text:
Calculate the number of moles of phosphorus in 15.95g of tetraphosphorus decaoxide
I've got this question and I cannot figure out how to get the correct answer.

Calculate the number of moles of Phosphorus in $15.95\:\mathrm{g}$ of tetraphosphorus decaoxide. ($\ce{P4O10}$)

I tried doing it like this:
Molar Mass of (element) = Number of atoms * Relative Atomic Mass = ### grams/mole
Mass of Compound in Grams = $15.95\:\mathrm{g}$
$M(\ce{P4}) = 4(30.97376) = 123.89504 \:\mathrm{g/mol}$
$M(\ce{O10}) = 10(15.9994) = 159.994 \:\mathrm{g/mol}$
$M(\ce{P4O10}) = 123.89504\:\mathrm{g/mol} + 159.994\:\mathrm{g/mol} = 283.88904 \:\mathrm{g/mol}$
$\ce{P}: (123.89504 \:\mathrm{g/mol})/(283.88904 \:\mathrm{g/mol}) = 0.4364207 \cdot (100\% = 43.64\% \ce{P})$
$15.95\:\mathrm{g} \cdot 0.4364207 = 6.9565\:\mathrm{g} \ce{P}$
$\ce{P} = 6.9565\:\mathrm{g}/ (123.89504\:\mathrm{g/mol}) = 0.056148 \:\mathrm{mol}$
But the answer in the back of the book is $0.2247 \:\mathrm{mol}$

You can go about it more directly as follows.  You have $(15.95/283.889)=\pu{0.0562 moles}$ of $\ce{P4O10}$.  From the molecular formula you see that for every mole of $\ce{P4O10}$ you have 4 moles of phosphorous.  Therefor you have $4 \times 0.0562 = \pu{0.2247 moles}$ of phosphorous.
Your error involved converting moles of $\ce{P4}$ to moles of $\ce{P}$.  If you multiply your answer by 4 you get the correct answer.


text:
Why is the para carbon in Benzaldehyde NMR more deshielded than the ortho and meta carbons?

Assuming this image is correct, why is the Carbon-3 more deshielded than Carbon-2 and Carbon-1? From what I understand, deshielding is the greatest when the Carbon atom is closer to electron withdrawer and it appears Carbon-3 is the farthest.

Do a search of 'electron withdrawing groups' and 'electron donating groups'. You'll see several cases where resonance forms are drawn out to illustrate which carbons are most or least shielded depending on the classification of the group attached to the aromatic ring.

Resonance effects come into play along with the inductive effects.  I've drawn a few of the resonance structures for benzaldehyde below, notice how resonance effects place some positive charge on carbons 2, 3 and 5 which results in their deshielding.  I might have guessed that the combination of resonance and inductive effects would place C2 below C3, but that's not the case.  At least C5 is at lowest field and C2 and C3 are close and both are below C1. The interplay of inductive and resonance effects is complex and estimating the magnitudes of these effects can get tricky.

source: Stack Exchange

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

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text:
Electrolysis of Strongly Basic Water
What happens when strongly basic water is electrolyzed? I know that the electrolysis of water in strongly acidic solution involves hydrogen protons; will these protons still be involved in strongly basic solution?

In a basic solution $\ce{H2O}$ will act as the $\ce{H}$-source for the hydrogen evolution reaction. The reaction will look something like:
\begin{equation}
 \ce{2H2O + 2e- -> 2OH- + H2}
\end{equation}

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


text:
Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid

If, in 2 M $\ce{HONO2}$, 2 mols $\ce{Al (s)}$ are oxidized to $\ce{Al(OH2)6^3+ (aq)}$, mols of reduction product hydrogen is
  (a) two;
  (b) three;
  (c) four;
  (d) five;
  (e) six.

Regarding the above question, how would one solve it?
Would it be valid to simply say that I know that 6 electrons are being transferred because we have this skeleton equation (balanced for charge):
$$\ce{Al -> Al(OH2)6^3+ + 6 e-}$$
And given that 6 electrons are being transferred and that hydrogen is a reduction product, we must obtain hydrogen from some source in this system. The source would probably be hydronium ion, since hydronium ion is less stable than water. Plus abstracting protons from water to form hydrogen gas wouldn't make much sense in strongly acidic solution (we'd be forming hydroxide ions as well)! And knowing the following reaction:
$$\ce{2 H+ + 2 e- -> H2}$$
Can we just say since we know that 6 electrons are being moved, and that hydrogen gas is a product, we know that the immediately above reaction is occurring, and according to its stoichiometry, 2 moles of electrons yields 2 moles of hydrogen gas?

You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.)
\begin{aligned}\ce{
(1)&& Al + 6H2O &-> Al(OH2)_{6}^{3+} + 3e^{-}&|\cdot2\\
+ (2)&& 2H3O+ + 2e^{-} &-> H2 ^ + 2H2O &|\cdot3\\\hline
(3)&& 2Al + 12H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\
(4)\equiv(3)&& 2Al + 6H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ \\
(5)\equiv(3)&& Al + 3H2O + 3H3O+ &-> Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\
}\end{aligned}
There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.
Hence, three mole of Hydrogen gas are formed by two mole of Aluminium.

Edit upon Dissenters follow up question "Why exactly would hydronium ion be the source?"
It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.
It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$

Thanks for the thorough response! I first tried using water instead of hydronium ion or hydrogen protons as the source of hydrogen ... why exactly would hydronium ion be the source? I have an inkling it's the source because it's less stable, right?

Also is your second equation balanced for mass? I see 6 H on one side and 4 H on the other side .. fixed it for ya!

(see edit) Nope you falsified it, left side $2\cdot3\ce{"H"} = (2\cdot1 + 2\cdot2)\ce{"H"}$.

@Jan I put it there, because I think it belongs there. (The new version of MathJax simply has a different display style of charges.) After all, the oxygen still is not positively charged, but the hydrogens are.

2 moles of electrons yields 2 moles of hydrogen gas

Doesn't your second equation say that 2 moles of electrons yields 1 mole of $\ce{H2}$?  Following your reasoning, wouldn't 2 moles of $\ce{Al}$ yield 12 moles of electrons, which in turn would yield 6 moles of $\ce{H2}$?  In any case, the balanced equation is $$\ce{12H2O + 6HNO3 + 2Al -> 2Al(H2O)_6^{+3} + 3H2 + 6NO3^-}$$ so 2 moles of $\ce{Al}$ would produce 3 moles of $\ce{H2}$ (answer B).


text:
Do buffers only work for species of the opposite variety? (I.e. weak acids buffer strong bases and vice versa)
I have looked at several sources including Khan Academy, ChemCollective and the answer to a similar question:
Can someone please explain buffers to me?
However, it is not clear whether buffers can only protect against the opposite species. I.e. a weak acid can only protect against the pH change caused by strong bases, or if it can also protect against the pH change caused by strong acids.
This is my thought:
If we have $\ce {H_2CO_3_{(aq)} + H_2O_{(l)}<=>H_CO^{-}_3_{(aq)} +  H_3O^+_{(aq)}}$
Then we add a strong acid $\ce{HCl + H_2O-> H_3O^+ + Cl^-}$
Then we have $x$ amount of protons in the solution and for every one of the weak acid's conjugate base, there is a proton. So, to me it appears that there is no way for the weak acid to act as a buffer for a strong acid. Is this correct or am I missing something?

Please consider that carbonic acid is actually not very stable in aqueous solution. $$\ce{H2CO3 <=>[\ce{H2O}] H2O + CO2}$$  For a buffer solutions it is better to use something along the lines of $$\ce{KHCO3 + H2O <=> KCO3^{-} + H3+O}.$$

@Martin I chose this simply because it is the pH buffer in our blood. Ref (http://www.chemistry.wustl.edu/~edudev/LabTutorials/Buffer/Buffer.html)

Actually, all 3 ionization states of carbonic acid are used to buffer our blood. Carbonic acid is unstable in solution and that it can be readily converted to carbon dioxide allows us to, effectively, exhale acid and maintain the buffer system.

Think of it in terms of the equilibrium constant:
$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$
Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is
$$Q\; =\; \frac{9\cdot 2}{1} = 18$$
Because $Q > K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have
$$Q\; =\; \frac{8\cdot 1}{2} = 4$$
and because $Q = K$ the reaction is now at equilibrium.
So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).

If you just add straight carbonic acid to water, the $\mathrm{p}\ce{H}$ will drop as it is ionized (as represented by your equation). Since it is a weak acid it won't ionize completely and you'll be left with carbonic acid, bicarbonate and hydronium in solution at equilibrium. Now, if you increase the concentration of $\ce{H+}$ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left, as protons react with bicarbonate to produce carbonic acid). Because the new protons are not free in solution, the pH will not drastically change until the bicarbonate is exhausted. Thus the solution is buffered.
Mathematically, the acid dissociation constant, which is unchanging for a specific species by definition, is given by $K_a=\frac{[\ce{HCO3^{-}}][\ce{H+}]}{[\ce{H2CO3}]}$. If you add more $\ce{H+}$, the ratio of $\ce{HCO3^-}$ to $\ce{H2CO3}$ must decrease for the system to regain equilibrium

So you're saying a weak acid can act as a pH buffer in a solution when a strong acid is introduced? The reason I don't understand is because for every conjugate base of the weak acid, there is a proton that it has dissociated from. [Conjugate base] = [H+]. When the strong acid is introduced (a supply of H+). The weak acid's conjugate base may react with the excess [H+], but for every conjugate base that reacts with an H+, Le Chatelier's Principle would require the weak acid to produce more conjugate base and thus more H+. In effect, I don't see how H+ is reduced.

