text:
How does chlorine form more than 1 bond?
How are perchlorate or chlorate or chlorite ions and their respective acids or compounds formed. $\ce{Cl}$ can't form more than one bond but still...
 $\rightarrow$'Perchlorate ion'
 $\rightarrow$'Chlorous acid containing Chlorite ion'
 $\rightarrow$'Chlorate ion'
Chlorine is making more than 1 bond... How is it possible?

In terms of valence bond theory Chlorine can have an expanded valence because it has an empty 3d subshell.

Hate hearing this shit cause every time they do a computational study on supposedly hypervalent main group elements they find minimal d orbital participation in bonding. Also that explanation doesn’t even remotely help when it comes to predicting the bonding behavior of main group elements, it’s just a bullshit hand wave explanation

The order of bonding, and so the valence state of Cl in $\ce{ClO_{x}-}, x>1$ compounds is very debatable. Generally, two models exist.

$\ce{Cl}$ atom, just like $\ce{S, P}$ and some others has unoccupied $d$-orbitals in the valence shell. It is possible to move some electrons from $p$-orbitals to $d$-orbitals, producing half-occupied orbitals that can participate in covalent bond formation
$\ce{O}$ atom has 6 electrons at the valence shell with four orbitals. It is possible to move one electron from chlorine atom to oxygen atom to form $\ce{O^{\dot{-} }}$ ion, that can than form a bond with newly formed half-filled orbital of chlorine. However, being electronegative and having high positive charge, chlorine partially pulls back the electron pair of newly formed bond.

Currently the second option is in favor of theoreticians, both by electron population analysis and relatively high charges in the oxochloric compounds, however, in school-level chemistry people traditionally resort to the first option. On the other hand, the second approach is the only option to adequately describe nitric acid and nitrates as nitrogen does not have $d$-orbitals on the valence shell.
Now, the stability of oxochloric compounds is a completely different matter. All of them are not very stable compounds, easily freeing their oxygen. Still, the process has significant activation energy, the state in between is even less favorable then oxochloric molecule. both oxygen and chlorine have electron-rich shells that repel potential reactants. Perchlorate ion is especially stable, at its surface is completely covered by outer negative shells of oxygen, so electrons cannot reach positively charged chlorine.

The Cl-O bond has a dissociation enthalpy of 269.1 kJ/mole in the gas phase. That's not very low. This suggests that yes, the Cl-O bond is stable. Similarly, the Cl-Cl bond has a dissociation enthalpy of 242.580 kJ/mole. Again, not very low, but not very high. For comparison, consider triply bound diatomic nitrogen - the third strongest covalent bond known. This species has a dissociation enthalpy of 945 kJ/mole - about three times higher than the aforementioned energies.
http://www.webelements.com/chlorine/bond_enthalpies.html
Chlorine can have multiple bonds, as you've shown in your Lewis structures. Try looking into molecular orbital theory.

I don't understand what is the big deal with bonds in acids or esters in higher oxidation states - I always thought of them as of single dipolar bonds. They shouldn't be written as double bonds. It's an archaism for me - now chlorine oxoanions aren't considered hypervalent. Even hypervalency of molecules like iodine heptafluoride has been criticized.


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why do we obtain a sigmoid curve in vapour pressure versus temperature graph
i have recently got a question in an assignment, which was somewhat like this

what would be the shape of curve obtained in a graph between vapour pressure & temperature of a binary solution in a closed vessel

generally we do all the experiments regarding vapour pressure in a closed vessel, so the volume must remain constant and from the formula $PV=nRT$, the number of moles remains constant in a binary solution, as volume is remaining constant we get the equation as $Y=mX$, which is a straight line passing through the origin,
but the answer was a sigmoid curve, how did we get that, can i get an elaborate answer for this quiery.

the answer already present over here is absolutely excellent but, as an extention i would add some points to it

the equation of $PV=nRT$ is made for gases and for a solution of gas and liquid in which the volatile solvent shows some vapour pressure on its solution

the curve sigmoid curve you get is through the Clausius Clapeyron equation which suits this condition [i.e. the situation of solutions]

after the whole solution is evaporated then you get pure solvent in the vapour phase after which your equation for an ideal gas $PV=nRT$ is applicable and you will get a striaght line there after

Above is a graph of the Maxwell Boltzmann distribution at several temperatures.  Imagine a point on the x axis being the point at which a liquid particle has enough kinetic energy to leave the solution.  As you can see, each increase in temperature results in a larger number of particles with enough energy to vaporize.  This generates an exponential graph for VP vs T.  However, eventually you reach the point where all of the liquid has reached the gas phase.  Then you have a linear graph of P vs T.

can i get the equation from which you've got this graph

The graph is a generic Maxwell Boltzmann distribution.  Look up the Clausius Clapeyron equation.  It gives the vapor pressure vs temperature for a given enthalpy of vaporization.

but where did my concept of the equation of $Y=mX$ went wrong could you please explain it

The graph is not linear while there is still liquid present.  Every successive unit of energy added by increasing temperature results in more particles having enough energy to vaporize.  Look at the v in the word velocity on the graph.  What is the area under the red curve vs the blue curve.  A doubling in T more than doubled the number of particles in the vapor phase.

yes now the thing stuck my mind, anyway thanks for your assistance

+1. This is a GREAT answer. No idea why it was -1 before! I see you followed a Materials Science proposal in the past. I really wish they allowed it to be launched. If you don't mind another attempt at a Materials proposal, would you be so kind to commit to this? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=OTA1YTI2NjQyNTU5ZjZkZmUxMGJjY2Q5NTQ1ZDJjYWI5ZWE1MTYzMTIwNjUzZGMwNjA5ZTU1M2Q5Y2VhMDliNFITayKhVuP2O5Uu3YycVIkNN2_sKy7FfIkiG5b9Kh7W0 It started slow but lately it's been booming with 5 commitments/day and very soon it will reach 200.

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Why are least electronegative elements usually in center of lewis structure
Many sources state that one of the primary rules for molecular structures is that the least electronegative element is in the center. However, none of them explain why this is the case.
Is this because it is most conducive to sharing electrons and forming bonds, or is it the most stable, or is there a better reason?
Also, why are many molecules symmetric. Is there a reason why symmetry in nature and in chemistry is preferred over non-symmetry?

Is this because it is most conducive to sharing electrons and forming
  bonds, or is it the most stable, or is there a better reason?

Your first rationalization is a good one. That is the likely reason the least electronegative element generally becomes the central element. That being said, there are exceptions. One being water. Oxygen is much more electronegative than hydrogen, which suggests that hydrogen should be the central atom, but the hydrogens are clearly peripheral atoms. This stems from the limitation that hydrogen can only form one sigma bond. That being said, hydrogen has also been found in unconventional bonding arrangements (see diborane for an example).

Also, why are many molecules symmetric. Is there a reason why symmetry in nature and in chemistry is preferred over non-symmetry?

Symmetrical conformations of molecules may or may not have lower energies than their asymmetrical counterparts. Consider the ternary interhalogen, $\ce{IBrF2}$. The symmetrical conformation is likely lower in energy than the asymmetrical conformation (draw out the Lewis structures to see).

As we can see, in the symmetrical conformation (left), there are only 90 degree fluorine/lone pair interactions, as opposed to the conformation on the right, in which there is a fluorine/lone pair interaction in addition to a bromine/lone pair interaction. The equatorially placed atoms have been left out of this consideration as are ~120 degrees apart from the lone pairs and thus their interactions are not as significant as the 90 degree axial/lone pair interactions.
That being said, there are myriad exceptions. Considered the eclipsed (left) and non-eclipsed (right) conformations of 2,3-dibromobutane. The eclipsed conformation is symmetric (it is also meso; all meso compounds are symmetrical but not all symmetrical compounds are meso). The non-eclipsed conformation is asymmetric. However, the non-eclipsed conformation is energetically preferred since eclipsing the bromines places them closer together, which creates more van der Waals repulsions.

Additionally, symmetry is a complicated topic in chemistry, especially with respect to more complex molecules and entities.

Why are least electronegative elements usually in center of lewis
  structure?

If there is a "central atom", then 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.  Also, 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.

why are many molecules symmetric. Is there a reason why symmetry in
  nature and in chemistry is preferred over non-symmetry?

The symmetric arrangements (linear, tetrahedral, octahedral, etc.) minimize destabilizing bonding electron - bonding electron repulsions.  Remember methane is a perfect tetrahedron - that is the lowest energy arrangement because all of the bonding electrons are as far apart from one another as possible in this geometry.

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Epoxide Opening with Sodium Hydride
I understand that in general, when in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons. And when in basic media, the nucleophile attacks the less substituted side of the epoxide due to steric reasons.
What if the hydride anion attacks an epoxide? I'd imagine the hydride anion to be quite small, and able to attack both the more substituted and less substituted sides of an asymmetric epoxide. What are the experimental results?
Also I have my doubts about hydride anion even being able to open an epoxide; I am told by Clayden that hydride anion never acts as a nucleophile because its 1s orbital is too small.

If I recall my organic hydride is such a poor nucleophile it won't open an epoxide or do many other organic reactions for the reason you stated.

Poor nucleophile because of its heavy solvation, right?

The hydride ion is actually gigantic. It's hard to get a reliable value because its poorly held electron cloud is very soft and deforms easily, but if I recall correctly it's theoretically the second largest common monoatomic anion known, at ~200 pm radius, losing only to iodide.

Huge because of its very poor ability to stabilize  electron density, right?

in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons

This is an $\ce{S_{N}1}$ like reaction. First we would protonate the epoxide oxygen.  If we examine the case of an unsymmetrical epoxide and draw the resonance structures for this situation (see below) we would expect the secondary carbocation structure to count more towards the actual structure than the primary carbocation structure.  Therefore, we would expect a nucleophile to react preferentially at the tertiary epoxide carbon, quite selectively.

in basic media, the nucleophile attacks the less substituted side of
  the epoxide due to steric reasons.

This would be a traditional $\ce{S_{N}2}$ reaction.  Attack at the least substituted (sterically most accessible) carbon would be expected.

What if the hydride anion attacks an epoxide? I'd imagine the hydride
  anion to be quite small, and able to attack both the more substituted
  and less substituted sides of an asymmetric epoxide. What are the
  experimental results?

