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Is Oxide Anion more stable than O^+
O likes to be negative if charged at all. Most metal oxides are of the form $\ce{A^{x+} O^{2-}_{x/2}}$ (where $x$ is the magnitude of the positive charge and $\ce{A}$ is a metal.) Here, despite the $\ce{O}$ preferring covalent bonds it most often forms ionic bonds because of how metals behave and EN difference.
There are some compounds that have $\ce{O^+}$ in them. Hydronium is an example of that. When the O is positively charged the ions form a cation complex which itself is in a complex of ions and so on and so forth. This is why $\ce{H3O+}$ isn't really $\ce{H3O+}$ as individual ions but rather a complex formed from $\ce{H3O+}$ ions spreading the positive charge across all positively charged oxygen atoms.
Is the oxide anion despite being really reactive more stable than the $\ce{O^+}$ ion since in the oxide anion the $\ce{O}$ already has an octet and it is negatively charged which is what $\ce{O}$ prefers?

You have a very rare view of this situation. Most people would say that $O^{-2}$ is very stable and unreactive, but will attach itself to positively charged ions due to the attraction of unlike charges. Reactivity is use to indicate how likely an atom (or ion) will react and change it oxidation state.

Well if you were to form from NaOH(aq) O^2-(which would require if you want it to stay aqueous superheating of the H2O(heating past boiling point without boiling))  than that would steal 2 protons from 1 water molecule forming another O^2- and so on until it reached the Na+ and then form Na2O which is the most common of all sodium oxides. Thus my saying that the oxide anion is very reactive.

$Na_2O$ would only form after all the water evaporated. But I don't see how you envision a water molecule losing an electron (or 2). I have never heard anyone mention $H_2O^+$.

I edited my comment to say "2 protons" once I realized my mistake. The Ion Pair effect is what makes some ionic compounds not completely disassociate in water. I think that Na2O is one of those that has a really strong Ion Pair effect(definitely stronger than NaCl) and so doesn't completely disassociate into 2 Na+ and O^2-. Calcium flouride also doesn't completely disassociate into Ca^2+ and 2 F- because of the Ion Pair effect.

and how do you know without the amount of positive charge in the metal known that in a metal oxide there would be the charge divided by 2 for the number of oxygen atoms?


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Position of nodal planes in molecular orbitals
From the attached picture (MO diagram of cyclopentadienyl anion) I can figure out where the nodal planes are, but why are they exactly at these positions?

Why is this nodal plane invalid?

The picture is not quite right. As these are all pi orbitals, you have to add one nodal surface (in this case a plane) to what it says there. This plane is identical for all orbitals.

I believe that the figure you added has a higher energy than any of the others. So if it is possible, it would be an anti-bonding orbital.

How do you know that?

First a descriptive explanation:
The Molecule is symmetric. Thus the resulting mocular orbitals are symmetric too. If you are constructing different MOs where the nodal planes are different than in the picture above they wouldnt be symmetric.
Mathematical Explanation:
These nodal planes result from a LCAO-MO consideration with hückel approximation. You have to form the secular determinant for such a problem at which the coulomb integrals are Alpha and the resonance is only dissimilar zero when the two considered atoms are neighbors. After solving this hückel-matrix you can determinate the coefficents of the several molecular orbitals. Doing that you will see that the nodal planes exist as you drew them above.
(More Information about the Hückel-Method: here)

I see more ways to put the planes .For example, a molecular orbital with one nodal plane can be constructed so that the nodal plane passes through two atomic orbitals. I don't understand how MOs in the image are symmetric and the one I proposed here isn't.

@Marko the molecule is of $D_{5h}$ symmetry, with a $C_5$ symmetry axis passing through the centre of the ring perpendicular to the carbon plane. This axis is accompanied by five vertical mirror planes. It can be shown, that these mirror planes are also the nodal planes for high lying orbitals. In the case you explain it would break this symmetry.

I attached the nodal plane I proposed. I still don't see how it isn't allowed while the others are.

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What is the difference between 'particles', 'molecules' and 'chemical species'?
I have read the Chem.SE question about the difference between 'compound', 'mixture', 'element' and 'molecule', but I would like to know the precise definitions for more terms (see below). From the article, these are the definitions:

Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).

Molecules are a group of atoms covalently bonded to each other.

Compounds are made up of two or more elements. Compound is often used interchangeably with the word substance.

Mixtures are collectives of different molecules or atoms.

How are compounds substances? Isn't 'substances' a general term for any chemical thing?
These are the terms that I would like definitions for:

chemical species
substance
particle
atom/isotope
salt

I am rather confused about these terms, because my tutor uses particles and molecules (and elements) as if they were the same thing. As for the rest, I want to know how to distinguish between general (such as chemical species, substance) and specific terms (such as particle, atom, isotope, salt).

Have you checked the IUPAC gold book for the definitions? IUPAC probably collects the most rigorous and well-defined terminology, to the extent possible. For example, you can find chemical species, but particle is too general of a term and has to be broken down into subsets if you want the word to have a more specific meaning.

You are already misstating a few things.

Elements are not just classes of atoms - all atoms of an element must have the same number of protons (atomic number). There are other classes of atoms that are not elements, such as the class of atoms that are radioactive, the class of atoms that are bosons, etc.

I see, perhaps you would also like to correct the answer from which I adopted these definitions?

That question was already answered, although perhaps the answer could be improved upon. Specifically, to add that a substance can be either an element or a compound. You are getting into the philosophy of science in addressing these definitions.

A chemical species consists of chemically identical atoms, molecules or ions (more generally molecular entities) and is identified as a single species in a spectroscopic experiment. The latter part of the definition is important, because, for example, a compound which is identified as a single chemical species by NMR spectroscopy can still contain different enantiomers (that are separate molecular entities) which cannot be distinguished by an achiral spectroscopic method.
A substance is matter which has a specific and distinguishable chemical composition and specific properties. From this follows that every pure element and chemical compound is a substance. For example, pure $\ce{Fe}$ and $\ce{CH4}$ are substances, while an aqueous solution of $\ce{NaCl}$ and gasoline are not. The latter two are a mixture of two ($\ce{NaCl}$ and $\ce{H2O}$) or more different substances and have therefore properties which depend on the composition.
Particles can be defined as microscopic constituents of matter, like atoms or nuclei. The term can also refer to even smaller constituents like electrons, protons, or neutrons, which are called elementary particles. It is also used in the definition of an atom, the smallest particle which is characteristic for a chemical element. Atoms with the same number of protons $Z$ (and equal number of electrons) represent the same chemical element. Isotopes are atoms of the same chemical element which have different mass numbers, i.e. the same number of protons $Z$, but different numbers of neutrons in the nucleus. In the three isotopes of hydrogen, for example, the nuclei contain different number of neutrons (0, 1 and 2).
A salt is a chemical compound which consists of cations and anions. It should be distinguished from a zwitterionic compound, which contains charged but overall neutral molecules as building blocks. An example for the latter is trimethylamine N-oxide.

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What does (heated) mean in a chemical equation?
I'm supposed to balance the following equation:
$$\text{Potassium chlorate (heated)} \rightarrow \text{Potassium chloride + Oxygen}$$
But I don't understand what is meant by the label "heated" in front of the first element. Do I ignore it, or am I supposed to add another element to the reactants of the equation?

In terms of balancing the equation, the fact that heat must be supplied can be ignored.  It is just providing information about the conditions necessary to run the reaction.

It doesn't have anything to do with a combustion reaction?

Usually it just means to supply thermal heat to a solution or a solid.  For example, heat the flask containing a solution of potassium chlorate.  A more detailed experimental procedure would specify how much heat and for how long (e.g. dissolve the potassium chlorate in water and then heat to reflux for 1 hour).

Energy the forgotten reactant in chemistry! Did not some famous scientist establish that mass-energy is what is conserved.  For this reaction to happen smoothly a catalyst, usually MnO2, is advisable. BTW boiling in water would be a different reaction if it happened, either oxidation of water or disproportionation of the Chlorate ion.

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Mercury vapor release from amalgam fillings
What would cause vapour release from elemental mercury? Would it be pressure, friction... could one stop the vapour release through cooling the mercury?

The release of mercury vapour from fillings has been studied and the results are always the same unless very unusual conditions apply: the amount of vapour released is insignificant to health. So reducing the rate is a waste of effort. Unless you work in a crematorium when you might want to take precautions.

All solids have a certain vapor pressure, and if this vapor pressure exceeds the atmospheric pressure, then evaporation of the solid occurs. Some of the solid turns into gas. More specifically, it's the uppermost layer of the solid that evaporates.

What would cause vapour release from elemental mercury? Would it be
  pressure, friction...

Friction could definitely increase the kinetic energy of the mercury particles and increase the solid's vapor pressure. An increase in pressure could also increase the temperature and therefore kinetic energy of the solid's constituent mercury atoms.

could one stop the vapour release through cooling the mercury?

Yes, that would decrease the kinetic energy of the particles and lower the solid's vapor pressure. Here's a chart relating temperature (in Kelvins) to vapor pressure. At physiological temperatures, we're still talking about an extremely low vapor pressure.

http://www.physics.rutgers.edu/~eandrei/389/NISTIR.6643.pdf

I have several amalgam fillings which I hope to have removed as soon as possible. In the mean time I am cooling my fillings with cold water after eating. I am avoiding foods and drinks which are hot also. After my amalgams are removed I hope to undergo some kind of chelation therapy. Thanks for your answer.

I really don't think that ice would make much of a difference. Is there a specific reason you are undergoing chelation therapy? Or is it out of an abundance of caution?

I will be undergoing chelation therapy to remove mercury from my body tissue and cells. Mercury interrupts healthy functioning in many ways. I am also seeking chelation agents which cross the blood brain barrier. I have read several papers outlining the dangers of mercury poisoning from amalgam fillings.

Do you know whether you have been poisoned? Frankly this sounds like you've read too much into disreputable sites on the safety of mercury fillings. Ironically too, the disposal of mercury amalgams is largely unregulated in the US and disposal itself by dentists accounts for over half of mercury emissions into the environment - such as the water you might be drinking.

