text:
Calculating entropy change of surroundings
I have seen many similar questions but have not found my answer. Why do we use the actual heat involved in the process to calculate entropy change of surrounding? The only answer I can think of is that we are treating the process as reversible with respect to the surroundings even if it is irreversible with respect to the system. But, this is an irreversible path for the system, and a reversible path will not involve the same heat, so now, the entropy of surrounding calculated must be different. This seems to contradict the idea that the entropy of surroundings must be a state function. So why is this reasoning incorrect?

See my answer in the following thread:  http://chemistry.stackexchange.com/questions/42681/thermodynamics-entropy-and-temperature/42694#42694

When you evaluate the change in entropy of an entity (in this case the surroundings) between an initial and final (equilibrium state), you separate that entity from everything else, and then you only look at how that entity has changed between the two states, disregarding the actual process path.  In the case of the surroundings (treated as an infinite reservoir), it had received a certain amount of heat between the initial and final states, and its internal energy and entropy increased as a result.  In the formalism of thermodynamics, changes in an  ideal infinite reservoir are always regarded as taking place reversibly, irrespective of other irreversibilities in the actual process.  So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir).
You may also find my response in the following link pertinent to this question:  Thermodynamics entropy and temperature

Yes, I understand that the process can be modelled as reversible with respect to the surroundings. But consider a reversible process between the same two states for the system, then this process is also reversible for the surroundings, but since it involves a different heat change, the entropy calculated must be different for the same process in two different ways.

As I said,  to get the overall change in entropy, you need to look at each of the entities separately between their actual  initial and final equilibrium states.  You don't run the entire process over again reversibly for the combination of system and surroundings to get the change in entropy.  You take each entity separately over a  reversible path that may or may not bear any resemblance to the actual path followed by the system in the irreversible process (or to the reversible paths selected for the other entities).

So now if I treat the surrounding as one entity, any process is reversible for this entity due to it being modelled as a large reservoir.Why should I calculate the entropy change using the q for the actual process. As you said, the entropy change for any entity must depend only on the amount of heat extracted reversibly, independent of  their resemblance to the actual process.What I claim is that there are infinite ways in which I can take the surrounding between the same states reversibly, involving different amount of heat extracted.

That is just not so.  If the surroundings is purely an infinite reservoir, then the change in internal energy of the reservoir is just Q for the actual process (no work is done by the reservoir).  Once the internal energy change is fixed (i.e., the final state is fixed) and the temperature  is fixed, the entropy change can only be Q/T.

OK, so we are actually defining the state of the surroundings by this Q, which becomes the internal energy change, just as we define the system using variables like pressure, volume and temperature.Is that right?

Sure.  The internal energy per unit mass is just another example of an intensive property.  And we know that it takes two intensive properties (in this case internal energy and temperature) to define a thermodynamic equilibrium state of a single phase system.


text:
What is the product of the chemical reaction between phenol and ferric chloride?
The chemical reaction between phenol and ferric chloride is a test for the presence of phenol. They react with each other to produce a violet complex. However, the reaction is given differently in different links:

$$\ce{3ArOH + FeCl3 → Fe(OAr)3 + 3HCl}$$
source 1 , source 2
$$\ce{6 ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$$
source

I guess the 2nd reaction is the appropriate one because it shows the formation of a complex, $\ce{[Fe(OAr)6]^3-}$. How could  $\ce{Fe(OAr)3}$ be considered a complex? It should act like a ionic salt like $\ce{Na^+C6H5O^-}$  or $\ce{NaOAc}$.
So, what is the correct product of the reaction; $\ce{[Fe(OAr)6]^3-}$ or $\ce{Fe(OAr)3}$?

Well you'd need complexation and deprotonation equilibria to describe it properly...

After doing some looking around, and going back to check what I thought I knew about coordination complexes, I think I understand why you're seeing two equations. I first recommend reading the pages on Chemguide that talk about coordination chemistry, particularly the first half of this page on acidity, and the pages on reactions with hydroxide and ammonia. Both were helpful to me in understanding an investigation I'm doing on copper (II) complexes, but the same ideas apply to 3+ compounds.
The important concepts to keep in mind are that coordination compounds (in situations like this, at least) will either undergo acid-base reactions or ligand-exchange reactions. Acid-base reactions result in neutralization and a precipitate, ligand-exchanges cause a color change. Phenol is a relatively strong acid (stronger than alcohol at least), and so is ferric chloride. I had to look around but this page says at the very bottom that all oxygen-containing compounds act as bases in the presence of Lewis acids (like ferric chloride).
$\ce{3ArOH + FeCl3 -> Fe(OAr)3 + 3HCl}$ is the acid-base reaction, and you get a ferric phenolate salt as a precipitate.
$\ce{6ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$ is the ligand-exchange. It's not immediately obvious, but you can rewrite the two reactions a different way and it becomes clear.
$$\ce{3ArOH + [FeCl3(H2O)3] -> Fe(H2O)3(OAr)3 + 3HCl}$$
$$\ce{6ArOH + [FeCl3(H2O)3] -> [Fe(OAr)6]^3- + 3H2O + 3H+ + 3HCl}$$
Those three $\ce{H+}$ ions are from the excess phenol. The ligand-exchange occurs only when the phenol is in excess and more water molecules can be replaced, making the iron compound ionic and soluble again.
So depending on the context, you'll see the one with three or the one with six. Both are correct, but $\ce{[Fe(OC6H5)6]^3-}$ is the new complex and $\ce{Fe(OAr)3}$ is an intermediate product.
Hope that helps.

source: Stack Exchange

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

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

sort: votes


text:
Why do displacement reactions occur?
Suppose we have a compound, say copper sulfate. I know that sulfate is a poly-atomic ion, having valency 2. And, in this case, copper also has valency 2. So, to fullfill their 'octet' requirements, they both react. The problem is: Suppose we take the metal zinc and bring it in contact with copper sulfate. Now, a displacement reaction occurs, in which zinc replaces copper because of its higher reactivity.
Copper sulfate is a compound that is neutral; both the cation and anion have completed their octets; why then does zinc replace copper? Or, in other words, after completing the octet, why it is necessary for sulfate to leave copper and react with zinc?

$\ce{Cu^2+}$ and $\ce{SO4^2-}$ ions do not react. Your description of the ionic bonding in copper(II) sulfate sounds like a description of covalent bonding, which it is not. 2) In aqueous solution, which you are probably talking about, the ions are separated and there are hardly any interactions between ions since they are surrounded by water molecules. The reaction is between Zn and $\ce{Cu^2+}$, not Zn and $\ce{CuSO4}$. 3) The octet rule is irrelevant in transition metal compounds like those of copper. 4) The octet rule does not, and cannot, predict whether a reaction takes place.

It is due to that zinc is more reactive than copper which is less reactive metal so, zinc can easily displaces copper from its solution. We can understand it as zinc ion, $\ce{Zn^2+}$ ion form more strong electrostatic attraction with sulphate ion, $\ce{SO4-}$ ion.
Also, we can also consider the electrochemical series or activity series to give a better reason for the same. In activity series, zinc has greater value of standard reduction potential than copper's standard reduction potential value. It means, zinc can easily reduce Copper and it get oxidised.
$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$$

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


text:
Iodine medical products
After an injury or getting a scare it is typical to use medical products containing iodine. They typically leave a huge red spot on the skin.
Why are creams and solutions containing iodine used in the medicine?

Iodine kills microorganisms, although the mechanism does not seem to be well-known: "Similar to chlorine, the antimicrobial action of iodine is rapid, even at low concentrations, but the exact mode of action is unknown." (Clin. Microbiol. Rev. 1999, 12(1), 147–179) I did see some newer research on the topic though. Some less reputable sources on the Internet write that iodine has multiple mechanisms of action, which makes it difficult for microorganisms to develop resistance.

@orthocresol looks like an answer to me!

This Wikipedia article gives a good, if brief, discussion of a slow-release form of iodine called "povidone-iodine" in which iodine is the only active ingredient.

Free iodine, slowly liberated from the povidone-iodine (PVP-I) complex
  in solution, kills eukaryotic or prokaryotic cells through iodination
  of lipids and oxidation of cytoplasmic and membrane compounds. This
  agent exhibits a broad range of microbicidal activity against
  bacteria, fungi, protozoa, and viruses. Slow release of iodine from
  the PVP-I complex in solution minimizes iodine toxicity towards
  mammalian cells.
PVP-I can be loaded into hydrogels, which can be based on
  carboxymethyl cellulose (CMC), poly(vinyl alcohol) (PVA), and gelatin,
  or on crosslinked polyacrylamide. These hydrogels can be used for
  wound dressing. The rate of release of the iodine in the PVP-I is
  heavily dependent on the hydrogel composition: it increases with more
  CMC/PVA and decreases with more gelatin.


text:
Growing crystals that are in a magnetic field or subjected to an electric field
I know different crystals resonant at different frequencies but is it possible to grow crystal at a certain frequency.  An example would be growing crystals using Epsom salt or Alum in a magnetic field or when it's subjected to an electric field while two electrodes are submerged into the saturated solution.
The reason I ask is I know they can "program" magnetic fields one company doing this is called Correlated Magnetics
A video that explains "programming" magnets
I would like to do something similar but with crystals.  I do know there are ways to test the resonance of crystal oscillators Video showing how to find resonance frequency of crystal oscillator
How to test Rochelle salt piezoelectric crystals voltage oscilloscope
Now I'm just trying to see and find out if I can "program" crystals at various frequencies by subjecting them to magnetic fields or electric fields.  Has anyone tried this or know of research being done?

I'm afraid your idea is bad. One can influence magnetic properties using magnetic field, but they are too weak to influence mechanical properties. Even if you'd get one strong enough it would be highly dangerous.

On second thought maybe you could make sth with piezoelectric or magnetostrictive materials.

