text:
without using electronegativity, which one of these molecules is polar?
If we are given 5 molecules without their electronegativity
BrF5
PBr5
CCl4
XeF2
XeF4
How to determine that one of them is polar or not ?
I know by using the difference in electronegativity we can determine the polarity
But again, electronegativity is not given  ?? /:

Hi Maher, Use the dissenter answer as a tip, but please show your effort and what you think according to our homework policy! Thanks!:-)

I think he has shown some effort - he says "How to determine that one of them is polar or not ?
I know by using the difference in electronegativity we can determine the polarity But again, electronegativity is not given ?? /:"
This shows that he understands the basic concept of polarity but is stuck when applying it to molecules.

@thomij Maher has been encouraged to elaborate a bit beyond just a sentence in his attempts.  As such, I'm leaving this closed for the time being.

@jonsca Maybe we should include in the guidelines a statement about the required length of the elaboration. Otherwise the standard is not very clear.

@thomij I think that's a good idea.  I'm addressing it on a case-by-case basis at this point, and going more by the spirit of the law than the letter of the law.

Also, in my reading of the problem statement (as portrayed in the title), the fact about electronegativity seems to have already been given.

sorry ! when I asked this question I have not taken geometries and this question is about geometries and at that time I didn't know that it is about geometries! today I took an introduction to geometries and we will continue to take more about it .. once I feel that I get the required knowledge I will come here to discuss it if it is still not clear.. Sorry again

When we say something is polar, it has a very specific meaning. It means that there is a spatial separation of charges, or a dipole moment. In the simplest case, this is given by:
$$
\mu = Qr
$$
Where $\mu$ is the dipole moment, $Q$ is the charge on each pole (equal and opposite) and $r$ is the distance between the charges.
Anything with charges can have a dipole moment, and therefore be polar. In chemistry, we typically worry about whether or not two things are polar:

Bonds
Molecules

Bond Polarity
Chemists are concerned with the polarity of bonds because polarity affects the character of the bond - the more polar the bond is, the more it behaves like an ionic bond. The less polar, the more it behaves like a covalent bond. This has implications for everything from naming rules to reactivity, but all you need to know for this problem is:

All bonds between two atoms are polar, unless the atoms are identical.

(This is including whatever else the atoms may be bonded to). The only question after that is "how polar is the bond," and that is where tables of electronegativity come in. Using an arbitrarily defined scale, we can measure the relative polarity of bonds by comparing the difference in electronegativity between the two elements.
Molecule Polarity
For molecules, polarity isn't quite so simple. We know that all bonds between dissimilar atoms are polar, but in a molecule, sometimes the dipole moments add up to form a net dipole moment of zero. This is a little bit difficult to explain if you don't have a mathematics and physics background that includes vectors and summations of forces/moments.
If you do, then when I say something like:

The net dipole moment for a molecule is equal to the sum of dipole moments over each bond.

It will make sense to you. If you don't, then you have to get a little more creative. I tell my students to imagine that there are ropes connecting each outer molecule to the central molecule, along each bond. Then imagine pulling each rope towards the side that is more electronegative (has a higher electron density). The more polar the bond, the harder you pull on the rope. Then imagine whether the molecule moves. If the forces balance, it stays put, so the net dipole moment is zero and it is not polar. If it does move, there was a net dipole moment, so it is polar.
This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).
Let's look at each one, using wikipedia's geometric pictures.
$\ce{BrF5}$

As you can see, there is a lone pair at the bottom (this is a square pyramidal geometry). The net dipole moment will be pointing "up", which makes this a polar molecule.
$\ce{PBr5}$

This one has a trigonal bipyramidal geometry, with each bond symmetrically opposed to each of the others. If you add up the vectors, they result in a net dipole moment of zero. Therefore, it is non-polar.
EDIT: As ron points out in the comments below, this is not really a binary molecular compound, and in reality forms an ionic crystal structure with $\ce{PBr4+}$ and $\ce{Br-}$. I don't think the author of the question intended you to worry about that, but if they did, the answer is still "non polar," although we wouldn't really call it a "molecule" any more.
$\ce{CCl4}$