Considering I have 8 fillings and that each can release as much as 40ug mercury vapour per day, and the lungs absorb as much as 80% of that vapour, and also considering the well established toxicity of mercury which has no use in body what so ever, I do think I have harmful levels of mercury in my system yes.

What paper supports your claim? Please link.

I will be having my fillings removed according to strict safety protocols.

http://europepmc.org/abstract/med/3481133   http://link.springer.com/chapter/10.1007/978-1-4613-0961-1_9

There will always be some molecules with enough kinetic energy to escape (evaporate) from the surface of a solid or liquid.  For each substance and at each temperature there is a specific "vapor pressure" that when reached, the solid or liquid and the vapor are said to be in equilibrium.  Again, there are always molecules escaping from a solid or liquid, but once there are enough of the these molecules in the vapor above the solid or liquid, and the equilibrium vapor pressure has been attained, then the number of molecules returning from the vapor to the solid or liquid will equal the number of molecules escaping from the solid or liquid.
The problem with mercury vapor in an open environment like the mouth is that the air in the mouth is continually exchanged with fresh air.  Therefor, whatever concentration of mercury vapor has built up will, to some degree, be decreased by air exchange.  Hence mercury will, more or less, continuously evaporate in such an environment. Cooling will reduce the equilibrium vapor pressure and consequently reduce the rate of mercury evaporation.


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removing mercury from the brain
I have conducted many searches on this subject with regards to agents which cross the blood brain barrier. I have read about lipoic acid. Is anybody on here aware of any studies or findings which show effective mercury binding and transportation from the brain?


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How to find out oxidation state when compound has multiple electronegativities
How do I solve for oxidation state in something like CH3-O-F? The fluorine is more electronegative than the oxygen which is more electronegative than the carbon which is more electronegative than the hydrogen. In this case would everything except the fluorine be oxidized?

http://chemistry.stackexchange.com/questions/62397/does-flourine-in-fno3-have-1-oxidation-number?s=1|1.2815

You can assign oxidation states with a method similar to assigning formal charges.

Draw the Lewis structure.
Assign each atom all of its lone pair electrons.
For bonding electrons assign all of the electrons to the more electronegative element.
For each atom perform the following calculation:
number of electrons in the atom in its elemental state - assigned electrons = oxidation state.

In your example, the oxidation number of each $\ce{H}$ is +1 and of $\ce{F}$ is -1.  The oxidation number of $\ce{C}$ is -2 and of $\ce{O}$ is 0.

To successfully assign oxidation state you should define what oxidation is. Oxidation is defined as the loss of electrons.
This, along with the fact that oxidation state is represented as a charge - i.e. something may have a "+2" oxidation state or a "-8" oxidation state, it follows that oxidation state is the charge that an atom possesses if all bonding were ionic. Ionic interactions involve the taking of electrons; covalent bonds involve the sharing of electrons.
Now, what "takes" the electrons? Clearly, the element which can best stabilize the electrons. This would be the most electronegative element.
So I would start by drawing out the Lewis structure of $\ce{CH3OF}$. Assign electrons to element based on relative electronegativities, and then think about the charges. I.e. if an element normally has 7 electrons in its valence (such as fluorine) - but is assigned 8 electrons, what do you think the charge on the fluorine will be? It has surfeit of electrons. This should suggest to you that the fluorine has a negative 1 oxidation state.

okay so if I start with methanol(CH3OH) and end with CH3OF(Methoxy Flouride I think is what it is) it occurs in 2 steps. Carbon starts off in a -2 oxidation state and O in a -2 oxidation state. The O gets deprotonated by a hydroxide anion. The oxidation state does not change at either the O or the C. Fluorination of the alkoxide oxidizes the O to a -1 oxidation state. The C does not change. It doesn't get reduced or oxidized.

of course the deprotonation could happen at the C and give you a carbanion which then gets oxidized by the flourine and the oxygen does not change in either the bonding or the oxidation state in this case.

and the difference between fluorination and fluoridation is that fluoridation means you use the fluoride anion whereas  fluorination means you use some source of fluorine which could be the fluoride anion or a compound with fluorine in it.


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Why does HCO₂⁻ exhibit delocalized bonding while CO₂ does not?
$\ce{HCO_2^-}$ exhibits delocalized bonding and $\ce{CO2}$ does not?
What I am confused about is, that both of them have resonance structures. Therefore, why doesn't $\ce{CO2}$ exhibit delocalized bonding?

Carbon dioxide does show delocalised bonding, there are two delocalised $\ce{\pi-O-C-O}$ bonds, see this question for more details.

Also this question. Bond delocalization in carbon dioxide is easily represented in valence bond theory by charge-separated contributing structures.

Here is an orbital diagram for $\ce{CO2}$.

Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be "orthogonal" to one another.  This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.
Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$).  Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.
For completeness, we can restate the drawings from above in terms of resonance structures.  In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$.  The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.

In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question).  Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$.  But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array.  This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon.  Just like the "particle in a box" problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron.  So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion.  Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens

$\ce{CO_2}$ does not show delocalization because the p-orbitals of the carbon atom are orthogonal. They can't interact. The hydroxymethylidene ion shows delocalization because the lone pair on the carbon atom is "upgraded" to a full p-orbital, just as (one) the lone pair on each oxygen because that would stabilize the negative charge. Now you have a system of three p-orbitals linked together. There is no need for $\ce{CO_2}$ to stabilize anything, so the energy required to "upgrade" to a full p orbital (on the oxygens) is too high of a barrier.
Please correct me if I'm wrong (then I'll remove the answer). This is a very simple analysis, there could be things in MO theory that I'm not aware about.

To the downvoter of me and Ron, care to explain please? (he'll probably never answer, but I'm polite so I ask it anyway)

I did not downvite, but the premise of both answers and the original question is wrong. Carbon dioxide shows strong pi-electron delocalization. See Martin's and my comments to the original question, which give references.


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Mercury binding agents which cross the blood brain barrier
The two substances I have come across are cilantro and lipoic acid. How effective would either of them be? Are there any other agents known which produce the desired results?

What are your desired results? Are you looking for something that will enter the brain, grab a mercury atom, then leave the brain carrying the mercury atom out?

Yes of course. I am interested in chelation therapy to remove mercury from my brain and then out of my body.

http://www.sciencedirect.com/science/article/pii/S0162013413003358


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Using water model in a computer program
I am writing a small simulator for situations where water is the medium of all reactions.
A quick googling confirmed that there are water models available in computer readable formats. Being a beginner in molecular chemistry, I was hoping someone can direct me to already mature libraries which I can use to read water models.
Following seems to be good candidates:

http://msl-libraries.org/index.php/Main_Page
http://ball-docs.bioinf.uni-sb.de

Can't get you correctly! Have you checked Gromacs and Ambertools?

Nope. Ambertools look very promising. Thank you. I was hoping for a small library which can read the water model and let me populate my own data-structures (CGAL).

Libraries which support Water model Simulation:

Assisted Model Building with Energy Refinement(AMBER)
Chemistry at HARvard Macromolecular Mechanics(CHARMM)
The Molecular Modelling Toolkit

Cool Site for Water Model Generation:
Water Models

Image Source: http://biomodel.uah.es/en/water/index.htm

I saw (what I think was) your recently deleted question posted automatically in this chat room. You may have seen How can I calculate the charge distribution of a water molecule? but this may be more helpful: How to “get my feet wet” in Density Functional Theory by simulating a water molecule using Python

@uhoh Thanks a lot! I was looking for this!


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Catalytic iodine in synthesis of phosphonium salts
From Clayden et al.

However, iodine is expensive and a way round that problem is to use a
  catalytic amount of iodide. The next phosphonium salt is formed slowly
  from benzyl bromide but the addition of a small amount of LiI speeds
  up the reaction considerably.

The iodide reacts as a better nucleophile than Ph3P and then as a
  better leaving group than Br–. Each iodide ion goes round and round
  many times as a nucleophilic catalyst.

In a private conversation with ron I asked him how this catalyzing effect worked, he said that question would be more suited here and I agreed with him. It seems a bit strange at first that a strong nucleophile can also be a good leaving group (he really wants that nucleus, right, why would he leave?).
This is the explanation I came up with (probably wrong, but it is always better to first think self): Iodine is a stronger nucleophile here than bromine because of its higher energy HOMO, which lies closer to the energy of the $\ce{C-X}$ $\sigma^{*}$ (here $\ce{X}$ = $\ce{Br}$) LUMO. So it will react more rapidly than bromine, resulting in an equilibrium.
Now iodine is also a good leaving group, firstly because the $\ce{C-I}$ bond is higher is energy than the $\ce{C-Br}$ bond (less stable bond). And secondly (quoting Ron from another answer. Well explained, no need to rewrite it again): the larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. So $\ce{I^{-}}$ will be more stable, making it easier for iodine to leave.
My thought was that iodine can easily knock of bromide but not remain in that position for very long (therefore the equilibrium). Now, if $\ce{PPh_3}$ happens to meet the benzyl iodide it is more eager to react with it than with the benzyl bromide (lower energy $\ce{C-X}$ $\sigma^{*}$). The resulting phosphonium salt is either to stable to be attacked by iodine again or precipitates out (or both?), so that benzyl bromide continues to be transformed into benzyl iodide.
Is this correct?

Iodine is a stronger nucleophile here than bromine

I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much.  As long as iodide is a strong enough nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen.  Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.
I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.
I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is unhybridized just like phosphine ($\ce{PH3}$).  If it is unhybridized than our lone pair on phosphorous is in an s-orbital.  Since the s-orbital has little "directionality", this lone pair on phosphorous will be a weak nucleophile.  On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile.  So now matter how TPP is hybridized, we expect it to be a weak nucleophile.
Iodide is, as you note, a much better leaving group then bromide.  In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.
The reaction is run with just a small (catalytic) amount of iodide.  As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP.  During this last step, iodide is regenerated and the cycle starts again.