The main influence on resonant frequency of a crystal is mechanical: the mass of the crystal and the restoring force of elasticity cause the crystal to vibrate at a specific frequency when disturbed. To fine-tune the crystal, bits may be chemically or abrasively removed or the mass may be increased slightly by rubbing with graphite.
That said, crystal orientation does influence resonant frequency because the modulus of elasticity of a crystal such as quartz is anisotropic, i.e. it varies significantly in which plane it is measured. Perhaps applying a strong magnetic or electrostatic field would change the resonant frequency -- but you would need to isolate the effect on the crystal from that of its mounting, particularly if there are metallic contacts.


text:
Determining Reaction Orders
Can you figure out the reactant's reaction order (First-order, second-order, third-order, etc.) JUST by looking at a chemical reaction? What if you were given a multi-step reaction? A single step reaction?
For example, can you tell what order $\ce{2NO_2Cl}$ is just by looking at this equation: $$\ce{2NO_2Cl -> 2NO_2 + Cl_2}$$
Or this multistep equation?
$$\ce{NO_2Cl -> NO_2 + Cl}$$
$$\ce{Cl + NO_2Cl -> NO_2 + Cl_2}$$

This is a common kinetics question... lots of websites discuss it. Here's chemguide's page on it. No, you cannot tell the order from the balanced overall reaction. You can tell the order if you are given the mechanism, though.

By given the mechanism, what do you mean? And how can I derive the order from it?

Do you mean the graph of reaction path and energy?


text:
Why is fluorine more reactive than iodine despite the weaker I-I bond?
The atomic radius of halogens increases as we go down the group due to the addition of new shells. As a result, the bond length of halogen $\ce{X-X}$  increases down the group. So, less energy is required to break the bond (bond dissociation enthalpy decreases). So, as per this assumption, iodine should be the most reactive halogen. But this is not true. Rather,  fluorine is the most reactive halogen. It reacts violently with almost all chemicals. Bond cleavage of fluorine is much easier that that of other halogens, which may be due to the repulsive force of the lone pairs of electrons.
$$\ce{2F2(g) + 2H2O(g) → O2(g) + 4HF(g) + heat}$$
So, what about iodine? This source gives the values of bond dissociation enthalpy. It shows that the B.D.E value of iodine ($\pu{151 kJ/mol}$) is less than that of fluorine ($\pu{158 kJ/mol}$). So, iodine should be the most reactive halogen and not halogen. But this does not happen. In most of the reactions of iodine, equilibrium is maintained. To push the reaction forward, a catalyst is used.
$$\ce{I2(l) + H2O(l) <=> OI-(aq) + 2H+(aq) + I-(aq)}$$
Another similar phenomenon is that acidity of hydrohalic acid increases on going the group. $\ce{H-X}$ bond length increases on going down the group. Bond dissociation enthalpy decreases. Bond breaks into corresponding ions. So, $\ce{HI}$ is able to furnish $\ce{H+}$ ion faster than $\ce{HF}$. So, $\ce{HI}$ is more acidic than $\ce{HF}$. But, the reverse is observed. $\ce{HF}$ is so reactive that it reacts with glass and has to be stored in wax bottles.
$$\ce{SiO2 + 4 HF -> SiF4(g) + 2 H2O}$$
$$\ce{SiO2 + 6 HF -> H2SiF6 + 2 H2O}$$
In most of the reactions of hydroiodic acid, equilibrium is maintained. To push the reaction forward, a catalyst is used.
$$\ce{H2(g) + I2(g) <=> 2 HI(g)}$$
My question: Why is it so?

A catalyst does NOT affect the position of equilibrium!

In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from reacting. Obviously, the higher this barrier, the less reactive the halogen is (in general).
However, unless you're trying to consider the relative reactivity towards homolytic cleavage of the bond, it isn't sufficient to look at the bond strength. The other bit is the driving force for the formation of the products – in other words, something that encourages them to react.
In this case, the formation of fluorine-containing products is generally much more thermodynamically favourable than that of the corresponding iodine-containing products. Some factors influencing this are:

The electron affinity of fluorine is much larger, which makes reduction to fluoride much easier
$\ce{E–F}$ bonds are stronger than $\ce{E–I}$ bonds, in both an ionic and covalent sense.
Fluoride has a larger solvation enthalpy than iodide

For example, factors 1 and 3 lead to the large differences in the reduction potentials: $E^\circ(\ce{F2}/\ce{F-}) = \pu{+2.87 V}$ versus $E^\circ(\ce{I2}/\ce{I-}) = \pu{+0.54 V}$.
Silica is readily attacked by $\ce{HF}$ not because it is a strong acid, but rather because of the strength of $\ce{Si-F}$ bond thus formed.
That deals with the thermodynamic parts of "reactivity", in the sense that it explains why some reactions with iodine require elevated temperatures for any significant conversion, whereas reactions with fluorine proceed readily at room temperature.
The same factors, though, apply to the kinetic aspects of "reactivity". The activation energy is largely dictated by bonds that one has to break or form, for example, and the rate of electron transfer processes usually increases with $E^\circ$. Exceptions to this arise when – for example – the reactants have solid state structures which must be broken apart before reaction can occur. However, that is not the case here.


text:
Food Concentrate For Baking Using Hard Alcohol ex. Jack Daniels
I am trying to find a way to use hard alcohol such as Fireball, Jack Daniels, Sailor Jerrys and so on and turn them into a baking concentrate. This means the flavor of the alcohol cannot burn off at a high heat. Will xanthan gum or gum Arabic used to make baking concentrates with fruit juice provide the same effect? Or is there a way to take the flavor of the alcohols and turn them into a food concentrate? I just need the flavor not the alcohol itself so if burning off the alcohol which is the actual part that seems to be burning off since it boils at a much lower temperature than water I can do that. Thank you in advance for your time and help.
Cheers,
Savannah

You might find better answers on Seasoned Advice.


text:
Why are discharge tubes filled at reduced pressure?
Why are the gas sample tubes of hydrogen filled at a reduce pressure when observing hydrogen's emission spectrum? In my HS chemistry lab, we looked at gas sample tubes of hydrogen through a spectrometer and I am wondering why they were filled at a reduced pressure?

Think of the distance that an electron travels before hitting an atom of hydrogen. If the gas were dense then that distance would be short. In other words it would require more voltage to get a discharge across the tube, and more voltage to maintain the discharge. With more voltage, you'd also get more current when the discharge does occur. That would mean more heat. // ;-) I'd guess part of the reason here too is safety. A high voltage with very, very low current isn't deadly. I got a nasty jolt and a small electrical burn from such an apparatus, but lived.

Googling 'Paschen curve' may prove enlightening (pun intended).

Not only does it take a high starting voltage and a current-limiting device (e.g. inductor) to maintain a continuous discharge at higher pressures, but the quality of the light emitted is changed by atomic (and ionic) interactions, tending to broaden lines into a quasi-continuous spectrum.
A good example is the sodium-vapor lamp, often used for street lighting. In the example cited, compare the line spectrum of a low pressure lamp with that of a high pressure one (though there is also a contribution from other gases in the second image) .
The low-pressure hydrogen glow discharge corresponds well with calculated values and is a useful tool in showing how theoretical calculation of orbitals can be translated into an observable spectrum. Calculation of high-pressure phenomena is much more difficult.


text:
Why are 1,3-diols protected with aldehydes but 1,2-diols are protected with ketones?
1,3 diols are usually reacted with benzaldehyde when protection of the OH groups is desirable, but acetone is used when the relation between the hydroxyl groups is 1,2. Why is it so?

I have never heard of that. Check this out: http://www.organic-chemistry.org/protectivegroups/carbonyl/dioxanes-dioxolanes.htm

Hmmm.. isnt acetone cheaper and more easily available?

Let's look at the protection of a typical 1,3-diol with benzaldehyde and acetone:

The theory is that the formation of the six-membered cyclic acetal is less favourable with acetone than it is for benzaldehyde, because with acetone you cannot avoid having one methyl group being axial.
The next-most stable thing after a six-membered ring is a five-membered ring, which can be formed by the protection of a 1,2-diol instead of a 1,3-diol. So, acetone happily reacts with those 1,2-diols to form five-membered cyclic acetals, whereas benzaldehyde is more selective for 1,3-diols since it can enjoy the stability of a six-membered ring.
When there is a choice, like in glucose, you get rather different results:

However, I am not sure how much this principle extends to protection of compounds with only two hydroxyl groups. I am guessing it should be possible to protect 1,3-diols with acetone too - after all, a methyl group is hardly the worst axial substituent to have.


text:
How to calculate hydrogen from SMILES string?
How to calculate the number of hydrogen atoms from SMILES string?
For example,
SMILES String: C1=CC(=C(C=C1C(CN)O)O)O

The known result is C8H11NO3

I think you wish to know how to add explicit hydrogens to a structure. The fact that your structure is originally a SMILES is not particularly important.

I think basically you just want to calculate a molecular formula from the SMILES?

@GeoffHutchison I just want to calculate a molecular formula from the SMILES. Because, there is no universal SMILES format handled by any chemical database. So, matching exact SMILES string is very difficult. Instead matching molecular formula is easier.

Other answers have done a good job of providing a description of the concepts and algorithms needed to answer the question.  Here I'll demonstrate the opposite: a black-box empirical approach.
In Python using rdkit:
from rdkit import Chem
from rdkit.Chem import rdMolDescriptors

my_smiles_string = 'C1=CC(=C(C=C1C(CN)O)O)O'
my_mol = Chem.MolFromSmiles(my_smiles_string)
print(rdMolDescriptors.CalcMolFormula(my_mol))

This shows the result C8H11NO3.  You could parse that further to get only the number of hydrogens, if you need to do so.  Alternately:
my_mol_with_explicit_h = Chem.AddHs(my_mol)
print(my_mol_with_explicit_h.GetNumAtoms() - my_mol_with_explicit_h.GetNumHeavyAtoms())

displays 11.  Slightly more cumbersome but it has the output you want.  If you take the latter approach you don't even need to import rdMolDescriptors.

Information on the system can be found here: https://en.wikipedia.org/wiki/Simplified_molecular-input_line-entry_system
http://www.daylight.com/dayhtml/doc/theory/theory.smiles.html
The correct number of hydrogen atoms to fulfill the bond count around the other elements is assumed unless placed in square brackets. The bonds can be read from left to right and assumed to be a chain, unless placed in braces which represent a branch.
So there would be 1H around the first C (going from left to right) this is because it is c1 indicating it's one half of a ring closure so will have an extra bond attached, there is one H on the second C. The third C has a branch and an O (the chain continues after the braces close) which would have a single H attached, the third carbon has no H as the branch is attached via a double bond. (so far we have 3H)
The first C in the branch has no H and attaches to another branch and another O with an H.
The first C in the second branch has a single H. The second C in the second branch is important because it has 1 after it like the first C. This indicates that they should have a bond between them forming a ring closure. It continues down the branch which now looks like a substituent on the ring. So it has no H.
The third carbon in this branch has an O (with an H) and a new branch, so it has a single H.
(So we have 7H so far)
The third (and final 'branch' has a C attached to a N and two H, the N has two H attached. which adds 4H making 11H.