No, because when the conjugate base is protonated, you get the original acid again.

@Klik I think you applied Le Chatelier's Principle incorrectly. The equilibrium shifts in the other direction. Because you added more [H+] in order for the equilibrium constant to remain the same the reaction must shift left and produce $\ce{H2CO3}$.

I'm saying that adding a weak acid to water produces it's conjugate base and that that system will buffer pH changes in either direction. If you increase the concentration of H+ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left). The acid dissociation constant, which is unchanging for a specific species by definition, is given by Ka=[A-][H+]/[HA]. If you add more H+, the ratio of A- to HA must decrease for the system to regain equilibrium.

@canadianer I understand what you're saying now. Could you include the details you just explained to me in your answer? I will mark it as correct. Thanks for the explanation.

You got it. How's that?

Sorry, user beat you to it. He's also provided a good illustration involving the calculations. I've up-voted your answer though.


text:
Why is N₂ stable but HCN and C₂H₂ unstable?
Compounds with triple bonds generally seem to be unstable. $\ce{HCN}$ and $\ce{C2H2}$ are high-energy, relatively short-lived molecules that will readily polymerise or react with other organic molecules. My naïve mental picture is that they don't really want to have a triple bond because the geometry makes it awkward, so they'll do anything to transfer an electron somewhere else and turn it into a double or a single bond instead.
But $\ce{N2}$ seems to be an exception. By my naïve reasoning, when an $\ce{N2}$ meets an organic molecule it should be eager to shrug off its triple bond and join the party. But in fact this doesn't happen, and nitrogen fixing organisms have to do quite a bit of work to get nitrogen to participate in organic molecules.
So my question is, what is it about $\ce{N2}$ that makes its triple bond so energetically favourable while $\ce{C#C}$ and $\ce{C#N}$ bonds are so unfavourable?
Note that this question isn't about reactivity with $\ce{O2}$ but rather with organic molecules. I'm interested in why it's so difficult for $\ce{N2}$ to react with organic molecules in general, given that other molecules with triple bonds seem to react with them very easily.
(Information to guide the scope of answers: I'm coming at this from a physics background, so thermodynamic concepts can be taken as understood, but I quickly get lost when it comes to electrons and orbitals and so on, hence the rather basic question.)

I guess the simplest explanation is, the bond energy difference between N$\equiv$N and N$=$N are about 500 kJ mol$^{-1}$, while C$\equiv$C and C$=$C are about 200 kJ mol$^{-1}$.http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles/Bond_Energies For HCN, there are other channel, such as HCN$\rightarrow$ H$^+$ + CN$^-$. The next question is, why the bond energy difference of N$_2$ is much larger than HCCH. The Kohn-Sham HOMO (I know it is problematic, but N$_2^+$ is a failure of Hartree-Fock) of N$_2$ and HCCH are $\sigma$ and $\pi$ orbitals, former is more stable

Part of the answer must be to do with the possible mechanisms available for a reaction with something to happen. There are no easy reactions starting with dinitrogen, but lots of easy ones starting with acetylene or hydrogen cyanide (but they don't all start with the triple bond).

I think this question can be answered reasonably well with thermodynamic considerations.  When we say something is stable, it is (or should be) assumed that the compound does not engage in a chemical reaction.  It is important when we discuss the stability of compounds to focus not only on the compound itself but what the products of the reaction are.
The three triple bonds you mention ($\ce{N#N}$, $\ce{C#N}$ and $\ce{C#C}$) all have very high bond dissociation energies (from various sources such as this one they are $\approx950, 940\text{ and }960\:\mathrm{kJ/mol}$) so all of these triple bonds require a large amount of energy to break.
Now, when we think about the stability of acetylene, I suspect most people are thinking about its stability towards ignition (which has been studied, by the way).  That gives us a reaction we can explore, namely the combustion of acetylene in the presence of oxygen to produce carbon dioxide and water.  WolframAlpha can calculate the thermodynamics of this reaction, giving a result of $-2600\:\mathrm{kJ/mol}$; clearly a very exothermic reaction.
Now let's look at the combustion of nitrogen in the presence of oxygen, which could form nitrogen dioxide.  Again, WolframAlpha (Note, if you happen to play around with WA, make sure you put your reaction in quotation marks or it won't be interpreted properly) gives us a result of +66 kJ/mol or an endothermic reaction.
What do we make of these results?  Focusing on the products we see that for the acetylene combustion, there is a large amount of energy released when creating the products.  The same is not true for the products of nitrogen combustion.
In summary, when considering stability, we must think about we mean by the word, since that meaning will lead to a chemical reaction, and the tools we use to describe chemical reactions help us quantify that stability.

Thanks. I'm sure that what you say is right, but as I hoped would be clear from the question, I'm not so concerned with the reactions with O2 as with other organic molecules. I'm not asking "why doesn't N2 burn" but "why doesn't it generally react with biomolecules?"

@Nathaniel you might want to edit the question to make this point clear (especially since you refer to organic molecules but not biomolecules).  The answer will be the same (you need to consider the formation energy of the products) only in the case of biomolecules (of which there are many) the reactions will be more complicated.  If you have a particular reaction in mind, that might help lead folks to a usable answer.

I specifically don't have a particular reaction in mind. I'm asking why, in general, is it difficult to react N2 with organic molecules, which is why I asked for help from chemists rather than just looking up the delta-G. I will edit the question to clarify that, although it does actually say it already.

I disagree on your use of stable, the meaning you are implying is the one of unreactive. Stable should be used in the sense of "it is an isolatable chemical species."


text:
Why can air accommodate more water when the air has a higher temperature?
The amount of water(g) that air can accommodate depends on temperature. The absolute humidity then is defined as the percentage of water to air and decreases when the temperature decreases. The relative humidity is the ratio between the current and the maximum humidity, which increases with decreasing temperature. The temperature at which the relative humidity reaches 100% is called the dew point.
Anyway, why can air keep more water when it is warmer?

For a substance to be a gas, the kinetic energy of the molecules that make it up must be higher than the electrostatic potential energies that hold it together. When we talk about air "holding" water we are really talking about the amount of water that would be in the vapor phase relative to the amount that would be liquid at equilibrium.
Since temperature is a measure of the average kinetic energy of molecules, as the temperature increases, the fraction of water that can exist as vapor increases. Since air is already a gas under any conditions where water vapor can exist, it looks like the air "holds" more water as temperature increases.

Ah, so it actually has nothing to do with any air-property.

Yes, the vapor-to-liquid ratio for water at equilibrium would increase with temperature for air or any other gas (even vacuum). The exact value would depend on the total pressure of the gas phase, and that would be a function of the composition of the gas. However, the general trend would be the same.

The liquid-vapor coexistence line of the phase diagram of water has a positive slope $\mathrm{d}P/\mathrm{d}T$. Humidity in air is a partial pressure of water vapor. If partial pressure of water is increased starting from $<100\%$ humidity, it will reach the vapor-liquid transition and any additional water vapor will simply condense into droplets ($=100\%$ humidity).
The first-order phase transition of condensation will force any water molecules added to a saturated system to condense into liquid, and so there is a high limit to the humidity at a given temperature.
An increase in temperature corresponds to an increase in the saturation vapor pressure, thus at warmer temperatures there will be a higher partial pressure of water vapor in the air at a given $\%$ humidity.


text:
Carbon-13 NMR of dimethyl 3,4,5,6-tetraphenylphthalate?

I am trying to assign the peaks here to dimethylphenylphatalate but all the benzene rings are throwing me off. All I can get is that the further most signal is the carbonyl carbon (168.71) and the one prior to that (143.27) might be the ester carbon (from the methyl). Can someone help?

Here's a numbered drawing of the title compound to help us keep track of things.  First thing we notice is that there are 13 different carbon atoms in the structure and we see 13 signals in the spectrum, that's a good start.

Let's start with the easy ones.  The upfield peak at 52.28 is in the expected position for C1, the ester methyl carbon.  Down at the other end of the spectrum, the peak at 168.71 is in the expected position for an ester carbonyl carbon (C2).
In the region for aromatic carbons we see 5 low intensity peaks between 143 and 132.  These belong to the (unprotonated) quaternary carbons 3, 4, 5, 6, 10.  Using diethyl phthalate as a model compound, I would guess that the peak at 132.19 is from C3.  The two peaks at 138.73 and 138.60 have the same height and are closely spaced suggesting very similar environments, I might guess them to be C6 and C10, just not sure which is which.  Noting that the peak for the carbon corresponding to our C5 is downfield of the peak corresponding to our C4 in the model compound, I would expect the same trend to hold here, just with each of the peaks shifted downfield due to attachment of a phenyl group.  So I would assign the peak at 143.27 to C5 and the peak at 139.29 to C4.
This leaves the protonated aromatic carbons for us to consider.  The 2 peaks at 125.97 and 126.83 are half the intensity of the other protonated aromatic signals, therefor assign them to C9 and C13 as there are only half as many of these carbons as compared to the other protonated aromatic carbons. It's not clear which peak belongs to C9 and which to C13.  This leaves us with the 4 signals due to C7, C8, C11 and C12.  Using the same resonance structure - delocalization arguments I made in your benzaldehyde cnmr question from yesterday, I would expect C7 and C11 to be downfield from C8 and C12.  Therefor, I would assign C7 and C11 to the peaks at 129.71 and 130.88, not knowing which is which.  C8 and C12 then belong to the signals at 126.88 and 127.41, not knowing which belongs to which.
Obviously there is some guess work in here and if it was an important project I'd do some further experiments and look for some additional model compounds in order to confirm these assignments and remove the discrepancies, but this is a good start.