Carbon tetrachloride has a tetrahedral geometry, and all the dipole moment vectors cancel. Therefore, it is non-polar.
$\ce{XeF2}$

This one is linear - no net dipole moment, so it is non-polar.
$\ce{XeF4}$

I could only find the ball-and-stick model for this one. It is square planar - there is a lone pair of electrons (not shown) on each "face" of the square on Xe. There is no net dipole moment, so it is also non-polar.
Out of all of these, the only polar molecule is $\ce{BrF5}$ (the first one).
To review the steps:

Draw the Lewis structure
Figure out the geometry (using VSEPR theory)
Visualize or draw the geometry
Find the net dipole moment (you don't have to actually do calculations if you can visualize it)
If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.

Isn't $\ce{PBr_5}$ ionic, with a tetrahedral $\ce{PBr_4}$?

Yes, but for the purposes of the homework, I doubt they were looking for students to know that. It's hard to find examples of trigonal bypyramidal molecules without using the same ones over and over, so this one ends up being in lots of assignments. At any rate, you would still call it non-polar (polar and ionic are different) and if you looked at the tetrahedral PBr4, that would also be non-polar. Usually the point of these types of questions is to test whether students can draw Lewis structures, know the geometries, and visualize them in space to decide if its polar.

Use molecular geometries. Some geometries are asymmetrical, so they necessarily are polar. Other molecules are symmetrical, so even though they contain elements of great differences in electronegativities, these molecules are non-polar.
Consider carbon dioxide - which has the elements carbon and oxygen in its composition. The EN of carbon is 2.5; the EN of oxygen is 3.5. Clearly, there is a big difference in electronegativity, which suggests that the oxygens are isolating electron density away from the carbon.

As we can see in the above picture, the two permanent dipoles cancel each other out since they are equal in magnitude and opposite in direction. So carbon dioxide might have elements of great difference in EN, but is still non-polar.
Now apply this knowledge to your problem. To determine molecular geometries, understand that both electrons and attached atoms influence geometry. Try looking up and applying the AXE method. I will leave my answer at here for now, since you have not shown much work in tackling this problem.


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How to create a buffer solution from NaOH and CH₃COOH?
In my chemistry book, they say that an approach to making a buffer solution of $\mathrm{pH}=5.09$ is by adding an appropriate amount of strong base $(0.052\ \mathrm{mol}\ \ce{NaOH})$ to $0.300\ \mathrm{l}$ of $0.025\ \mathrm{mol/l}\ \ce{CH3COOH}$. But, I haven’t been able to figure out how they got those numbers. If I am not mistaken the relevant equation is,
$$\ce{CH3COOH +OH- <=>CH3COO^- + H_2O}$$
where the weak acid and conjugate base are respectively $\ce{CH3COOH}$ and $\ce{CH3COO^-}$. My concern is that this equation above goes nearly to completion $(K=1.8\times10^9)$ so I don’t see how this can even be a buffer solution. If not, how do they get the values? I would appreciate any help.

Try considering reaction stoichiometry. You know the extent of the reaction. Now consider stoichiometry.
EDIT: If these are really the numbers your textbook gave you, then the textbook is wrong; you cannot create a buffer system with these numbers specifically. I suspect that you didn't transcribe the numbers from your textbook correctly. In any case the general idea, however, is correct - you can indeed form a buffer solution using sodium hydroxide and a solution of acetic acid.
All you have to do to realize this is to know what a buffer is - a buffer has significant concentrations of both an acid and its conjugate base (or base and its conjugate acid) - and consider reaction stoichiometry.

If the reaction goes to completion then what would happen to the buffer when you add a base, since there is no weak acid left. Can you please be more concise? I copied the answers verbatim, if they are wrong can you suggest correct answers.

"what would happen to the buffer when you add a base, since there is no weak acid left."
Do you always run out of acid when you add base? What if you add less strong base than you have acid? That was my point. Consider reaction stoichiometry.

So, the amount of OH-(~NaOH) should be half that of the weak acid?