Thank you for your information. You have given an explanation for constructing the SMILES string from the chemical structure. But, I need to calculate atoms from the SMILES string (ie., parsing the SMILES, especially H). The problem is -- in SMILES representation, Hydrogen (H) is omitted.

But you know the valency of carbon and nitrogen (and other organic components) You just need to work out how many other elements are attached to the carbon and whether they are attached via double bond or single and take them away from the overall valency. It effectively displays the skeletal formula in text form.

Wherever the chemical formula comes from, it looks quite wrong.

You have to convert the SMILES to a graph model. Most everything you want to do with SMILES requires you construct the graph model, i.e., nodes (atoms) and arcs (bonds). Once you have that, you can go through all nodes, and see how many of their sites are occupied by explicit bonds, the rest you hydrogenize.
If you do chemical informatics, you really need to learn these principles and learn to program them yourself.
But taking this obvious solution aside for a moment, with SMILES the question is always: can you simplify the solution so as to work purely text string based? Might you be able to use regular expression replace pattern matching to come up with your solution? And the answer is categorically no! Because regex patterns do not work for nested parenthesis as you use in SMILES to write out branches.
Could you use a parser to do it? May be. But if you go through this effort, you might just as well reconstruct your nodes and arc model, which then you can use to do so much more than to just hydrogenize.


text:
Why enol is unstable?
Alkynes undergo acid-catalyzed addition of water across the triple bond in the presence of mercuric ion catalyst. The products are enol and keto. I want to know why is keto more stable than enol.

Try a Google search... "keto-enol tautomerism"

http://chemistry.stackexchange.com/questions/18904/which-is-the-more-stable-enol-form

It's is basically because in keto form the C=O bond is more stable than the C=C bond in enol. And the difference is probably close to 12 kcal.

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


text:
Why doesn't hydrogen displace metals from their oxides?
As I understand it, a displacement reaction is where a more reactive element kicks out a less reactive element from a compound.
For example, chlorine is more reactive than iodine, so chlorine will displace iodine from potassium iodide, yielding potassium chloride and iodine.
If magnesium reacts with water you get magnesium oxide and hydrogen. But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards? The above paragraph predicts that hydrogen would displace the magnesium from magnesium oxide.
So why does it actually happen the other way round?

‘But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?’ — no.

Magnesium is much more reactive than hydrogen.

Geez, couldn't we simply edit out all mentions about "displacement" ? You're asking about redox reactions.

"Reactive" is a rather unfortunate choice of word in the given situation. You are asking about electropositivity, not reactivity.

If magnesium reacts with water you get magnesium oxide and hydrogen.

True. So you can't put out a magnesium fire with water. It just burns hotter.
$$ \ce{Mg(s) + H2O ->T[aqueous] MgO(s) + H2(g) ^}$$

But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?

No, in the above reaction you're reacting water with magnesium. The hydrogen is given off as a gas, though a tiny amount will stay dissolved in the water.

So why does it actually happen the other way round?

It will under suitable conditions. So with heat and under a pressurized flow  of hydrogen gas the reverse reaction will occur.
$$ \ce{ MgO(s) + H2(g) ->T[heat, pressure] Mg(s) + H2O(g) ^} $$

Hydrogen definitely will reduce metal oxides in the right environment.  Metal oxides tend to be very stable compounds, and hydrogen, though it is flammable, is not terribly reactive stuff either at room temperature.  Simply combining the two substances is unlikely to yield any real results.  Most metal/metal oxide combinations won't react vigorously at room temperature either.
However, at higher temperatures, when a reasonable fraction of the collisions between reactants involved will overcome the activation energy of the reaction, hydrogen readily reduces some metal oxides.  A fairly common lab involves the reduction of copper(II) oxide by hydrogen.  When my students do this, they first use zinc to reduce the hydrogen in sulfuric acid to hydrogen gas, then that is piped to a tube containing copper(II) oxide powder that is being strongly heated.  The reaction is strongly exothermic; it can be thought of as a thermite reaction, so this isn't particularly surprising.  Water and metallic copper are produced.
$\ce{H2 + CuO -> H2O + Cu}$
The activity/reactivity series of metals can be used to predict which metals will be reduced by hydrogen.

Thanks, I'm still a little confused -- magnesium burns if you light it with a blue flame, and hydrogen explodes at an orange flame. Care to comment on where my understanding of reactivity is going wrong?

@spraff Are you confused about why magnesium is emitting higher energy visible photons than hydrogen when it burns?  It's difficult to make generalizations about reactivity, but we can say that magnesium donates electrons much more readily than hydrogen does.  Magnesium and hydrogen will even react with one another to form magnesium hydride, in which hydrogen has a -1 oxidation state rather than its usual +1.  In this type of reaction that is what matters, which species wants to donate electrons more.


text:
Half-equations for the combustion of sulfur dioxide
Given the following redox equation:
$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$
$\ce{SO2}$ is being oxidized to $\ce{SO3}$ and $\ce{O2}$ is being reduced to $\ce{SO3}$.
I figured a way to write the half equations:
Oxidation half reaction: $$\ce{SO2 + H2O -> SO3 + 2H+ + 2e-}$$
Reduction half reaction: $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$
After balancing and summing, I get: $$\ce{2SO2 + 2H2O + O2 + 4H+ -> 2SO3 + 2H2O + 4H+}$$
By cancelling $\ce{H2O}$ and $\ce{H+}$ we obtain the desired redox equation.
The used half equations necessitate the presence of water and acidic medium at first, but in reality we need neither water nor an acidic medium.
Is there a better way of constructing half-equations that do not require any extra materials at the beginning? I mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it.
In the "preparation" section here they didn't mention water or acid.
In addition, I guess oxygen must have sulfur trioxide as the corresponding reduced form instead of water.
Am I mistaken?

If you wanted oxygen to have sulfur trioxide to be the reduced form, then you need a sulfur-containing compound on the LHS. You then have two choices: 1) take SO2 as the sulfur-containing compound, which gets you back to the starting reaction anyway 2) introduce a new sulfur-containing compound that doesn't appear in the overall reaction

Orthocresol already sufficiently pointed out, that half-reactions are a book-keeping tool for electrons. When talking about electrolysis there might be coincidental overlap between half-reactions and the overall mechanism, but as the wording is supposed to indicate, this is not necessarily true.
Many reactions look simple at first glance, but on the second look they appear as highly complex systems of different equilibria. Actually the most simple looking reactions are often the ones that are painstakingly complex.

I actually mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it. Am I mistaking?

Since you have been asking about "the real half-reactions" I thought I give a little more insight into the mechanism.
I found a fairly old publication on the mechanisms of the formation of sulfur oxides during combustion.[1]
For the purpose of illustrating the complexity this shall be enough, but I am fairly certain there are more modern approaches and kinetic and mechanistic studies available today.
The reaction itself proceeds only at moderate temperatures (400-600 °C) and is slightly exothermic.
$$\ce{SO2 + 1/2O2 <=> SO3 (g) + $99.0$~kJ}$$
Because the reaction proceeds quite slowly, catalysts are necessary. Here comes into play what you find in Orthocresol's answer: you need a species that provides you with oxygen(-2) ions. In praxis you use $\ce{V2O5}$ catalysts, e.g. in the synthesis of sulfuric acid.[2]
Sulfurtrioxide is very hydroscopic, which indicates, that the reactions that you proposed cannot happen, since it would hydrolyse immediately in the presence of water.
$$\ce{SO3 + H2O -> H2SO4(l)}$$
Sulfurtrioxide has many modifications, most commonly it exists as $\ce{(SO3)3}$ units in solid phase. Condensation is also one of the driving forces of this reaction, in the gas phase the equilibrium between $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ would not be entirely on the product side.
There is also a variety of different other sulfuroxides that are possibly involved in this complex equilibrium. This is due to the fact, that sulfurtrioxide is a powerful oxidiser.
Let's have a glance at a few reactions that are involved in the formation. The reaction itself most likely proceeds via a radical pathway in the gas phase, with $M$ being a catalyst of some sort (it can be the wall or some inert molecule):
$$\begin{align}
\ce{SO2 + O2 &<=> SO3 + O}\\
\ce{SO2 + O + $M$ &<=> SO3 + $M$}
\end{align}$$
One important reaction that disturbs the equilibrium is
$$\ce{SO_{$n$} + O <=> SO_{$n-1$} + O2}.$$
I also found a publication about the mechanism in the troposphere and things get even more complicated there.[3] The reason for that is dilution, a reasonable amount of molecules in excited states, and various other species involved. This topic became interesting because of air pollution and acid rain.
I am only giving a few examples and leaving out a lot of details, as stated previously, this is for illustrative purposes.
$$\begin{align}
\ce{SO2 + O2 &<=> SO4}\\
\ce{SO4 + O2 &<=> SO3 + O3}\\
\ce{SO2 + SO2 &<=> SO3 + SO}\\
\ce{SO2 + O3 &<=> SO3 + O2}\\
&\text{etc.}
\end{align}$$
As soon as you consider species that are usually available in the troposphere, e.g. $\ce{N2}$, $\ce{NO}$, $\ce{H2}$, $\ce{CO}$, $\ce{CO2}$, etc, things get really complicated.
Conclusion
You cannot write half reaction equations for this equilibrium. Half reactions in general do not represent the reaction mechanism; they are a just a tool to generate overall reaction equations.
The mechanism of the reaction is significantly more complex than only two half-reactions.
You are not mistaken, you don't need water to perform the combustion, however, you need some sort of catalyst, that is able to absorb the excess bonding energy.

References

Arthur Levy, Earl L. Merryman, William Thomas Reid, Environ. Sci. Technol., 1970, 4 (8), 653–662.
Arnold F. Holleman, Nils Wiberg, Egon Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. De Gruyter: 2008. (in German)
English version: Nils Wiberg, A. F. Holleman, Egon Wiberg  (Eds.): Holleman-Wiberg's Inorganic Chemistry. 1st Edition. Academic press: 2001. (Amazon, Google books)
Jack G. Calvert, Fu Su, Jan W. Bottenheim, Otto P. Strausz, Atmospheric Environ., 1978, 12 (1-3), 197-226.

+1 for "Sulfurtrioxide is very hydroscopic" this clearly indicates that my proposed half reactions do not expose reality.