COSY should help...

I wouldn't trust the signal intensity in a 13C NMR, unless it is explicitly set up to be a quantitative experiment.

Your right that you can't integrate the signals like you can in proton nmr, but in my experience quaternary carbons are always distinguishable by their low intensity, and for protonated carbons when one signal is, say, half the size of another and you know that matches the ratio of the number of carbons for these two signals, you're usually OK.

Thanks a lot ron, I got a good feel for analyzing this kind of NMR's.


text:
Does this activation of carbonyls usually involve a SN1 or a SN2?
The process I have seen in class to activate a carbonyl function is acetalisation.
One of the steps involved is the following :

So I was wondering, is that a $\ce{S_{N}1}$ or a $\ce{S_{N}2}$ ?
(PS : drawing molecules with paint is awful, anyone knows a good free molecules drawer ?)

It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction.  It is commonly referred to as an "addition-elimination" reaction.  You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate.  Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction.  This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl.  Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously.  Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product.  You "add" a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you "eliminate" one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.

Here's another, more specific drawing for the case at hand.

What would be the formula of the tetrahedral intermediate ?

See the picture I've added to my post.

where does the =O come from ? where is $R_2$ gone ? (and btw what do you use to draw those molecules?)

You had the word "carbonyl" in your title, so I assumed you were starting from a carbonyl compound - so I drew one in my diagram, 2) I replaced R2 with X, my drawing is a little more general, If R2 is an alkyl group, then it can't leave and we would just have an equilibrium between the ketal (the tetrahedral intermediate) and the starting ketone, but it is still an example of an "addition-elimination" reaction, 3) I use ChemSketch freeware for my drawings

Here is the acetalisation i'm talking about http://i.gyazo.com/351823bdf8c4b8437ae81ec6302ee407.png

Since you're starting with a ketone, the reaction is called ketalization.  The final product, the ketal, is in equilibrium with the starting ketone.  For most ketones the equilibrium lies far to the ketone side (not much ketal present).  Still, for the reasons I presented above, it is an addition-elimination reaction that proceeds through a tetrahedral intermediate.

I read that the IUPAC wanted us to stop using 'ketalization' and use only 'acetalization' for both, am I wrong ? Btw could you update your image with something matching my reaction please? It would help me :)

Yes, you are correct, the term "ketal" is now out of favor.  I've added a second drawing does it help?

http://i.gyazo.com/47a470c98228b6f8e46897aafac7e4c0.png whete is the second O gone ? Isn't there some H2O missing ?

Good catch, error in my drawing; fixed it. Does it make sense now?

Yes, thank you ! I just have two more question. 1) why is this way better (energy/stability wise) to do it this way rather than using a carbocation ? 2) I've made some researches before posting my question here and didn't find any answer, where do you know it from ? (or is it just the answer you naturally came with when you thought about it?). It's not that I don't believe you, but I always feel better when I have additional sources other than just Q&A websites, especially in chemistry. :)

In the top line of my drawing, the second structure from the left is an oxonium ion.  It has a resonance structure with positive charge on the carbon.  So it has some carbonium ion character, but it is not a full-fledged carbonium ion.  No carbonium ion rearrangements are observed - maybe because much of the positive charge resides on the oxygen. 2) I understand.  It is the way I learned things.  You've seen addition-elimination reactions with carbon-carbon double bonds.  Well in this case the only difference is that it involves a carbon-oxygen double bond.

Oh wow.  I totally misunderstood the original question.

This is an SN1-like process, the intermediate is an alyoxonium ion (R1R2C=O+R), which, having a double-bond resonance form is a somewhat better intermediate than a free carbocation, as in the tradition SN1.
There is too much steric hinderance for this to be a SN2 reaction with a reasonable reaction rate.  That's a definite no-go.

The carbocation is (R1)(R2)C-OR with a lacunary central C, how is that equivalent to a (R1)(R2)C=OR structure ?

Resonances forms involving donating a lone pair from the oxygen to the carbocation forms a double bond

I got it now with @ron 's drawings, thank you :)


text:
How do I determine whether it is the *ferrous* or *ferric* ion that is implied in this displacement reaction?
I'm supposed to write a balanced equation for:
$$\ce{Iron(s) + copper(II) sulfate(aq) -> ?}$$
If I use Iron(II)/ferrous I get
$$\ce{Fe + CuSO4 -> FeSO4 + Cu}$$
and if I use Iron(III)/ferric,
$$\ce{Fe + CuSO4 -> Fe2(SO4)3 + Cu}$$
(Before balancing).
Is there a way to determine which one I am supposed to use?

Tip: Fe(III) salts are used in etching of copper for making custom electric boards by enthusiasts.

Yes, it is. Generally, you should split the reactions into half-reactions and then consult standard electrode potential table. In case differences are large, results are obvious. In case differences are small (<0.2-0.5 V), some calculations involving Nernst equation must be performed. This, of course, have limited use, since kinetic differences are not considered, but for metal substitution for salts such procedures usually works. For details consult electrochemistry chapter of your favorite physical chemistry textbook. I'm too lazy to write a proper answer here.

@permeakra it's good you are not at least not lazy in writing answer in comment. I am even lazy in that. That's why i normally give some useful links :)

You can determine which reaction takes place by the standard electrode potentials of the half reactions. The relevant standard reduction potentials for the possible reactions (source of the values) are:
$$\ce{Cu^2+ +2e- ->~ Cu} \quad E_0=+0.342\mathrm{V}$$
$$\ce{Fe^2+ + 2e- ->~ Fe} \quad E_0=-0.447\mathrm{V}$$
$$\ce{Fe^3+ + e- ->~ Fe^{2+}} \quad E_0=+0.771\mathrm{V}$$
The overall potential of the electrochemical cell is defined as the difference between the standard reduction potentials of reduction and oxidation reaction.
$$E_{cell}=E_{red}-E_{ox}$$
For a spontaneous reaction to occur, $E_{cell}$ must be positive. With the potentials of copper reduction and oxidation of iron to $\ce{Fe^2+}$, you get
$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^2+/Fe}}=+0.342\mathrm{V}-(-0.447\mathrm{V})=+0.789\mathrm{V}$$
It is obvious from the values that $\ce{Cu^2+}$ cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ because $E_{cell}$ is negative for this reaction.
$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^3+/Fe^2+}}=+0.342\mathrm{V}-(+0.771\mathrm{V})=-0.429\mathrm{V}$$
The balanced equation for the redox reaction between iron and copper(II) sulfate is therefore
$$\ce{Fe + CuSO4 ->~ FeSO4 + Cu}$$


text:
Endothermicity of Reaction

Regarding this question, how can one solve this without access to thermodynamic data?
I suppose we should keep the following information in mind:
1) Double bonds are not double the strength of single bonds. Same goes for triple bonds. Reason has to do with orbital overlap with regard to sigma and pi bonds.
2) In order of increasing stability: $\ce{O_2, O_3, O}$. We know this because diatomic oxygen is what usually exists in large quantities at ground level; ozone is a pollutant because it is quite reactive and can thus harm tissues; we don't usually hear of monoatomic oxygen.
Therefore:
a) Cannot be the answer. Going from ozone to diatomic oxygen is exothermic. At least part of this reaction is exothermic.
b) Definitely not the answer. Going from ozone to diatomic oxygen exclusively.
c) This is endothermic, but perhaps not as endothermic as creating monoatomic oxygen.
d) Looks pretty endothermic; we are splitting oxygen.
e) Definitely not the answer, this is fundamentally exothermic.
How can we definitely pick out the right answer? What reasoning would you employ?

Your reasoning is correct, that's the way to approach the problem,

I believe that you would have to know at least these things to start with:

All enthalpies of bond dissociation are endothermic
Complete dissociation of triple bonds is more endothermic than double bonds, which is more endothermic than single bonds (when compared for the same elements).
The difference between going from a triple bond to a double bond is smaller than double to single, and that is smaller than going from single to complete dissociation (you referenced this in your question)

We can summmarize these facts with the following figure:

Here D(O-O) represents the bond dissociation enthalpy of an oxygen-oxygen single bond. The numbers inside of circles represent "steps" between each successively stronger bonds, and are there just for easier reference.
With that in mind, let's look at the options again, and count how many of each type of bond are broken and formed:
$ a) \space O_3(g) \rightarrow O_2(g) + O(g) $
Ozone has a resonance structure, which means that we need to consider the bonds as being somewhere between a single and double bond in strength. We could imagine them being about halfway up step 2.
Broken: 2 x single/double resonance bonds
Formed: 1 x double bond
So we are looking at an overall reaction that would be slightly more endothermic than the breaking of a single bond. To make that easier to see, think about taking these energetic "steps" using the figure above:
(+1.5, +1.5) - breaking the two resonance bonds
( -1, -2) - forming the double bond
$ b) \space  2O_3(g) \rightarrow 3O_2(g) $
Here we are breaking two sets of resonance bonds, but forming 3 double bonds. So our steps look like:
(+1.5, +1.5, +1.5, +1.5) - breaking six resonance bonds
(-1, -2, -1, -2, -1, -2) - forming three double bonds
There are three -1, -2 steps, each of which is larger than a +1.5 step. However, they are not much larger, and the total of the difference may or may not be greater than the fourth 1.5 step. This means we can't say for sure whether it is exo- or endo-thermic, but we can at least say it would be less endothermic than breaking a single bond.
$ c) \space  3O_2(g) \rightarrow 2O_3(g) $
Now we are doing the opposite of (b). We still don't know the sign, but we know the magnitude is the same as (b), which was less than (a), so either way it can't be (b) or (c).
$ d) \space  O_2(g) \rightarrow 2O(g) $
Here we are breaking a double bond, and forming nothing. This is the clear winner so far, with steps +1, +2.
$ e) \space  O_3(g) + O(g) \rightarrow 2O_2(g) $
In the final case, we break two resonance bonds, and form two double bonds. Our steps are:
+1.5, +1.5, -1, -2, -1, -2
Since the formation of each double bond releases more energy than the dissociation of each resonance bond, this is clearly exothermic, as you stated in your question.
Taking all of this together, we have to choose (d) as the correct answer.