If you want a one to one acid to conjugate base concentration buffer, yes. Not all buffers must have such a ratio.

So, I am missing one more equation $\ce{CH3COOH +H2O <=> H3O+ +CH3COO-}$, where the concentration of $\ce{CH3COOH}$ is defined by the excess. After this its like any other buffer, we use the $K_a$ of the equation above and $[\ce{H3O+}]=K_a \frac{[\ce{CH3OOH}]_{\text{excess}}}{[\ce{CH3COO-}]}$

I believe they had done a typo error it should be 0.25 M CH3COOH.

I think that would be a more reasonable molarity for acetic acid in this case.

Awesome! Feel free to mark my answer as accepted then, so this question can be closed.

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Reduction and Oxidation - Alternate definitions
Why is reduction defined as the gain of hydrogen?
Is it because hydrogen has a very low electronegativity and easily donates its electrons to other elements?
From what I understand reduction is formally defined as a decrease in oxidation state; oxidation state assignment involves supposing that all bonding is ionic and that electrons are therefore parceled to the most electronegative atom in a bond. And if something gains H - an atom of exceedingly low EN - then that element is reduced because that element will likely be more EN than H and therefore "take" the hydrogen atom's electron.

What book is this from?

organic by brown. Note that this definition isn't exclusive to this book.

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Definitions_of_Oxidation_and_Reduction

I see - it's also in my copy of Wade's organic chemistry. I suppose it's a shortcut method used by organic chemists. I'll update my answer.

The "official" definition I've always used is that reduction is the gain of electrons. IUPAC seems to use the same definition. It looks like organic chemistry books often use the addition of oxygen to identify oxidation, and the addition of hydrogen or the removal of oxygen to identify reduction. This is in line with the historical definitions (at least according to wikipedia) that defined reduction in terms of the change in mass on heating (if oxygen was lost, the weight would be reduced).
When using these shorthand rules, I think it's important to distinguish between the purposes of the definitions. The more formal "official" (according to IUPAC) definition is intended to get as close to the underlying physics as possible. The less formal "shortcut" definition used here is intended to quickly allow chemists to identify electron transfer just by looking at the structural formulas. In fact, in my copy of Wade's Organic Chemistry (4th ed) it says

Most organic chemists habitually use the following simple rules, based on the change in formula of the structure:

It then goes on to define redox reactions in terms of gain or loss of oxygen and hydrogen. This implies that the author doesn't intend it to be read as a formal definition - just a shortcut you can use in a pinch.
This is a case of "it works well enough" being used to make life easy - but it is important to remember that reduction isn't really being defined as gain of hydrogen in the formal scientific sense. It is still defined as a gain of electrons - it just so happens that in organic reactions, it is very difficult to add electrons to something without also adding hydrogen, and so the shortcut works (due to electronegativity differences between O, C, N, P, and H, as you pointed out).
Here's an example of a redox reaction where this rule won't work, but looking at electrons will:
$$
\ce{Li(s) + 1/2H2(g) -> LiH(s)}
$$
We are adding hydrogen to lithium, and the oxidation state of Li goes from 0 to +1 - it is oxidized, not reduced.
I have been looking for an example of an organic reduction that does not involve the addition of hydrogen, but haven't been able to find one - which is probably why this shortcut rule works.

Going to address your concern in my post above shortly.

" it just so happens that in organic reactions, it is very difficult to add electrons to something without also adding hydrogen" - why is that so?

because you are limited to O, C, N, S, and P - all of which are more electronegative than H.

@thomij  Examples of organic reductions not involving hydrogen include halogenation of an olefin, naphthalene reduction to the radical cation by NH3/Na, R-X -> R-M where M is Mg, Hg, Li, etc.

@ron - Those are good ones! I made my answer a wiki - feel free to add them with the mechanism if you wish.

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Dangerous chemicals from plastic
When the plastic of water bottles is exposed to light some atoms come off from it and blend with the water, so my question is which are the chemicals and its effects

Do you have a reference for this?