That's a very interesting question and shows some good thinking on your part, so kudos on that first.
The simple answer is, no, there is no way you can construct half-equations that don't involve "extra species". Just consider the oxidation half-reaction. Since $\ce{SO2}$ and $\ce{SO3}$ have different numbers of oxygen, you need some way to balance the oxygen:
$$\ce{SO2 + $\text{oxygen(-2)-containing species}$ -> SO3 + 2e-}$$
The problem is that there are no oxygen(-2)-containing species in the reaction, apart from $\ce{SO2}$ and $\ce{SO3}$, of course! So the "closest" you could get to a pure set of half-reactions would be:
$$\begin{align}
\ce{2SO2 + 2O^2-} &\ce{-> 2SO3 + 4e-} \\
\ce{O2 + 4e-} &\ce{-> 2O^2-}
\end{align}$$
but even that involves some weird oxide ions that the combustion strictly doesn't need.
The actual problem is that they are only a formal way of representing redox reactions. The way you are taught to balance half-reactions, which involves adding $\ce{H2O}$ and $\ce{H+}$, only applies to the case of an electrochemical cell in aqueous solution.
Those half-reactions that you are trying to construct do not represent what is going on at all in the reaction. So, the issue is not that you cannot construct "pure" half-reactions. The issue is more of that you are trying to interpret the half-reactions wrongly. Half-reactions are merely a tool to 1) count electron transfer 2) construct a correct overall equation after cancelling out duplicate species - nothing more than that.


text:
Chair conformation. Wedged lines vs no wedged lines?
As the title suggests, I have been wondering why some chair conformations utilize wedged lines for some of the bonds while others have regular thin lines for all bonds, you can take cyclohexane as an example.

That's not really an example... I still have no idea what you mean. Can you draw a diagram?

https://www.google.it/search?q=chair+conformation&client=ms-android-sonymobile&biw=360&bih=205&prmd=ivsn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwig29zPqpfKAhXDPBoKHRFKBvgQ_AUIBigB#imgrc=WH1NU69XoV8aJM%3A (wedged lines) and https://www.google.it/search?q=chair+conformation&client=ms-android-sonymobile&biw=360&bih=205&prmd=ivsn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwig29zPqpfKAhXDPBoKHRFKBvgQ_AUIBigB#imgrc=6KPI7N1GlWFpaM%3A (no wedged lines)

Oh, it's really just the same, there's nothing special about them.

Mant thanks orthocresol! I was kind of going crazy haha.

Cyclohexane is often drawn as a regular hexagon even though its prevailing chair conformation is not planar.

Perspective diagrams are preferred when discussing conformation (since there is no other reasonable way to depict boat and chair forms) and acceptable when it is just desired to emphasize configuration. (For the depiction of carbohydrates and structurally related molecules, Haworth projections can be also preferred.)

The use of wedged and bold bonds in such perspective diagrams is discussed in Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006):

ST-3.3 Use of bold bonds to emphasize perspective
Since perspective diagrams imply a single absolute configuration, it is not necessary to use solid wedged or bold bonds to emphasize that configuration further. In some cases, such bonds may help an inexperienced viewer to understand the perspective used. (…)

However, note that

Only solid wedged and bold bonds should be used for the purpose of emphasizing perspective. Hashed wedged bonds and hashed unwedged bonds should not be used, either in isolation or in combination with bold bonds.


text:
What is Q-carbon? Does it exist?
Glowing press releases and news articles in 2015 proclaimed a new allotrope of carbon. However, even the journal article is light on chemical detail (e.g. no structural formula).

What is its bond structure?
Why is it ferromagnetic? fluorescent? harder than diamond? etc?
Or is it an erroneous result?

Update: as of late 2020, Bhaumik & Narayan have published further papers about Q-carbon, but I still have not seen any reports of replication by others.

Ref: Jagdish Narayan and Anagh Bhaumik J. Appl. Phys. 2015, 118, 215303.

At this point, it is my personal belief that the experimenters at NCSU are either self-deluded or outright lying.

There's no structural formula because it's what we call an amorphous solid. That is, it's like glass. It has no clearly defined crystal structure, and is likely to be extremely hard, but also extremely brittle.
As to why it's ferromagnetic or fluorescent, I'm unsure. It's likely they don't quite know, either. I suspect it has something to do with the curious bond structure of Q-carbon, which is a mishmash of both 3-way and 4-way bonds. In comparison, graphene is uniformly 3-way bonds, while diamond is a fairly uniform composition of 4-way bonds.
I speculate (don't quote me on this, I can't find a reliable source) that the way the electrons interact across the carbon atoms has something to do with why it illuminates (as photon emission is not an uncommon effect of electrons changing energy levels) and with why it's magnetic (as ferromagnetism explicitly refers to the quantum mechanical properties of electrons).
Either way. Please take this explanation with a grain of salt, but, if you're able to find more reliable sources, feel free to tack it on.

Umm . . . I'm skeptical of this answer. Sources: http://google.com/search?q=amorphous-carbon and http://google.com/search?q=diamond-cubic

@FooBar I too think much of this answer is kind of weak. But, giving google search terms as sources for why you disagree with it doesn't really give any info as to why you are skeptical of the answer. For one thing, the same search terms will give different people different results. Then it's hard to say which result was of interest to you. Anyway, this answer is old and the poster doesn't seem to be around any more so it probably doesn't matter anyway ;)

@airhuff the original answer said that diamond is amorphous (see the edit history).


text:
Peak Wavelength of Emission Spectrum of a Flame vs. Actual Flame Temperature
I am looking at a lab test of two variants of mixed hydrocarbon gas (57.8% $H_2$, 16.9% $N_2$, 12.4% $CH_4$, 1.11% water vapor, and 11.8% "unknown" gases)  burning in oxygen which shows a discontinous emission spectra. The authors are claiming that since their graph shows a peak wavelength at 471 nm, the gas therefore has a peak flame temperature at 6132 K / 5859 °C / 10578 °F according to Wien's Displacement Law, which is absurdly higher than the adiabatic flame temperature of any known hydrocarbon gas.
Now, I know that the spikes shown correspond to the $C_2$ and $CH$ Swan Band. Can someone provide a detailed explanation as to why this is not an accurate measurement of flame temperature?
I should also mention the graph only ranges from 200 nm to 1100 nm, and that they fitted the emission spectrum to a blackbody curve. Could the peak wavelength actually be further into the infrared spectrum? Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?

Your skepticism seems very well founded. Wikipedia has "an acetylene/oxygen flame burns at about 3,773 K" https://en.wikipedia.org/wiki/Oxy-fuel_welding_and_cutting // Fitting to a peak in the spectra seems totally weird // Could you provide a reference to the paper?

http://www.magnegas.com/docs/MG-Flame-report.pdf

My suspicion is the authors chose an arbitrary peak and fitted it to that point. There is no rationale or equation they used in justifying this action.

I'm not an expert on measuring flame temperatures, but the method used seems totally wrong. You don't curve fit the whole spectra but rather take background measures (i.e not on a peak) on the high energy side and extrapolate to zero intensity. At least that is what I did for x-ray fluorescence and it worked well.

Another possible issue is perhaps they are confusing "flame temperature" with "color temperature" due to emission lines which I think is more likely. Just because a blue flame is blue doesn't mean it burns over 6000 K.

@MaxW I'm having trouble trying to explain that all in my head since I'm no expert on flames myself, but I still find it disingenous and sloppy for a serious laboratory to present data like this.

I'm not an expert but surely you need to exclude strong emission lines before fitting (it doesn't look like they have done this). Even then it seems like a stretch to assume the rest of the emission is black body (which is the only thing an emission curve fitting would make sense for).

There seems to be a number of problems here.

spectrometers need to be calibrated in intensity on a wavelength-by-wavelength basis. Each part of the system has an efficiency which can vary with wavelength, but the grating response tends to be one of the strongest. Here is a random example of a grating efficiency curve from a Thorlabs catalog :

You can see right away that a grating efficiency curve can impose it's shape on the measured spectrum, generating a shape superficially similar to a blackbody, but totally instrumental. Without calibration for efficiency, you just have ADC counts vs wavelength raw data!

Fitting of a Plank shape should be done to the spectrum without ignoring the peaks. In other words, a good fitting program will fit the shape you are looking for (Plank) PLUS fit the peaks at the same time. And don't forget to calibrate intensity first (see 1.)!

" Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?" Probably not unless you really know what you are doing. Flames are beautiful and complicated. There's thermal emission from gas in addition to peaks, but stronger is the blackbody-like emission from hot soot particles. Since there will be a distribution in temperatures within the 3D volume of a flame, you have a spectrum recorded from a significant fraction of the flame volume will include a variety of Plank spectra corresponding to different temperatures.

So the answer is: you have to do fitting, remove peaks by fitting them simultaneously, and more importantly, you must calibrate the spectrometer intensity, because grating efficiency curves look strangely Plank-like, althouth they fall to zero much faster above and below their useful wavelength range. Also have to deal with considering the sample volume of the flame and whether or not it's true black-body emission or not.
Those "peak" values, are closer to peak efficiency wavelengths of the spectrometer, and not really related to Wein's displacement law at all!

I calculated a small table of values using the formula for Wien's law: $$ \lambda_{max,nm} = \dfrac{2898000}{T}$$
where T is the temperature in Kelvin and the wavelength is in nanometers.
I can't reconcile my table, the graph, nor the calculated temperature. For the top graph it looks like the cutoff is something like 350nm which would be rough 8250 Kelvin which is much greater than 6132 Kelvin which the authors cited, and much hotter than an acetylene/oxygen flame ($\approx$3773K).
  T(Kelvin)   W(max) in nanometers
  2000        1449
  2250        1288
  2500        1159
  2750        1054
  3000         966
  3250         892
  3500         828
  3750         773
  4000         725
  4250         682
  4500         644
  4750         610
  5000         580
  5250         552
  5500         527
  5750         504
  6000         483
  6250         464
  6500         446
  6750         429
  7000         414
  7250         400
  7500         386
  7750         374
  8000         362
  8250         351
  8500         341
  8750         331
  9000         322
  9250         313
  9500         305
  9750         297

This is Ref. 1 used in the report:
http://www.tempe.mi.cnr.it/zizak/tutorial/cairol06-flame-emission.pdf
Fig. 7 shows "Emission spectrum of a rich pre-mixed butane-air flame." By the logic of fitting to a blackbody curve, then the peak flame temperature of butane-air becomes 9500 K, which is definitely not the actual flame temperature. I think my assertion over the confusion between "color temperature" and "flame temperature" is probably correct.