The hydride anion is not free, but complexed, so it has reasonable steric bulk - enough for stereoselective reactions to occur.   Further, the reaction between LAH and epoxides is quite facile, no doubt accelerated by the release of steric strain in the 3-membered epoxide ring system. A traditional $\ce{S_{N}2}$ reaction takes place.  In the figure below, the alcohol shown is formed in ca. 90% yield.  Because of the "backside" attack nature of the $\ce{S_{N}2}$ reaction axial attack is strongly preferred in cyclic systems.

That's interesting. My professor commented that the hydride anion was so small that the opening of the epoxide is not stereoselective.

"Lithium aluminum hydride usually delivers hydride with excellent selectivity to the less substituted carbon of an epoxide", see page 358,  http://www.wavefun.com/products/books/Chapter4.pdf    a few more examples showing high stereoselectivity in http://depts.washington.edu/chemcrs/bulkdisk/chem531A_win06/notes_Epoxide_opening.pdf

I was thinking of sodium hydride; would that be the same thing?

No, each form of hydride  ($\ce{LiAlH_4, NaBH_4, NaH}$, etc.) has noticeably different properties.

Oh okay. I was referring to NaH exclusively.

Sorry, somehow I missed that.

It's okay at least I gleaned a lot about the nature of LAH!

https://chemistry.stackexchange.com/questions/156054/can-nah-open-the-epoxide-ring-to-form-alcohol-if-so-how

There's much detail to it beyond the scope and nit necessary but the final result is:

Bases (or ions with negative charge) attack the sterically unhindered positions
Acid catalyzed ring opening happens from the position best to stabalize positive charge even though it doesn't form


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Why does blood get further oxidized than Iron oxide?
Oxygenated blood is bright red and deoxygenated blood is dark red or brown.
If you take oxygenated blood and leave it in the air it will turn dark red, then brown, then finally a bluish green from exposure to atmospheric oxygen.
Why does it oxidize this far to get a copper oxide color to it after it has passed the phase where it is the same color as iron oxide? Why doesn't it stay at the iron oxide color?
This often happens during the last few days of my period that the bleeding is extremely light and the color of it when it first comes out is reddish brown (sort of like iron oxide) and after a few hours, it starts turning bluish green (sort of like copper oxide). However, this happens very slowly and after 24 hours of exposure to atmospheric oxygen still only a fraction of the blood is further oxidized. I don't take medications or drugs that have a lot of sulfur so I don't think it is from $\ce{H2S}$ or $\ce{SO4-}$ (which would give it a dark green color, not light bluish green) and I don't have excess copper in my blood either.
I think that it is Iron(II,III) oxide with the Iron(III) released and the autoionization of water (which happens in both the liquid and gas phase) forming Iron(II) Hydroxide which can be further oxidized to Iron(III) hydroxide which has that bluish green color along with a small amount of Copper oxide forming (Due to the fact that our bodies need some copper and like Iron there is always some copper in the bloodstream).

It's extremely unlikely copper has to do with anything. According to Wikipedia, there is less than 100 mg of copper in the entire body. The amount present in any reasonable quantity of lost tissue would be measured in micrograms.

So it is more likely to be Iron(III) hydroxide causing that green color than than copper oxide because Iron is so abundant and copper is much less abundant?

If it is not in inert (or pure oxygen) atmosphere, the change of colour can also be caused by bacteria.

There are two misconception behind your question:

You assume you can directly compare the colors of different compounds and derive oxidation states e.g of a heme (porphyrin) comparing to iron-oxide. No, it is not true.
You assume this whole story has anything to do with copper. No, the blueish-green is nothing to do with copper, it is a (porphyrin) degradation product.

The color of transition metal compounds mostly can be derived from the crystal field (ligand field) around the central metal ion. In inorganic iron-oxo/hydroxo compounds have different structure and hence different field than heme compounds. In heme, the N coordination of porphyrin ring is generating a very strong planar field, and the oxygen coordinates as molecule on the top of it. Charge transfer to the ring can also be an important factor in heme derivatives. As you see it is a whole complicated mass, and not much direct connection with different inorganic salts you see in lab.
The colors you see are coming from oxo-hemoglobin (which you have in your healthy blood), and most probably several steps of enzymatic degradation of heme. The porphyrin ring is oxidized and cut open by an oxygen on the the iron. Several colorful compounds are produced, including the iron complexes themselves and biliverdin (green), bilirubin (yellow). These are the same colors you see under your skin, of you hit yourself hard and have some bleeding in your tissues. First the injured area is deep red/blue, than go to greenish, finally yellowish patch. This also happens continuously in your body, when your blood cells get old, and heme is enigmatically degraded so the whole thing can be recycled:  http://www.elmhurst.edu/~chm/vchembook/634bilirubin.html

This often happens during the last few days of my period that the
  bleeding is extremely light and the color of it when it first comes
  out is reddish brown(sort of like iron oxide) and after a few hours it
  starts turning bluish green(sort of like copper oxide).

Well, period blood isn't necessarily just blood; it's also sloughed off tissue. That would complicate the analysis; the green color could arise not necessarily from what's in your blood but also what's in your tissues, and I wouldn't be surprised if there were copper in your tissue. Copper is after all an essential trace element for health.
http://en.wikipedia.org/wiki/Copper_in_health

but tissue I think comprises most of the heavier ones and those I only see going to dark red before I have to change my pad. In the extremely light ones I think it is mainly deoxygenated blood flowing out(sometimes not constantly)

I could simplify the analysis by looking at Iron(II) oxide, Iron(II, III) oxide, Iron(III) oxide, Copper oxide, Iron(II) hydroxide, Iron(II, III) hydroxide, and Iron(III) hydroxide concentration changes in oxygenated blood(from the blood vessels themselves) over time and see if any increase as it goes from the reddish brown Iron Oxide color to the bluish green Copper Oxide color but how could I get enough oxygenated blood to test that without causing severe bleeding from arteries?

Try a flame test. Use a piece of nichrome; stick in blood, then stick in open flame. Note the color of the glow.

how will that tell me the amount of Iron oxide, Iron hydroxide, and Copper oxide there is?

Well, it might alert you to the presence of copper in the first place. Green could be a lot of things other than copper.

yeah like H2S or SO4- or Iron(III) hydroxide. Those could all be what is causing the blood to oxidize further from reddish brown to bluish green except that I think the sulphur compounds would lead to a dark green, not a light bluish green.


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How to determine oxidising and reducing agents in the reaction between ammonia and hypochlorous acid?
How do I figure out what is the oxidizing agent and the reducing agent in the following reaction:
$$\ce{\overset{-3}{N}\overset{+1}{H}_3 + \overset{+1}{H}\overset{-2}{O}\overset{+1}{Cl} <=> \overset{-1}{N}\overset{+1}{H}_2\overset{-1}{Cl} + H2O}~?$$
I assigned known oxidation numbers for the elements, but I'm not sure how to proceed with the rest.
Is $\ce{NH3}$ the reducing agent? Or nitrogen? Going from $-3$ to $-1$
and $\ce{HOCl}$ the oxidizing agent, chlorine going from $+1$ to $-1$?

You're on the right track with assigning oxidation numbers to each element. And you've assigned them successfully.
To successfully complete the task you need to define a few terms:
What's oxidation?
What's reduction?
What's the oxidizing agent? Try thinking of this in terms of what oxidation is, and then try figuring out what an oxidizing agent would do.
What's the reducing agent?

And what is a Redox Reaction?

^Good question to be asking oneself.

Thank you! Oxidation would be the loss of at least one electron, and reduction is the gain of at least one. Oxidizing agent is the one that is reduced because it lets the other one get oxidized.
So if Nitrogen in NH3 is going from -3 to -1, it is losing electrons (since charge is increasing) meaning it is oxidized, making it the reducing agent?

As Dissenter had already pointed out, your assignment of oxidation states of each element is spot-on. Through your comments you seem to know that-

Oxidation refers to the loss of electrons or an increase in the
oxidation state of an atom by another atom, an ion, or a molecule.
Reduction refers to the gain of electrons or a decrease in the oxidation state
of an atom by another atom, an ion, or a molecule.
Substances that have the ability to oxidize other substances (cause
them to lose electrons) are known as oxidizing agents, as they remove
electrons from another substance, and thus itself get reduced. As it
accepts electrons, it's also called an electron acceptor. Vice versa for a
reducing agent.

(For more details you can check out the Wiki page on redox reactions.)
For deducing which is the oxidizing agent and why is it so we come to the actual picture. The change of oxidation state of $\ce{N}$ from -3 to -1 indicates it being oxidized as it loses 2 electrons (and hence the negative charge on it decreases).
Thus by the definition above, $\ce{NH3}$ is the reducing agent as it gets oxidized itself (as $\ce{N}$ goes from -3 to -1).
While change of oxidation state of $\ce{Cl}$ is from +1 to -1, this shows that it has been reduced as it has gained 2 electrons (and hence the charge on it changes from positive to negative). Thus $\ce{HOCl}$ is the oxidizing agent as it gets reduced itself (as $\ce{Cl}$ goes from +1 to -1).

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Why does charcoal absorb smells from the air, and where do those smells 'go' once absorbed?
I've read that using blocks of charcoal can remove some odors from the air by placing them in the vicinity of the odors. (Maybe not as efficiently as other forms of carbon, as pointed out in the comments of that link, but something nonetheless.)
What is the chemical process by which charcoal can do that, and once the odors are 'gone', where do they 'go'? In the case of 'toxins in gases', what typically is done to them upon 'deodorizing' by the charcoal block - are they collected, neutralised / transformed upon contact, or something else?

The scent molecules are adsorbed by the charcoal.   The higher the surface area of the charcoal the more smell it can absorb

I know, but I'm wondering what chemical process this absorption is :). I've even read that they can be re-used after some 'airing out' so, if that's true, it's interesting to know what the process is! 'Stuff' (including odors and toxins in gases) can't just disappear, something happens to them! I'm wondering what it is in this case.

It is a form of carbon that has very large surface. The odors and impurities are held to the surface with Van der Waals forces. The capability of removing notable amounts of these impurities or odors is thanks to the extraordinarily large surface.