Yes, I would second @jonsca, it would be helpful if you could provide a reference to a specific example. Photodegradation of polymers is a real phenomenon, but this is distinct from dissolution of the polymer. As far as I know, a bigger problem is the leaching and/or off-gassing of more volatile polymer additives (e.g., plasticizers). Thermal and acid/base-induced degradation are at least equally significant (depending on the polymer in question) as well.

Incidentally, and separate from your question, there are plenty of companies that manufacture bottles and other products free of plasticizers. You can also minimize any possible exposure by taking simple and obvious precautions: avoid heating the plastics, keep products refrigerated when possible, avoid exposure to light (especially UV), do not use harsh cleaners on the plastics (i.e., anything beyond simple detergents and water), do not store highly acidic or highly basic liquids for extended periods of time.

It all depends on the polymer used in making the plastic bottle. I have heard that some of these chemicals you refer to are carcinogens. I won't pretend to know the formula nor the name of these carcinogens. However, the polymer must ve, to some extent, photodegradable given that the breakdown that you speak of occurs after being subjected to light. This would indicate that the polymer chain has a carbon oxygen double bond as part of the main "backbone" of the polymer and will have been formed by condensation polymerization. This may help to indicate the type or chemical that you're asking about.

I agree that it's crucial that the OP be specific about the polymer in question, but it need not contain carbonyl groups to be vulnerable to photodegradation. Polypropylene, polybutadiene, polystyrene, and low-density polyethylene are some examples -- there need not even be any $\pi$-bonds, chromophores, or heteroatoms at all.

authors: ا. د. سحر الملا


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What affects bond angles?
I have three factors in mind:
1) vdW repulsions
2) Bond length
3) Electron-electron repulsions.
I know that vdW repulsions between ligand atoms push the atoms apart. And bond length obviously influences bond angle; if the bonds were very long, then the bond angle can shrink without bringing the ligand atoms too close together.
Are there any other factors or are these it?

This is not a full answer, so I'll leave it as a comment. You also need to consider various electronic effects distinct from sterics. Overlap of $\pi$-orbitals, for example, introduces planarity constraints. There are also phenomena such as the anomeric effect and the gauche effect, which are often explained in terms of hyperconjugation. These are factors in determining lowest-energy conformers for given molecules, and sometimes have subtle effects on bond lengths and bond angles (and I mean beyond the obvious impact on dihedral angles).

These are just a few of the many complex electronic phenomena that may be relevant. In organometallic chem, there's also $\pi$-backbonding, for example, which shortens the bond between the metal and ligand, and can therefore slightly alter bond angles in asymmetric complexes.

Thank you. I was curious as to the statement my prof made. It was something along the lines of that bond angles can be rationalized using vdW repulsions alone.

I don't know what to call it, but ring formation could influence bond angles, such as cyclobutane.

@Dissenter, yeah, I would charitably describe that as an oversimplification on your prof's part. I suspect your prof may have wanted to avoid getting into certain gray areas (which are numerous in chemistry); I've certainly seen this plenty of times with my own profs. Steric effects (which is kind of a catch-all term mostly referring to Pauli electron-electron repulsion) are certainly quite dominant in many cases, especially in simple and highly symmetrical structures, but they're not the whole story.

I think the key word is "rationalized" - you can't predict bond angles with VSPER, except for the simplest cases, but you can rationalize them once you know what they are. For example, water has a bond angle of about 104.5 - you can't predict it with electron repulsion alone, but you can say "the electron domain geometry is tetrahedral, and there are two lone pairs on oxygen, so we would expect the bond angle to be something less than 109.5 degrees."

Bond angles mainly depend on the following three factors:

Hybridization: Bond angle depends on the state of hybridization of the central atom
Hybridization: $\ce{sp^3}$, Bond angle: $109^\circ$, Example: $\ce{CH4}$
Hybridization: $\ce{sp^2}$, Bond angle: $120^\circ$, Example: $\ce{BCl3}$
Hybridization: $\ce{sp}$,   Bond angle: $180^\circ$, Example: $\ce{BeCl2}$
Generally s- character increase in the hybrid bond, the bond angle increases.
Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.
Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.

