}3p^{6}3d^{5}$ in its outer electron layer.

It's the first Google result for "electron pushing" and it's from the U. Wisconsin, Madison website (not that I go there; I actually happen to be a few thousand miles away).

A "sextet" atom has a lack of electrons and is eager to get 2 additional valence electrons to fill its octet. By a Lewis acid-base reaction (between ether and $\ce{BF3}$) or lone-pair donation from a neighboring atom (the mesomeric structures of the two carbocations), another bond with 2 electrons to the sextet atom is made and its octet is completed.


text:
Correct Electron Pushing with HCl and Water
My book indicates that the following electron pushing is wrong:

So what would the right electron pushing diagram be?
I have the following in mind. My rationale is:
1) Electrons attack protons, not the other way around. This is just convention right? Is this just a conceit in chemistry; just a standardization? On a related note is this why one person described organic chemistry as:
"The foundation of every reaction in organic chemistry can be found in one statement: Nucleophile Attacks Electrophile."
Rather than electrophile attacks nucleophile?
2) Water is the base here; water is the Lewis Base (nucleophile) so water will attack the hydrogen cation (which is a bare nucleus).
3) HCl is the acid here; H+ is the Lewis Acid (electrophile, so it's going to go toward the lone pairs on the water).
So:
Would this be right?

If the electrons are to go somewhere, they should be attracted to that place. Consider the incorrect diagram you posted. The oxygen atom in water is partially negatively charged; the bonded electrons are drawn to the oxygen atom due to its high electronegativity. So it would not make much sense for the electrons in HCl to "attack" the oxygen atom, because the negative electrons would not be attracted the negative oxygen atom - in fact, they should be repelled.
Also, keep in mind that protons are rather stationary, at least where chemical reactions are concerned. It is the electrons that move around and form bonds to whatever compound is willing. Since "attack" is a verb of action, it helps to think of something mobile to carry out that action. The protons do not attack; the electrons do. Even in "electrophilic attacks", such as electrophilic addition, it is the electrons that move around and form bonds, even though the "disturbance" is provoked by something electrophilic.
My experience in organic chemistry is much like your quotee's. Just find out where the electrons go, and draw the schemes.

Ah, I understand. Electrons are the mobile species. Electron flow characterizes reactions (and also conductivity). Electrons form bonds. Electrons are found in clouds and move around at will while protons are bound to the nucleus. I should have thought of the chemistry first and have attempted to answer my own question; like you say, the proton is quite stationary relative to the electron.

Yes, there you have it! The essence of chemistry :)

1) Nucleophiles often have excess electron density while electrophiles are characterized by a lack thereof. The electrons flow in the direction of lower electron density (and not the other way around), because they are "attracted" by the positive partial or formal charge of the electrophile. This is the reason why the first scheme from your book is wrong, because it shows a nucleophile attacking another nucleophile.
2), 3) and the reaction scheme you have drawn are correct.


text:
What are the factors that affect the corrosiveness of an acid?
Acids are usually corrosive but what determines this? Is it the concentration of the acid or strength of the acid ($K_a$ value)?

Without saying against which substance/surface the corrosiveness is asked for, the question does not make much sense. And an answer is even less thougtful.

Corrosiveness of an acid is a more complex concept, of course the concentration of the acid and the strength of the acid are two main factors that determinate the success of the corrosion reaction. For example this is the Pourbaix diagram of iron both concentration and $K_a$ determine the pH of the solution so if this is higher then about 9 the metal is passivate and you can't have any corrosion.
I add to these another important factor:  temperature.
In some case the corrosiveness is due to a more complex interaction that is directly linked to the material you are dealing. One example is acqua regia in this case the strong corrosiveness is due to the couple effect of two different acid that are able to de-passivate the metal allowing  an enhanced corrosive effect on noble metals. However acqua regia is not the best choice for an organic material in this case Piranha Solution has a greater corrosive effect.


text:
What is formed when hydrogen peroxide reacts with titanium?
I just got night contact lenses, and you are supposed to use a "one-step" cleaning solution to clean it over a period of six hours. At the bottom of the lenses case, there is a "titanium plate neutralizer", which causes the liquid to bubble. What is this gas?
Edit: it was platinum, not titanium