Its simple adsorption. The smells/ chemicals are stuck at the surface, and once you expose the activated charcoal to sunlight for a few hours, it release the smells and is again activated and charged, and good to use.
A good quality, with high adsorbent surface area activated charcoal kit can work good for a period of about 1 year, however you need to keep it in sunlight at-least once a month.

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Is BF3 an electrophile?
Boron has an empty $2p$ orbital.
But is it an electroplhile?
I know it is a Lewis acid.

Lewis acids are by definition electrophiles. Electrophiles love electrons, or negative charge. Boron has an empty 2p orbital and there exists a strong partial positive charge on the boron due to the extremely electronegative fluorine atoms covalently bound to boron. This strong partial positive character, coupled with a vacant orbital, makes BF3 a potent Lewis acid and thus an electrophile.
That being said, note that acidity and basicity are both thermodynamic properties, while electrophilicity and nucleophilicity are both kinetic properties. That is, while you can label a Lewis acid an electrophile, it would be incorrect to say, for example, that because a certain Lewis acid is a "strong" Lewis acid, then it must be a strong electrophile.
Formation Constants; Lewis Acidity


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Proteins and their solubility
I have 2 statements , and i am not sure, whether they are true or false.

Proteins like albumin and haemoglobin are insoluble in water

(in my view , almost all proteins are soluble- so this is a wrong  statement)

Proteins which are soluble in water have complex 3D structure and they are globular molecules. - (in my view I think this is also wrong)

Please explain these statements.

I think 2nd is correct, but have no idea  about 1st. Take a look at image of proteins soluble in water

If 2nd is correct, it sense only globular proteins are soluble? I am not sure ?

All proteins have complex structures. There are three broad classes: globular, membrane and scleroproteins. Globular proteins are generally water soluble while membrane and scleroproteins are not.
Albumins and haemoglobin are globular and water soluble.


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Need a pair of compounds that is similar in formula but different in structure
I would like to explain chemical formulas is not representable in terms of the molecular structure of the compound.
Any good suggestions of examples?

What about carbon? Diamond, graphite, graphene, nanotubes, fullerenes...

The classical example is dimethyl ester $\ce{CH3-O-CH3}$ and etanol $\ce{CH3-CH2-OH}$ pair. They have exactly same brutto-formula, but different structure and very different properties. The first is a gas, the second is a liquid easily reacting with sodium metal.
Carbon is also interesting example. It has two common forms with exactly same chemical formula $\ce{C}$, but very different properties: a very hard and transparent diamond and soft black graphite.

Examples from geology abound, and are probably more interesting to everyday folks.

Titanium dioxide, $\ce{TiO2}$ is commonly found in a rutile form that has a body-centered cubic tetragonal strucutre, but anatase has tetragonal pyramidal crystal structure.  Rutile has very unusual optical properties that anatase does not have; and conversely, anatase enjoys catalytic properties that rutile does not.  This system very well studied by chemists and materials scientists.
Hydrated magnesium silicate, $\ce{Mg3Si2O5(OH)4}$ comes in a bewildering array of polymorphs.  Several (chrysotile, tremolite, anthophyllite, amosite, crocidolite, and actinolite) are collectively known as asbestos.  Their unusual fibrous crystal structure led to their use in a variety of fire-resistant applications until it was realized that inhaling fine shards of these minerals was a major health hazard.  In contrast, lizardite is also a hydrated magnesium silicate with formula $\ce{Mg3Si2O5(OH)4}$, but does not have a fibrous crystal structure and is not hazardous.

Some examples from biochemistry are also very interesting because they matter to folks' everyday lives.

Trans fats like elaidic acid, $\ce{C18H34O2}$ were previously abundant in the food supplies of many Western countries because of partial hydrogenation of plant-derived vegetable oils.  Dietary consumption of trans fats is now believed to be a major health hazard.  However, the cis isomer of elaidic acid is called oleic acid.  It is the major component of olive oil and other plant-derived oils, and is not believed to be hazardous.  It's formula is also $\ce{C18H34O2}$.


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Energetic Placement of Atomic Orbitals in the HCl Molecular Orbital Diagram
How are the 3p orbitals of chlorine lower in energy than the 1s orbital of hydrogen?

I'd say there are two important things to consider here: Firstly, $\ce{Cl}$ is more electronegative than $\ce{H}$. And secondly, as you proceed from the left to the right in any row of the periodic table, the s and p orbitals go down in energy. This is a consequence of the fact that the valence electrons do not screen each other effectively. Thus, the addition of one proton to the nucleus and one electron does not cancel; instead, the valence electrons "feel" an increased nuclear charge.
The electronegativity difference and the reduced screening among the valence electrons seems to be enough to shift the $3\ce{s}$ and even the $3\ce{p}$ orbital of $\ce{Cl}$ below the energy of the $1\ce{s}$ orbital of $\ce{H}$. The experimental state averaged ionization potentials (IP) could serve as a good guide for the placement of starting orbitals in an orbital interaction diagram. For the $\ce{H}$ $1\ce{s}$ orbital you have an IP of $-13.6 \, \mathrm{eV}$. For the $\ce{Cl}$ $3\ce{s}$ and $3\ce{p}$ orbitals you have IP's of $-25.2 \, \mathrm{eV}$ and $-13.7 \, \mathrm{eV}$, respectively.$^{[1]}$ So, this indicates that the $3\ce{p}$ orbital of $\ce{Cl}$ does indeed lie slightly below the $1\ce{s}$ orbital of $\ce{H}$.
Update:
Here is a table of the experimental state averaged ionization potentials, in electron volts, for the main group atoms. However, those $\ce{p}$ atomic orbitals for groups 1 and 2 and for the $\ce{s}$ and $\ce{p}$ orbitals of the sixth row are not experimentally known and hence, calculated values (indicated by an asterisk) have been used.$^{[2,3]}$ The latter include relativistic corrections. The table is taken from the 2nd Edition of the book Orbital Interactions in Chemistry by T. A. Albright, J. K. Burdett, and M.-H. Whangbo.

[1] L. C. Allen, J. Am. Chem. Soc. 1989, 111, 9003 ; L. C. Allen, J. Am. Chem. Soc. 1992, 114, 1510; L. C. Allen, Acc. Chem. Res. 1990, 23, 175.
[2] A. Vela, J. L. Gazquez, J. Phys. Chem. 1988, 92, 5688.
[3] J. P. Desclaux, At. Nucl. Data Tables 1973, 12, 311.

You can also test that theory with the bohr model for hydrogen like atoms. It is nothing really accurate, but it draws the picture quite good: $$E= {1\over 2} m_\mathrm{e} v^2 - {Z k_\mathrm{e} e^2 \over r} = - {Z k_\mathrm{e} e^2 \over 2r}.$$ With $Z$ as nuclear charge and $r$ as the radius, $k_\mathrm{e}$ is a constant and $E$ is the energy: $E\propto -Z$.

Thank you very much. Do you have a table with the averaged ionization potentials (IP)?

@Marko It's your lucky day ;) I added one for the main group elements. But unfortunately, I don't have any data for transition metals.


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How to figure out which phase on a phase diagram will be the most dense?
If I am given a phase diagram, how can I determine which phase will have more density?
What should I consider to solve these kinds of questions?
Does the answer have to do with the area that each phase occupies?  Or the more area a phase occupies, the more the density?

The solid phase is more dense than the liquid phase. The line that separates solid and liquids bends right.
If the solid phase is less dense than the liquid phase, the line that separates solid and liquids bends left.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site.

This could really do with some explanation of the Clapeyron equation, it is technically an answer, but not much of one

Look at the lines on the phase diagram.  If you go to a line and then increase pressure you move to a new phase, the denser phase.  Edit:  Add the below.

Go to the dotted green line.  Now, go straight up (increase in P).  What happened?  You went from solid and liquid being in equilibrium to just liqui.  The liquid is more dense than the solid.

could you be more clear !

The phase which has the higher density is the phase which exists under higher pressure and lower temperature. Thus, in a phase diagram showing pressure as a function of temperature, look for the phase which lies in the upper left region.

Usually the solid phase is the densest. To be sure, look at the slope of the solid-liquid line. If it is decreasing, the liquid phase is denser, if it is increasing, the solid phase is denser. For example, the phase diagram of water has a negative solid-liquid line; the liquid phase of water is denser.
The densest phase exists at the highest pressure and lowest temperature. This comes back to looking at the top left corner of the graph.


text:
Difference between actual position of electron and Radial Distribution Probability
Its known that the radius of maximum probability of 2s orbitals is more than that of 2p orbitals. It means that the maximum probability of finding an electron in an 2s is further away from electron than that of 2p. This means 2p electron are mostly found nearer to atom than the 2s atom. Well! thats totally wrong.
So I want to know the factors due to which 2p orbital lie further to nucleus than the 2s one.