Good Side note:
Triple bonds repel other bonding-electrons more strongly than double bonds.
Double bonds repel other bonding-electrons more strongly than single bonds.
Heres a nice read into the different components (extra)

Hybridization is a man-made concept that we use to try and approximate reality.  Being a concept, hybridization has nothing to do with the actual energetics of why a molecule adopts a certain geometry.  In fact, there are many cases were hybridization fails to explain reality, the water molecule being a case in point.  It is often taught that the lone pairs residing on the central oxygen in water are ~sp3 hybridized, while in fact these two orbitals are largely unhybridized.  P.S. Could you elaborate on what you mean by "repel" when you write, "Triple bonds repel other bonding-electrons more"

@ron you're exactly right about hybridization. But what it allows us is classification. It allows us to associate particular arrangements with particular geometries. And with Chemistry, there are a ton of exceptions, so it's definitely not perfect. But when thinking about bond angles its helpful to associate like things so we can see patterns.
And by Repel, I mean simply to push away, as in more often a more significant change in bond angle.

I am strongly agreeing with @ron. Not only is hybridisation, in this way of using it, a man-made concept (of purely mathematical considerations) - it is also always only the result of a geometrical arrangement of atoms in a molecule. And as this is true, it can only be applied to molecules composed out of the first and second period. A popular example for the failure of this concept is $\ce{PH3}$, which has bonding angles close to $90^\circ$. (Apart from the obvious water example ron already mentioned.)

The extra reading link is not available..

Downvoting because angles dominated by geometric constraints and dispersion forces are absent, consider paracyclophanes https://en.wikipedia.org/wiki/Cyclophane and  http://dx.doi.org/10.1002/anie.201304674

Could u explain more clearly how electronegativity affects bond angle?


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What is an effective fabric-friendly solvent for aluminium chlorohydrate?
I have recently found out that the white/yellowish "sweat" stains one  occasionally finds on t-shirts might be aluminium chlorohydrate polymers.
While talk about aluminium and antiperspirants is all over the press and there are lots of DYI-instructions for removing them, they all sound quite pseudoscientific.  I find it strange, for instance, that some guides recommend both vinegar and/or baking soda as a remedy...
Any ideas what else I could use to dissolve such stains without also dissolving much of the pigments?

I am surprised that there still hasn't been a single chemistry-based answer to this! I've started a bounty to attract more attention to this question

Bleach and baking soda are no use for removing deposits of aluminum chlorohydrates. These deposits are made of $\ce{Al(OH)_{}Cl_{(3−)}}$ and they are only soluble in somewhat concentrated $\ce{HCl}$ or $\ce{NaOH}$ solutions, whose pH are either $<1$ or $>13$. $\ce{HCl}$ solutions are not always available in domestic settings. $\ce{NaOH}$ solutions are available in specialized stores selling strong liquids for cleaning oven windows darkened by soot. This liquid is a rather concentrated solution of $\ce{NaOH}$. It should be diluted at least $10$ times with water before using it on a T-shirt. If not, it could hydrolyze your T-shirt. And take care ! It is corrosive for the skin. Use gloves ! It may also change the color of the T-shirt.

Thank you! In 17 hours I will be able to award the bounty; I've set an alert to do so. Your precise answer is greatly appreciated!

to be precise with dilution: should I look at getting a 10% solution of NaOH, or do I need less than that for the purpose of treating fabric?

The concentrated solution I was mentioning is probably $50$% $\ce{NaOH}$. A 10% dilution is about $5$% in$$\ce{ NaOH)$.. if you use it for destroying aluminum chlorohydrate deposits, do it quickly. The dissolution should be done in a couple of seconds. Rinse without delay.

I would assume however, that an $\ce{NaOH}$ solution at $\ce{pH\geq13}$ would not really be fabric or pigment-friendly. Are there any alternatives, and if I have the means, could I expect better results from $\ce{HCl}$? In personal anecdotal experience from chemistry courses strong non-oxy acids took a lot longer to burn into my skin and clothes than strong bases.