Hydrogen peroxide will damage your eye tissue; that's why there are enzymes in our cells actually to catalyze the conversion of $HOOH$ to water and oxygen gas. The enzymes are actually contained in peroxisomes (name is self-explanatory).* The neutralizer serves the same function of catalyzing the otherwise relatively slow decomposition of $HOOH$ into water and oxygen gas; hydrogen peroxide will decompose in the presence of light and that's why bottles of hydrogen perioxide are opaque - generally black or brown (but not transparent like bottles of which witch hazel or isopropyl alcohol). I have seen platinum used as a catalyst in contact cleaning solutions but I suppose titanium can be used as well.
*Edit: peroxisomes actually create $HOOH$ upon catabolism of cell by-products but can also break down its own toxic byproduct.

What happens to the platinum after it is used for a certain amount of time? The box says to replace the neutralizer disc after 30 days. Also, what happens to HOOH when it decomposes (what results)?

You get water and oxygen gas when the HOOH dissolves. The platinum is actually consumed through HOOH decomposition. I've found out the hard way (from using my contact neutralization case for way over 30 days ... ouch!)

What does the platinum become after the chemical reaction? (just curious)

So if after 30 days, I take out the neutralizer somehow and slice off the outer layer of platinum (which would be very hard indeed), would the reaction still work?

I'm thinking there is a very, very thin layer of platinum on there so if you scrape off what was consumed, then you won't have much left. Also now I'm unclear on the catalysis of the decomposition of HOOH by Pt. By definition, a catalyst is NOT consumed in a reaction, but by all empirical means, it appears that Pt is consumed. Also, HOOH can be both reduced and oxidized, depending on pH. I'm thinking that HOOH is oxidized because that's when it creates oxygen gas. When it is reduced, it produces reactive radicals (not good for your cells!).If HOOH is oxidized then PT must be reduced; gains e-

If it were pure platinum, would the electrons from the inside go to the outside to participate in the reaction?

No clue; I'm lost. Better consult someone better versed in redox chemistry than I am.

@Jason The decomposition of hydrogen peroxide is a disproportionation reaction, hence one oxygen is reduced and one is oxidised. The platinum/ titanium disk should not be compromised by this reaction as they are only catalysts. There might/will be reaction paths (sideways) that wear down every catalyst. Depending on the thickness of the Pt layer is might be oxidised by air. Mechanical forces might also be at work. And there could be many more effects at play...

If I was cynical I would think the neutraliser is consumed on purpose at a rate that 50 doses of cleaning solution will consume.  I also suspect that using your own platinum catalyst (thin and cheap loop of wire from reputable manufacturing jeweller) would last forever.


text:
Does corrosion follow the law of conservation of mass?
For e.g .Due to rusting , iron becomes weak .So it should become weak only if the amount of iron is reduced. Even if we say the reduced iron is now the rust ,the amount of rust practically is very less comparing to iron.

The answer to the question in the title is yes.
Iron oxide flakes away from the iron so just looking at the iron would make you think that the mass has disappeared; it's actually on the surface below, unless someone has moved it.

In chemistry we assume the total mass of reactants to be conserved in after reaction (except in the nuclear reactions, of course). It is a fundamental law and hold true everywhere.
Even in your question the Oxide layer formed above the iron makes the Ferrous Oxide which generally get removed, hence you think the mass had reduced.

+1. This is a GREAT answer. No idea why it was -1 before! I see you followed a Materials Science proposal in the past. I really wish they allowed it to be launched. If you don't mind another attempt at a Materials proposal, would you be so kind to commit to this? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ZTUyZWY5OTQ1NDI4ZGVlMmU5MzM0ZThiZjNhOTRkYWFlMTA1NjM0OWEzOTZhYzlmMTBlNWQ4ZmIzYWE3MWFkN-4At28O9gekwyERVbYPhFIPwyHiajfB5BeMyrZKJaYV0 It started slow but lately it's been booming with 5 commitments/day and very soon it will reach 200.


text:
What is the correct temperature probe for a wet etch solution?
What type of probe or probe coating should one use to measure the temperature of a roughly 100c aqueous solution of cupric chloride?
A stainless steel probe seems incorrect. Is a Teflon coated probe sufficient or is something else more appropriate?
Is there a standard way of protecting metal probes in corrosive liquids?