Ok, let's start out by looking at the hydrogenic atom. The wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0
\end{equation}
with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$.
The probability of finding the electron in the hydrogen-like atom, with the
distance $r$ from the nucleus between $r$ and $r + \mathrm{d}r$, with angle $\theta$ between $\theta$ and $\theta + \mathrm{d}\theta$, and with the angle $\varphi$ between $\varphi$ and $\varphi + \mathrm{d}\varphi$ is
\begin{align}
  | \Psi(r, \theta, \varphi) | \, \mathrm{d} \tau = \left[R_{n,\ell}(r) \right]^{2} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} r^{2} \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\varphi \ .
\end{align}
To find the probability $D_{n, \ell}(r) \, \mathrm{d}r$ that the electron is between $r$ and $r + \mathrm{d}r$ regardless of the direction, you integrate over the angles $\theta$ and $\varphi$ to obtain
\begin{align}
  D_{n, \ell}(r) \, \mathrm{d}r &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \underbrace{\int_{0}^{\pi} \! \! \int_{0}^{2\pi} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} \sin \theta  \, \mathrm{d}\theta \, \mathrm{d}\varphi}_{= \, 1} \\
  &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \ .
\end{align}
Since the spherical harmonics are normalized, the value of the double integral
is unity.
The radial distribution function $D_{n, \ell}(r)$ is the probability density for the
electron being in a spherical shell with inner radius r and outer radius $r + \mathrm{d}r$.
For the $2\ce{s}$ and $2\ce{p}$ states, these functions are
\begin{align}
  D_{2, 0}(r)  &= \frac{1}{8} \left( \frac{Z}{a_{0}} \right)^{3} r^{2} \left( 2 - \frac{Z r}{a_{0}} \right)^{2} \mathrm{e}^{-Zr/a_{0}} \\
  D_{2, 1}(r)  &= \frac{1}{24} \left( \frac{Z}{a_{0}} \right)^{5} r^{4}  \mathrm{e}^{-Zr/a_{0}} \ ,
\end{align}
where $a_{0}$ is the Bohr radius.
The graph of those two function can be seen here:

The most probable value $r_{\mathrm{mp}}$  of $r$ is found by setting the derivative of $D_{n, \ell}(r)$ with respect to $r$ equal to zero. From the graph above it can be seen that $r_{\mathrm{mp}}(2\ce{p})$ is smaller than $r_{\mathrm{mp}}(2\ce{s})$ (and if you calculate it you can check that for yourself).
But $r_{\mathrm{mp}}$ is not the quantity you are actually interested in. Because although $r_{\mathrm{mp}}$ gives you the most probable distance of the electron from the nucleus you want to know the average distance $\langle r \rangle$ between the electron and the nucleus as that is the distance you will get when you measure $r$ repeatedly and average over your results.
The radial distribution functions may be used to calculate the expectation values
of functions of the radial variable r. You can get them via
\begin{align}
  \langle r \rangle_{n, \ell}  &= \int_{0}^{\infty} r D_{n, \ell}(r) \, \mathrm{d} r \ .
\end{align}
If you evaluate this integral for the $2\ce{s}$ and $2\ce{p}$ orbitals you get
\begin{align}
  \langle r \rangle_{2\ce{s}} = \langle r \rangle_{2, 0}  &= \frac{6 a_{0}}{ Z } \\
  \langle r \rangle_{2\ce{p}} = \langle r \rangle_{2, 1}  &= \frac{5 a_{0}}{ Z } \ .
\end{align}
You find that the $2\ce{s}$ electrons are on average further away from the nucleus than the $2\ce{p}$ electrons.
So, for the hydrogenic atom your claim that $2\ce{p}$ electrons are further away from the nucleus than the $2\ce{s}$ electrons is wrong. But in real atoms you have more than one electron and those electrons will influence each other. One important effect here is, that the inner electrons (or electron density) screens nuclear charge from the outer electrons. Thus the outer electrons feel a lower effective nuclear charge $Z^{\mathrm{eff}}$ than the inner electrons. The extend of this screening is distance dependent, so $Z^{\mathrm{eff}}
= Z^{\mathrm{eff}} (r)$.
Now, if you look back at the formula for $\langle r \rangle_{n, \ell}$ things will certainly change since $D_{n, \ell}(r)$ will have a different form and the additional $r$-dependence of $Z^{\mathrm{eff}}$ will make things more difficult. Assuming that the general form of $D_{n, \ell}(r)$ will be similar in the real atom, it can be expected that the parts of $D_{n, \ell}(r)$ for small $r$ now have a higher weight in the integral of $\langle r \rangle_{n, \ell}$ than before since these parts of the electron density suffer less from the screening effect. Since $2\ce{s}$ electrons have a small portion of the electron density that lies very close to the nucleus ("penetrates deep into the core region") and $2\ce{p}$ electrons don't this will lead to the observation that $2\ce{p}$ electrons are on average further away from the nucleus than the $2\ce{s}$ electrons.
A more involved explanation of the effects at work in multi-electron atoms can be found at this answer of mine. The important bit starts at the paragraph "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?"

This means 2p electron are mostly found nearer to atom than the 2s atom. Well! thats totally wrong.

Why is this idea totally wrong?
Let's consult a radial probability graph. Here's one for hydrogen.

This graph shows that the 2p electrons have a good probability of getting closer to the nucleus than 2s electrons.
You might be wondering why then s-electrons are lower in energy than p-electrons of the same subshell. Notice the little bump in the pink line on the left end of the graph? The 2s electrons can penetrate closer to the nucleus than the 2p electrons can.

ya! you r right... but again max probability of finding a 2s-electron in further from the nucleus or behind the 2p one... "Maximum Probability" it should have some importance.Acc. to me, most of the time electron would be find in the region of max RPD... And it spends a small time near the nucleus.. :/

Try integrating to find the area under the radial probability curves. In Physical Chemistry by Atkins there is an example of such integration, and the author finds that the area under the 3s curve is larger than the area under the 3p curve at a set point from the radius.


text:
Oxide Anions in Redox
Consider the following unbalanced redox equation:
$\ce{NO_2^- +e^- -> NO3^-}$
This occurs in acidic media.
My professor would note that an negative two oxidation state oxygen "disappears" from left to right.
Oxide anions are strong bases, and since this process occurs in acidic media, and since an oxide anion has "disappeared," it must have been consumed by the strongest acid present - hydronium ion. Specifically, two equivalents of hydronium ion. These two equivalents of hydronium ion react with oxide anion to produce three waters.
Thus we balance the equation as such.
$\ce{NO_2^- +2e^- +2H_3O^+ -> NO3^- + 3H_2O}$
Now, what is the validity of this method, given that redox reactions likely don't involve an oxide anion intermediate? I can find no mention of any $\ce{O^2-}$ intermediates in redox literature.
If that is the case, how can this method still yield a redox equation that is valid in terms of mass and charge? Is there some likely intermediate that is close enough in character to the oxide anion?

Your equation is wrong. In acidic medium, water is the one which gives an oxygen to the nitrite. And that should be an oxidation half-reaction (as nitrite oxidizes to nitrate), but you wrote a reduction where two electrons are spent, not released.


text:
Peroxide Linkages
Why are peroxide linkages generally weak?
Is it due to Coulombic repulsions? I can't see how placing two oxygens with high degrees of partial negative character next to each other is a good thing in terms of bond strength. I suppose a way to test this hypothesis is to compare the relative stabilities of HOOH and FOOF, as the hydrogens are electron donating while the fluorines are electron withdrawing.
Or does this have to do with promotion of a bonding electron to an antibonding orbital? I know that peroxide linkages are easy to break with application of light/heat ... this suggests to me that we can easily promote an electron to anti-bonding status and weaken the O-O link. Still, this does not answer why it might be easy to promote an electron to antibonding status.

Dissenter, I've been thinking about this question since you posted it yesterday.  My first thought was that this question has a straightforward answer based on lone pair repulsions.  However, I ran into problems when I tried to correlate this hypothesis with physical properties.  For example, look at the following Table.  As we move from hydrogen peroxide with 2 lone pairs on each of the adjacent oxygens to hydrazine with
\begin{array}{|c|c|c|} \hline
%Header
\text{Molecule} &
\text{Bond Strength}~/ (\pu{kcal/mol}) &
\text{Bond~Length}~/ \pu{pm}\\ \hline
\ce{H3C-CH3} & 85 & 154\\ \hline
\ce{H2N-NH2} & 60 & 145 \\ \hline
\ce{HO-OH} & 51 & 149 \\ \hline
\end{array}
one lone pair on each of the adjacent nitrogens, the bond strength increases and the bond length decreases as one would expect if lone pair repulsions were the dominant controlling factor.  However, as we move further on to ethane with no lone pairs, the bond length increases.  This suggests that if lone pair repulsions are playing a role, then, at least with regard to bond length, some other variable is also playing a significant role and obscuring the "expected" trend.
Next I examined an an approach based on MO analysis. Let's examine the MO diagrams for $\ce{HO-H, H2N-H}$ and $\ce{H3C-H}$, they are pictured below.

We see that water has a (roughly) p-hybridized lone pair (the $\ce{1b_1}$ orbital) that is non-bonding.  For ammonia, the lone pair is in an (roughly) $\ce{sp^3}$ hybridized orbital (the $\ce{2a_1}$ orbital) that is lower in energy than the non-bonding atomic p-orbitals.  In the case of methane, there are no lone pairs and all MOs are bonding (lower in energy than the p AOs).
Now, imagine what would happen to these MOs if we removed a hydrogen atom from each of them and  remixed the resultant MOs with themselves (e.g. dimerize $\ce{HO{.},~ H2N{.}~ and~ H3C{.}}$).  My (simple) analysis suggests the in forming hydrogen peroxide the $\ce{1b_1}$ orbital would mix and split to form 2 new MOs, one bonding and one (slightly more) antibonding, both filled with 2 electrons, for a net destabilization relative to water.  Similarly when we form hydrazine, two $\ce{2a_1}$ orbitals will mix to produce two new MOs, one lower in energy than a non-bonding p AO, and the other either slightly lower or slightly higher (depending on the degree of splitting) than a non-bonding p AO.  Energetically speaking, the hydrazine case shouldn't be as destabilized as hydrogen peroxide.  Finally, when we form ethane, all orbitals will be more stable than a non-bonding p AO.
This analysis, based on the relative energetic positioning of the lone pairs in these molecules, appears to be in accord with the observed bond strengths of these molecules.

The effect is not limited to peroxide linkages. Let's look at energy of single bonds.
\begin{array}{|c|c|c|c|} \hline
%header
\text{Bond} &
\text{Energy}~/ \pu{(kJ/mol)} &
\text{Bond} &
\text{Energy}~/ \pu{(kJ/mol)} \\
\hline
\ce{C-C} & 347 & \ce{Si-Si} & 340 \\ \hline
\ce{N-N} & 160 & \ce{P-P} & 201 \\ \hline
\ce{O-O} & 146 & \ce{S-S} & 206 \\ \hline
\ce{F-F} & 154 & \ce{Cl-Cl} & 240 \\ \hline
\end{array}
We can clearly see, that the effect manifests between $\ce{C/Si}$ and $\ce{N/P}$, strengthens at $\ce{O}$ and somewhat reduces at $\ce{F/Cl}$. How $\ce{C-C}$ and $\ce{N-N}$ bonds differ? The answer is simple: in $\ce{N-N}$ bonds both atoms have unbound electron pairs. So, the reason for the effect is interaction of electron pairs.
Now, rationalization for how the interaction affects the bond energy is more open to debates. It is well known, that in VSEPR electron pairs take more place than atoms, so it may be said to be purely steric. It is known from MO theory that antibonding orbitals unbond a bit more than bonding bond, so it may be attributed for details of MO structure. Finally, according to Pauli principle, electrons cannot be in same 'place', so electrons with same spin in two electron pairs are repelled from each other, forcing them from each other.
How this effect is partially compensated in $\ce{F-F}$ and $\ce{Cl-Cl}$ bonds is more interesting, I would say that it is because of their smaller orbitals, resulting in less effective π-interactions and stronger p-p σ bonding


text:
Acid catalysed mechanism of the addition of alcohols to alkenes

I believe that this mechanism is false.
The protonated ether would in my opinion give a proton to methanol rather than $\ce{H2SO4}$ because the protonated $\ce{CH3OH}$ is more stable than $\ce{H2SO4}$.
Can someone confirm my thought?

pKa of H2SO4 is -9
pKa of CH3OH2+ is -2.5

Dissenter's analysis is pretty much on target. The one thing I'd add, you are correct that the proton is probably given to back to methanol rather than bisulfate because there is probably a drop of acid and a lot of methanol.  But thermodynamics will win out and eventually the proton will be transferred from protonated methanol to the bisulfate anion to regenerate the acid which, as Dissenter noted, is a catalyst - not to be consumed.