I just did the calculation, $\ce{PPh3}$ gives a nicely directionalized $\ce{sp}$-lone pair on DF-BP86/def2-SVP. $\angle(\ce{Ph-P-Ph}$ is about $102^\circ$. The $\ce{C-P}$ bonds are $38%~\ce{sp^{$4.5$}~P}$ and $62%~\ce{sp^{$2.6$}~C}$. I would argue, that this is due to steric effects. But as you pointed out, this has little to no effect on your explanation.

Is there much delocalization of the lone pair into the benzene rings?


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Nitrogen dioxide dimerization
$\ce{NO}$ can't dimerize and that is explained because the free electron is in an antibonding orbital. The same is true for $\ce{NO2}$, but it can dimerize, how?

Having an electron in an antibonding orbital means that the electron exists in a high energy orbital.  If anything, this means the electron will be "motivated" to react and form a bond, particularly if bond formation lowers the energy of the electron.  An electron in an antibonding orbital can form a bonding (as opposed to antibonding) bond with positive overlap.

Why doesn't oxygen for a dimer then?

Good question!  Remember that NO only dimerizes at low temperatures.  Apparently molecular oxygen does similar things.  Since molecular oxygen has 2 unpaired electrons, maybe we would expect a polymer or a cyclic oligomer?  It turns out that one form of solid oxygen actually exists as $\ce{O_8}$.  Read about "red oxygen" or $\ce{O_8}$ in this Wikipedia article  http://en.wikipedia.org/wiki/Solid_oxygen#Red_oxygen  Here's a link to an article where $\ce{O_4}$ is reported  http://onlinelibrary.wiley.com/doi/10.1002/1521-3773(20011105)40:21%3C4062::AID-ANIE4062%3E3.0.CO;2-X/abstract

Are you sure that nitric oxide can't dimerize? It has a rather high bond order - 2.5 - and we can see that from molecular orbital theory. Therefore even if its free electron is in an antibonding orbital, that wouldn't decrease the bond order to 0 or below.
The nitric oxide dimer has actually been extensively studied too.
http://pubs.acs.org/doi/abs/10.1021/ja00381a052

How can it dimerize if antibonding electrons residue in HOMO?

http://chemistry.stackexchange.com/q/13537/5084

I don't think that hyponitrite is the same as a NO dimer.


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Nitric Oxide Dimerization
This is the molecular orbital diagram for nitric oxide:

Now, we know that nitric oxide can dimerize. However, why does nitric oxide dimerize as to form an $\ce{N-N}$ bond? Why not an $\ce{O-O}$ bond? Wouldn't the lone electron be more likely "possessed" by the oxygen since oxygen is more electronegative (if we were to step away from molecular orbital theory for a second).
Is this a Coulombic effect? Oxygen, by nature of being more electronegative, attracts more electron density to itself. This in turn would weaken any hypothetical $\ce{O-O}$ bond due to electron-electron repulsions.

This is the molecular orbital diagram for nitric oxide || It shows oxygen having orbitals with higher energy, which is obviously wrong. Please, check. As for the question, the $\ce{ONN}$ angle in N2O2 is ~100, so I would say the nature of the bond is debatable. While antibonding orbital obviously is shifted towards atom with higher initial orbitals, the angles suggests partial O-O or O-N interactions as well.

What does bond angle suggest about the nature of the bond, exactly? How can you conclude that there are O-O interactions based on angles?

Oh wait nvm I see it after I drew out the diagram; the 100 degree bond angles deviate from the usual sp3 hybridized bond angle; this suggests that the oxygens may be interacting with each other and closing up the bond angle from 109.5 degrees. Are there really available orbitals on the oxygens to do this/

Something is not correct with your MO diagram.  Here is a correct MO diagram for NO

Note that in this diagram the oxygen atomic orbitals are lower in energy than the nitrogen AOs.  This makes sense since the AOs of the more electronegative element should always be at lower energy than the AOs of the more electropositive element (in your diagram the relative energies of the oxygen and nitrogen AOs are somehow reversed).  The energetic ordering of the starting AOs is important as we'll see in a minute.
The closer in energy a MO is to an AO from which it was formed, the more character of that AO it shows.  So let's examine the nature of the single electron in the $\ce{\pi_{2p}^{\ast}}$ orbital since this electron will be involved in forming the new bond when $\ce{NO}$ dimerizes.  This MO is formed from both the nitrogen and oxygen p atomic orbitals, but it is closer in energy to the nitrogen p AO.  Therefore, we would expect this electron to have more nitrogen and less oxygen character (in other words it will reside more around nitrogen than oxygen).  Because this electron has greater density on nitrogen, dimerization to form an N-N bond is more likely than dimerization to form an O-O bond.

Are the AOs of the more EN element necessarily always lower in energy? Aren't there cases in which size trumps EN (i.e. with the binary halide series of acids)?

Also by "more likely" you're not excluding the possibility of the formation of a dimer with a peroxide linkage?

I think it would be "always", but I'm reluctant to use "always" so let me say "generally" the more EN atom will be at lower energy.

"more likely" addresses the kinetics of the situation.  There is a larger coefficient on N than O in the $\ce{\pi^{\ast}}$ orbital.  So initial bond formation involving N is more likely, but certainly (assuming there is some coefficient on O) oxygen bond formation could occur too, just to a lesser extent.  Now enter thermodynamics - no matter what predominates initially (kinetically), since the reaction is reversible, eventually we will wind up with a product equilibrium distribution matching the free energy difference between the isomers.

Cool. But I don't see any papers exploring the isomer of the nitric oxide dimer with an O-O linkage. Why not?

Using standard bond energies calculate the heats of formation of the 2 isomers.  If the difference is 3 kcal/m or more,  then the equilibrium mix will be > 99:1

So we can safely ignore entropies of formation?

For a back of the envelope calculation - yes.  Further, in this case, I would expect the two reactions to be very similar entropically.

Using this table (http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm) it appears there is a difference of about 17 kcal/mole. Huge!

How are you relating enthalpies of formation to product mixtures? I'm guessing you are using dG = -RTlnK in some form or fashion?

Exactly, so with a 17 kcal/m difference K ~ 10^12 !!!

Why would a peroxide linkage be weaker than a N-N linkage?

That would make another good question.

And how are you using dG=-RTlnK to arrive at K? This is my train of thought: TdS is ~ equal for both the formations of ONNO and NOON, so dG for the two should differ by 17 kJ/mol. Then 17 kJ/mol = -RTlnK, and for K I get 0.99.

Could this be because I forgot to convert R = 8.31J/(molK) to a kJ/(molK) value?

17 kcal/mol or kJ/mol?

It is 17kJ/mol. http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm. So 17kJ/mol is ~ 4kcal/mol and this suggests a >100:1 ratio in favor of the ONNO isomer versus the NOON isomer. I see where I screwed up!

"are you using dG=-RTlnK to arrive at K"? Yes, I get K ~ 1200 if delta G ~ 17 kJ/mol

you can always to a quick check: 1 power of 10 for each 1.4 kcal/mol at RT

You might want to continue the discussion in [chat].


text:
Analytically combining Dynamic Equilibria that have a two way affect?
Lets say I want to model a water system as the following set of equilibria:
$\ce{H2O <=> OH- + H3O+}$ | $\ce{A=[OH^{-}][H3O+]}$
$\ce{CO2 + H2O <=> H2CO3}$ | $\ce{B=\frac{[H2CO3]}{[CO2]}}$
$\ce{H2CO3 + H2O <=> HCO3- + H3O+}$ |  $\ce{C=\frac{[HCO3-][H3O+]}{[H2CO3]}}$
$\ce{HCO3 + H2O <=> CO3^{2-} + H3O+}$ |  $\ce{D=\frac{[CO3^{2-}][H3O+]}{[HCO3-]}}$
The whole system starts off as just $\ce{H2O}$ and $\ce{CO2}$ from the atmosphere. Now consider that calcium carbonate ($\ce{CaCO3}$) from limestone is going to react with carbonic acid in the water ($\ce{H2CO3}$) to cause the dissolution of calcium ions, increasing the water hardness (We will ignore the small affect of calcium carbonate dissolving on its own very slowly). Thus the following equilibrium also exists:
$\ce{CaCO3 + H2CO3 <=> Ca^{2+} + 2HCO3-}$ |  $\ce{E=\frac{[Ca^{2+}][HCO3-]^2}{[H2CO3]}}$
Using basic math, I can find the relationship between hydronium and calcium concentration:
$\ce{\frac{(E[H3O+]^2)}{(C^2 B[CO2] )}=[Ca^{2+}]}$
This can be linked to pH:
$\ce{\frac{E}{(C^2 B[CO2 ](10^{2pH}) )}=[Ca^{2+}]}$
My question is, can I change this to take into consideration the change in pH over time caused by the change in Carbonic Acid etc. without a computer?

If you have one equation for each unknown, you can use this approach.  http://chemistry.stackexchange.com/questions/14436/4-compounds-in-equilibrium-with-one-another-determine-their-equilibrium-concen


text:
Whats the effect of pressure on water electrolysis of sea water with increase in depth?
I’m currently working on an underwater robot and was hoping to use the gas bubbles produced by water electrolysis for buoyancy control. The operating depths range from 200–1000 m.

The higher pressure itsels doent make it much more difficult. But...
For water electrolysis you need high overpotential, preferably large electrodes with small electrode gap between them. You want clear, filtered water and no fishes electrocuted, burning down the whole submarine due to an accidental short.
And you will need plenty of gas for buoyancy control, especially at $200{-}1000\ \mathrm{m}$ depths, therefore you need a lot of electricity:
According to Faraday's law $$m=(Q/F)(M/z)$$ where $F = 96485\ \mathrm{C/mol}$ you can make like $4.5\ \mathrm{l}$ of hydrogen gas (1 atm, 25 °C) with a $10\ \mathrm{A}$ current in 1 hour! That is like $0.04\ \mathrm{l}$ at $1000\ \mathrm{m}$ ($100\ \mathrm{atm}$) of a potentially explosive gas.