The lack of a control gas in the testing such as butane or acetylene make it difficult to accept their results. They referenced a butane spectrum off Wikipedia and it is nearly identical to the second graph I posted.

Revised comment: I'd add that 350nm or so is probably getting near to the UV cutoff for air. The oxygen in the air will absorb short wave UV, so UV radiation shorter than 350 would be attenuated quickly by air.

The authors stated that their spectrometer wasn't sensitive enough to detect emissions in the UV range. Mind you, the Ocean Optics HR2000 spectrometer is the same model I used when I took introductory physics and chemistry to analyze discharge tube spectra. You would think a dedicated optics lab would have access to better equipment. I also found different experimental setups for measuring flame temperatures but none of them involved using a single spectrometer.

I'm really stumped as to why the first spectra appears to cutoff at about 350nm. About 8200K seems way to hot. If you are trying to look for the minimum wavelength then as you go from visible to UV you'd have to consider the detector optics as well as the air absorption problem.

So am I, unfortunately.
"Combustion Measurements" by Norman Chigler gives several methods of measuring temperature:
https://books.google.com/books?id=aZGBBs7moYgC&printsec=frontcover#v=onepage&q&f=false

Is the background luminosity (not on a spectral "line" of some sort) of a flame really black body radiation?!?


text:
Why is carbon monoxide a neutral oxide?
Carbon monoxide ($\ce{CO}$) is considered to be a neutral oxide, not acidic or basic. But why? It reacts with hot concentrated sodium hydroxide solution to give a solution of sodium methanoate:
$$\ce{NaOH + CO -> HCOONa}$$
Doesn't the reaction of carbon monoxide with the basic hydroxide ion show that it must be acidic?

The idea of acidic/basic/neutral oxides is hardly used anywhere apart from high school, so I would say that the classification of CO as a "neutral oxide" has more to do with the particular syllabus you are taught rather than chemistry as a whole.

Well, carbon monoxide can be created from formic acid by adding sulphuric acid which will dehydrate said formic acid:
$$\ce{HCOOH ->[\ce{H2SO4}] CO + H2O}$$
Therefore, we can imagine the reverse reaction theoretically, which would make carbon monoxide an acidic oxide. However, the forward reaction does not proceed easily and it needs both the high acidity of sulphuric acid and its strong dehydrative properties to actually work. And your question mentions using hot, concentrated sodium hydroxide to make the reverse one work.
Most oxides that are classified as acidic or basic either have a very electrophilic central atom (e.g. $\ce{CO2}$) which can be attacked by the weak nucleophile water (which in turn can then release an acidic proton), or they have a high charge density on the oxygen which allows it to abstract a proton from water directly. Carbon monoxide is neither. If you check out its molecular orbitals, you will notice that even though carbon is partially positive it has the largest HOMO contribution, meaning a proton would be more likely to attatch to the carbon side — which doesn’t want one at all. The LUMO is, luckily, also more carbon-centred, meaning nucleophilic attacks on carbon are possible. However, it is also degenerate due to the double bond so that an attack is not favoured.
Thus, the carbon monoxide molecule is one that won’t react with water at all and totally defies the concept of acidic/basic oxides.
Abbreviations:

HOMO is a widely used abbreviation for the Highest Occupied Molecular Orbital, i.e. the one with the highest energy that still contains electrons. It is usually the orbital that will attack nucleophilicly or that will be attacked electrophilicly.
LUMO is a widely used abbreviation for the Lowest Unoccupied Molecular Orbital, i.e. the virtual (unoccupied) orbital that has the lowest energy. When considering a nucleophilic attack, the attacking electrons will usually interact with the LUMO. Electrophiles attack with other molecules’ HOMO with their LUMO.


text:
How do valence shell configurations affect attraction and repulsion forces between atoms?
It seems clear that the electrons of one atom are attracted to the protons of another and that this is the source of attractive forces between atoms. However, it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms. Are the electrons not attracted to protons in the same way? Is it because the repulsion between the electrons wins out over the electron-proton attractive forces. Furthermore, why do valence shell arrangements with few electrons bring about attractive forces?

it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms - It is, in some cases.

Possible duplicate of Why are noble gases stable

Firstly, why should covalently bonded atoms be incapable of bonding? Bond changes are the very basis of quite a huge amount of chemical reactions. Oxygen occurs in the $\ce{O2}$ form, which is covalently bonded, but readily breaks its O–O bond in order to form other bonds.
Now, to your question. Why are elements in group 18 (generally: think of the xenon fluorides) inert? Why is sodium so willing to give up its electron, while neon doesn’t form any compounds at all, when they only differ by one proton in the nucleus?
The answer lies in the orbital energies. You can find a table listing the outer shell orbital energies page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.). Let us compare the outer orbital energies of Ne and Na. Neon has two 2s electrons with an energy of $-48.47\ \mathrm{eV}$ and four 2p electrons with an energy of $-21.59\ \mathrm{eV}$, while the 3s electron in the sodium atom lies at $-5.14\ \mathrm{eV}$. The difference is overwhelming. When a ionic compound is formed, an electron is removed from the less electronegative atom and added to the more electronegative atom, forming a cation and an anion. Now, the Na electron is much easier to remove than the Ne electron – you just need $-5.14\ \mathrm{eV}$ to do the task, while eliminating the Ne electron is much more difficult. The electron is then placed into an atom such as fluorine, whose outer 2p subshell lies at $-18.65\ \mathrm{eV}$: you needed energy to remove the Na electron, but energy is released when you place it in the F atom (since it is put into a low energy orbital). Placing an electron from F to Na wouldn’t release energy because it is much more stable in the fluorine atom, due to their orbital energies (if you want to simplify things my anthropomorphizing the electrons, you could say they’re “happier” in the fluorine atom). Removing an electron from the neon valence shell is very difficult because it has very low energy; adding one doesn’t bring much because since the second shell is full, you would have to add it to the third shell – its energy is similar to the Na 3s shell, so it’s not convenient.
I have to note here that while the first ionization energy corresponds more or less to the orbital energy (Na: $495.8\ \mathrm{kJ/mol}$ which corresponds to $5.139\ \mathrm{eV/atom}$), the electron affinity doesn’t (F: $-3.399\ \mathrm{eV/atom}$), because of electron-electron repulsion in the fluorine p subshell. That’s why an isolated $\ce{Na+ - F-}$ compound cannot exist: the additional energy released when forming the lattice is needed to form a stable entity.
Having said this, we can move on. You may ask: Yes, but why are the F electrons more stable than the Na ones, when Na has more protons? The reason is that the outer Na electrons are in the third shell, and thus they have higher energies. In a hydrogen-like atom (an atom with only one electron, which is clearly an oversimplification, since it doesn’t involve screening) the orbitals have the following energies:
$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$
You can see that raising the $Z$ (atomic number) value from 9 (F) to 11 (Na) isn’t enough for lowering the energy, since $n$ (the principal quantum number, i.e. the shell) changes from 2 to 3.
So, this is why sodium forms positive ions, fluorine forms negative ions, and neon doesn’t form any ions at all (chemically). Nota bene: as I said, the formula above doesn’t take screening into account – when going down a group, it (usually: see d-block contraction) prevails over the growing nucleus. That’s why Ar is less inert than Ne, even though the formula above would suggest the contrary.
But what about covalent compounds? According to VB theory, bonding occurs through orbital overlap. Lower energy orbitals mean smaller overlaps, which is inconvenient if you want a stable bond: Carbon forms bonds while neon doesn’t because its orbitals are less stable (see table) and can overlap with other orbitals. Its orbitals are less stable because neon has more protons with the same principal quantum number – see the equation above to understand the concept.
In short: noble gases, especially those positioned higher in the periodic table, don’t bond because their outer orbitals have the lowest energies. Ultimately, this is the result of having the largest number of protons possible for a given outer shell in the ground state.

If you consider Hydrogen bonding to itself, the bond is stabilized both by the electric attraction between proton and electron. As well the 1s orbital of both atoms merge into one bonding orbital, this single orbial is more energetically stable because electrons like to be paired off (spin 1/2 and -1/2) in orbitals.
The electron configurations of the noble gas feature valence shells with completely filled orbital pairs. According to the Pauli Exclusion Principle (PEP), no two electrons may inhabit the same quantum state in the same system. If all available electronic states for the atom are filled then there is no stabilizing phenomena for either the loss or gain of a new electron, and the formation of bonds is not favored energetically.
An analogy commonly used for the PEP is that there is an additional repulsive "force" between elections beyond the electric repulsion, it is precisely this "force" that prevents the noble gasses from bonding. Though it's not so much a force as it's simply the inherent fermionic behavior of the electrons at play. However, exotic noble gas compounds do exist, but are not terribly stable.

Spin pairing has a net energy cost associated with it, not an stability gain. The lower energy of bonding orbitals has nothing to do with spin pairing.

I don't understand "If all available electronic states for the atom are filled then there is no stabilizing phenomena for either the loss or gain of a new electron" - there is no such thing as running out of electronic states

There are noble gas compounds that are almost stable as a rock.


text:
Negative LUMO Energy
During the (DFT-)computations that I have done over the last years, it happened quite often that I got negative energies for the LUMO and many times also for a bunch of virtual orbitals.
Is there a common explanation what negative energies for LUMO means?
I was told that it might have to do with the overestimation of correlation energy within DFT. But what would it mean if there is no error?

Negative compared to what?
Most chemical system have several unoccupied orbitals still in the negative energy range, meaning lower excited states are still bound systems (they don't fall just apart).

Compared to nothing ... negative absolute energy. Your answer would support what I would assume intuitively.

Positive LUMO energies are far more uncommon. It essentially means the molecule disintegrates upon any electronic excitation.

Professor Zipse from the LMU Munich (who does computational chemistry among other things) has no problems in presenting negative energies of virtual orbitals for the allyl cation system on his webpage. I think there is really no problem.
(The same would be true for the benzyl cation when taking the benzyl anion as a starting assumption, since the HOMO would be emptied, however there is also a change in orbital energies involved when changing the charge do I carried on looking until I found the allyl cation, too.)


text:
What is the order of stability of the following carbocations?

What is the order of stability of the following carbocations?