Teflon is resistant against corrosive chemicals and stable up to temperatures of about 200 °C, so it is suitable for use in hot aqueous salt solutions. Using a coating or casing made of plastic which is unreactive and resistant against heat should be a good way to protect temperature probes.


text:
The Acidity of the Hydrated Iron (III) Ion
Why is $\ce{Fe(OH_2)_6^{3+}}$ fairly acidic? This iron has six water molecules coordinated to it.
In other words, water itself is a very weak acid. But when water is coordinated to iron, it becomes more acidic. I can think of a few reasons; can you think of more? Also are my reasons valid?
1) Induction. The negative charge generated through loss of the hydrogen proton is stabilized by the coordinated water molecules. Plus the center of the square planar molecule is greatly positively charged (and therefore can handle the negative charge, despite the fact that iron itself is actually electropositive rather than electronegative - at least in iron's ground state).
2) Size of the ferric ion. The ferric ion is rather big (it is, after all, able to coordinate six water molecules, as opposed to only four for some other metal ions). So whatever positive charge it has after being ionized once is spread out over a large surface area and that makes entire compound relatively stable.
3) Oxidation state of the central ferric ion - positive 3. This indicates a high degree of ionic character in the coordination bonds. Withdrawal of electron density from the water molecules makes the hydrogens even more positively charged and thus even more electrophilic.
4) Positive charge density of the central ferric atom. Coulomb's law is in (significant) effect here. Positive and positive. Like charges repel. Kicking off a hydrogen proton - no problem.

When a water molecule coordinates to $\ce{Fe^3+}$, one non-bonded $sp^3$ hybrid orbital (with one of the 2 lone pairs) of oxygen overlaps with an empty $d$ or $p$ orbital of $\ce{Fe^3+}$. Electron density is transferred from oxygen to iron, and due to the electronegativity of oxygen, the polarization of the $\ce{O-H}$ bond in the water ligand is increased (more electron density is "pulled" from $\ce{H}$ towards $\ce{O}$). The hydrogens are more positively polarized than in a free (uncoordinated) water molecule, and this increases their electrophilic character and acidity to the point that they are readily abstracted by solvent (non-coordinated) $\ce{H2O}$ in aqueous solution:
$$\ce{Fe(H2O)6^3+ + H2O \rightleftharpoons  Fe(H2O)5(OH)^2+ + H3O+}$$
$$\ce{Fe(H2O)5(OH)^2+ + H2O \rightleftharpoons  Fe(H2O)4(OH)2^+ + H3O+}$$
Further deprotonation results in the neutral complex $\ce{Fe(H2O)3(OH)3}$, from which $\ce{Fe(OH)3}$ precipitates.
The aqueous solution of $\ce{Fe^3+}$ aquo complexes is fairly acidic, wth a pH of up to 1.5 depending on iron concentration. The first dissociation equilibrium lies mainly on the product side, which can also be seen in the color of the solution: Pure $\ce{Fe(H2O)6^3+}$ is pale violet and found in crystals of some hydrated $\ce{Fe(III)}$ salts, while aqueous solutions get their orange color from $\ce{Fe(H2O)5(OH)^2+}$.
You have mentioned ion charge (oxidation state) and charge density (which is proportional to ion size) of the metal center as potential factors of influence. One can expect that with increased charge and/or charge density, the polarization of the water ligands also increases, and thus the aquo complex should be more acidic. This is true for many aquo complexes; however, there are also some deviations from this rule, like $\ce{Cr(H2O)6^2+}$ and $\ce{V(H2O)6^2+}$ (reference).

Thank you. I think you mean charge density is inversely related to ion size.

@Dissenter Yes, that's correct.

Also, is it true that we can use a simple size/charge model to rationalize acidities of the metal ions are isoelectronic with the noble gases?