Does the overpotential increase with water pressure?

Would the use of sea water facilitate the gas production process ???

@user8697 The better conduction (i.e. the salts in water) help the electrolysis. You cannot electrolize pure water, you have to add some ionic compound to it to improve conduction. I have no idea how big change of overpotential can be observed, but I would not expect much.

Does the salinity increase with depth  ? Would increased salinity facilitate the electrolysis ?


text:
How to calculate which ionic compound has the highest melting point?

Which of the following compounds has the highest melting point: $\ce{PCl5}$, $\ce{H2O}$, $\ce{NaCl}$, $\ce{SrCl2}$, $\ce{CaF2}$?

In general, ionic compounds will have higher boiling points than those of dipole-dipole interaction compounds.
So, the compound above that will have the highest melting point will certainly be an ionic compound, either $\ce{NaCl}$, $\ce{CaF2}$ or $\ce{SrCl2}$.
I thought that I should calculate the molar weight but that won't work, because the correct answer is $\ce{CaF2}$, and $\ce{SrCl2}$ has a greater molar weight.
Between ionic compounds, how can the one with the highest boiling point be predicted?

The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is
$$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$
$k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions.
So the larger the charge the greater the force, and the smaller the radius the smaller the distance between the ions (smaller $r$ in the equation) again leading to a stronger force.
So, first you should calculate the charge on each ion in a molecule. As can see, the charge on strontium ($\ce{Sr^2+}$) and calcium ($\ce{Ca^2+}$) ions is more than the charge on sodium ions ($\ce{Na+}$). Also, calcium ions are smaller than strontium ions. Fluorine ions are also smaller than Chlorine ions. So $\ce{CaF2}$ will have higher bond strength which means it will have higher boiling point since stronger bonds need more energy to break.

You can deduce it from the Coulomb law. As you can see, the bigger the charge of the respective ions, the electrostatic interaction between them is stronger. The charge of calcium and strontium cations is twice as that of sodium. r is the distance between the charges, so for smaller ions the distance would be smaller i.e. the interaction would be greater. Calcium cation is smaller than strontium cation, therefore the correct answer is indeed calcium fluoride.
It can also be explained with more sophisticated equations: Kapustinskii equation or Born-Lande equation. I will let you analyze them.


text:
Conformational analysis determining forces
So the determining forces in the conformational analysis (what will be the more stable state and what will be the more excited state) according to my textbook are:

Steric interaction
Electron repulsion
Interaction of bonding and anti bonding molecular orbitals (for example $\sigma$ and $\sigma^{*}$ in ethane)

Electron repulsion is quite clear. Filled (molecular) orbitals repulse each other. I have no clue however what is meant with steric interaction, my textbook makes a small remark about it saying it is a repulsive force when atoms come to close to each other. It notes that in the conformation of ethane this hardly plays a role because the hydrogen atoms are to small to interact. But isn't this the same as electron repulsion? What else makes (neutral) atoms want to avoid each other?
Then the last part of bonding and anti bonding orbital interaction (overlap) stabilizing the staggered conformation of ethane is also not so clear to me. How does partially putting electrons in a higher energy molecular orbital (anti bonding) make a molecule's conformation lower in energy?

what is meant with steric interaction
isn't this the same as electron repulsion?

I agree with you Jori, the basis of steric interactions is electron repulsion.  When anyone stops to think about "why are adjacent bulky groups destabilizing" they'll soon arrive at the destabilizing interpenetration of electron clouds.  However, the use of the term "steric" to denote interactions due to size and bulk has been around so long that it has simply become common usage - the way things are said. The following figure shows trans-1,2-di-t-butylethylene to be more stable than its cis counterpart.  It's easier to say and understand that the bulky t-butyl groups produce steric strain in the cis isomer, rather than to say electronic clouds interpenetrate in the cis isomer.  The word "steric" immediately creates a picture of sorts in a chemist's mind.  Tetra-t-butylethylene has yet to be synthesized; attributing this to steric bulk around the double bond provides a clear image of the problem.

Then the last part of bonding and anti bonding orbital interaction
(overlap) stabilizing the staggered conformation of ethane is also not
so clear to me. How does partially putting electrons in a higher
energy molecular orbital (anti bonding) make a molecule's conformation
lower in energy?

First off, let's recognize that the orbital interaction you are referring to is a way to understand the effect of hyperconjugation in various conformations.  Also, we are not putting electrons in a higher energy molecular orbital.
Here are some diagrams that may help explain this.  The first diagram on the left doesn't relate to hyperconjugation, but since it is part of the picture, let's explain it.  The Pauli Exclusion principle says that we can't have electrons with the same quantum numbers in the same orbital. In ethane, each pair of electrons in one C-H bond have the same quantum numbers as the pair electrons in any other C-H bond.  Therefor, the closer we bring 2 C-H bonds, the closer we are to violating the Pauli Exclusion principle.  This argument tells us that the eclipsed conformation should therefor, be higher in energy than the staggered conformation.
The figures in the middle and the right show the effects of hyperconjugation upon the eclipsed and staggered conformations of ethane, respectively.  According to Frontier Molecular Orbital theory we examine the interaction between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) to see if stabilization or destabilization results. In the hyperconjugated resonance structure for ethane we have a proton and a pair of electrons in the carbon $\ce{sp^3}$ orbital, that $\ce{sp^3}$ orbital with two electrons is our HOMO.  The only LUMO for it to interact with is the LUMO for the C-H (the $\sigma^{*}$ orbital) bond on the adjacent carbon atom.  We can mix the $\ce{sp^3}$ and $\sigma^{*}$ orbitals to produce 2 new MOs, one bonding (no nodes between these 2 orbitals) and one antibonding (1 node between these two orbitals).  What is pictured below is the bonding (in-phase, no nodes) combination of these two orbitals.  We see that in the eclipsed (middle) conformation the interaction in this orbital is destabilizing, there is no constructive overlap.  However, in the staggered (right) conformation the overlap within this orbital is now constructive - there is positive overlap.  The hyperconjugation argument also favors the staggered conformation being at lower energy.

All right, finished reading now. Two remarks: firstly, what force is exerted in the Pauli exclusion principle (i.e. what is it that prevents the orbitals from closing in at each other, I guess it is more than just simple electron repulsion. Can this force be calculated)? Secondly, the explanation you give was exactly the same I had in my head, but I completely agree, but I can't wrap my head around the following: by letting electrons move out their orbital and into an anti-bonding orbital aren't those plain C-H bonds weakened in such way that it neglects the effect of the hyperconjugation?

@Jori  Part 1 - "electrons of the same spin are kept apart by a repulsive exchange interaction, which is a short-range effect, acting simultaneously with the long-range electrostatic or coulombic force"  see http://en.wikipedia.org/wiki/Pauli_exclusion_principle#Consequences

@Jori Part2 - "by letting electrons move out their orbital and into an anti-bonding orbital"  Like I said above, they don't move into an antibonding orbital.  The HOMO (sp3) and LUMO (sigma*) combine to produce two new MOs, one more bonding, one more antibonding.  It is this new bonding MO that the electrons are in, so the electrons are now more stable.  This is hyperconjugation described using MOs.  The energy of this new MO is also dependent upon ethane conformation as the pictures suggest; the staggered being lower energy than the anti.

Yes I see that, but this new molecular orbital that is lower in energy then the sp3 or sigma* must be occupied by electrons. These electrons must come from the 1sp3-1s C-H bond, right? And they also move around in the area that was previously the sigma* of the C-H bond on the right side of your diagram, so mustn't they produce an destructive wave pattern for the sigma C-H bond there? I'm sorry if it are unintelligent questions, but the book (Clayden et al.) really doesn't explain molecular orbitals at all. They more or less define them as a postulate and leave detailed information out.

I understand you at a conceptual level, but the details confuse me.

"this new molecular orbital that is lower in energy then the sp3 or sigma* must be occupied by electrons" Yes; 2) "These electrons must come from the 1sp3-1s C-H bond, right?" Yes (2sp3-1s), the bond we're hyperconjugating, the one that now has an H+; 3) "mustn't they produce an destructive wave pattern for the sigma C-H bond" no, this has nothing to do with the sigma C-H bond, we didn't mix that one in, we were looking only at the HOMO-LUMO orbital interaction; 4) no such thing as an unintelligent question, and more to the point, your questions are well thought out.

Aha, a H+ arises. I was already wondering about this. So this H+ is now attached as an ion to the electron density of our new MO? Wouldn't that require much energy to ionize this bond? 2sp3-1s of course, excuse me. No we did not mix the C-H sigma bond (the right on the picture) in, but our new $\ce{sp^{3}-}\sigma^{}$ MO (lacking a better name) partly contains the old sigma, right? So putting electrons in it must partly fill the area where we would otherwise have our destructive interference. What other counter MO would there be left to destroy the right sigma bond else?

I think the main problem is, is that I don't understand MO theory, because it is poorly explained in my textbook. See for example: https://physics.stackexchange.com/questions/121671/orbital-plot-to-explain-bonding-and-anti-bonding-molecular-orbitals. I never really understood how anti bonding orbitals "break" bonding orbitals. This isn't explained and they assume you just roll with it, and as a result you are left with these kind of frustration in later chapters haha! :)


text:
Boric Acid - Mechanism of Acidity
I was wondering how one could test the hypothesis that boric acid behaves not as a Brønsted acid but as a Lewis acid in that boric acid cleaves water molecules as to make hydrogen protons.

rad
Could one run an experiment involving boric acid in heavy water ($\ce{D_2O}$)?
Thus, if boric acid behaves as a Brønsted acid, the concentration of deuterium cation should stay the same, right? Boric acid might release hydrogen protons but those wouldn't affect the deuterium protons, right? Wouldn't heavy water still undergo auto-ionization and create a certain equilibrium concentration of the heavy equivalents of hydronium and hydroxide ions?
And thus, if boric acid behaves as a Lewis acid, couldn't we somehow precipitate the $\ce{B(OH)_3(DO)^-}$, and test whether the precipitate has D rather than H? I.e. introduce sodium chloride into the system and evaporate to dryness. Wouldn't that leave behind a precipitate consisting partly of sodium ion and partly of the borate? On the other hand what would happen to the excess chloride ion?