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Why does calcium carbonate decompose into calcium oxide?
Why does calcium carbonate, when heated strongly, decompose into calcium oxide and carbon dioxide?
What is the mechanism? Can someone illustrate the means by which this happens? What bonds/associations are broken?

I'm guessing that one of the calcium ion has a strong intermolecular association with one of the partially negative oxygens, and this changes the bond character of the carbonate ion. Upon applying heat, this bond remains while carbon dioxide is off-gassed.
In other words, the positive charge on the calcium ion isolates the electron density to the nearest oxygen atom on the carbonate ion. This isolation of electron density away from the carbon and toward the calcium weakens the oxygen-carbon bond while strengthening the calcium ion-oxygen intermolecular association force.

It is a matter of entropy $\ce{CaCO3 -> CaO + CO2 ^ }$.

Yes, I agree Martin, but why not carbon monoxide? That would also yield an increase in entropy.

I think you approach this question in a very "organic-chemistrish" way. In this reaction the oxidation state of Ca and C do not change!

Some quick googling did not turn up a definitive reaction mechanism, and most of the papers I did find seemed to instead be focused on the overall reaction rate and controlling factors - probably because this is such an industrially important reaction and (at least according to this article) gas transport is the rate-limiting factor - this is common for heterogeneous reactions.
However, it is not too hard to propose a plausible mechanism. I would suggest the same mechanism you did - what is likely happening is that the oxygen near one calcium ion can pull some electron density away from the carbon, leaving $\ce{CO2}$ which diffuses away as a gas. The calcium ion stabilizes the new oxygen anion and forms a new, different lattice structure.
As @Martin points out, the reason that this process is thermodynamically favorable is that the products have a higher entropy. The standard Gibbs free energy of reaction (using data from Tro's Chemistry) is:
$$
\begin{align}
\Delta G &= \Delta H - T\Delta S \\
&= 179.2\space kJ/mol - T(0.1602\space J/mol*K)
\end{align}
$$
Note that $\Delta H$ is positive, but so is $\Delta S$.
At 298 K (room temp) $\Delta G = 131.3 \space kJ/mol$ - the reaction is not favorable.
The equilibrium temperature (ignoring temperature corrections) should be around 1120 K.

What do you mean by temperature corrections?

The enthalpy change is a function of temperature, so the real equilibrium temperature would be different.

The real equilibrium temperature would be he calculated temperature plus the starting temperature - is that what you're saying?

No, when you solve the equation for change in Gibbs free energy to find the equilibrium point, you are making an implicit assumption that change in enthalpy is independent of temperature. Since that is not true, the temperature you get is not exactly right. It is close in many cases, and can give a rough idea of the true equilibrium temperature.

In otherwords you are assuming that the enthalpy function is isothermic? Dunno if I'm using the term correctly but I would love to learn!

isothermic means "same temperature" and usually refers to a process that occurs at constant temperature. This isn't really a process - we are just changing T in the $\Delta G$ equation to see where $Delta G$\ becomes zero. Probably the best way to state the assumption is "we are assuming that the enthalpy change of reaction is temperature-independent."

Okay! What if we say that the enthalpy change of reaction is isothermal? That just means the same thing right as isothermic?

Let us continue this discussion in chat.


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What exactly is the "oxygen reduction reaction"?
Is it
$\ce{O2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)}$
and are there multiple types of ORR?
See here for context: http://www.nature.com/srep/2014/140122/srep03821/full/srep03821.html

Your reaction is perfect.

According to your cited paper the reaction they care most about is $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$ as you are suggesting. This should also be the total reaction and the most interesting one for applications, e.g. in fuel cells.
There are various suggestions for the mechanistics of the Oxygen Reduction Reaction. A quite recent review was published by Nagappan Ramaswamy and Sanjeev Mukerjee. A very early general scheme can be found at electronicstructure.wikidot.com, where they cited Halina S. Wroblowa et.al..


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I am trying to picture how electrons move around in atomic orbitals
Are they thought to continuously pop in and out of existence at various points inside the orbital defined by probabilities or do they follow definite paths that are made fuzzy by the Heisenberg uncertainty principle?
I sometimes find myself as thinking of the electron cloud as a gas with varying charge density corresponding to the shape of the atomic orbital.

Take look HERE I hope you will get complete idea

if you don't like to read then search it on YOUTUBE

Don't 'picture' it, your intuition will not work here. Quantum mechanics does not include such word as 'trajectory', meaning that common understanding of movement is absolutely useless in understanding of electone movement. The most understandable concept applicable here is idea of 'electronic density', i.e. how much electrons is located in region of space, i.e. your concept is not all that wrong.

Related post http://physics.stackexchange.com/questions/115188/how-does-an-electron-move-in-an-s-orbital

http://chemistry.stackexchange.com/questions/42032/do-really-the-electrons-could-only-orbit-in-specific-orbits-with-a-fixed-radius

I think you miss an important notion to get a better understanding of the behavior of the electrons: wave-particle duality. I believe that a better visualization takes into account that you have to deal even with a wave. From what you say it seems that your picture is shifted only to the particle-side,  keep meditating for a half an hour trying to visualize the electron as a wave. For me is a very interesting mind experiment. Then move to the orbital concept.
At Lund University researches took a picture of a moving electron. This image maybe could be helpful for your meditation:

Lund University. "Electron Gets Film Debut In First-ever Video Of Its Kind." ScienceDaily. ScienceDaily, 25 February 2008. www.sciencedaily.com/releases/2008/02/080222095358.htm.

The interpretation of quantum mechanics is an unresolved issue in physics and philosophy.

I sometimes find myself as thinking of the electron cloud as a gas with varying charge density corresponding to the shape of the atomic orbital.

This was Schrodinger's original interpretation when he developed/postulated the Schrodinger equation, that the wave function should be interpreted as charge density.  However, the Copenhagen interpretation soon became most popular. The Copenhagen interpretation associates the wavefunction with a probability distribution rather than a charge distribution.

Are they thought to continuously pop in and out of existence at various points inside the orbital defined by probabilities or do they follow definite paths that are made fuzzy by the Heisenberg uncertainty principle?

The electrons of atoms do not pop in and out of existence.  However, virtual electrons pop in and out of existence and can affect the energy levels of atoms slightly, for example contributing to the Lamb shift.
According to the Copenhagen interpretation, the electrons do not follow definite paths; however, there are other interpretations which have not been disproven, such as Bohmian Mechanics where definite paths are followed.

I hear you. I have been struggling with trying to visualize it as well. The problem is that when we get down to this world of quantum effects, stuff stops being easy to visualize just on the basis of pure weirdness. Whomever in the comments said "don't visualize it" is basically correct. But here goes anyway, following good advice is optional =)
I did however have some epic a-ha moments when I saw the models of vibration modes of circular drums. Especially as you go up in modality - it is fascinating to see the vibration mode $u_{21}$ and then think of the 3d orbital shape.
The Heisenberg uncertainity principle basically arises when we need to for instance pinpoint the exact location of the electron (how can momentum be preserved when we reduce the drums' surface area down to a point) - For me it is always easier to think of the electron as smeared out over a surface (or, better yet a space) until we actually try to locate it - at which point it will coalesce into a particle.
Disclaimer: All of this is factually B.S. - the only model that can accurately represent any single part or property of the universe - is the universe.

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What is the reason for anomalous expansion of water?
What is the reason for anomalous expansion of water? Why doesn’t it simply expand on heating or contract on cooling? Why it shows anomalous behavior at 4 degrees Celsius? Why not on 4.6 or 10 (or whatever) degrees Celsius?

So, the first question to ask would be: how "anomalous" is change in the thermal expansion coefficient?  It turns out that for most real materials the coefficient of thermal expansion is not a constant, and takes on both positive (expansion) and negative (contraction with increasing temperature) values.  For silicon as an example, see ioffe,  Above room temperature Si has a positive (but not constant) CTE, below about 100K silicon's CTE is negative.
The second question is why would this occur?  Taking a liquid or a solid (ignore gases for the moment), an individual atom (or molecule) sits in some kind of potential well which is what is causing the material to be bound together. At absolute zero, these atoms sit at the very bottom of the potential well.  As the temperature increases, heat is added and the atom starts rattling around in the potential well, and is able to explore more of the well.  There is no particularly good reason to believe that the shape of the well, across the entire temperature range of existence of the phase, will be such that a single linear coefficient will describe where the atom will most likely sit in the well at a given temperature. On top of that in a solid is the equilibrium concentration of point defects that is dependent on temperature.  As an example, Google for Simmons and Balluffi and their experiments combining x-ray measurements of the lattice parameter (where the atoms sit in the potential well) and thermal expansion (change in length of a heated bar) to determine the point defect concentrations vs temperature for various solids.

@OscarLanzi - well, that is not supported by experiment or theory. One place to start would be https://journals.aps.org/prb/pdf/10.1103/PhysRevB.94.174305

Not a place to go for me.  Paywall.  Can you summarize a few key points?

Ioffe "not found" when I click.