Protonating HSO4- in the presence of methanol is thermodynamically favorable?

Ron, which thermodynamic property are you stating? From my current perception of things, I think the protonated methanol would NOT transfer the proton to make H2SO4.

Isn't HSO4- a stronger base than neutral CH3OH?  How would you regenerate the catalyst H2SO4?

pKa of HOSO3- is ~1; pKa of CH3OH is ~16. This suggests to me that methanol is a better base.

I agree that HSO4- is a stronger acid, but is it also a stronger base than methanol? How do you regenerate the catalyst?

I don't know, could you please clarify? If HSO4- is a stronger acid than methanol that means HSO4- is a weaker base than methanol.

I just fall back to show me how you regenerate the catalyst?

What is the catalyst? I don't know if it's really sulfuric acid.

The figure says that the catalyst is H2SO4.  HSO4- has another ionizable proton so it is an acid, but what its pKb?

Why pKb if we're in methanol solvent? I was only using pKb as a rule of thumb; as a tool to rank the basicities on a relative basis. Also yes the figure says sulfuric acid is the catalyst but I don't necessarily agree.

HSO4- is an acid and a base

It is important to realize, that $\ce{H2SO4}$ is not the catalytically active species. It is also important to know that sulfuric acid always comes with water. The catalytic species is therefore either $\ce{H3+O}$ or $\ce{H2+OMe}$ as these are present in equilibrium. The scheme is only a sequence of reaction steps, it is no reaction mechanism as it completely lacks transition sates. (It is also omitted that the product has to be reprocessed in aqueous solution.)

What do you mean reprocessed in aqueous solution? And what if the problem meant liquid/pure sulfuric acid?

@Dissenter (If you are referring to a comment, then you should tag the person with @ so that s/he gets the message.) A reaction never stops with what is proposed with by the 'reaction mechanism', the product has usually to be quenched or otherwise reprocessed and after that purified. Pure sulfuric acid is almost never used since it would decompose almost any organic compound straight away. The autoprotolysis is even stronger in $\ce{H2SO4}$: $$\ce{2H2SO4 <=> H3+O + [HS2O7]^{-}}$$ It always comes with water.

@Martin what is the auto ionization constant for sulfuric acid?

@Dissenter http://en.wikipedia.org/wiki/Sulfuric_acid#Polarity_and_conductivity

The protonated ether would in my opinion give a proton to methanol
  rather than H2SO4

I don't think sulfuric acid is gaining any protons here. You mean hydrogen sulfate ion, correct?

Can someone confirm my thought?

The mechanism is acid-catalyzed dehydration, so the acid must be regenerated. Catalysts are by definition not consumed over the course of a reaction. Now, whether the acid is actually sulfuric acid is debatable, and in my mind, unlikely, given the leveling effect and that the solvent is methanol.
The leveling effect states that the strongest acid that can exist in high concentration in a particular solvent is the conjugate acid of the solvent. The conjugate acid of methanol has a pKa of ~ -2.5, while the pKa of sulfuric acid is ~ -9. Thus, sulfuric acid can't exist in high concentration in methanol. Sulfuric acid would immediately protonate the strongest base in the system. Therefore, it appears to me that the acting acid is not sulfuric acid. Perhaps it is protonated methanol ion.
Or perhaps it is the case that sulfuric acid reacts directly with the alkene. Whether sulfuric acid would be regenerated in this particular solvent, however, is questionable at best, given sulfuric acid's extreme thermodynamic instability in methanol. If it were regenerated, then the sulfuric acid would again protonate the strongest base in the system.
Of course, thermodynamic considerations are important, but so are kinetic considerations. Do remember that there is likely a lot more solvent than there is sulfuric acid and substrate (the alkene). So it seems most likely that the acting acid is protonated methanol - it is a lot more likely for sulfuric acid to bump into a methanol molecule rather than an alkene molecule.

Sorry for the mistake. Yes, I meant HSO4-

Can you explain why the hydrogen sulfate ion would attack instead of methanol?

That's a good question. What's the solvent?

My textbook oversimplifies again :9

I think so; is this Wade? This mechanism doesn't seem to make too much sense. We know that protonated methanol is a weaker acid than sulfuric acid. And we know from the leveling effect that the strongest acid that can exist in high conc. in a solvent system is the conjugate acid of the solvent. This suggests that the actual acid isn't sulfuric acid but rather hydrogen sulfate ion. Or protonated methanol.

I'll upload the question.

http://imgur.com/fn2yxyU
Solomon's, not Wade's.

I think matters have been oversimplified.

Do you agree that in case where methanol is the solvent, it would be CH3OH that would attack? I could discuss it with my teacher and come back. But I think we make more sense than the book.

You mean would methanol do the deprotonation?

I read your edited answer. We seem to agree. :)

I mean it might be the case that the alkene gets the proton from sulfuric acid before sulfuric acid protonates the solvent. I don't know. I'm just speaking from a thermodynamic standpoint.

I believe the sulfuric acid actually attacks the alkene first. This would make more sense because H-OR are more acidic than alkenes. So, adding a proton would be easier on an alkene.

Hey looks like we agree again.

Great, I'll check your answer. I think we made our point ;)

Weird: look at the following problem from the same textbook.
http://imgur.com/dcqpZqR

Which problem? Can you show me?

They show the protonated methanol.

I don't quite agree with that either because it should be the alkene getting protonated, not the methanol. In any case at least the end product is the same regardless of which pathway taken!

Very confusing!

do you know that the triple equality sign showing the hydride shift means in the last uploaded picture?

It could be the fact that there is a whole lot more solvent than there is substrate (alkene) so the methanol could indeed be protonated to a significant degree.

The three lines just means "equivalent" in chemistry. The triple lines are not indicating a hydride shift; it is just indicating that the two pentagons are the same (one is just purposely distorted as to make clear the hydride shift; the other has been adjusted to look pretty when printed).

@yolo123 When it comes to reaction mechanisms you have to understand, that these are only the most likely steps that are going to happen. Also the depiction of most reaction mechanisms is only a reaction sequence. A mechanism has always to include the transition states. At the same time, there are many reactions happening simultaneously, and there are also quite many equilibria involved. A mechanism is never as simple as it is depicted in most organic chemistry textbooks. It is only a guideline to reactivity of certain species.

The short version: You are correct, the schematic drawing is not the correct mechanism and it does also not provide the most likely path. For a training exercise it is close enough though. It shows the most important steps and it conserves stoichiometry.

The depicted scheme does not show a reaction mechanism in the sense of the definition given in the goldbook. A reaction mechanism always also has to include transition states, that lead to the respective products. It is rather a reaction sequence, that shows the most important steps. In this sense, the stoichiometry has to be conserved and in this sense it is as correct as it could be.
Noteworthy is, that the catalytically active species is not sulfuric acid, but instead a mixture of hydronium and protonated methanol, $\ce{H3+O/H2+OCH3}$. Sulfuric acid is such a strong acid, that it is capable of protonating any solvent that has a lone pair, even itself via autoprotolysis (in concentrated $98\%$ solution). $$\ce{2H2SO4 <=> [H3SO4]+ + [HSO4]- <=> H3+O + [HS2O7]- <=> H2O + H2S2O7}$$
The use of concentrated sulfuric in organic synthesis is usually not advisable, since its dehydration property can quickly decompose many organic compounds that contain hydroxyl groups. The use of diluted sulfuric acid (either in water or alcohols) is therefore much more common.
It is only logical, that sulfuric acid will protonate methanol, which can be considered a preceding equilibrium. $$\ce{H2SO4 + HOCH3 <=> H2+OCH3 + [HSO4]- }$$
Since methanol is also the solvent and the acid is only available in catalytic amounts, it is feasible to assume, that the equilibrium lies entirely on the right.
It is also important, that the last step must be also in equilibrium, since it is only the protonation/deprotonation of an ether. Here again we have much more methanol than product, so that the equilibrium will most likely be entirely on the product side.
A more descriptive mechanism would be the following:

For simplicity and sufficiently accurate bond formation and breaking is indicated by arrows, omitting the transition states. Resulting a slightly simplified picture. However, the preceding equilibrium - as you noted - should not be omitted. This formulation of the mechanism is particularly important for many simple organic reactions, as it focusses entirely on the reactants and products (and to a minor extent on the reactive intermediates).

When it comes to more complex reactions, e.g. with asymmetric catalysts it is often much better to include the transition states.
From this it can also be seen, that the choice of sulfuric acid is somewhat arbitrary, as most acids would do the same. However, sulfuric acid is often used, since bisulfate and sulfate are very weak nucleophiles.

Additional note
In all schemes only the most likely path is shown. Any possible side reactions, like concerted [2+2] cycloaddition, have been omitted.

Why hydronium ion? Where does that come from? I understand that we're not dealing with pure sulfuric acid but your equilibria do not show where the water can come from.