There will always be exchange of hydrogen and deuterium, even for highly covalent bonds like carbon hydrogen bonds, resulting in a wild mixture according to a shared Boltzmann statistics.

Why does this exchange happen? Kinetics?

It would be helpful if your diagram better reflected your question.  Also, deuterium is not radioactive.

As for acidity of boric acid you can find on https://en.wikipedia.org/wiki/Boric_acid that it dissociates in both ways, but is very weak. However it's acidic enough that it's only a matter of time until it would become almost completely deuterised if dissolved in D2O. It happens because in mixture there would be much more deuterium then normal hydrogen. The acid is too weak to capture or detect its anions. Maybe kinetics of deuterisation would give us some info about properties of its anions.


text:
Enthalpy of the reaction between hydrogen and oxygen

The reaction between hydrogen and oxygen to yield water vapor has
  $\Delta H^\circ = \pu{- 484 kJ}$. How much $pV$ work is done, and what is the value of $\Delta E$ in kilojoules for the reaction of $\pu{0.50 mol}$ of $\ce{H2}$ with $\pu{0.25 mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-\pu{5.6L}?$
$$\ce{2H2(g) + O2(g) -> 2H2O(g)} \qquad \Delta H^\circ = \pu{- 484 kJ}$$

I use the formula $\Delta E=\Delta H - p\Delta V$ to determine $\Delta E.$ However, when determining the enthalpy, the solutions manual does this:

$$\Delta H = \frac{\pu{-121 kJ}}{\pu{0.50 mol}~\ce{H2}}$$

Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-\pu{242 kJ}$. Or do we take into account all four hydrogen atoms? That would give us $\pu{-484 kJ}/4 = -\pu{121 kJ}.$

Is ∆E or ∆H change in enthalpy?

$\Delta H$ is change in enthalpy

The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.
In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water:
$$\ce{2H2(g) + O2(g) -> 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$
(By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)
Thus, the molar enthalpy relating to the amount of hydrogen is
$$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$
i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.
However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is
$$\begin{align}
0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt]
&= -121\ \mathrm{kJ}
\end{align}$$


text:
In which of the following compounds does Cl have the most positive partial charge?
The problem is :
In which of the following compounds does Cl have the most positive partial charge?
SCl2 ,, HCl  ,, BrCl  ,, OCl2 ,,
before I go through how I will solve this problem .. There is a question : Can Cl have a partial positive charge?
Well, all of them are polar, so there is a partial charge on each one of them
but how to compare or specify the partial charges?
The only thing I can think about is electronegativity
is it like .. the greater the difference in electronegativity the more the partial charges ?
Or .. ?

The greater the difference in electronegativity the more the partial charges is true. Compare the electronegativities of the atoms bonded with chlorine and you can easily determine in which compound chlorine has the biggest partial charge. Chlorine can have a partial positive charge when it's bonded with a more electronegative element like oxygen or fluorine.

$\ce{Cl}$ can have a positive charge if it is bonded to something more electronegative like oxygen for example.
You specify the partial charges by looking at electronegativity AND how many are bonded to 1 atom. For example take Chlorine Heptoxide($\ce{Cl2O7}$):
Each $\ce{O}$ has a partial negative with the 1 in the center having the largest of the negative charges since it is bonded to 2 chlorines whereas the other 6 have smaller negative charges since they are each bonded to 1 chlorine.
Each $\ce{Cl}$ has a partial positive and they are equal since both of them are bonded to 4 oxygens.


text:
Why alcohols don't behave as acids?
If I understand everything correctly, carboxylic acids behave like acids because oxygen is much more electronegative than hydrogen, so it takes his electron and releases H+. The problem is that alcohols have an OH group too, so why don't they behave like acids?

The figure below illustrates these two ionization reactions.  The key difference between the two is that when the carboxylic ionizes the carboxylate anion is formed.  The carboxylate anion is much more stable than the alkoxide anion formed when the alcohol ionizes.  The carboxylate anion is much more stable because we can draw resonance structures to describe it.  This resonance stabilization of the carboxylate anion explains why carboxylic acids are so much more acidic than an alcohol.

Although the answer is fine, there is still something weird when we got into inorganic acids. How would we justify that $\ce{HClO4}$ is so acid when it has no resonance structures (hibridisation is $sp^3$)?

@santimirandarp  The $\ce{ClO4}$ anion is highly resonance stabilized (just google for drawings). This enhanced stability of the perchlorate anion explains why the acid is so highly ionized.

There is no resonance without using $d$ orbitals. That's my point..

@santimirandarp Sorry, perhaps I'm not following you.  You asked why $\ce{HClO4}$ is so acidic and I said because of resonance stabilization of the perchlorate anion.  Hybridization $\ce{sp^3}$ for chlorine is OK, no need for d-orbitals.

No problem. Perchlorate anion has four simple bonds, so it is $sp^3$ and all oxygens are with same partial charge $-1$ there is no resonance...just one structure. Am I wrong? If we say there is a double bond, the only way possible is by means of $d$ orbitals...

@santimirandarp I think you are asking how is the tetrahedral geometry found in the perchlorate anion consistent with $\ce{sp^3}$ hybridization and resonance stabilizaton.  Permeakra's answer in this earlier Q&A might be helpful.  If not you might want to post a new question.  It is a very different question than the one above.

thanks, yes it was. So there is no resonance, at least it doesnt follows from that answer. In any case, it should be interesting to see some calculations probably...

@santimirandarp Yes, I suspect this is one of those cases were traditional hybridization approaches may break down and calculations would be informative.

hmm... yes but not that strange. $\ce{H2SO4}$ is another example...i.e no stabilization by resonance. (it is not a downside of your answer, I'm just trying to understand the whole picture).


text:
How can I find the heat that will be released in kJ per gram for the reaction of Al with Fe2O3?

Find the heat that will be released in kJ per gram when aluminium ($\ce{Al}$) reacts with $\ce{Fe2O3}$ as follows:
  $$\ce{2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)}$$

The information given is:

$\Delta H = 12.40 \space \rm{kJ/mol}$ (for $\ce{Fe}$)
$\Delta H = -822.2 \space \rm{kJ/mol}$ (for $\ce{Fe2O3}$)
$\Delta H = -1669.8 \space \rm{kJ/mol}$ (for $\ce{Al2O3}$)

If it asked me to find the heat released in kJ/mol that would be easy:
$$
\Delta H^\circ = \sum \nu_{i}\Delta H^\circ_f (\rm{products}) - \sum \nu_{i}\Delta H^\circ_f (\rm{reactants})
$$
I assumed that there is one mole and I found the mass in grams by multiplying the molar mass of each compound by one mole.
Then I divided the $H^\circ_f$'s by those amounts to get the new values "expressed in kJ per gram"
Then I used the previous equation and I got an answer around 10 but the correct answer is 15.24 kJ/g.
The question is: is the way I used wrong, or is there something else I didn't pay attention to?

When you converted the heats to kJ/g, you inadvertently "broke" the Hess's law equation that you plugged them into.
Each of the enthalpies of formation in that equation is multiplied by the number of moles of the matching species, and so the units of the enthalpy of formation are kJ/mol.
If you assume one mole of product and convert that to grams, you then would have to convert all of the other species to grams (two moles of Al, two moles of Fe, etc).
It's easier to just do the whole calculation in moles, then divide the final answer by grams. However, you need to specify which species is the gram basis. In other words, you are asking for the heat per gram, but per gram of what?
Whatever the answer is, then find the total heat output in terms of kJ/mol of that substance, then divide it by the molar mass of the substance (in g/mol).
One way to avoid these kinds of problems is to always explicitly include the units in any calculation that you do. In this case, you would have quickly seen that the Hess's Law equation would not give you the correct units.

I'm guessing that you're looking for the heat released per gram of aluminum.
You're correct to here
ΔH° = Σ(ν × ΔHf°) (products) - Σ(ν × ΔHf°) (reactants)
Since there are 2 moles of aluminum, you need to divide this ΔH° by 2, then you have heat released in kJ/mol of aluminum.
To convert this to kJ/gram you need to now divide by the atomic weight of aluminum (26.98 gm/mol).
When I did this I got an answer of 15.25 kJ/gm aluminum heat released.

it is not specified for any reactant or product .. the problem exactly says : " Calculate the heat released in kilojoules per gram when aluminium (Al) reacts with Fe2O3" ..?

So I think it is fot the reaction  !


text:
Why did Dalton assume that atoms are spherical?
One of the primary components of Dalton's theory is to "express" an atom as an unbreakable spherical shape filled with whatever. The question is, why didn't he suggest any other shape, like for example, cubes?

I don't know what Dalton was thinking exactly, but I do know that for a very long time philosophers and scientists have assumed that things are spherical in the absence of any indication that they are not.
The reason for this was originally related to religious beliefs, and the idea that God or (The Gods) would only create perfect objects, and the sphere (being perfect) was the obvious choice for heavenly objects and subatomic particles.
The assumption that things are spherical is still used very often, although it is not usually directly related to religious beliefs anymore. These days, the idea is that you should always use the simplest possible explanation that works. The simplest geometric shape is a sphere - you only need one dimension to describe it, and it has some other interesting properties like minimized surface area-to-volume ratio. Mathematics can also be greatly simplified by using spherical coordinates when things are spherical, which is another big advantage.
Therefore, when we imagine particles that we can't see, a sphere is a natural choice, and I would guess that Dalton was using one or both of these lines of reasoning when he came up with the theory.

What exactly do you mean with "you only need one dimension to describe it [a sphere]"? Clearly you're not talking about spatial dimensions, so I assumed you might mean measure (i.e. a sphere is defined solely by its radius). However, that's not unique of spheres, since cubes and many other polyhedra can be determined by their side length.