I think that it is clear thatt the first two are more stable than the last 2 as they involve resonance in ring, i.e. the mesomeric effect. Also as the first one has a bigger carbon group attached to it so it will show a larger inductive effect, therefore $1>2$. The third one has a tertiary carbon which is stabilized by hyperconjugation, while the fourthth shows resonance.
As resonance is given more priority the fourth should be more stable, but this doesn't match with the answer.

It's very hard to actually just state the sequence, unless you have the relative stability of benzyl/allyl/tertiary carbocations committed to memory. However what I can say is that it's very dangerous to simply make generalisations like "resonance is stronger than inductive effect" - these are the "rules" that will lead you to incorrect conclusions.

The actual order is given in the question here: http://chemistry.stackexchange.com/questions/27086/carbocation-stability-order

This is in part addressed here: https://chemistry.stackexchange.com/q/74943/4945 and https://chemistry.stackexchange.com/q/72825/4945


text:
Are these two structures of 1,3-dichlorocyclobutane enantiomers?

According to me they are not enantiomers because the mirror image can be rotated by  $180^\circ$ to get the same compound.
Giving it a little more 3-D look:

But the book says they are enantiomers. I don’t understand how?

The book is wrong.

Related: Does trans 1,3-dichlorocyclobutane have zero dipole moment?

They are certainly mirror images, but since they are superimposable on each other, they are not enantiomers. http://goldbook.iupac.org/E02069.html

Like stated in the question Loong linked, there is also a mirror plane present.

The molecules are identical (i.e. the same substance)  not enatiomers.

@Loong

“The books says” is somewhat better than “my book says” (to know that you didn't write it yourself), but it's still too incomplete bibliographical info.


text:
Understanding the mechanism of Blanc chloromethylation
I have seen this question: Mechanism for chloromethylation of benzene.
The answer there uses a picture from wikipedia which confuses me. It gives $$\ce{H2C=O <=>C[HCl] H2C+-OH <-> H2C=O+H}$$ and the benzene attacking the $\ce{C}$ from $\ce{H2C=O+H}$, not the $\ce{C}$ from $\ce{H2C+-OH}$ as I would have expected.
Is it only a quirk in the picture or do I have some fundamental misconceptions?

I found the question this is a duplicate of. http://chemistry.stackexchange.com/questions/37870/resonance-forms-in-electrophilic-aromatic-substitution Different reaction, same question, although it is a poor answer. Maybe mark that as a dupe of this

@orthocresol I agree, but my resonance structure is easier, as it has only two parts ;-)

Resonance is a concept which expresses a more realistic bonding situation through a set of conformations. Therefore the following can only be treated as one structure:
$$\Bigg[~\ce{H2\overset{+}{C}-OH <-> H2C=\overset{+}{O}H}~\Bigg]$$
It is therefore irrelevant in which structure you indicate the nucleophilic attack since it is the same carbon. Usually you would rather draw it the way that the arrow points to the structure where the carbon carries the positive charge, since you would expect this structure to have a higher contribution to the ground state.

So my misunderstanding is that resonance is not some equilibrium state but rather a quantum mechanic "dead and alive" at the same time, like Schroedingers Cat?

@GyroGearloose yes that is pretty much the case. There would be a couple of more resonance structures, but they are not too important, so most of the times you neglect them. Whenever you see resonance structures think of them as one state.

The result is the same no matter which resonance form you use. Resonance forms are only formal representations, extreme forms of the true structure which is somewhere in between.

Note that to get from the blue resonance form to the red resonance form, you'd need to push an arrow:

So there is no real difference which resonance form you use - in the end you would draw exactly the same arrows. In the blue reaction, you draw two arrows. In the red reaction, you only draw one arrow. But, to get from the blue resonance form to the red resonance form, you have to draw one arrow anyway.
Don't read too much into the number of arrows or whatever - the only point I am trying to make is that from a mechanism-drawing perspective, it does not matter which resonance form you use, because all roads lead to the same product anyway. All those arrows are just formal bookkeeping.

I could (naively?) interpret $\ce{H2C+-OH}$ as Lewis acid, but not so $\ce{H2C=O+H}$.

@GyroGearloose No, since those are the same thing, if one is a Lewis acid, the other must be as well.

Bookkeeping... You like that word, don't you. Nice pictures! +1


text:
The meaning of the sulfite anion attached to the benzene ring
I needed to work with PEDOT:PSS solution, so I also looked up it's chemical composition and the wikipedia page threw up this pic.
I need to understand how the sulfite anion is connected to the benzene ring in the structure on the right. I have studied it in chemistry in my school days but I have not really used it afterwards, hence I have forgotten. Your explanations will be appreciated.

If your wondering about the bond, the line going to the center means that the sulfonyl group is attached at any available position (2, 3, or 4) on the ring, and no particular position is given.

Do you think it is possible to use this compound mixture to displace the Se/S ion from CdSe or CdS

I suppose it* could*. To what extent I cannot say. I can say It will have similar acid/base chemistry to sulfate ions or better yet similar to p-Toluenesulfonic acid


text:
Why is water's IR spectra indistinguishable from the IR spectra of a solution of glucose in water?
Anhydrous glucose has an absorption spectra (glucose IR spectra) that is quite different from the spectra of water (Liq. water absorption spectra). However, when a solution of glucose is prepared in water, the absorption spectra is almost indistinguishable from the absorption spectra of water alone. Is this because glucose forms additional bonds when in water? Pointers to links that explain this in detail is also appreciated. Most links I searched on Google just stated this as an observation without explaining the underlying reason.

In order to answer accurately more details would be needed on exactly how the spectra were taken, the concentration of the glucose solution and so on. // The gist of the problem is signal to noise. There is a lot of water and little glucose. So you have to detect a relatively small glucose signal in a large water signal.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works. As Max said, this question would greatly benefit from an IR spectrum of the actual solution.

@MaxW I am referring to the plots in this link. So if I am getting this right, we would need large amount of glucose in water to be able to see peaks from glucose? At 4.4mol/lt, the difference in absorption spectra is very little.

The article is doing in-vivo monitoring, so you can't just dissolve more glucose. So yes, there is a poor S/N ratio.

Thanks! Is there a website that holds data for uv-visible-IR absorption spectrum of common chemicals? The NIST database lists only the IR spectrum.

Why use IR? Its not even quantitative. I would suggest polarimetry as is typically used for measuring glucose content and is quantitative.

Like @MaxW said, in this case it is probably a problem with the SNR ratio of the glucose to water. Of course there could be some differences in a solid state spectra to a spectra of a liquid. This could be caused by interactions with the solvent.
An additional database for IR-spectra (and more) is SDBS. There are more databases like reaxys, but it is not for free. Maybe you university or library owns an account.

I see the spectra in the following way:

The sample and measurement: glucose in water and IR absorption measurement

Considering free glucose molecule, we should expect the vibrations from $C-H$, $O-H$, $C-O$ and $C-C$ bonds and other associated vibrations.
The approximate vibrational frequencies are :
a) $C-H$ = ~2800 - 3300 $cm^{-1}$ (stretching)
b) $O-H$ = ~3800 - 3200 $cm^{-1}$ (stretching)
c) $C-O$ = ~1200 - 1030 $cm^{-1}$ (stretching)
d) $C-C$ =  usually not good for interpretation
e) $C-H$, $CH_{2}$ and $O-H$ bend(in plane) = 1500-1200 $cm^{-1}$
and now considering water,
the $O-H$ stretching band  spectrally overlaps with the bands of glucose (~3750 - 3200). The combination band (~2300) and scissor mode (~1600) are also relatively broad and strong. Due to this spectral overlap and also larger infrared  absorption cross-section of water, glucose is not clearly identifiable in solution.

Instrument and spectral quality : For obtaining a high resolution spectra with good s/n ratio the instrument should be good (FT-IR) and long exposure is needed.
The spectra shown by @user1155386 has the $x-axis$ as nanometers (not wavenumbers  ! ), indicating that the spectral resolution is low. It is noisy showing that exposure time can be improved (the exposure time shown there is only 10 milliseconds).

However, still it would be difficult (but not impossible) to differentiate glucose solution and water. The main hurdle would be get a good spectra and the reason 1. mentioned above.
Additionally, I should mention Raman spectroscopy is better for aqueous solution measurements.

Take a good spectrum (not too high resolution, more than a few scans if your spectrometer is not in top shape) of your solution, and an evenly good spectrum of pure water, and subtract the two from each other. The glucose spectrum (i guess looking quite different from the spectrum of the solid) should appear as plain as day.
Try to scale the relative intensity of the two spectra so as to minimise the residual difference. The absolute intensity signal intensity in IR spectrometers tends to drift quite heavily (compared to the very small signal difference you want to observe) on a timescale of minutes.


text:
Why is H bonded to C5 drawn below the plane in Haworth formula for D-glucose?
I have always seen the hydrogen atom bonded to C5 drawn below the plane as in the following figure. Shouldn't it be drawn above the plane since it is drawn on left side in Fischer projection?

Think again about C-5 in the Fischer projection, it has the OH on the right side, not CH2OH.

@orthocresol Sorry, I don't understand how does that affect the location of H atom?

The Haworth projection depicts the pyranose (or furanose, as the case may be) as a planar ring. The Fischer projection depicts the open chain carbohydrate with the carbon backbone in a single plane. Importantly, the groups on the horizontal bonds are coming out of the plane and the vertical bonds are going into the plane. The result of this is that the carbon backbone is curving toward itself, and the compound is partially poised to close a ring. However, the ring of the pyranose is not all carbon. In glucose, it's the oxygen at C5 that adds to the aldehyde to close the ring. We can manipulate the Fischer projection so that this oxygen at C5 is coplanar with the rest of the atoms that will make up the ring. Rotating around the C4-C5 bond places this oxygen on the Fischer projection's vertical axis. The Haworth projection is essentially a closed version of the Fischer projection tipped on its side with C1 at the right. Forming the hemi-acetal gives the pyranose form of glucose.

You imply in the question that groups that appear on the right side of the Fischer projection should appear on the bottom of the Haworth projection and groups that appear on the left side of the Fischer projection should appear on the top of the Haworth projection. That idea is true. However in the process above the -CH2OH is rotated to the the left side of the Fischer projection. Consequently, it appears on the top of the Haworth projection.

The reorientation of a Fischer projection to a Haworth projection is explained in Subsection P-102.3.4.1.2 of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows. The relevant step that affects the orientation of the hydrogen atom at the carbon atom C-5 is the turn (c) that places the oxygen atom in the ring.