@Dissenter, even assuming the reagents were classified as anhydrous, the autoprotolysis and related self-ionization equilibria (which Martin described) of sulfuric acid would result in a hodgepodge of species. Greenwood & Earnshaw note the following species present in pure sulfuric acid (in order of decreasing abundance, with $\ce{H2SO4}$ itself being the solvent): $\ce{HSO4^{-}, H3SO4^{+}, H3O^{+}, HS2O7^{-}, H2S2O7, H2O}$. Of course, when added in catalytic quantities, they won't be the dominant species.


text:
How to determine the pressure to output a gas to achieve a particular gas flow rate?
I have a 420-liter (14.9 cubic ft) helium cylinder and a pressure regulator with a gauge supporting up to 160 psi. My requirement is to ensure a flow of 10 to 15 liters / min. How do I determine the pressure setting on my pressure regulator to fulfill my flow rate requirement?

Well, since the pressure in the tank will decrease with usage, you can't ultimately do what you seem to want to do.
At a given backing pressure, there are equations to determine the orafice size to get a flow rate - google 'choked flow'. Downstream of the regulator, your orafice should be good for quite some time.
For an experimental setup, one would use either a mass flow controller (MFC) to actively control the flow rate, or (much cheaper) one of those little flow indicators with the ball floating in the He (air) stream. Note the second option here is basically an adjustable orafice with integrated flow indicator.


text:
Is an SN1 reaction an elementary reaction/one step reaction?
I need some help. From what I understand, elementary reactions are reactions which only has one transition state. Another definition that I have come across is that elementary reactions obey the rate law.
I believe that $S_N1$ are NOT elementary reactions because they proceed through more than one transition state. However I am not so sure about the other definition (the rate law one). Could someone explain to me what exactly an elementary reaction (or a one step reaction) is? And can someone explain to me if an $S_N1$ reaction is an elementary reaction or not.

From http://goldbook.iupac.org/E02035.html:

A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.

So you're right that elementary reactions occur in a single step/have a single transition state. I'm not sure what you mean by "obey the rate law", but necessarily, a reaction that has only a single transition state is first order in each reactant. Because there is only a single step, the concentration of every molecule involved in that step effects the rate.
In practice, because the probability of three molecules colliding together in a way that can cause a reaction is very low, all elementary reactions involve one or two molecules. A single molecule reaction is just a dissociation, e.g. $A_2 \rightarrow A + A$, with a first order rate in A and an overall order of one. A two molecule reaction could be something like $A+B\rightarrow AB$, with still a first order rate in each of the reactants for an overall order of two.
In your question, you've identified that SN1 reactions have multiple transition states, so SN1 cannot be an elementary reaction. In fact, the 1 in SN1 means that the overall order of the reaction is one, but since it involves two reactants, the order would be two for a single step reaction. Naturally, an SN2 reaction has an overall order of two and is an elementary reaction, proceeding through a single transition state where a bond between the incoming nucleophile and the carbon centre is formed at the same time as the leaving group's bond is broken.

You are absolutely correct. In SN1 we can observe coupled elementary reactions, the two you already described. Since the dissociation in SN1 is so much slower than the following association, we observe overall first order reaction. Obeying the rate law simply means, that a fractional order is not possible. Elementary reactions are alway first or second order reactions.


text:
Normal and heavy water mixing
This question seems silly, but this just came in my mind while i was learning about hydrogen.
Means will there be any reaction when $\ce{H2O}$ will be mixed with $\ce{D2O}$?

Yes, a reaction known as "hydrogen exchange" will take place.  When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture.  Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.

Statistically, you would get that ratio, but unfortunately the $\Delta H_f$ for $D_2O$ is about 9 kJ/mol different from $H_2O$ so that the equilibrium is not that easy to determine. http://www1.lsbu.ac.uk/water/data.html

@ LDC3 Excellent point LDC3, I'll edit my answer.

@LDC3 the relevant quantity is not the difference vetween D2O and H2O, but the difference between their average and HDO. We would expect HDO to fall nearly halfway between H2O and D2O in enthalpy and free energybif formation, so the statistical model would likely be accurate in any fluid sample.


text:
Mercury brain half-life
If have read conflicting reports on the half-life of mercury in the brain. I would like to hear some estimates from this community.
I am also wondering how long chelation therapy would take to remove mercury from the brain.

I don't think chelation therapy will remove mercury from the brain. Chelators are good at grabbing inorganic mercury, but the brain usually takes up organic mercury. Chelation therapy for mercury poisoning tends to be reserved for acute mercury exposure, not chronic exposure, as the chelators often have side effects. Why are you concerned with mercury in your brain?

I have amalgam fillings and I am concerned that the mercury vapour they release - 80% of which is taken up by the lungs - has crossed into my brain and is causing mild symptoms such as concentration problems.

Since you're concerned about mercury from fillings, you may have already taken a look at the wikipedia article about dental amalgam toxicity. It would be worth reading through some of the sources listed, too.
I assume you're an adult, I don't know if you're male or female. Children are more sensitive to mercury than adults because their brains are actively developing. If you're female, you may become pregnant at some point and there would be some concern about fetal mercury exposure.
However, according to a 2003 WHO report, the amount of mercury ingested from dental amalgam is low, with exposures estimated from 1ug per day to 12.5ug per day, but the vast majority of people are exposed to less than 5ug per day. The mercury from fillings is elemental mercury, which is poorly absorbed, less than 0.01% is taken up, with smaller amounts reaching the brain.
Removing your existing fillings probably isn't worth it, as the removal process would expose you to more mercury than just letting the filling sit there.
I don't know if this will make you feel better or worse, but coal fired power plants are probably a larger source of mercury exposure than your fillings. The mercury from the coal plants is methylmercury, which is much more efficiently absorbed than elemental mercury. This bioaccumulates in the environment, so that fish have relatively high levels of mercury in them.
As far as your perceived concentration problems, we really can't say if it's caused by the mercury or not. If it concerns you, you should see a doctor. And it may be more efficient to treat the symptoms through ADD medication than trying to remove mercury.

I have read reports of amalgam fillings producing as much as 50 µg per day of vapor with a 80% uptake through the lungs. This is what worries me. There is evidence that the half-life of mercury in the brain is many years.

It would be better if you put this as comment

This is fine as an answer as long as you summarize the evidence that's in the paper.  This question overall is a bit borderline for our scope, though, since it's more of a health issue than something that we can answer from a chemical safety perspective.

If there is no information on the chelation aspect in the publication, this is probably best left as a comment as several users have indicated.

The study below (2014) suggest the half life in the brain (where it counts for mercury poisoning), may be much longer than thought: on the order of years or decades.
The retention time of inorganic mercury in the brain--a systematic review of the evidence. (https://www.ncbi.nlm.nih.gov/pubmed/24368178)
Also: article about this study (http://www.speciation.net/News/A-new--study-finds-Inorganic-mercury-stays-in-the-brain-for-years-if-not-decades-;~/2013/12/29/7042.html).

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text:
Why do negative oxidation states not extend to -8?
There are oxidation states from -4 to +8
Why do the negative oxidation states not go all the way down to -8? I mean if an atom is hypervalent meaning that it can have more than an octet than that means it can have potentially 8 bonds(counting a double bond as 2 bonds and a triple bond as 3 bonds respectively) to other atoms more electronegative than itself and thus Xenon can have a +8 oxidation state and some metals can as well.
Doesn't that also mean that a hypervalent atom can have 8 bonds to other atoms less electronegative than itself and thus have a -8 oxidation state?

Actually it's now -5 to +9.

Why do the negative oxidation states not go all the way down to -8?

Let's think about what we would need for this to occur - i.e. an oxidation state of -8.
We would need a central atom that has considerable electronegativity - one that is at least more electronegative than its attached atoms. This already excludes numerous molecules; most molecules' central atoms are, on a relative basis, more electropositive rather than electronegative.
In addition, this atom would need to have one of the following configurations; "allocated" refers to the assignment of oxidation state - since oxidation states are assigned on the premise that all bonding is ionic - the more electronegative element in a bonding configuration is "allocated" all the bonding electrons in the calculation of oxidation states. I.e. in H2O, the oxygen would be allocated both pairs of bonding electrons in the H-O bonds, and oxygen would then "have" 8 electrons, and its oxidation state would be -2.
1 valence electron and be "allocated" 9 electrons
2 valence electrons and be "allocated" 10 electrons
3 valence electrons and be "allocated" 11 electrons,
and on the other extreme:
6 valence electrons and be "allocated" 14 electrons,
7 valence electrons and be "allocated" 15 electrons,
8 valence electrons and be "allocated" 16 electrons.
As we can see, the first set of criteria are basically incompatible. Elements only having 1 or 2 valence electrons are going to be small (on a relative basis). These elements would also much rather be oxidized than reduced. Oxidation allows these elements to easily become isoelectronic with a noble gas.
The other set of criteria for a negative 8 oxidation state is also hard to fulfill. None of the group 6 elements except the possibility of sulfur and elements below it are hypervalent - i.e. they can accommodate more than 8 valence electrons. Even if we consider sulfur as able to accommodate 12 valence electrons (as some depictions of sulfuric acid suggest) we must remember that sulfur is not very electronegative and unlikely to be allocated any electrons.

Sulfur is more electronegative than quite a few elements, especially metals, but also the oxygen group elements below sulfur, Nitrogen group elements below N, Carbon group elements, Boron group elements, Xenon, Iodine, and Astatine.

Right, but I can't think of any molecules or ions in which sulfur is the central element and metals are ligands, the other way around yes. Such as silver sulfide.

Doesn't the possibility of a -8 oxidation state only require that a hypervalent atom has 8 bonds(covalent(with a double bond counted as 2 and a triple bond counted as 3 respectively) or ionic) to 8 atoms less electronegative than itself, not that it is the central atom?

Well, you also have to consider how many electrons the element has in its valence.

right but it has to have 8 bonds to something less electronegative than itself(with a double bond counted as 2 and a triple bond as 3 respectively) whether they are covalent, ionic, or both(even though it is rare to have both types of bonds in the same compound unless you are talking about a hydroxide or something along those lines.)


text:
How hazardous are peroxide-able ethers if stored properly?
This question is directed towards laboratory scientists and technicians who actually deal with ethers that can develop peroxides upon concentration (e.g. diethyl ether, THF, dioxane). I am already very familiar with the literature, the normal precautions (dispose of after 1 year, test for peroxides, etc.), how to remove peroxides (shake with ferrous salt or metabisulfate).
My question is how hazardous are peroxides in reality if stored properly. If a peroxide-able ether upon concentration is stored in a cool dark cabinet, is rarely opened, has an inhibitor (BHT or ethanol), and argon is used above the headspace, it is likely that the ether can be used for longer than a year?
Again, I need the input of an actual practitioner. Simply quoting written material will not help me, as I have consulted LOTS of written material and I believe I understand the theoretical issues. I am looking for practical experience.  I have heard informally that some ethers are available in European countries as OTC solvents, and so it leads me to think that the standard warnings may be too severe. Thanks!