@OscarLanzi - yeah, kind of weird - my autolink at work didn't go there directly either, but going through the PRB main site got it just fine. As the abstract says, 'The phonon modes' contributions to the thermal expansion are analyzed and ... shown to be dominated by negative mode Gruneisen parameters at specific points on the Brillouin zone boundaries.' Applies to both Si and Ti.

The question "Why exactly 4.6 grad" has no simple answer. The question 'wtf happens with water' may have reasonably simple answer. I think about it this way: water has surprisingly low-density crystal structure with lot of holes, meaning that many defects putting a molecule into such hole would result in increase of density. Thus, the actual form of density-temperature dependency for water is result of two opposing forces: various defects, that increase its density and thermal expansion that decrease its density.
This, however, is not the only way for such dependency with extremum to occur. Another reason for negative thermal expansion coefficient would be fact, that with thermal oscillation increase and increase of mean distance between atoms, the position between mean position of some atoms may decrease. A simplistic model would be this simple physical experiment: http://youtu.be/rpnoq9hz1gw with increased speed of rotation of the chain, mean distance between its rings becomes less and less, but their mean position are slowly drageed to one point, and if chain is rotated fast enough, its vertiacal size becomes a bit more than size of one ring. IRL similar mechanism is the reason for some pulled polymers to contract at elevated temperatures. It is possible for some compounds to have negative thermal expansion for similar reason, though I can't give an example at the moment.

When water is heated, breaking of tetrahedral structure having an ordered arrangement of H-bonds continues up to 4 degree centigrade and the contraction caused by the close proximity of water molecules is greater than thermal expansion (molecular agitation due to vibrational and kinetic energy). Consequently, volume continues to decrease up to $4 \, ^\circ$C. Therefore, water has its highest density at that temperature and it decreases with increasing temperature.


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Solubility of p-nitrophenol in various substances?
So I had an organic chemistry lab where we had to dissolve p-nitrophenol in water, dilute $\ce{NaHCO3}$, dilute $\ce{NaOH}$, and dilute $\ce{HCl}$ to test for solubility.
The p-nitrophenol dissolved in everything but water. I just want to make sure if these results are theoretically correct. From what I understand, the nitrophenol should have also dissolved in water because its a polar substance and it shouldn't have dissolved in $\ce{HCl}$ because I think the Hydrogen in the phenol is itself acidic due to electron withdrawing effects of the benzene.
Can someone please explain what should happen?

p-Nitrophenol is actually modestly soluble in water at neutral pH (16 g/l at 25 °C) because of its weak acidity ($p\mathrm{K_a}=7.16$ at 22 °C) and the formation of intermolecular hydrogen bonds between the phenolic OH group and water molecules. The latter would also explain why the compound dissolves in diluted $\ce{HCl_{aq}}$ despite being protonated at lower pH values. The importance of intermolecular hydrogen bonds is also demonstrated by the significantly lower solubility of the ortho isomer, which forms intramolecular hydrogen bonds and therefore appears to be less polar to the solvent (source):

In basic solutions, p-nitrophenol is deprotonated to p-nitrophenolate. The anion is stabilized by delocalization of the negative charge into the benzene ring and the electron-withdrawing nitro group (only two of the possible resonance structures are shown):


text:
Can entropy increase during bond formation?
Recently the question was asked, "Is bond formation “strictly” exothermic?"  As a corollary to that question, what is the relationship between entropy and bond formation?  Does the entropy of the bond formation reaction always decrease when a chemical bond is formed?

“Although in defolded state protein has grater entropy, greater degree of disorder, it folds into original conformation (lower entropy), and this seems to be in collision with the II law of thermodynamics. But by refolding molecules of water surrounding him increase its own entropy by forming maximum number of hydrogen bonds (they’re squeezing, pushing in hydrophobic regions from protein surface inside of protein, allowing hydrophilic regions to come out to surface to form H-bonds with water), so net change in entropy of solute and protein is than increase of entropy, and process is spontaneou

Entropy of what? Isn't the entropy of the universe always increasing?

Yes, entropy of the system could increase. E.g., Chelation http://en.wikipedia.org/wiki/Chelation

@user26143 Chelation describes a process of multiple bond formations and breakings. I think the OP is asking about the entropy for the formation of one bond.

Are you talking about gas phase? In solution, solvent organization is going to make things very complicated for this question...

@Zhe Just the bond formation itself, neglecting any external effects

I do not think that an entropy decrease is a requirement to form a bond.
Think about a "normal" case: $\ce{HCl}$ (a gas) and $\ce{NH3}$ (another gas) react together to form $\ce{NH4Cl}$, which is a solid. This is an example of bond formation where entropy clearly decreases.
But now think what would happen if two liquid or solid compounds react together to form a gaseous compound. An increase of entropy has taken place. Can't think of a concrete example, but I think this might be the case of, for example, an esterification between a liquid alcohol and a liquid carboxylic acid that form an ester where, due to the lose of $\ce{-OH}$ groups, intermolecular interactions are lower and it can result with the product molecule being more volatile than the original compounds.
It's all about the number of micro states in which the compounds can exist, and gas formation from solids or liquids importantly increases entropy.
(PS: These are just some thoughts, really, I'm no physical-chemist.)


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What free software exists to view molecule pictures within a file explorer?
I am testing my small api library that generates lots of chemical files such as MOL and SDF. I was wondering if there is a software that can show me the pictures of chemical structures inside each file so that I don't have to open each of these separately. would prefer something free for Windows and Mac.

I found this tool named thumbfish (Windows only). It can display thumbnails for variety of formats and is free

I wish thumbfish could also handle *.cif and *.mrv files.

@andselisk you should try to convert them with free openbabel software

On the Mac, there's ChemSpotlight. The most recent version includes a QuickLook component that will allow previews of most molecule files, including interactive 3D depictions using HTML/JavaScript.
You can also batch generate previews of MDL, SDF, etc. using Open Babel to export previews to SVG or PNG formats.

What about JMol ?
It opens .Mol and a slew of other formats in Java which is fairly widely compatible. Open-source too.

jMol has an application that can show structure pictures but then I have to open each of my files to view them. I am looking for something that will generate a thumbnail for my structures.

JMol is scriptable, and it can generate image files, too.

There is Free Maestro or Maestro Academics, but you have to read EULA before, maybe you are going to use it for commercial purposes.
This is software for creation and opening of the databases, can read almost all formats. And also has 2d drawing tool and is cross-platform.


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Protonation of Guanidine
Why is the $\ce{sp^2}$ hybridized nitrogen atom in guanidine the most basic?

I know that as a general guideline, $\ce{sp^3}$ hybridized atoms are more basic than $\ce{sp^2}$. Bases are electron pair donors; having the electrons further away from the nucleus makes donation easier.
On the other hand, protonating the double bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule. Protonating the other nitrogens does not allow for the same degree of resonance stabilization.
And finally, in one textbook, I found mention of a +M effect - what is the +M effect? And while we're on that, what's +/- I? I understand that I stands for induction, but what's +I? What's positive induction?
One more note, since the double bonded nitrogen is $\ce{sp^2}$ hybridized, does that mean it has a greater negative charge density than the other nitrogens by nature of having greater s-character? Could this + charge density be giving the double bonded nitrogen more nucleophilic character?

There is no sp³ nitrogen in guanidine.

+I stands for inductive effect and compounds less electronegative than carbon(attached to carbon chain) can be identified as +I For eg. Alkyl groups,CH3...
They can be described as donating groups
Therefore it can be considered that these group generally carry negative charges..

For the record, I think it's clearer if you represent lone pairs as dots instead of lines. It's a bit easier to read in the giant mix of lines that already exist. At least you circle your charges, so you can't confuse the line-lone-pair with a negative charge...

This may be another example of kinetics vs. thermodynamics. It is generally the case that $\ce{sp^3}$ hybridized lone pairs are more basic than $\ce{sp^2}$ hybridized lone pairs because the more s-character in an orbital, the more stable (unreactive) the lone pair.  However in guanidine, ground state resonance structures (see picture below) tend to make the $\ce{sp^3}$ nitrogens less basic.  Whether they are actually more or less basic than the $\ce{sp^2}$ nitrogen I don't know.  But let me start by assuming that the $\ce{sp^3}$ nitrogens are more basic.
To whatever extent the $\ce{sp^3}$ hybridized nitrogens are more basic, they will protonate more rapidly than the $\ce{sp^2}$ nitrogen.  However, due to resonance stabilization, the product formed by protonation of the $\ce{sp^2}$ nitrogen is thermodynamically more stable than the product formed upon protonation of the $\ce{sp^3}$ nitrogen.  All of these protonations are simple equilibria between starting material and protonated product.  In this case with guanidine there are two protonation equilibria occurring in parallel.  Even though protonation of the $\ce{sp^3}$ nitrogen is kinetically favored due to its increased basicity, the product we observe is from protonation of the $\ce{sp^2}$ nitrogen due to its greater thermodynamic stability.  Because both of these processes are in equilibria, thermodynamics controls the outcome.  If the $\ce{sp^3}$ nitrogens are actually less basic than the $\ce{sp^2}$ nitrogen, then kinetics and thermodynamics favor the same outcome, kinetic protonation of the $\ce{sp^2}$ nitrogen and formation of its thermodynamically favored products.

If I am interpreting your reaction coordinate correctly, you are saying that the protonation of the sp3 nitrogen has a lower activation energy but results in a thermodynamically unfavorable product; the protonation of the sp2 nitrogen has a higher activation energy but a more thermodynamically stable product?

Correct, and since they are all equilibria, thermodynamics, not kinetics, controls the outcome.

What do you mean they are all equilibria? As in, we observe both the reactant and product to actually exist, so the activation energy barrier can be overcome, and thus we can generally ignore the activation energy and just say that the thermodynamically favored product is the likely product?