@Dissenter As you can see in the reference, there is a possibility to use such a model, but it has its limitations. This starts with the definition of the radius of an ion. As you can see in the 2 plots ($pK_a$ vs. radius), there is not a single definition of ionic radius, since the values vary significantly.


text:
Can a s orbital overlap with any p orbital to form a sigma bond?
In internet resources, only $p_z$ orbitals are shown to overlap with $s$ orbitals to form sigma bond. But my teacher says that according to Valence Bond Theory, $s$ orbital overlaps with all three $p_x$, $p_y$ and $p_z$ orbitals to form sigma bonds. I am really confused about this. I heard that the concept changes when it comes to molecular orbital theory. Only similar orbitals (like $1s$-$1s$,$2p_x$-$2p_x$,$3p_y$-$3p_y$) form bonds but ($2s$-$2p_z$) do not.
Please clarify if there is any difference in concepts of overlapping between 'Molecular orbital theory' and 'Valence Bond Theory'.

The axis of orientation is used so we can designate one orbital from another. But since each orbital is equal, the assignment of the axis is arbitrary, so it doesn't really matter which is $P_x$ and which is $P_y$ or $P_z$. So if there can be an overlap of an $S$ and $P$ orbital, it is us who decide if it is $x$, $y$ or $z$.

Certainly. An example of a s-p sigma overlap is the bonds between hydrogen and the halogen in hydrogen halides.

$\ce{Px}$, $\ce{Py}$, and $\ce{Pz}$ orbitals can all overlap with an $\ce{S}$ orbital to form a bond.  It simply depends upon along which axis (X, Y or Z) the bond will be directed. These concepts of overlap do not change between Molecular Orbital theory and Valence Bond theory.

As an additional note here: The choice of the coordinate system is completely arbitrary. Orbitals can also be rotated, linar combined, ..., to form bonds.

Of course there is only one concept of bonding which never differ in either cases.
Now that here you have a doubt about the bonding between s and p orbital. First of all you should know that s orbital can form bond between with p orbital on any axis (x,y or z axis) and all forms bond in sigma bond.


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Cyclic Voltammogram - How to determine Overpotential
http://youtu.be/1f92vGOridg
Clicking on the link above will take you to a video with a diagram of a voltammogram.
Through my own experiments I have managed to obtain a similar diagram. However, I am unclear on how to determine the overpotential from such a diagram.
I know that the overpotential is simply the potential required over and above the amount required to get the reaction going thermodynamically.
If you could, please be explicit about what points I should choose exactly to make my calculation.
Thank you !


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Mechanism of Acidity - Boric Acid
How does boric acid behave as an acid? Is it a Lewis acid or a Brønsted-Lowry acid?

what are your thoughts/ideas on the question?

Wikipedia says boric acid is a Lewis acid. It later states that it is a tribasic Brønsted acid.
A Brønsted acid is a compound that is able to provide $\ce{H^+}$ ions. A Lewis acid is a compound that can accept an electron pair; it would need to be electron poor. Some molecules have both characteristics and either feature could be used in explaining the reactivity of the compound.
Tribasic means that there are 3 hydrogens that can be used. I'm believe that the author used the term basic acid to mean that the hydrogen is attached as a hydroxyl group instead of hydrogen bonded. Phosphorous acid is always written as $\ce{H3PO3}$, although it could be written as $\ce{P(OH)3}$.

It also says it's a "tribasic" Bronsted acid. I'm not clear on what that means and it appears there is quite a bit of debate still about its acidity. I just want the best supported answer.

Yes, but what's a tribasic acid? What's a basic acid?

The term "basicity" of an acid refers in this context to the number of conjugate bases. $\ce{H3BO3}$, when reacting as a Brönsted acid, is tribasic because the 3 conjugate bases $\ce{H2BO3^-}$, $\ce{HBO3^{2-}}$ and $\ce{BO3^{3-}}$ are formed in the dissociation steps. The term does not imply that the acidic protons are bound to an oxygen (OH group). $\ce{H2S}$, for example, can be described as a weak dibasic Brönsted acid.

Ah, thank you. There are the HS- and S^2- conjugate bases.

Boric acid is a Lewis acid, therefore when in water it accepts the lone pair from oxygen and becomes $\ce{(OH)3B^{-}-OH2^{+}}$.
From this compound a proton can be donated and this is how boric acid behaves as a Brønsted acid.