I believe, that when I was still a student in university, we kept and used DEE and THF for more than a year, especially when dried and kept in inert atmosphere. I think most safety concerns apply to quantity. There are probably other reasons why you should not use solvents after a year or so...

One of may previous labs used some diethyl ester and THF, and no accidents occurred. As long, as they are properly stored (no direct sunlight, tightly sealed bottle, a tiny bit of $\ce{KOH}$ on the bottom of the bottle) they are reasonably safe. Of course, I would be a bit ... concerned in case of work with diisopropyl ester, but not so much.

@permeakra: Thanks! Can you answer these two questions I have: 1) do I need to pre-dry (I would assume yes with something not so water reactive), and 2) how much to use (I would assume you don't need much)? You seem to agree with my answer to 2). And I have no diisopropyl ether and agree that it is a higher order of storage danger!

DEE and THF dissolve quite a bit of water, so boiling and then distillation of marketed solvent  over excess of sodium or potassium hydroxide is common. In case of extra large quantities, another pre-drying stage may be economically justified, but in common labs it is usually dried with alkali hydroxide directly. The amount of hydroxide required for drying is very dependent on quality of the raw solvent, so no suggestion here. I would use roughly 1-2 table spoons per liter unless it is extremely wet.

As far as I remember, it was a small amount. Again, 1 table spoon per liter should do.

@permeakra: Thanks again. But my concern is that using such a water reactive drying agent first may produce a fire hazard. Might it not make sense to pre-dry with something less water reactive? I have a list of such agents.

@trb456 DEE and THF are fire hazards on their own.  Both DEE and THF have flash points below zero, bp at 35 and 66 Celsium respectively and autoignition temperature at roughly 160 and 320 C respectively. Given that, room temperature is already above flash point for both of them, but since they boil well below autoignition temperature, they are not likely to ignite as far as you heat liquid solvents. Even if heating source (i.e. energetic reaction) will have high temperature, the temperature of the vapors will be equal to bp, which is not enough to autoignite. Granted, THF is safer.

@trb456 after some thoughts I wouldn't use sodium metal on large quantities without pre-drying or testing, though.

Peroxides are no fun.  There have been explosions when ether solutions were concentrated to dryness.  What you really want to do is store your ether over potassium hydroxide - hydroperoxides are slightly acidic, and the base forms hydroperoxide salts and pulls them out of solution.
Also, if you keep solvents for more than a year you clearly aren't turning them over fast enough.  Keeping flammable material that you don't need just increases the danger in case of fire.

Thanks for this. I've tried looking into this but could not find hard procedures. Two questions I have: 1) do I need to pre-dry (I would assume yes with something not so water reactive), and 2) how much to use (I would assume you don't need much)?

Having accepted Abel Friedman's very sound answer, I wanted to append a coda of sorts. After posting this question, I tested all my opened peroxide-able ethers (diethyl, THF), and there were no measurable peroxides despite being well past the 1 year mark. I double-checked that the test was valid with hydrogen peroxide, which produced a strong positive for peroxides.
I conclude that the standard recommendations for peroxides are similar to the warnings on MSDS sheets, in that they err on the side of extreme caution. As Abel Friedman says, peroxides are no fun! So testing and careful storage (full airtight containers, argon in the headspace, dark cool location, inhibitors) are essential. But if you do these things, the "store-by" dates do seem to be excessive.
As to the valid point that I might not be turning over my solvents fast enough, I realized that the converse could be true in a working lab. That is, solvents are used frequently, and thus are left out in the light and opened frequently. If these are not used quickly enough, there is a real risk of peroxides. But it would seem that dark airtight storage (and inhibitors!) does retard peroxide formation well beyond the recommended limits.


text:
How to figure out if the reaction proceeds via SN1 or SN2 mechanism?
I have been assigned with the task of figuring out whether a reaction is $\ce{S_{N}1}$ or $\ce{S_{N}2}$. The information that I have is, that the yield is dependent on solvent polarity. More specifically, the higher the solvent polarity, the higher the yield.
The problem is that both $\ce{S_{N}1}$ or $\ce{S_{N}2}$ products are the same. Both result in charge separation (therefore both are favored by polar solvents).
I proposed to repeat the experiment. I proposed to make the site of substitution chiral. That way it is possible to check for $\ce{S_{N}2}$ inversion.
Another proposal was to use the fact that the rates of $\ce{S_{N}2}$ and $\ce{S_{N}1}$ were different. However, I am not sure how to use this fact and perform an experiment.
What should I do in this situation? The people that I am working for do not want to repeat the experiment at all. Using the rates approach seems to be the best option since there won't be a need to obtain new materials. But I have no idea on how to do this. What are your suggestions? How can I figure out whether the reaction proceeded through $\ce{S_{N}1}$ or $\ce{S_{N}2}$ by using rates approach?

In praxis there is no such thing as a purely $\ce{S_{N}1}$ or $\ce{S_{N}2}$ mechanism, therefore it is hard to figure it out in the first place. Why do you need to know this? Is it of any implication for the continuance of your study?

As the other answers state, kinetic analysis is one way to go, especially if you want to avoid the hassle of inserting a chiral label. If you're interested at looking at the relative rates, your best bet for a practical technique is probably NMR, depending on your reactants and solvent, so long as the reaction isn't too fast. It's easy enough to get 30 s resolution for arrayed proton NMR experiments.
ESI-MS may also be an option if your compounds ionize well and you're using a compatible solvent. A reaction in a 500 µL syringe should be good for an hour or two of measurements.
Depending on your exact reaction, optical techniques could potentially be used, as well. If you're lucky, it may be as simple as absorbance measurements.

As you have the information that the yield depends on solvent polarity. It is the key point to be considered in analysis of type of reaction. In case of a $\ce{S_{N}1}$ reaction, the polar solvent favors the reaction. $\ce{S_{N}1}$ is a two stepped reaction, in which the 1st step is carbocation formation, such that there appears the dipole dipole interaction between the carbocation and polar solvent which favors the reaction.
Whereas in the case of $\ce{S_{N}2}$ reaction, the dipole dipole interaction develops between polar solvent and the nucleophile rather than with intermediate product. It is because the charge is dispersed in two centres in intermediate while charge is centered in single center in nucleophile. This deactivates the $\ce{S_{N}2}$ reaction and the reaction cannot proceed forward smoothly.
Since you have another specific information that the higher the solvent polarity, the higher the yield. Since higher solvent polarity favors the $\ce{S_{N}1}$ reaction, the higher will be the yield through a $\ce{S_{N}1}$ mechanism. So the reaction with tendency with higher solvent polarity and higher yield is a $\ce{S_{N}1}$ reaction.

Good argument. In addition, if the leaving group is an anion (which is likely) then this will be also stabilised by a more polar solvent, as it is favouring the dissociation of carbocation and leaving group.

Yeah. The polar solvent makes the carbocation more stable by creating dipole dipole interaction which will  favor the formation of carbocation and as the formation of carbocation is the rate determining step , use of polar solvent also increases the rate of SN1 reaction.

You should be very careful about changing the reagents or solvent, because then you are looking at a different reaction than the one proposed which may go by a different mechanism.
Your idea of looking at rates is a good one. Since an SN2 reaction depends on the concentration of nucleophile, while SN1 does not, set up two experiments exactly the same (same concentration of electrophile, same solvent, same temperature, etc) but double the amount of nucleophile in one of the experiments. If the reaction rates differ (by a factor of two), then the mechanism is SN2. If the rates are the same, then the mechanism is SN1.
Another method that chemists use to learn about reaction mechanisms is by isotopic labeling. Depending on the exact electrophilic substrate, you could consider labeling a proton as deuterium (or other atom as necessary), rendering the substrate chiral. Then you could take advantage of the difference in stereochemical outcomes of SN1 and SN2 mechanisms. However, whether this can be utilized depends on the exact substrates of the reaction being considered.

Another proposal was to use the fact that the rates of SN2 and SN1
  were different.

SN2, as its name suggests, is a bimolecular mechanism and its rate depends on not just substrate concentration but also nucleophile concentration. SN1, however, is solely dependent on substrate concentration, as carbocation formation has a high activation energy barrier, and overcoming this barrier takes time.

I proposed to repeat the experiment. I proposed to make the site of
  substitution chiral.

How would the results of this proposed experiment apply to the results of your previous experiment with the achiral substrate?

The information that I have is that the yield is dependent on solvent
  polarity. More specifically, the higher the solvent polarity, the
  higher the yield. The problem is that both SN1 or SN2 products are the
  SAME. Both result in charge separation (therefore both are favored by
  polar solvents)

Well, try thinking about the intermediates, and try thinking about the nucleophile. Nucleophiles can be aided or hindered by the solvent.

How would the results of this proposed experiment apply to the results of your previous experiment with the achiral substrate? - In SN2 the stereochemistry is inverted, in SN1 a racemic mixture is to be expected. Based on %ee you can propose the mechanism to be more like SN1 or SN2.


text:
Explaining Ionization to kids
I have a quest to explain a physical, chemical theory to kids of age 8-12.
The topic is "Ionization". We all understand from early experiments of Physics and Chemistry (the Millikan oil drop experiment) how a neutral atom in gas becomes a charged atom where radiation knocking the electrons out of the neutral atom.
Here's the story line I came up with. Can you guys please help me to make this a success and correct the wrong impressions, expressions and examples in the narrative?
I am open for any better examples.