(…)
  Two reorientations are necessary from the standard Fischer projection to prepare the acetalisation or ketalisation procedure: the first reorientation, step (a), consists in placing the nonterminal hydroxy groups vertically; the second one: step (c) is the reorientation of carbon C-5 to place the oxygen atom in the plane of the ring. (…)

I apologize for asking this on an old post, but is this reorientation just a conceptual convenience or does it occur in actuality?


text:
what are the methods to reduce ticl4 to ticl3
I need the methods for reducing TiCl4 into TiCl3. I have searched for the methods ,but I couldn't find the answer. Please suggest me the methods

What have you searched for? Just saying "I have searched" isn't helpful at all.

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Please elaborate

Various methods are described in detail in Handbook of Preparative Inorganic Chemistry; 2nd ed.; Brauer, G., Ed.; Academic Press: New York, NY, 1965; Volume 2, pp 1187–1192. (You may want to look up the details, such as the used equipment, temperatures, and reaction times, in this book or similar literature.)
In most methods, titanium(IV) chloride is heated while a stream of hydrogen is passed through:
$$\ce{2 TiCl4 + H2 -> 2 TiCl3 + 2 HCl}$$
If pure titanium metal is available, titanium(III) chloride can be prepared by separate placing of titanium(IV) chloride and titanium in stoichiometric proportions in a thick-wall quartz tube and heating of the tube using a temperature gradient. (Note that an explosion shield is recommended for this method.)
$$\ce{3 TiCl4 + Ti -> 4 TiCl3}$$
A simple method is the reduction of titanium(IV) chloride with freshly prepared antimony by heating in a melt-sealed bomb tube.
$$\ce{3 TiCl4 + Sb -> 3 TiCl3 + SbCl3}$$


text:
Electrochemistry
My book says:

"At each electrode-electrolyte interface there is a tendency of metal
  ions from the solution to deposit on the metal electrode trying to make
  it positively charged. At the same time, metal atoms of the electrode
  have a tendency to go into the solution as ions and leave behind the
  electrons at the electrode trying to make it negatively charged. At
  equilibrium, there is a separation of charges and depending on the
  tendencies of the two opposing reactions, the electrode may be positively
  or negatively charged with respect to the solution. A potential difference
  develops between the electrode and the electrolyte which is called
  electrode potential."

My doubt is whether electrode potential  develops after we connect the reduction and oxidation half cells or it develops as soon as we dip the metal in the solution of it's electrolyte before connecting it to other half cell?
And if the electrode potential develops before connecting it to the other half cell why can't we measure the absolute potential difference(without using standard hydrogen electrode as reference)?

When you dip the electrode into the solution, it is indeed possible that electrolyte may interact with the surface of the electrode and there may be a limited amount of "charge flow" in the form of the creation of surface bound species. Without a bias created with the addition of an other electrode, however this reaction will be self terminating and it would resemble something similar to a space charge layer that can be observed with semiconducting materials, i.e. charge would flow until their fermi levels equalized. The extent of this flow is minuscule and it essentially would create a surface dipole which can be thought of as a kind of potential, but the electrode potential is defined with respect to its counter electrode so they aren't quite the same thing.
Does that answer your question? Potentials are always with respect to something else so by the definition of your question, you would need the other electrode in there, but we can still have the potential of a certain subspecies of the electrode with respect to a different part of the electrode.


text:
How to find the molar volume of 1 gas in a 2 gas system using PV_m=ZRT?
Using $PV_m=ZRT$, would I need to use the total pressure of the system or the partial pressure of the gas I am interested in to work out the molar volume?

Partial pressure.

If you want to get the partial molar volume of a species in a non-ideal gas mixture, you can treat it as an ideal solution.  In this approximation, the partial molar volume of the species is equal to the molar volume of the pure species at the same temperature and pressure as the solution.  Another method is to evaluate the molar weighted averages of the critical pressures, critical temperatures, and ascentric factors for the various species in the mixture, and then treating the mixture as a pure gas.  You then take the partial derivative of the volume with respect to the number of moles of the species under consideration.  I would try it both ways to see how the results compare.


text:
Comparing pKa of hydrocarbons
We have to compare their acidic strength, and we know acidic strength is directly proportional to stability of the anion.
So the first compound is more acidic due to its being aromatic after deprotonation. The second most acidic is the fifth compound, because of $\mathrm{\% s}$ character of $\ce{C-H}$ bond. And the least acidic is sixth compound because its anion is antiaromatic.
But how would we compare the second, third and fourth compounds?

Source: MS Chouhan; Advanced Problems in Organic Chemistry 11th Edition; Chapter 1 General Organic Chemistry problem 27

Nice work! You clearly understand how factors like aromaticity/antiaromaticity and % s-character in a $\ce{C-H}$ bond can affect acidity.
You can use Coulson's Theorem to estimate the % s-character in the other molecules. In its general form Coulson's Theorem describes the relationship between the angle $\ce{X-Y-Z}$ in a molecule and the hybridization of the $\ce{Y-X}$ and $\ce{Y-Z}$ bonds according to the following equation:
$$\mathrm{1+\lambda_{XY}\lambda_{YZ}cos(\Theta_{XYZ})=0}$$
(where $\mathrm{\lambda}$ represents the square root of the hybridization index of a bond, e.g. $\mathrm{\lambda=\sqrt{3}}$ and the hybridization index $\mathrm{=3}$ for an $\mathrm{sp^3}$ hybridized bond)
In molecule III where the two $\ce{C-H}$ bonds are equivalent, Coulson's Theorem simplifies to
$$\mathrm{1+\lambda^2cos(117)=0}$$
and we find that $\mathrm{\lambda^2 = 2.2}$, or the $\ce{C-H}$ bond in III is $\mathrm{sp^{2.2}}$ hybridized.
As the hybridization index gets smaller, there is more s-character in the $\ce{C-H}$ bond which results in a more stable carbanion and consequently increased acidity. This is consistent with our thinking that, in general, acidity follows the order $\mathrm{sp > sp^2 > sp^3}$; or, the larger the $\ce{X-C-H}$ bond angle, the more acidic the $\ce{C-H}$ bond.


text:
Turns in beta-alpha-beta loops (protein motifs)
In the context of protein secondary structure, do the loops between strand and helix in a beta-alpha-beta motif form into the conformation known as a beta turn?

Why is this on hold as unclear? Both Beta-Alpha-Beta motifs and beta turns are well defined concepts. This is also quite a frequent source of confusion, as the names are similar enough that many students assume that beta-turns are only associated with BAB units. Of course, it's on the edge of chemistry, and more structural bioinformatics...

I thought I vtc'ed as homework, but I can't remember any more. I will try and edit later to improve the quality of the question.

@gilleain It's easy to close this question as off-topic. The explanation you gave should probably be in the body, so that people who are not familiar with proteins would at least have an understanding as from where any confusion is coming from. In its current form the question is only helping people who already know the answer, hence don't need the question. // I downvoted, but I will not dequeue the question. Please [edit] the question to give some more context.

So a beta-alpha-beta structure will have two loops that could be turns - between the first beta strand and the alpha helix, and the helix and the second strand. They might be beta turns, but they might not.
Although there are different definitions of beta-turns, one common one is the presence of a hydrogen bond between the carbonyl of residue (i) and the amino of residue (i+3). I'm not actually sure how frequent this is, but you can see this from a search on PDBeMotif. There is also Motivated Proteins, which gives nice 2D (and 3D) images of motifs.
However, you could get other conformations of these loops that are not beta-turns, especially if they are quite long.


text:
Heat of combustion of stereoisomers of 1-tert-butyl-2,4-dimethylcyclohexane

Which of the compounds below has the smallest heat of combustion?

I know that the heat of combustion is indirectly proportional to the stability of a molecule. And here both $\ce{CH3}$ and $\ce{C(CH3)3}$ are +I groups (inductive effect) but apart from that, I am stuck on how to compare these compounds.

I guess, you are expected to recall energy difference for equatorial and axial positions of substituent in chair conformation of cyclohexane ring. However, I'd say that IRL this effect is very small and dispersion interactions in liquid phase may trump it, but are much harder to predict in this case.

@permeakra  so how could we compare in this

"you are expected to recall energy difference for equatorial and axial positions of substituent in chair conformation of cyclohexane ring" - that's literally the entire answer

Which of the compounds below have the smallest heat of combustion?

The question is asking, which of these 4 compounds is the most stable.  The most stable compound will have the lowest heat of combustion.
In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position.  This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can be in an axial or equatorial position.  In the axial position, it is destabilized by the steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

These interactions are nicely described by cyclohexane A-Values.  The A-Value measures the energy difference (in kcal/mol) between having the substituent in the axial and equatorial positions. The larger the A-Value, the larger the preference the substituent has for the equatorial position. The A-Value for a methyl group is 1.7 kcal/mol. The A-Value for a t-butyl group is so large (>4 kcal/mol) that the t-butyl group is effectively locked into the equatorial position. t-Butyl substituted cyclohexanes do not undergo chair-chair interconversion - this would place the t-butyl group in an axial position, hence they are conformationally locked into the chair with the t-butyl in an equatorial position.
It is this last piece of information that helps simplify our problem.  Here are the structures of the 4 cyclohexanes redrawn in the chair conformation.

Each structure has the t-butyl group in the equatorial position.  Compound 1 has one methyl group in an axial position; this adds 1.7 kcal/mol to the energy of the molecule. Molecule 2 has two axial methyl groups, therefore it is 3.4 kcal/mol higher in energy.
I'll leave molecules 3 and 4 to you to figure out.  Once you determine their relative energies, which one of the 4 has the lowest energy relative energy?
Note: There are exceptions to the additivity of A-Values. See near the bottom of this page for a few examples.


text:
Why does ozone have such a large solubility in water?
At $25~\mathrm{^\circ C}$, the solubility of ozone is $109~\mathrm{mg~L^{-1}}$. This is $13$ times higher than the solubility of diatomic oxygen. What causes this?

It is polar. That would be my guess.

If you look at the molecular structure of ozone below, you can see that it is polar.  Diatomic oxygen is non-polar.  This mild polarity of ozone greatly enhances the solubility of the molecule in a polar solvent like water.


text:
can anyone suggest me good chemistry video lectures
Can anyone please suggest me some websites from where i can get free chemistry video lectures?it must be relevant to iit jee syllabus.

What type of chemistry?

@A.K video lectures on organic,physical and inorganic.it must be of iit jee level,if you haven't heard about iit jee then take it to be undergrad level.