Also, I should note for historical interest that atoms have previously been thought of as cubes, and that according to quantum mechanics atoms are not in general spherical.

@Nicolau - What I meant is: you only need one number to describe a sphere. For a regular polyhedron, you need two numbers - the number of vertices or sides, and the length of each side.

@NicolauSakerNeto Just for the record, the conclusion of that topic was the atoms are spherical..

@Greg Ah yes, a few months after I wrote this comment here, DavePHD corrected me in comments of the very question I linked. Indeed, atoms are spherically symmetric, and I was wrong previously. Good thing that you pointed it out here, as I'd already forgotten about this comment. Thanks for helping keep the record straight.


text:
The highest melting point
Inspired by this question, what is the substance or compound with the highest melting point? At standard atmospheric pressure and ever-increasing temperature which substance a) eventually becomes liquid b) does so at the highest temperature?
I don't know much about glasses or related semisolids, so perhaps a reference viscosity? Glass transition temperatures are kind of like melting points, but some materials have a glass transition temperature and a melting point.
Some initial casting about on the internet finds that rhenium and tungsten have melting points at about 3400 °C, and hafnium carbide claims to be 'one of the highest' at 4100 kelvin. Does it actually melt or just decompose and then sublimate? 4100 kelvin is its 'refractory temperature', but I'm not certain how that's defined.

there are reports, that crust of neutron star is solid, composed of iron nuclei and degenerate electrons. The surface temperature of such star is about $10^6$ Kelvin.

Well, according to this Wikipedia article, it might be an alloy of tantalum carbide and hafnium carbide, specifically $\ce{Ta4HfC5}$.  A melting point of 4215 °C (4488 K) is listed.  But who really knows the answer to your question with so many new, proprietary alloys with complex compositions being fabricated these days.

It seems like 4-5 thousand kelvin is where the high-performance ceramics melt. I wonder if there would be interest in a melting-point competition, with teams sponsored by Dow, Du Pont, etc. Like Darpa's self-driving car thing they do in the desert.

It's human nature to compete.  Set up a competition and people will be there.

It seems this mixed carbide isn't all that good after all and hafnium carbonitride is probably higher melting, but it's virtually impossible to even make solid data about that https://doi.org/10.1103%2FPhysRevB.92.020104


text:
Elemental Potassium as both an oxidizer and a reducer
On this table I realized that elemental potassium could adopt oxidation states of -1, 0, and +1.
http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
Does this mean that elemental potassium can function as both an oxidizer and a reducer? I understand that oxidizers are reduced (gain electrons), and that reducers are oxidized (lose electrons). How can potassium anion be formed? The potassium anion must be extremely unstable due to potassium's extremely low electronegativity, correct?
Also in what compound does potassium actually have a negative oxidation state?

http://www.nature.com/nature/journal/v317/n6034/abs/317242a0.html  This abstract seems to suggest the ion is made in a solution of liquid crown ether.  I'm assuming the affinity of the crown ether for K+ is so strong that a potassium atom can take an electron from another potassium atom.

Which table are you referring to? @brinnb This might be a similar approach like solvated electrons from sodium or potassium $\ce{Na/K (s) + (n + m) NH3 (l) ->[][-78~^\circ\mathrm{C}] [Na(NH3){n}]+/[K(NH3){n}]+ + [e- . (NH3)_{m}]}$.

Martin, thank you, and I have included the link to the table on the OP.

Take a look here

Did you realize that the K anion existed a priori?

yes I knew it, but it was just because my book had question which of the following can act as both oxidizing and reducing agent and answer was K:)

Which book is that? I must take a look.

It's Modern fully solved MCQ's semester 1 class 11. Publisher Kautulia. These book helps to prepare for JEE(joint entrance exam)

Is this for the JEE?

Also in general how does one realize that the K anion exists, a priori?

@Dissenter Ya this book help to prepare for JEE but just for class 11 when you don't know to much stuff. I don't think anyone can realize about it unless he/she have read it in  some books or using stack exchange ;)

AFAIK, sodium in presence of 0.5 equivalent of proper crown ester disproportionate into complex of positive ion and crown ester and negative ion, and the solid salt was obtained, but I can't find the reference, sorry. Likely same for potassium. Surprising enough, negative potassium ion is more stable, that double oxygen anion. AFAIK, all or most double negative anions are unstable to dissociation into monoanion and free electron. Still, we have compounds with oxygen in -2 oxidation state.

Compounds having potassium anions are referred to as potassides and are rare, usually involving K- within a cyrpt such as:

Crystalline Salts of Na- and K- (Alkalides) that Are Stable at Room Temperature J. Am. Chem. Soc. vol. 121, pages 10666-10667 describes preparation of K+(aza222)K- , where "aza222" is the compound in the above figure, but with the 6 H's replaced with methyl groups.  The compound is formed by dissolving potassium metal in methyl amine solution of the methylated aza222.


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Are forming coins starting with sodium chloride possible without the clay melting from the molten metal?
Is forming metal coins (specifically ones with copper, nickel, and zinc) possible with these steps:

Carve a sheet of sodium chloride so that it is big but not too big and so that the design sticks out of the sheet of sodium chloride.
Cut the sodium chloride so that you have 1 big circle with the design
Form a mold out of clay from the sodium chloride design
Scale it down without losing detail using a carving machine(instead of hand carving like with the sodium chloride) that carves the design on each side on a separate piece of clay(which is much easier to carve than sodium chloride).
Once the clay is in the leather hard stage(usually takes about a day) put a piece of clay of uniform thickness and hardness around each side of the coin so that you can hand carve the rim(make sure the rim is not too wide) This will make there be space for the metal
Once the rim is in the leather hard stage attach the 2 sides to a base, again made out of clay
Fire the clay
Smelt the metal
pour the metal into the clay mold
get the metal coin out of the clay mold
Heat the metal again but don't melt it
Cool the hot metal
Wash and dry the metal
reuse the clay mold so that you don't have to start all over again from hand carving sodium chloride but rather only from step 8 which is smelting the metal

One of my big concerns of this process is that a copper, nickel, zinc mixture might melt the clay as soon as the metal is poured in but at the same time I don't want to use a metal that you can bend with the pressure from your hand like you can with aluminum.

Melt not likely- more likely cause disintegration due to thermal shock. Why not use a green sand process- Wikipedia describes Sand Casting reasonably well.

so your saying that even with molten metal at thousands of degrees that the clay mold won't melt? I mean clay, yes it does have a high melting point and so does sodium chloride but those are not nearly as high as with a copper, nickel, zinc mixture, especially sodium chloride compared to metal.

and I was thinking of starting with a sodium chloride cast because you can carve it reasonably well(its mohs hardness is only 2.5) and going to a clay mold because clay is much more flexible than sodium chloride(If you try to bend sodium chloride it just breaks) and then finally metal which is hard like sodium chloride but flexible like clay because the majority if not all coins are made of a metal of some sort and not clay or an ionic compound(both of which dissolve in water)

Sodium Chloride isn't a bad idea but it will be hard to shape to get much detail. Wood would be better for detail. You're title should be a question. Another element is that the ration of mold (clay in this case) to casting material is typical large, the process happens quickly and its protected by a layer of steam- all that protects the mold from damage. Need to consider expansion of the metal when casting and other factors.

Why would Sodium Chloride be hard to shape to get much detail if it is hard and you can carve it easily. It seems like if it is hard and you can carve it easily than it would be easier to get the fine details.

I now noticed you intend to obtain a solid sheet- it'll probably crack up a bit when you try to carve it up- leading to bigger and more jaggered holes than you intended.

Let us continue this discussion in chat.

What you are describing is a process similar to lost wax casting. The problem of course is that the clay mold can only be used once. You have to crack the clay mold open to recover the casting. https://en.wikipedia.org/wiki/Lost-wax_casting


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Synthetical formation of diamonds
As everyone knows a diamond won't last forever. It will eventually turn into a graphite over a long period of time, as the free energy of graphite is lower than that of diamond, hence making the reaction energetically favorable.
However, is it possible to do the reverse, by using graphite and the applying energy onto graphite so as to synthesize diamond. If so, how is it done and can it be done in a lab by a chemist? Does that mean that people who managed to do it can become rich? As a diamond costs so much more than a graphite. Shouldn't we synthesize diamonds in this way? Can I become  rich by synthesizing diamonds and selling them?

Yes it possible to synthesis diamond in lab, i am sure it can be done by synthesis of charcoal but don't whether it is possible by graphite. You might have heard that to cut some metals cutters with diamond blade are used, this type of blade are made up of laboratory- created-diamond. See this

They are not as expensive as real diamonds made by nature

A simple web search 'synthetic diamonds' into the all knowing google and: http://en.wikipedia.org/wiki/Synthetic_diamond :)

Why same link is added 3 times?

@Freddy : Same link thrice O.O Where?

@ashu first in my comment, second in user2353082 and third one in your comment

Natural Diamonds are formed under high pressure and temperature at depths of 140 to 190 kilometers (87 to 118 mi) in the Earth's mantle. Carbon-containing minerals provide the carbon source, and the growth occurs over periods from 1 billion to 3.3 billion years (25% to 75% of the age of the Earth). Diamonds are brought close to the Earth's surface through deep volcanic eruptions by a magma, which cools into igneous rocks known as kimberlites and lamproites.
According to Wikipedia,

The formation of natural diamond requires very specific conditions—exposure of carbon-bearing materials to high pressure, ranging approximately between 45 and 60 kilobars (4.5 and 6 GPa), but at a comparatively low temperature range between approximately 900 and 1,300 °C (1,650 and 2,370 °F). These conditions are met in two places on Earth; in the lithospheric mantle below relatively stable continental plates, and at the site of a meteorite strike.

So yes, Diamonds can be produced synthetically if you can simulate almost similar conditions as in Earth's Mantle. (See high-pressure high-temperature process; Synthetic Diamond) An alternative, and completely different growth technique is chemical vapor deposition (CVD).