Yes, in protonation reactions the activation energy is usually low, so assume both pathways are in operation with guanidine.  Then there are 2 parallel protonation reactions both starting from guanidine and leading to their respective protonated species.  In such cases, no matter which of the 2 products is kinetically preferred (and therefor formed first), whichever product is thermodynamically preferred is what we will observe in our beaker of product.

Cool! This leads me to a corollary question which I will be posting about soon.

Do you have a source for this graph? The HOMO of guanidine has the largest coefficient at the doubly bonded nitrogen - this would contradict your theory, hence protonation would be thermodynamically and kinetically favourable at this position.

As I say in the first paragraph, "Whether they" (the sp3 nitrogens) "are actually more or less basic than the sp2   nitrogen I don't know."  In the second paragraph I examine both possibilities, sp3 more basic than sp2 (I drew the graph to illustrate this case), and sp2 more basic than sp3 (where I say, "kinetics and thermodynamics favor the same outcome").  There is no single "theory" to contradict.

There are no sp³ nitrogens in guanidine.

@ron What about arginine? it is also basic than lysine. Can the same be applied to arginine?

Your misconception is directly at the beginning of your question and turns up multiple times. There are no $\mathbf{sp^3}$ nitrogens in guanidine.
Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced.
This is also why the $\ce{-NH2}$ nitrogens are not basic. Their lone pair is not sitting in an orbital all alone waiting for a proton. It is actively resonating with the π system of the other two. The only available lone pair guanidine has is that of the $\ce{=NH}$ nitrogen sitting in an $\mathrm{sp^2}$ hybrid orbital. Thus, a second protonation requires somewhat superacidic media.

The basicity of the double-bonded nitrogen is chiefly a result of the delocalization of charge permitted by the two single-bonded nitrogens with allylic lone pairs, as the resonance contributors you've drawn suggest. Notice that, upon protonation, the three nitrogens become chemically equivalent, with the positive charge shared equally among all three. If either of the single-bonded nitrogens is protonated, the resulting cation does not benefit from charge delocalization, which is a comparatively less stable arrangement.
Since the basicity is a function of the stability of the resulting guanidinium cation, it's a thermodynamic phenomenon. However, it may be worth mentioning that kinetic factors also likely favor protonation at the double-bonded nitrogen, since the lone pairs on the other two nitrogens are somewhat delocalized. This is the meaning of the $+M$ notation, which refers to the amine groups being electron-donating by the mesomeric effect (basically synonymous with resonance), while $-M$ would indicate groups which are electron-withdrawing. This is an effect active within the $\pi$-electron system of a molecule, which is probably better understood when considered from the standpoint of MO theory. The concept of resonance/mesomerism is a convenient abstraction that usually works well qualitatively to approximate the electron distribution in molecular orbitals (especially frontier orbitals). In reality, MO theory probably provides a more accurate rendering of the electron distribution, with $\pi$ orbitals that span multiple atoms. I say this to emphasize that resonance is not a physical process, but just a very expedient way of representing electron delocalization in $\pi$ orbitals.
Similary, $+I$ indicates a group being electron-donating by induction, while $-I$ denotes groups that are electron-withdrawing via induction. In essence, an electronegative group creates bond polarization, which can act to draw electron density away from an atom, while an electropositive group can do the opposite. This effect operates principally through the $\sigma$-bond skeleton of a molecule, and weakens as distance from the polarized group increases.

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Why does diazo coupling preferentially occur at the para position?
When the diazonium ion couples with dimethylaniline, why is the para position favored?

Is this only due to steric factors or do electronic factors (such as ortho-para directivity) play a role here too? In particular, I suspect ortho-para electronic factors might be significant because some of the resonance structures have a cation directly next to the electron-donating group.

Both steric and electronic factors play a role in this reaction.  Resonance structures such as those shown below favor electrophilic attack (the diazonium ion is an electrophile) at the ortho and para positions. Since there are twice as many ortho positions as para positions in N,N-dimethylaniline, on a purely statistical basis we might expect a 67:33 ortho/para product distribution if only electronic effects controlled the reaction.  Since diazonium coupling usually produces a majority of the para coupled product, steric factors must be playing a role as well.  The somewhat bulky dimethylamino group does make it more difficult for an electrophile to approach the ortho position and the diazonium ion is relatively large.  These two steric factors combine to reduce the amount of the ortho isomer and the para isomer predominates.


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Activation Energy in Acid/Base reactions
For simple acid/base reactions - i.e. those involving inorganic compounds - does activation energy ever play in reactions?
Or can we ignore kinetics entirely and just consider thermodynamic factors such as conjugate stability?
I'd be very interested if there were some book or article examining inorganic molecules' acid-base behavior from a kinetic standpoint.

Proton transfer, especially in aqueous media, has an activation barrier that is easily surmounted at room temperature or higher.  Such reactions are said to be diffusion controlled (the reaction happens at the rate of proton diffusion through the media) or spontaneous.  In such cases thermodynamics will control the process.   That said, I'm sure there are cases where the base the proton is transferring to, is, for example, surrounded by bulky ligands substantially increasing the activation energy for the process.

What about electron transfer?

Generally electron transfer is also very rapid.  It occurs often in biological systems, and you wouldn't want to spend a lot of energy (to get over a barrier) for a common process in a biosystem.  Electron transfer in other systems (such as redox) is also usually very fast.  However, because of the mass difference between a proton and electron, the mechanism for electron transfer may differ from proton transfer in that tunneling through the barrier (rather than passing over the barrier) is more common.


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Which argon compounds exist, supported by experiment or computation?
Just as $\ce{HArF}$ was synthesized...Could something like $\ce{Li-Ar-Li}$ exist with AXE geometry $\ce{AX2E3}$? Any other argon compound proposals out there?
I also know that there are some posible excimers like $\ce{LiAr}$ and $\ce{KrF}$ (maybe $\ce{KrF2}$). Can they be considered covalent compounds, even if explained via excited dimers/excimers or the OM theory (recall the technical name for lithium-doped argon matrices in the chemical literature)?

Related: http://chemistry.stackexchange.com/q/31845/15489

That depends on what you consider a compound. Well, so called "disodium helide" was made, so Li2Ar perhaps isn't out of question, but it neither Na2He isn't molecular, nor  would Li2Ar (probably) be.

ArBeO and ArBeS are experimentally known.
ArAuF, ArAgF and ArCuF are also experimentally known.
CUOAr$_n$ and related complexes are known.
ArBeNCN and ArBeNBO are predicted by theory.

But, could be Li-Ar-Li (AX2E3, linear molecule, trigonal bipyramid orbital) possible or not? Theoretically, at least...

Lithium is very electropositive, so it tends not to form covalent bonds.

Hydrogen is also electropositive...And we have HArF and, as well, excimer-like Li-Ar (but unstable) compounds. IN the Pauling scale, H is 2.3 and lithium 0.9. I mean, couldn't it be possible? I am not asking if it is hard (likely), but even if electronegativity is LOW, I can not see why it could not be possible in certain extreme conditions such a bond... Of course, lithium doped argon excimers are something different to my idea.

And I understand, even if they have special features, ionic bonds like "very polar" covalent bonds...Theoretically, what would forbid a Li-Ar-Li "molecule"? Example: http://journals.aps.org/pr/abstract/10.1103/PhysRev.179.19

Or http://www.physics.wisc.edu/~tgwalker/002.Spin-Rotation-Noble.pdf

And even if lithium is very electropositive, you also have the dilithium molecule $Li_2$, two lithium atoms COVALENTLY BOUND in a single dimer molecule...So, I have no reason to neglect a Li-Ar-Li compound by electronegativity, at least...

And finally, if you search, you can find this adduct with sodium http://scitation.aip.org/content/aip/journal/apl/100/11/10.1063/1.3695064 I agree the problem is that the compound can be very unstable in normal conditions, but I can not see a good objection if we allow certain extreme conditions...

If the 2 elements being considered have an electronegativity difference of >1.7, a covalent bond will not form.  The Allen electronegativity of Ar is 3.24 and of Li is 0.91.  The references to LiAr you cite are van der vaals molecules not covalent molecules.  Li-Li has an electronegativity different of zero.  The sodium reference is Na+, not neutral Na, and says the same type of cluster forms for Li+ with Ar.

I see. Let me explain why I am asking this question. In some acces exam to the university in my country, it was asked the following question:

Argue what kind of "bonding" will POSSIBLY form between   =3 and Z=18. Of course, the normal question IS that argon does not bind with lithium since it has a complete valence shell with eight electrons. But, since Argon is in the 3th period, I was wondering if it could be admissible an answer with hybridation sp³d ...I agree with you in the fact that lithium is very electropositive, but seeing other compounds I don't understand yet why Li-Ar-Li is not possible...

In summary, I want to understand what forbids argon to hybridize sp³d and "bond" two electrons with the single electrons of two lithium atoms...

I have found this http://twitpic.com/e99l3k here ab initio Calculations on Alkali Atom - Rare Gas
Van Der Wads Clusters
Bosoon Lee and Sungyul Lee*
College of Environment and Applied Chemistry, Kyunghee Universi~, Kyungki-do 449-701, Korea
(Received March 29, 2000)... Of course, I guess that a Van der Waals complex IS NOT a covalent compound itself, but it seems that it can be "understood" as a week aggregate compound due to Van der Waals forces...