There's a high-class (=high voltage) shopping mall area (=radio activity area metaphorically) in the country I am living right now. One particular shopping mall called ION. If you enter that shopping mall, you will shop and get excited (=charged) and spend your pocket money (=losing electrons) by the temptation (=radiation) from fancy shoppers around. Shopping means getting exposed to radiation.
Some shoppers are highly influential (=alpha rays with +2 charge) and there's a high tendency you will spend money (=lose electrons). e.g. Movie theater. You get excited even looking at it from far.
Some other shoppers are less influential (=beta rays with -1 charge). e.g. Junk food, a piece of cake you like to eat. You have to go near to see the food and then spend your money on it. The excitement of eating the food lasts for shorter period compared to the movie you watched. So the impact from movie is higher. You will talk about it in school for a week or two. Unlike the food you had and forgotten by the next meal.
The last type of shoppers are least influential (=gamma with 0 charge). e.g. Manga shop. You don't find it as exciting as watching a moving or eating favourite food. But you may go there as a matter of need (for a manga copy) or peer-pressure from your friends to buy manga and spend your money. You are kinda forced to get excited (=charged) to spend a buck on it.
So it's quite dangerous for you (=neutral atom) to enter or go near a shopping area (= radiation area) where you will be losing your money and get excited for nothing...

I do not have a very good feeling about this approach. Anyone that can make sense of your shopping mall explanation surely is capable of understanding ionization itself.

@iad22agp sorry I don't follow your comment mate. Do you mean it's suitable for the kids or not? If shopping mall-metaphor is hard for you to comprehend, I don't think the kids would do...what do you suggest?

If you enter that shopping mall, you will shop and get charged (=excited) ... should be ... excited (=charged) ...

Feel free to edit :) much appreciated. And our intention is to make a positive impact on kids to be more conversant with science. So a simple, catchy, memorable and fun example would make that happens... @LDC3 there's another meaning to charge. e.g. charge, a fee. So I need to find a better way to indicate that excitement is like charging a battery.

@iad22agp do you think I should swap the "Food" with "Manga". Means food refers to Gamma and manga refers to Beta? To have an impact from food one has to go quite near and consume it. Plus food is kids don't really like per se...

It might just be best to use the Bohr model of the atom and explain that the electrons and protons cancel out the charges in a neutral atom. When an electron is removed or added, the charge changes. Metaphor will only complicate things. The Bohr model isn't correct, but for illustrating basic concepts like ions to children the simpler the better. Once the concept of ions is understood, you can move on to ionization through radiation by incoming particles stripping away or adding electrons.

I am sorry that I am not able to contribute more meaningfully to this effort. I think the best approach to teaching kids about chemistry is to let them experience the behavior of matter itself. Without that experience, a concept like charge or ion really does not mean anything. You could use a magnet to affect the behavior of electrons in a fluorescent tube. Or observe a salt dissolving or crystallizing out of solution (salts are made of positive and negative ions). Or run a current of electricity through a salt and see a metal be plated on one electrode.

I think the key concepts that are explainable at this age level are:

Atoms are made of protons, neutrons, and electrons
Protons have positive charge, electrons have negative charge, neutrons have no charge
Opposite charges attract
Atoms can gain or lose electrons, but not protons or neutrons
Atoms hold onto their electrons more or less tightly depending on which atom it is (you won't be able to really explain why, but you could just show them a periodic table and explain that non-metals (the right side) like electrons more than metals (the left side))
Atoms can "steal" electrons from each other
Electrons can be "knocked off" of atoms by high-energy photons (light or EM radiation)

The difficulty with explaining these things by analogy to regular-life experiences is that there is no regular-life experience that captures the whole picture. This means you are limited to activity-type demonstrations, videos/pictures, or analogies for specific pieces of the puzzle.
I would recommend a mixture of all three approaches. Kids like hands-on stuff, so maybe you could explain the basic concepts with very simple analogies, then do an activity where the kids are given "electrons" (tennis balls, maybe) and are assigned to be metals or non-metals. Then you could have them give each other electrons and count charges as they go. If you had containers that could hold 8 electrons, you could even represent valence shells and the octet rule - metals could start with two or three electrons, non-metals with 5, 6, or 7, and the metals would need to empty their shells while the non-metals would need to fill them.
This would demonstrate a very important concept that is hard to teach in the abstract: Electrons don't disappear.
Since you are also interested in teaching about ionization by radiation, you could assign each student a number representing their ability to "hold on" to an electron. Here is where you might be able to use a shopping-type analogy - only instead of students giving up electrons as money, they would give them up in exchange for money. In this case, the money represents the energy, and you would be giving them the energy (money) needed to remove (sell) an electron. Atoms that charged a higher "price" (require more energy) would wind up with more money at the end of the exercise.
There are many variations on these ideas that you could use, but I think anything that covered the seven points I mentioned above while simultaneously developing an intuitive understanding that energy and electrons are conserved could be called successful.

@downvoter : I don't see a reason to downvote this answer. I was taught about ions by real life analogy and I misunderstood most of the things. Better understand nothing than to misunderstand things. :)

+1 I am taking a good look at your concept @thomij :) And appreciated all the you guys here making the contribution for our next generation little scientists =)

@bonCodigo SE has always strived to help the next generation scientists :)

thomij has given an excellent answer. I'd like to add a few more. He told:

Atoms can "steal" electrons from each other

You could do an arrangement to explain why atoms need to steal electrons. Get some marbles and a four-bladed boomerang like this (or any other medium-sized radially symmetric object):

(source: dxcdn.com)
Now, balance the boomerang on it's center. Add marbles to the corners of the 3 blades. The kids can see that the boomerang will remain balanced if all the four corners are placed with marbles. This is a real-life analogy to the Octet-Rule, although real atoms need 8 for balancing while our analogy was with 4.
I hope this analogy is worthwhile.

Pritt, cheers! This was 3 years ago ^^ kids (not mine) have grown up and I moved ahead a nanometer or so.

@bonCodigo Oops I totally forgot to see the date.


text:
What are some of the canonical texts for polymers? Is materials science more related to polymers than chemical engineering?
I am a freshman in college. My major is chemical engineering. I would like some recommendations for a textbook that's an introduction to polymers.
I am wondering whether a major in materials science is more related to polymers than chemical engineering.

I don't know of any undergraduate level texts that would be suitable for a freshman, unless you have already taken physics, statistics, and chemistry up through organic chemistry.
I can tell you that Fundamentals of Polymer Science by Painter and Coleman is a good introductory text, but I used it in the final year of my Chemical Engineering program. There could be better resources as well - that's just the only one I am personally familiar with.
It also depends a lot on which specific aspects of polymers you are interested in. This actually matters a lot for your second question as well. The three main branches of polymers studies are:

Polymer chemistry
Polymer physics
Polymeric materials

There are sub-fields in each of these, and a lot of overlap, but essentially they focus on:

Polymer chemistry - How are polymers made? (this would include engineering and production of polymers)
Polymer physics - How does the behavior on a molecular level affect the macroscopic properties?
Polymeric materials - How can we use the unique material properties of polymers to make useful stuff?

At the undergraduate level, you won't get too specific on any one of these areas, no matter which degree you choose. If you majored in chemistry, you would learn a little more about polymerization reactions. If you majored in physics (and took electives related to polymers) you might learn more about the statistical mechanics of polymers. If you majored in materials science and/or engineering, you would learn more about how we use the polymers, and less about how they are made and why they act the way they do. Chemical engineers learn a little bit of everything, but they don't get too deep into any one particular branch.
The real differences show up if you go on to graduate school or work in a particular role in industry. There, you will definitely focus more, and your degree can make more of a difference. So I would say, for your second question, major in whatever aspect interests you more - if you want to know the details about how polymers are made in a lab, then maybe chemistry is a better option. If you really like math and physics and want to understand how they relate to the behavior of polymers, then physics is good. If you are more interested in how to characterize and use polymers, then materials science/engineering is a good bet. And if you like a little bit of everything, or are interested in industrial-scale production of polymers, then chemical engineering is a good choice.
Keep in mind as well: it is hard to predict now what you will really want to do in four or more years - so if you make a choice based on which aspects of polymer science you are interested in, and then later decide that polymers are actually not what you really like, you will still be studying the "right" stuff for you.


text:
What is the name for the equation ΔG = ΔG° + RT ln Q?
As the title states, what is the name of the equation
$$\Delta G=\Delta G^{\circ}+RT \ln Q$$
My textbook does not explicitly state the name and I could not find anything on google.

I've never heard that this equation has a special name.

I'm fairly sure there is no "name" for that specific equation. It's just one (of very many) relations involving Gibbs free energy.

this might help.


text:
Why is there sulfur in black powder / gun powder?
Black powder is composed of charcoal, sulfur and a nitrate salt. The charcoal and sulfur serve as fuel and the nitrate is the oxidizing agent.
However, wouldn't it be cheaper to leave out the sulfur, thus having only charcoal as fuel?
Wikipedia says about this (emphasis added):

Black powder is a granular mixture of

[...]
sulfur (S), which, while also serving as a fuel, lowers the temperature required to ignite the mixture, thereby increasing the rate of combustion.

I guess after this initial research my question boils down to: Why is sulfur required as an ingredient in gun powder, and how does it lower the ignition temperature of the mixture?

Black powder is a mixture of solids. As solids, they are not particularly inclined toward fast reactions between each other. Elevated temperature generally liven things up, because it allows diffusion of solids into each other. However, both carbon (charcoal) and potassium nitrate are relatively stable compounds, so even when potassium nitrate is melted, they still do not react immediately,  further heating is required.
Sulfur, on the other side, is quite easy to melt. Nitrates are not very powerful oxidizers, but potassium nitrate in particular is a low-melting compound. And at the moment both potassium nitrate and sulfur are molten, their interdiffusion becomes quite easy, so they ignite, and charcoal follows.
Summing it up: charcoal is almost impossible to melt and potassium nitrate is a relatively weak oxidizer and also hard to melt, so the mixture is hard to ignite. Sulfur is easily flammable, easy to melt, so it is much easier to react with potassium nitrate solid, and its presence makes mixture much more ignitable.

Nice answer. You have any references for that?

Yes, but it's a rare specific firework-related book, and it will take some time to dig for it. Are you still interested?

Yes. It is a damn good chemistry question I've never seen properly explained before. I'd love to see the detail.

@matt_black john a. conkling, chemistry of pyrotechnics, chapter 5, ISBN 0-8247-7443-4. Some pdfs are googlable.

Thanks for the link. I will look it up to fuel my understanding and pyromania.