Yup i know it could be done but im more interested in the economic aspect of it. Wouldnt ie possible for someone to synthesize it and made large profits. If so why has it not been done?

Read Synthetic Diamond article from wikipedia. It will provide you with all information you need.

@yswong Yes, there are several manufacturers of synthetic diamond. But it requires lot of investment and difficult technologies as it is described in the answers, so not a one man gig.


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Why are paramagnetic materials attracted to ferrimagnetic materials(such as iron, nickel, and cobalt)?
Diamagnetism exists in every molecule or ionic compound. It is the ability to be repelled by a magnetic field.
So you might think from this, that liquid oxygen would either be repelled by the magnet or not affected at all, yet it is attracted to it.
$\ce{O2}$ is paramagnetic, but how can a paramagnetic material like oxygen be attracted to a ferrimagnetic material like Iron other than the fact, that neutral attracts both positive and negative charges and $\ce{O2}$ is neutral.
Is it just this, or is there another part to it, like charged iron atoms inducing a dipole in $\ce{O2}$ or something?

First of all, in your question i feel a confusion about the nature of magnetism and paramagnetism. These magnetic interactions have nothing to do with attraction/repulsion of electric charges, and magnets do not attract or repel Coulomb charges (when we discuss the magnetic part of the interaction). In short static, magnetic and electric properties are two different stories.
Second, iron is ferrOmagnetic and not ferrImagnetic (capitalisation is by me). Forget ferrimagnetic materials for a second, and let's just talk about ferromagnetic, paramagnetic and diamagnetic materials:

Diamagnetic materials have no free electronic spin in the system, therefore they do not interact with magnetic field as a first approximation. There is a very small repulsive interaction coming from the polarisation of paired spins, but it is very small compared to any other interactions. Most organic compounds are in this category.
Paramagnetic materials have free electronic spins, which can flip / rotate. When put the sample in magnetic field, this field polarises the spins which turns into an attractive interaction. This is analogous with the electric field induced dipole moments, but this is a magnetic interaction, not an electronic interaction. The alignment of spins are constantly changing, as simple thermal excitation can rotate them. Therefore paramagnetic materials has no permanent magnetic moments like magnets. Typical examples are molecular oxygen, transition metal complexes, etc.
In ferromagnetic materials the free electronic spins are having a strong interaction with each other, and aligned parallel in strict order. This alignment is so stable that the magnetic moment of spins adds up, and there can be a macroscopical magnetic field around the sample (i.e. it is a magnet).

When you see the attraction between paramagnetic and magnetic materials, you see the polarisation of the paramagnetic materials in the magnetic field of ferromagnetic materials, which is an attractive itneraction.

does that also caused an induced dipole since it makes a nonpolar molecule like O2 polar? And magnetism can be simplified to attractions and repulsions between charges where neutral charge(like O2 which is a neutral molecule) attracts any charge, positive or negative. This is also the basis for static electricity.

@caters As I said you mixing up badly magnetic and electric properties. The dipole moment, polarity/nonpolarity are all electric properties. Static magnetic field has no interaction with charges themselves, and do not induce electric dipole moments, O2 do not turn to polar.


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What is the Solubility of Common Noble Gases in Concentrated Phosphoric Acid, 85% w/v?
By Henry’s Law, the concentration of a dissolving gas in a liquid is proportional to the partial pressure of that gas in contact with the liquid, mathematically,
$ p = k_\mathrm{H} c$,
where $p$ is the partial pressure of the dissolving gas, $c$ is the concentration of the dissolving gas in the liquid (solution), and $k_\mathrm{H}$ is a constant (Henry’s constant?) which depends on temperature.
What is the value of $k_\mathrm{H}$ for common noble gases (in particular, xenon) in concentrated phosphoric acid, at, say $298~\mathrm{K}$? Can the solubility at other temperatures be calculated using an extrapolation formula?

Value seems to be $k_H^{\ominus}=4.3\times10^{-3}\;M/atm$ for Xenon in water.For it in conc. phosporic acid seems not to be available as probably no one has carried the experiment in this solvent since water is most common and used commonly.Maybe you would?
Also, A simple way to describe Henry’s law as a function of temperature is:
$$\LARGE k_H=k_H^{\ominus}\times e^{\left(\frac{-\Delta_{soln}H}R\left(\frac1T-\frac1{T^{\ominus}}\right)\right)}$$
where $\Delta_{soln}H=$enthalpy of solution. Here, the temperature dependence is:
$$\frac{-d\ln k_H}{d(1/T)}=\frac{\Delta_{soln}H}R$$

Useful Links:
Table of constants, The NIST Chemistry WebBook, The Pesticide Properties Database (PPD)


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What can we say about the Gibbs free energy if it decreases or not?
[Sorry if the question looks trivial but I have thought about it for days and red a number of books all with no success.]
In almost every thermodynamics book that has something to say about chemical reactions there is a proof for the fact that at constant pressure and temperature the process is spontaneous in the direction in which the Gibbs free energy decreases. In some of the books it is also remarked that this conclusion is for when there is no "additional work" (beside the work of pressure in expanding the system), if there is an additional work then except for $dG=0$ the reversible process will yield into that additional work gaining its maximum value:
\begin{align}
\text{The only work is expansive:}\qquad & dG\le 0\\
\text{There are some additional works present:}\qquad & dG\le \delta W_{\text{additional}}
\end{align}
In this regard I have a question:
 When we have some additional works there in the problem (like electric work in an electrochemical cell), so that the second inequality holds and the maximum work is assigned to $dG$ as $dG=\delta W_{Add,Max}$, why still we feel free to use the first inequality and claim $dG$ should be negative?

In the case of electrochemical cells we have: $\delta W_{E,Max}=-n\,F\,E_{cell}\,d\zeta$ wherein $\zeta$ measure the advancement of the reaction. So we would obtain in a reversible (i.e. infinitely slow) reaction $\Delta_rG=\frac{dG_{p,t}}{d\zeta}=-n\,F\,E_{cell}$. But all the books then state that since the reaction is spontaneous in the direction in which $\Delta_rG\le 0$ so we would obtain $E_{cell}\ge 0$, while we know the inequality $\Delta_rG\le0$ is obtained from $dG\le0$ which doesn't hold here.

My question in other wording reads like this. Does the electrochemical cell problem involve an electric work or not? If "yes" then from where do we know the Gibbs free energy should decrease? And if "no" then from where do we obtain the equality $\Delta_rG=-n\,F\,E_{cell}$?

I think the confusion may be because you are conflating two ideas about Gibbs free energy.
The first is that Gibbs energy is a thermodynamic potential - this means that on a given surface in phase-space (the NPT surface), it is the "driving" force for, well, everything. Including chemical reactions.
The second idea is that the net change in Gibbs energy when going from reactants to products indicates the direction that the reaction will (eventually) go, and that its magnitude tells you the maximum amount of work that could be done by that reaction.
The second idea follows from the first - this becomes easy to understand if you just picture the thermodynamic potential as a gravitational potential, since we all have a good intuitive understanding of gravitational potentials.
Imagine a hill representing the thermodynamic potential (in this case Gibbs energy):

The rock represents a reaction coordinate - say, the concentration of a reactant. What will happen to the rock? It will go downhill. What will its maximum energy be when it reaches the bottom? Whatever the difference in potential energies is. What is the minimum energy it would need to get back the top? Again, whatever the net difference is in potential energies.
When you ask about $dG$ being negative, you are asking about the differential change - in other words, what is the slope? Obviously, if the reaction proceeds, the slope must be negative - "things roll downhill"
When you ask about $\Delta G$ being negative, you are asking whether the net change is positive or negative. The system can never do more work than the net change in $\Delta G$, because that is the maximum change in thermodynamic potential energy.
For your problem, the reaction involves moving an electron against a potential - the cell potential. The reaction can provide $\Delta G$ amount of energy with which to move that electron.

Dear @thomij, thank you for your great explanation, only one point remains in the dark side for me, that the reaction itself involves the transfer of electron so moving the electron is not a work done by the reaction on the system's surrounding, am I wrong?

The electrons are moving against an external potential, which is the cell voltage. If all of the reactant molecules were in the same side of the cell, there would be no external potential and the total $\Delta G$ would be released as other forms of energy (heat mostly). Since in a Galvanic cell the reactants are separate, but electrons can flow through a circuit, then the voltage of the circuit is the external potential that the reaction is doing work against.


text:
Is sucrose made of 2 glucose molecules or 1 glucose and 1 fructose?
I have read from several sources about sucrose and some say it is 2 glucose molecules and some others say it is 1 glucose molecule and 1 fructose molecule.
I know that both of these are related disaccharides since they have the same molecular formula of $\ce{C12H22O11}$
So which one is actually sucrose and what is the other disaccharide.

This was discussed recently here on Chemistry SE  http://chemistry.stackexchange.com/questions/14562/why-do-other-sugars-melt-whereas-sucrose-decomposes   I checked a number of other links and all agree that sucrose is glucose + fructose

I don't think it is a duplicate because it is about the structure of 2 related sugars, not the melting point of sugars.

Yes, but the link shows the structure of sucrose and identifies the two monosaccharides.

but several sources I have read said that sucrose is glucose + glucose.

maltose is glucose+glucose

@ron - it is true that the answer can be found in other places, but that doesn't make this a duplicate question. OP is asking directly about the structure, so to be a duplicate, the other question (not answer) would have to be about the structure of glucose as well.

@caters - if you would link or reference your sources, someone can probably help clear up your confusion. At this point all anyone can say is "if your source says sucrose is two glucose molecules, then your source is wrong."