@DavePhd about that electronegativity rule you assert ... this listing gives F=3.98 and H=2.2, so no covalent bond forms between H and F. //Hydrogen halides are generally considered covalent molecules unless they are dissociated in a solvent

Wikipedia now lists dozens (hundreds?) of known argon compounds, so the concept is nothing new.  Most are a bit contrived, but one compound that could have a real existence on Earth -- or rather, in Earth, is a nickel-argon compound:

At 140 GPa and 1500K nickel and argon form an alloy, NiAr.[1] NiAr is stable at room temperature and a pressure as low as 99 GPa. It has a face-centred cubic (fcc) structure. The compound is metallic. Each nickel atom loses 0.2 electrons to an argon atom which is thereby an oxidant. This contrasts with Ni3Xe, in which nickel is the oxidant. The volume of the ArNi compound is 5% less than that of the separate elements at these pressures. If this compound exists in the core of the Earth it could explain why only half the argon-40 that should be produced during the radioactive decay that produces geothermal heating seems to exist on the Earth.[2]

There is an implication in the above passage that argon can assume a partial negative charge, despite ostensibly having full valence subshells, a feature not evident in the more usual noble gas compound choices such as oxides and fluorides.
Cited references:
1.
Dalton, Louisa (30 October 2019). "Argon reacts with nickel under pressure-cooker conditions". Chemical & Engineering News. Retrieved 6 November 2019.  Electronic link goes to Ref. 2.
2.
Adeleke, Adebayo A.; Kunz, Martin; Greenberg, Eran; Prakapenka, Vitali B.; Yao, Yansun; Stavrou, Elissaios (15 October 2019). "A High-Pressure Compound of Argon and Nickel: Noble Gas in the Earth's Core?". ACS Earth and Space Chemistry. 3 (11): 2517–2524.  Link


text:
Are 6-hydroxyhexan-2-one and hexanoic acid isomers?

Are the following two compounds isomers?
(A) $\ce{CH_{3}-CO-CH_{2}-CH_{2}-CH_{2}-CH_{2}-OH}$
(B) $\ce{CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-COOH}$

Although they have the same molecular formula ($\ce{C_{6}H_{12}O_{2}})$, I think that this two compounds are not isomers.

Isomers are compounds having same molecular formula, but a different structural formula.
As they satisfy the above condition, they are isomers and they are structural isomers. Structural isomers have a few variety based on connectivity of constituent atoms.

Chain : which has a different arrangement of carbon molecules
Position : different position of same functional group
Functional : different functional group for the same atoms
the others (tautomerism, metamerism and ring-chain isomerism)

They are functional isomers as they have different functional groups. First one has a ketone and an alcoholic group, while the second one has a carboxylic acid functional group.
You will come to see that it is easier to recognise structural isomers compared to the rest (stereoisomerism).
When speaking about isomerism, a picture is worth a thousand words:

They are certainly isomers since they have same molecular formula but differ in structural formula and their chemical properties are also different.
Isomers can be classified into structural isomer/constitutional isomer and stereo-isomer. The above compounds are structural isomers and in depth, they are functional isomers which is sub division of structural isomers in which the isomers differ in functional groups.
The compound A has two functional groups ketone and alcoholic groups (priority order ketone> alcoholic group) and compound B has carboxylic acid as functional group and they have same molecular formula.


text:
how does electronic conductance of a metal change with density?
The electronic conductance of a metal depends on its density.Does this mean that if we increase the density of the metal,its electronic conductance will increase?How does it happen?

You cannot change the density of a metal without also changing other properties.
Brass is a mixture of several elements. It can have different densities depending on the mixture of the elements. I haven't researched it, but the electronic conductance would likely be different for each type of brass.
Do you have a source for your statement?

The electronic conductance of a metal depends on its density.

It's given in a reference book of chemistry.

@user47568 I'd be curious what the book is. On the one hand, I can imagine that making a less dense metal (i.e., adding impurities, voids, filling with air, etc.) would likely create a less conductive material, I don't think compressing a metal and increasing the density necessarily makes a more conductive material. Maybe this is true in some cases, but recent experiments and quantum calculations have shown complicated effects with high pressure materials.


text:
Why does the solubility of group II hydroxides increase and the solubility of sulfates decrease down the group?
I know that solubility of alkaline earth metal hydroxides increases down the group and solubility of alkaline earth metal sulfates decreases down the group. What is the reason for these trends?

A few weeks ago I would have given the hard soft acid base theory (HSAB) explanation.
I have heard this "trend" explained in this way:  Hydroxide is hard and sulfate is soft.  So, the harder ions are, the more soluble with sulfate and less with hydroxide.
About a week ago I was playing around with some DH, lattice energy, and Ksp data and I found that barium seems to go against the "trend".
I found these posts that show that many of the "trends" we are taught in chem class aren't as trendy as we are told.
Look at this:  http://www.chemguide.co.uk/inorganic/group2/solubility.html
Is this a homework question or a curiosity question?  If the former, the "trend" isn't that trendy and the teacher should read the links above.  If the latter, do you know that barium salts tend to go against the "trend"?

This answer could be improved by adding more information from the hyperlink. If the link breaks, then this answer isn't helpful.

Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy.
Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy.
Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy.
Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group.
Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.


text:
NCl₃ + H₂O → NH₃ + HOCl , the resulting solution is acidic, basic or neutral?
When $\ce{NCl3}$ is dissolved in water, $\ce{NH3}$ and $\ce{HOCl}$ is formed.
Can you please explain me whether formed solution is acid or basic or neutral? thank you!

Try balancing the equation. What do you notice?

You should specify initial concentrations. generally, your question has no simple 'qualitative' answer, as both acid and base produced are weak. However, analytical chemistry textbooks should have an entire  section about calculation of pH of similar solutions.

Here's our initial reaction, $$\ce{NCl3 + 3H2O -> NH3 + 3HOCl}$$ $\ce{NH3}$ is a weak base and $\ce{HOCl}$ is a weak acid, but there is more of the acid, so at this point the solution would be mildly acidic. However $\ce{NH3}$ and $\ce{HOCl}$ are known to react as follows $$\ce{NH3 + HOCl->NH2Cl + H2O}$$ and in aqueous solution the resultant chloramine can decompose as shown in the following equation $$\ce{3 NH2Cl ⟶ N2 + NH4Cl + 2 HCl}$$  Clearly, an acid is generated in this last reaction.  So although the initial products are neutral, eventually the solution will be strongly acidic.

What's the extent of these reactions?

I suspect they all move further to the right with time - the ammonia escapes or is consumed, the chloramine decomposes or reacts further, just a guess.  Open or closed containers would make a difference in the rates.

I was wondering because the extents can affect whether the solution is actually mildly acidic, strongly acidic, or not acidic at all. Strongly acidic implies that the third reaction goes forward to a large extent.

Good point.  But you'd be a tough professor.  The OP's teacher just wanted to know acidic, basic or neutral; you're drilling down to not, mildly or strongly acidic :) I would guess that the third reaction goes far to the right as the nitrogen leaves the liquid phase, just don't know the timescale.

What about this reaction? NH3 + HOCl -> NH4+ + ClO-? This reaction would be large extent, and would probably consume a fair bit of the ammonia and hypochlorous acid.

K for aforementioned reaction is 52 btw.

Since you give a K value, I assume its an equilibrium.  Since the product doesn't react further, it will revert to reactant as the ammonia is siphoned off in other reactions.  See problem 4 (scroll down to the answers) at https://engineering.purdue.edu/~jafvert/ce%20554/CE%2059700%20008%20S2010%20HW%204.pdf  Looks like the reaction shown above is the major pathway for $\ce{NH3 + HOCl}$

Aren't all chemical reactions in equilibrium at least to some extent? Also, you can get the K value easily. K for the reaction = Ka(reactant acid)/Ka(conjugate acid).

Yes, at least to some extent.

@ron no that can;t be. My teacher explained me the correct answer.Your answer is wrong. Anyway thank you very much!

@YomalSamindu It would be helpful if you posted the correct answer.

@YomalSamindu So, what is the correct answer? Can you at least shortly explain it here in a comment?

@YomalSamindu As the others say , it would be nice if you give your perception over here in the comments or as an answer .

It is less but acidic since in HOCl, OH bond dissociates to form proton.Though NH3 is a lewis base,it is not  strong enough to neutralize the acidic HOCl.

Consider coming over here. You might be interested since there are Indian highschool students out there.


text:
Hydrogen halide acidity
Why is it that
$$\mathrm{p}K_\mathrm{a}(\ce{HF}) < \mathrm{p}K_\mathrm{a}(\ce{HCl}) < \mathrm{p}K_\mathrm{a}(\ce{HBr}) < \mathrm{p}K_\mathrm{a}(\ce{HI}),$$
even though the electronegativity decreases down the group? The more electronegative the atom accompanying hydrogen, the lower the energy of the σ* bond. The lower the energy of the σ* bond, the easier it is for nucleophiles to attach to it (i.e. hydrogen is more easily removed from the compound, thus making it more acidic).
Obviously, something is wrong with this reasoning. I could guess it is because hydrogen binds to different $\mathrm s$ orbitals with the different halogens. For $\ce{F},$ $\ce{Cl},$ $\ce{Br},$ $\ce{I}$ these are $\mathrm{2sp^3},$ $\mathrm{3sp^3},$ $\mathrm{4sp^3},$ $\mathrm{5sp^3},$ respectively.

There are a few things to keep in mind. The $pK_a$ values you typically see depend on approximations for thermodynamic activity that only hold for dilute solutions, and are solvent-specific (typically water). Concentrated solutions of $\ce{HF}$ are actually extremely acidic, stronger than concentrated $\ce{HCl}$. $\ce{HF}$ is quite a strange beast, and its reactivity is very complex and concentration-dependent. Finally, the consideration of the acid's LUMO energy is probably more relevant to reaction kinetics, while acidity (as measured by $pK_a$) is a thermodynamic parameter.