This channel has a lot of great videos on physical chemistry.
https://www.youtube.com/user/TMPChem/

I find Free Lance Teach to be a quality in-depth source for most chemistry concepts.


text:
Alkyl shift vs Hydride shift in hydroboration-oxidation reaction
A part of the hydroboration oxidation reaction in alkenes is as follows:

but in the 4th step, why is alkyl shift happening instead of hydride shift (from Boron atom)?

Related: Oxidation Of Trialkylborane - Doubt In Mechanism

You get B=O -boron doesn't like to have coordination equal 2


text:
Is Tollen's test safe?
Tollen's test is an important test to distinguish between aldehyde and ketone.
$$\ce{2 [Ag(NH3)2]+ + RCHO + H2O → 2 Ag(s) + 4 NH3 + RCO2H + 2 H+}$$
But is this reaction safe? Is there any threat in the formation of silver nitride (fulminating silver)? It is formed when the products of most of the reaction of silver($\ce{I}$) salts are kept aside undisturbed. It act as a potential explosive or a  contact explosive. If there is any threat in the formation of silver nitride, what precaution must be taken to prevent it?

Great question - In general I'd say that not enough attention is given to safety. A good chemist has a plan to dispose of excess reagents as well as the various waste reaction products.

The Tollen's test is safe, but silver amine solutions should be prepared as needed just before use.  The formation of explosive silver nitride is a slow process and can be prevented by treating the solution with mineral acid after use.

I agree, but it would be nice if you add an equation or link and elaborate a little

See here for a more detailed answer.


text:
How does changing volume affect concentration of solutions?
My teacher performed an experiment and made us watch.
He prepared 3 conical flask containing 50 mL mixture of sodium thiosulphate, potassium iodide and acetic acid.
He also had 3 measuring cylinders containing 12.5 mL, 25 mL and 50 mL of hydrogen peroxide of the same concentration.
He added few drops of universal indicator to each of the flask. Then he added the $\ce{H2O2}$ to them as well and started timing.
The general idea is that the indicator is going to change colour from orange to green, which is the colour when he stopped timing. The flask added with 50 mL $\ce{H2O2}$ had the shortest time recorded.
It is intuitive that higher volume of $\ce{H2O2}$ added will increase the speed of reaction but I don't know how to explain it. In kinetics we learn the effect of concentration on rate of reaction, but i cannot use this idea to explain because the concentration does not change! All three solutions of $\ce{H2O2}$ added has the same concentration because they are undiluted (He did not add distilled water to the $\ce{H2O2}$ solutions to make them the same volume).
So how can I explain why the use of higher volume (50 mL) will make the reaction go faster?
An example of the experiment can be viewed through youtube :
https://www.youtube.com/watch?v=Tlj5RvCSpLw

It is $\ce{H2O2}$ in the reaction solution, not the concentration in the reagent. The reaction volumes were 62.5, 75 and 100 ml.

All three solutions of H2O2 added has the same concentration because they are undiluted. So how can I explain why the use of higher volume (50mL) will make the reaction go faster?

As Max said, it doesn't matter and how concentrated is reagent before using it or how much is it, but what is the concentration in reaction mixture.

Consider the reaction from this perspective: you have three solutions of H2O2, 12.5mL 25mL and 5omL. You dilute each of them with 50mL of your thiosulphate solution.  Which flask now has the highest H2O2 concentration?

The concentration of $\ce{H2O2}$ doesn't change but when you have more of it in the reaction (50 ml vs. 12.5 ml) there are more moles of $\ce{H2O2}$ meaning more collisions speeding up the reaction.


text:
Why does the reduction of a Weinreb amide give an aldehyde instead of an amine?
The reduction of a regular amide with $\ce{LiAlH4}$ yields an amine. However, with a Weinreb amide, the product is an aldehyde. How can this be justified? I can't find the mechanism for the Weinreb amide reduction - is it known?

A dimethylamide anion is a highly active nucleophile, which makes it a bad leaving group. Remember that the two alkyl groups are electron-donating, thus destabilizing the negative charge on N. (blue mechanism favored)

However, in the case of a Weinreb amide, the oxygen atom adjacent to the nitrogen atom decreases the electron density on nitrogen, so that the attack from the nitrogen atom becomes unfavorable. The reaction intermediate enclosed in the box is stable due to the oxygen atom of the methoxylamide coordinating to the aluminium atom. After hydrolysis, the methoxylamide anion becomes a much better leaving group, due to the oxygen atom stabilizing the negative charge. (red mechanism followed)

Where did you get such nice pictures?

You'll have to ask my chemistry teacher for them.

If the methoxylamide anion leaves in the course of the reaction, then the aldehyde would be formed in the presence of LiAlH4. The aldehyde is much more reactive than an amide, so it would be reduced to an alcohol. This is precisely what the Weinreb amide tries to avoid.

Good point.
Due to the good affinity of oxygen with metals (for instance, Al), the oxygen atom next to the nitrogen may actually form a O - - - Al - - - -O complex, which stabilizes the intermediate state. It is after the hydrolysis that the aldehyde is formed, thus preventing the over reduction to an alcohol.

It would be excellent if you could include that last comment in your answer somehow.

Reaction of acyl carbonyl compounds with strong nucleophiles (hydrides, organometallics) is a very common and useful reaction, although it has some limitations. One of the simplest cases is the reaction of an ester with a strong nucleophile. As one would expect, the strong nucleophile attacks the electrophilic carbonyl carbon, giving a tetrahedral intermediate. In most cases, the tetrahedral intermediate breaks down readily to give a new carbonyl compound. In this case, that carbonyl intermediate will be a ketone (from a Grignard, e.g.) or an aldehyde (from LiAlH4). Both of those functional groups are more reactive than the ester starting material, so the next equivalent of nucleophile will react faster with the aldehyde/ketone, eventually giving a primary (from LiAlH4) or tertiary (from a Grignard) alcohol product.

The situation at the tetrahedral intermediate changes slightly when an amide is the electrophile. As cfcief points out, the less electronegative nitrogen will kick out the more electronegative oxygen (stabilized by aluminum) to give an iminium intermediate. Again, this intermediate is more reactive than the starting material, so a second nucleophilc attack takes place.

The key feature of the Weinreb amide is that after the nucleophilic attack at the carbonyl, the tetrahedral intermediate is stabilized as a five-membered cyclic chelate. Without decomposition of the tetrahedral intermediate, the only active electrophile is more of the Weinreb amide starting material. On workup, the strong nucleophile is quenched, the metal washed out, and the tetrahedral intermediate is allowed to decompose to the aldehyde/ketone product.

Although the N-methoxy-N-methylamide was specifically designed by Weinreb to behave this way (ref), it wasn't until more than 20 years later that mechanistic evidence was observed (ref).


text:
Does the van der Waals equation remain valid when repulsive intermolecular forces dominate?
The van der Waals equation for a real gas is:
$$RT =\left(p+\frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m}-b)$$
We have understood this formula by saying that $a$ is the term which is for force of attraction between the gas molecules and thus the pressure of a real gas will be less than that of an ideal gas.
Therefore some factor must be added to the pressure of a real gas to make it equal to ideal gas pressure so that it can be used in $p_\mathrm{ideal}V_{\mathrm{ideal}}= nRT$.
Also, if we see the compressibility factor graph , suppose for $\ce{N2}$:

(Source: TutorVista)
If $Z < 1$, then attractive forces are dominating. We see that after a particular temperature, the value of $Z$ becomes greater than $1$, implying that the repulsion forces are dominating in $\ce{N2}$.
Now, since repulsive force dominates, the pressure of the real gas will increase and be greater than that in ideal case.
To find the $p_{\mathrm{ideal}}$ now, we will need to deduct some value from the $p$ of real gas, which means $a$ should be negative. However, this is not so as $a$ can never be negative.
No value can ever be deducted from $p$ in the van der Waals equation.
Then how does the van der Waals equation still work and give the correct data?

If you think too hard about it you will get a headache. The real truth is that such equations provide extra coefficients to fit the experimental data and can thus be used to adjust the relationships. Wikipedia lists 10 different equations and depending on the range for P, number of moles and V (and how the coefficients were fit), some equations work better than others. https://en.wikipedia.org/wiki/Real_gas

Have you tried plotting up a z factor  chart using the Van der Waals equation, in terms of reduced temperature and reduced pressure to see how it compares to the chart for real gases?  You might find it very instructive.

You need to consider the $b$ term, as well as $a$.
Rearranging the van der Waals equation to solve for $p$ gives
$$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$
You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result.
But a positive $b$ will act to increase the pressure compared to the ideal gas result.
At high pressures (small molar volumes), the effect from finite molecular volume ($b$) outweighs the effect of intermolecular attraction ($a$).


text:
High Viscosity of Noble Gases
Why are the viscosities of noble gases higher than almost all other gases? This seems counter-intuitive since they have much weaker inter-molecular interactions and lower boiling points.
Gas Viscosity Table:
http://www.engineeringtoolbox.com/gases-absolute-dynamic-viscosity-d_1888.html

I wonder if this is at all related to the unintuitive fact that gasses increase in viscosity as temperature is increased, unlike the majority of liquids. Noble gasses and higher temperatures both provide situations which behave closer to an ideal gas.

It is very related.  Higher gas temperature means higher particle velocity.  Higher velocity means more collisions.  More collisions means higher viscosity.

In the case of temperature particle velocity is increased by higher temperature as opposed to lighter particles (He, Ne, Ar.)

Viscosity in gasses is generally different than liquids.  Like liquids exhibiting fluidic properties, the viscosity is related to internal friction; however, the friction mechanism is different than inter-molecular interactions experienced by closely packed molecules. The molecules in a gas are far apart resulting in the influence of an entirely different friction mechanism.
The friction mechanism dominating gas viscosity is related to the number of collisions between gas particles. More collisions means more internal friction. In gasses, two dominant factors affect the number of collisions: (1) the mean distance a gas particle has to travel before it collides with another particle and (2) the average velocity of the particle within the gas.
Some noble gasses (He, Ne, Ar) are monatomic particles with a lighter mass than the diatomic elemental gasses. The light mass results in a much higher velocity particles at a given temperature (consider  $E=\frac{3}{2}kT=\frac{1}{2}m \bar{v^2}$) resulting in more collisions. Xenon has a high molecular diameter, so the shear increase in size of the molecule increases the probability of a collision, thus reducing the mean distance it travels before a collision.
More information is available from the following link.