Sucrose is composed of one glucose and one fructose monosaccharide.  Two glucose molecules form maltose.


text:
O=Cl=O and O-Cl-O in a ring are not resonance structures?
Resonance structures are about moving electrons to form anions, cations, and new bonds. Structural isomers can be resonance structures.
For example take phenyl ketone. Of course you can form resonance structures by moving the pi electrons from benzene. However you can also move the electrons from the carbonyl double bond to the oxygen forming phenyl oxide. From there you can again move the electrons from the benzene ring. All of these are resonance structures and the ones with the alkoxide anion are structural isomers as well as resonance structures.
In the same way you can move the pi electrons from the $\ce{O-Cl}$ double bond to the oxygen to form $\ce{O-Cl-O}$(I don't know how to make the negative charge on the left oxygen not look like a 4 e bond(which unless you are talking about a double bond is nonexistent)) which can then form a peroxide bond($\ce{O-O}$ single bond) to make a cyclic peroxide. Resonance structures are all about moving electrons and in the process forming anions within molecules(like the alkoxide anion), cations within molecules, and new bonds.
So how are chlorine dioxide($\ce{O=Cl=O}$) and this cyclic peroxide that forms not resonance structures and just structural isomers?

Resonance structures are about moving electrons to form anions, cations, and new bonds. Structural isomers can be resonance structures.

No, resonance structures do not form anions or cations. Resonance structures should all have the same overall formal charge.
Also moving electrons around when drawing resonance forms does not create new bonds. These bonds always have existed; it's just that one picture is inadequate to represent them all. Hence the resonance forms.
Structural isomers also cannot be resonance structures. Structural isomers differ fundamentally in terms of the bonding between or among their constituent atoms. Resonance structures are just things we draw on paper to better represent the true nature of bonding. For one chemical entity, there may be multiple resonance forms, but that doesn't mean that each is a discrete form that is able to be isolated.

In the same way you can move the pi electrons from the O−Cl double bond to the oxygen to form O−Cl−O(I don't know how to make the negative charge on the left oxygen not look like a 4 e bond(which unless you are talking about a double bond is nonexistent)) which can then form a peroxide bond(O−O single bond) to make a cyclic peroxide. Resonance structures are all about moving electrons and in the process forming anions within molecules(like the alkoxide anion), cations within molecules, and new bonds.
So how are chlorine dioxide(O=Cl=O) and this cyclic peroxide that forms not resonance structures and just structural isomers?

I'm not sure what you mean by this.

I mean by anions and cations those WITHIN molecules. And phenyl oxide is a resonance structure of phenyl ketone or even benzaldehyde and is a negatively charged molecule because of the alkoxide anion. Also it does form new bonds(like as I already said the C-O single bond).

That "new bond" isn't a "new" bond in the sense that none of the resonance pictures you can draw for the phenoxide anion actually exist discretely; the actual structure of bonding is a weighted average of the resonance forms. I'm guessing that the phenoxide anion is what you're talking about when you say "phenyl oxide."

Charged atoms within molecules are not anions or cations.

Oxyanion: Negatively charged oxygen atom within a molecule or negatively charged molecule containing oxygen, Oxycation: same thing except positively charged, Carbocation: positively charged carbon within a molecule, Carbanion: same thing but negatively charged.  These are ions because often 1 or the other occurs(like in the enolate anion where either the carbanion occurs or the oxyanion does) and in lots of them the molecule is charged. and yes that is what I mean by phenyl oxide and not the other interpretation of it(Benzene ionically bonded to the oxide anion(neutral attracts - and +)).

as for the movement of electrons in chlorine dioxide I mean that because of how pi bonds are much weaker than sigma bonds that you can move the pi electrons from the double bonds to the oxygen atoms making both O's negative and the Cl positive. This in turn can form a peroxide bond. Similar things happen with covalently bonded metal oxides forming metal peroxides in a ring.


text:
Is it possible for a reaction to have an entropy change of zero?
If the question seems rather "unscientific", be kind to accept my apology. My textbook is being highly unclear about the topic and I need to know the answer to this question before I go on with my studying.

Sure, in any trivial reaction involving the reformation of the reactants. Entropy is a state function and if you reform the reactants then the change in entropy is 0.

So then zero entropy change can hardly be found in biphasic mixtures, huh?

Yes, entropy change of reversible process is zero.
This will also help

Yeah, but there's no such thing like reversible process in real world...


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Why is it O-Cl-O and not O-O-Cl?
I'm currently preparing my exam and while I did never think about this question during the course, it begins to confuse me right now...
Why is it the structure of $\ce {ClO_{2}-}$ like $\ce {O-Cl-O}$ and not like $\ce {O-O-Cl}$?
I know how to place the covalent bonds and the electrons, but I don't know how to figure out the basic structure of the molecule. Is there a relatively simple rule to solve (and understand) such relatively simple molecules?

Is there a relatively simple rule to solve (and understand) such
  relatively simple molecules?

The most electropositive atom is usually at the center of a molecule because

a "central atom" implies that there are multiple (more than 1) bonds to the central atom. If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would.
electronegative atoms tend to carry multiple lone pairs of electrons. If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.

Try to draw the structure of $\ce{ClO2-}$. It will like this :

If you see the structure it is $\ce{O−Cl−O}$  and not $\ce{O−O−Cl}$.
Chlorine is in the center because Least electronegative atom will be at center. You can refer this links: YouTube and  chemistry education

The right O and Cl are connected by a covalent double bond? But Cl would then have 5 electron pairs which is wrong. (Furthermore, my textbook says, that the atoms are connected by single bonds only.)

@Michael this is $\ce{ClO2-}$ Lewis Structure, if you want to know how to draw structure you should see this video.

Doesn't $O-O-Cl$ have lower formal charges than $O-Cl=O$? Why is having Chlorine as the central atom preferable to having oxygen as it?

@user2612743 Least electronegative atom will be at center.

@Freddy Why is it the least electronegative? Is it most conducive to "sharing" electrons/forming bonds?

@user2612743 See the rules for drawing Lewis structures of mlecules here

@Freddy Unless I missed something that doesn't explain why the least electronegative is in the center

@user2612743 Frankly i don't know why but thats the rule! Yes and i also know Hydrogen is exception.

@Freddy: The Lewis structure in your answer says that there's a double bond between the right O and the Cl, right? That's not what my book tells and that's also not what I get. I get only single bonds.

@Michael I have given youtube link in second comment to Michael you can refer that if you have any doubt about how the structure came.


text:
Converting cyclic compounds to linear compounds (possible ?)
Can one convert from cyclic convert to linear compound?
Like
$Cyclohexane$ $+$ $?=Hexane$
Even in High Temp and Pressure?
I know it is very tough since we are going from a stable compound to a relatively more unstable compound.But still...???
And can we also convert $Benzene$ to some $Unsaturated Compounds$ ?
And if possible can anyone please tell the reaction mechanism of it?
Thanks

Yes, this can be done.  Unsaturated cycloalkanes can be ring opened using fairly straightforward methods.  For example, heating cyclobutene will produce 1,3-butadiene in an orbital symmetry allowed reaction.  But since your example involved cyclohexane, I'm guessing that you are more interested in the ring opening of saturated cycloalkanes, so that's where I'll focus my answer.
Catalysts are needed to ring-open cycloalkanes.  The catalyst complexes with a cycolalkane bond and then in a second step, hydrogen is added across this bond.  Mechanistically, it is basically the reverse of the hydrogenation reaction (e.g. catalytic hydrogenation of cyclohexene to cyclohexane).  Some of these catalysts are very indiscriminate, so the initially formed linear alkane will rebind to the catalyst and undergo subsequent isomerization and fragmentation reactions.  More recently catalysts have been developed that can open cycloalkanes to the corresponding linear alkane in fairly high yield and relatively free of other hydrogenolysis products. These reactions are run in solution at temperatures slightly above ambient.  Here is a link to a full text article on one such process that uses Lanthanide\Aluminum complexes;  Terbium seems to be the preferred lanthanide.  Yields are typically in the 65-80% range (cyclohexane, 67%; cyclopentane, 77%; cyclodecane, 77%).

Can we also convert Benzene to some linear unsaturated compound?

Yes, there are many known catalytic hydrogenation methods that reduce benzene to cyclohexane (see  http://orgchem.chem.uconn.edu/2444f2011/2444-111611.pdf).  Once you have cyclohexane, then you could use the Tb\Al method described above to reduce it further to the alkane.


text:
Inventor of CRT?
Let's cut to the chase. I've been researching through numerous, reliable and unreliable, resources. According to the CRT official site and wiki, the first cathode ray tube (CRT) was invented by Carl Ferdinand Braun, year 1897. Yet undeniable proofs of CRTs being used in experiments in 1860s have been collected.
So this leaves the question unanswered. Noting that one of the fundamental progresses in chemistry (discovery of electron) is based on this useful tube, I will be thankful to anyone willing to share some info.

Invention of CRT
In the year 1855, German inventor Heinrich Geissler developed mercury pump which is later modified by Sir William Crookes who became the first to produce cathode rays.

Figure 1: Geissler's Vaccum tubes(1862)
Since, then vacuum tubes are being used to study cathode rays, Braun's tube was actually an improvement to the these.
Otherwise, all of them have used some kind of Vacuum tube which produces cathode rays.

In 1858 Julius Plücker showed that cathode rays bend under the influence of a magnet.
In 1865 H. Sprengel improves the Geissler vacuum pump.
In 1869 J.W. Hittorf establishes that "rays" from the cathode travel in straight lines.
In 1871 C.F. Varley is first to publish suggestion that cathode rays are composed of particles.[Fig 2]
In 1874 George Johnstone Stoney estimates the charge of the then unknown electron.
In 1876 Eugen Goldstein introduces the term cathode ray to describe the light emitted.
In 1881 Herman Ludwig von Helmholtz suggested the idea that there is a smallest unit of electricity.
In 1890 Arthur Schuster calculates the ratio of charge to mass of the particles making up cathode rays (today known as electrons) by
  measuring the magnetic deflection of cathode rays.
In 1894 J.J. Thomson announces that he has found that the velocity of cathode rays is much lower than that of light.

Source:

http://www.chemteam.info/AtomicStructure/Disc-of-Electron-History.html
http://www.chemteam.info/AtomicStructure/Disc-of-Electron-Intro.html

Figure 2: First published paper on cathode rays.