Also, I'm not sure your reasoning about the LUMO energies is correct. See my discussion with ron in the comments to his answer. I could well be mistaken, but I think the LUMO energy of $\ce{HF}$ is higher than $\ce{HCl}$, for example. Keep in mind also that the extent of ionic character in a bond depends on a lot of factors, but generally you can expect that larger energy differences between interacting atomic orbitals (which tends to correlate with electronegativity difference) favors greater ionic character. Still, one also has to consider atom size, orbital overlap, polarizability, etc.

@GregE. I think I finally see my mistake now. I was assuming that if the $\sigma$ bonds rises in energy, the $\sigma^{}$ would also do, but the opposite is actually happening: when the σ bond rises in energy, the $\sigma^{}$  lowers, making it more easy to remove the proton in our case. I looked at http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf and it seems that the energy of the hydrogen $1s$ orbital does not differ that much from the energies of the $2sp^{3},3sp^{3},4sp^{3},5sp^{3}$ orbitals of our halogens, which if you come to think of it isn't very strange (continued)

because, although the orbitals with a higher principal quantum number have higher energy, they are also more shielded of from the nucleus by the electrons from the lower shells. This seems to cancel each other out mostly for atoms in the same group, leaving the energies of the $3p, 4p, 5p$ orbitals a bit higher for $\ce{Cl, Br, I}$ (higher bond energy, lower bond strength, more acidic). There are of course other factors to consider, but since I'm not studying quantum mechanics, but organic chemistry, I'll leave it up to you guys to explain the remaining granularities.

Last thing I noticed is that fluorine's $2s$ orbital has significantly lower energy than the corresponding $ns$ orbitals of $\ce{Cl, Br, I}$. Also note that the table I listed doesn't say anything about the $d$ orbitals, I don't know what effect these have on the hybridized orbital energy, but I think not much as they will be very high energy.

So to conclude, it seems that the binary halogen acids have a significantly covalent bond, especially $\ce{Cl}$ and $\ce{Br}$, so ionic bonding is only an issue with fluorine. The decreasing electronegativity down the halogen table is not large enough (doesn't pull the $\sigma^{*}$ bond energy far enough up) to overcome the increasingly more energetic bond between the halogens down the table and hydrogen, reducing their bond strength and thus increasing acidity.

You cannot assign an energy to a hybrid orbital. Hybridisation is a concept that can only be applied to a molecular bonding pattern - it is a mathematical, but not at all physical representation. All of these molecules are linear and hence only sp hybridisation is possible. Apart from that, s p splitting in Br and heavier is too big to hybridise at all. The s orbital becomes more and more inert.

@Martin Wait what? Are they linear? But they have three lone pairs? Can't you say: well $sp^{3}$ hybridized is energy of s orbital plus three times energy of p orbital divided by four (i.e. they don't have energy by themselves but get it from their components)? It's all so dang difficult to understand because this organic chemistry book (i.e. Clayden et al.) seems very shy about explaining these things is more detail.

The only possible arrangement of a diatomic molecule is to be linear. The symmetry of the molecular orbitals has to represent the nuclear geometry as well. Using hybrid orbitals will give you the wrong bonding picture - as they do not follow these constraints. Hybrid orbitals itself have no physical meaning whatsoever - they are just a mathematical interpretation (they do not exist). Org. Chem. often uses hybridisation as a law, but that is just wrong - one has to be much more careful in these terms.

Thermochemistry help; http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/HXacids/HXacids.html

Yes, you are on the right track, let's look at the situation in more detail.  In the haloacid equilibrium $$\ce{HX <=> H+ + X-}$$ anything that stabilizes HX will push the equilibrium to the left and make the $\mathrm{p}K_\mathrm{a}$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make the $\mathrm{p}K_\mathrm{a}$ less positive. Here's a table that summarizes some of the haloacid data:
$$\begin{array}{cccc}
\text{X} & \text{H-X bond strength / kJ mol}^{-1} & \text{Electronegativity of X} & \mathrm{p}K_\mathrm{a}(\ce{HX}) \\
\hline
\ce{F} & 565 & 4.0 & 3.1 \\
\ce{Cl} & 427 & 3.0 & -7.0 \\
\ce{Br} & 363 & 2.8 & -9.0 \\
\ce{I} & 295 & 2.5 & -11.0
\end{array}$$
There are 3 main factors that influence this equilibrium: bond strength, electronegativity and polarizabilty.
In the "Bond Strength" column we see that the $\ce{HX}$ bond strength decreases as we move down the column. This is because, as you noted, overlap between the hydrogen $\mathrm{1s}$ orbital and the halogen orbital is most effective with the fluorine $\mathrm{2sp^3}$ orbital (in fact, overlap is so good in the fluorine case that the bond has a significant covalent nature) and the effectiveness of the overlap decreases as we move down the column. Therefore, $\ce{HX}$ bond strengths should make the $\mathrm{p}K_\mathrm{a}$ decrease (less positive, more negative) as we move down the halogen column.
Your analysis of electronegativity is correct. The more electronegative the halogen, the more stable the halide anion. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to increase (more positive) as we move down the halogen column.
The larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to decrease (more negative) as we move down the halogen column.
So we have two factors, bond strength and polarizability, that work together to push the $\mathrm{p}K_\mathrm{a}$ lower as we move down the halogen column; and one factor, electronegativity, that works in the opposite direction. Based on the observed $\mathrm{p}K_\mathrm{a}$'s, the first two factors win out over the electronegativity.
BTW, I like your $\sigma$, $\sigma^*$ reasoning. Here's how I think of it in those terms.  As you bring two atomic orbitals closer together, they will split and form two new molecular orbitals. The degree of splitting is dependent upon the overlap of the two atomic orbitals.  Overlap is a measure of electron density between the atoms.  An electronegative atom will share its electron less readily decreasing the splitting.  If there is a strong bond, then the electrons are well shared and the splitting will be larger.

Oops, I forgot the lone pairs of the halogens, they will be hybridized into $sp^{3}$ of course :)

What do antibonding orbitals have to do with the binary acids of group 7? I don't think any of the antibonding orbitals are populated in the binary acids ...

HF has a significant covalent component to its bond.  The covalent nature of the HX bond drops off rapidly going down the column.  As the bond becomes less covalent (more ionic) there is less electron density between H and X causing the energy of $\ce{\sigma^{\ast}}$ to decrease.  HX acidity depends on a number of factors as I mentioned above.  No one factor alone can adequately explain the trend.

Are you saying that the HX series becomes more ionic as you go down the column? That doesn't make sense because the halogens get less electronegative as you go down the column.

and also less capable of forming a covalent bond with a small 1s orbital

Where does sigma star come from? It's not populated in any of the HF MO diagrams I see.

It's not populated, if a nucleophile were to attack it would interact with sigma star, in the HX case the splitting is a measure of bond strength

@ron, I'm interested in your thoughts on this. As $n$ grows, so does the energy of the $X_{np}$ orbital, becoming closer in energy to the $H_{1s}$. This should result in greater energy lowering/raising for the HOMO/LUMO, respectively. However, the overlap and resonance integrals decrease in magnitude with increasing $n$, which should have opposite effect. I'm fairly sure I've seen calculations showing the LUMO energy of $\ce{HF}$ to be highest in the series, but can't find a reference now. I'd expect it to be, per HSAB theory results if nothing else. Do you happen to know if that's correct?

@Greg E  As n grows wouldn't the energy difference between X(np) and H(1s) increase?

@ron, not to my understanding. The $2p$ of fluorine is lower in energy than the $3p$ of chlorine, for example. To be certain, I confirmed the numbers in Fleming's book on MO theory, which indicates a difference of $-4.9$ eV. Both are lower than the hydrogen $1s$.

If that's correct, then everything in your comment hangs together and makes sense.

The chlorine $3p$ is very close to the hydrogen $1s$, however. Assuming the trend continues, the bromine $4p$ would actually be higher energy. In any case, I don't think it's especially important in determining the acidity of the halides, since rarely would acid-base reactions involving them be kinetically controlled.

I'm not sure, but can you even talk about $\sigma$ and $\pi$ bonds in an ionic bonding structure? The atoms/molecules are rather attracted to each other in total. No orbital interaction. But perhaps what Greg says could be some truth in. I assumed that a halogen $5sp^{3}$ would be much, much higher in energy than a hydrogen $1s$. Does anyone have some online table for this?

@Jori, to the best of my knowledge, I would expect the $4p$ and $5p$ orbitals of bromine and iodine, respectively, to be higher in energy than hydrogen $1s$. That said, the calculated values given in Fleming's book Molecular Orbital Theory and Organic Chemical Reactions indicate that fluorine $2p$ and chlorine $3p$ are both lower in energy, the former significantly and the latter only marginally ($-4.9$ and $-0.1$ eV, respectively). Finally, bonding is a continuum, with totally ionic and totally covalent being extremes. AFAIK, the binary haloacids all have substantial covalent character.

You argued that the more the electronegativity, the more it pushes pka value upwards (increases). However, couldn't I make the opposite claim: as the electronegativity increases, the ionic character increases, and the compound tends to more readily dissolve (dissociate) in water?  I know that I'm wrong; could you please show me how I'm wrong in approach?

@trying You misread my argument.  I said the more electronegative the halide, the more dissociated the acid and consequently the more negative the pKa (push downwards); or, as we move down the series the pKA is pushed upwards. Your argument that the more electronegative the halide the more dissociated it is, is fine. It's just saying what I said a bit differently.  Hope this helps!