I can confirm that this works. Used on a black t-shirt (with several years worth of aluminium!) precisely as instructed above, and had great results - fabric and colouring were not damaged.

My undershirts have turned yellow in the armpit area without the benefit of aluminum antiperspirants. I assumed the stains were from carbonaceous exudants from my pores - or the biogenic products of bacterial life on my skin.
So I bleached the undershirts, soaked them in a diluted bleach (sodium hypochlorite) solution (about 1 in 10, I think). BIG improvement. Not perfect, but very good. One of the side effects was that the undershirts turned blue in the bleach because of the whiteners in the surfactant used to wash them. But that faded as the bleach continued to oxidize the stains.
You can go overboard and use more concentrated bleach: this will perk up the entire undershirt! Interestingly, it seemed to not only clean the underarm stains, but also attack/weaken the fabric there, so the stains went away, but small holes started to appear there, which grew larger and larger until the undershirts became nice dust cloths.
Next time, I think I will try an oxygen laundry bleach, even if it takes longer. Or H2O2.
Addendum: the supposition that the brown deposits in the armpit area are due to aluminum compounds is widespread. Even undershirt makers suggest it - but of course, take no blame for providing the substrate that shows the stain. One manufacturer (https://thompsontee.com/blog/the-science-behind-your-pit-stains-and-what-to-do-about-it/) even has a link to fixing it. What do they suggest? Change your antiperspirant; take proper care of your clothing; wear a sweatproof undershirt. (They offer the best sweatproof undershirt: 20% off!) But how to get rid of armpit stains?
#1. Vinegar, baking soda and HYDROGEN PEROXIDE.
#2. OxiClean or Raise Stain Remover (mixed reviews)
Nivea makes deodorants and has a site that recommends ways to remove armpit stains: try an acid. For example: #1: HYDROGEN PEROXIDE #2. Baking soda (?) #3.White vinegar #4.Lemon juice. https://www.nivea.com.au/advice/skin/removing-deodorant-sweat-stains
The desire to use environmentally safe washing agents in certainly admirable, but when you have to do a job, it is quicker to use the powerful tool. Of course, there is a learning curve. I love bleach because its activity is so adjustable and controllable - from an additive at low concentration to eradicate gingivitis by waterpicking to clothes cleaning to mildew removal from house siding, and probably a million more uses.
But I suppose hydrogen peroxide will be adequate. Or OxiClean.

Bleach and baking soda are no use for removing deposits of aluminum chlorohydrates. These deposits are made of $\ce{Al(OH)xCl{3-x}}$, and they are only soluble in somewhat concentrated $\ce{HCl}$ or $\ce{NaOH}$ solutions, whose pH are either <$1$, or > $13$.

Hi @Maurice, I think your comment is very close to the answer I was looking for! If you post it as an answer and perhaps suggest how one could access the solutions in domestic settings I would like to award the bounty!

@Maurice: Aluminium chlorohydrate is a group of water-soluble,[1] specific aluminum salts having the general formula AlnCl(3n-m)(OH)m. (Wikipedia)

I was telling my wife about this issue at dinner tonight. She said, "Have they tried Spray and Wash?" She does have a chemistry degree, but even more valuable, she does laundry!

Have you tried baking soda or washing soda?  It may sound pseudoscientific but many people have had success.
Even more successful is using baking soda as a deodorant, avoiding the aluminum altogether.

Baking soda is dirt cheap so there's no money to be made from it.  Baking soda is a base and the malodorous compounds in sweat tend to be acids so the baking soda deprotonates them, making them less volatile and unable to reach the nose.

Tried it, to no avail, sadly! Tried just the soda and soda with vinegar

I have been looking for a solution to two related problems:

yellow armpit stains on white cotton t-shirts
antiperspirant gunk buildup on polyester gym t-shirts.

Oxiclean 24hour soak helps with light yellow staining on white t-shirts but does nothing to dissolve antiperspirant buildup. Oxiclean is a mix of washing soda and some sodium percarbonate. Didn't have any luck with pure sodium percarbonate.

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Why do ionic substances with higher lattice energies tend to be less soluble in water than substances with lower lattice energies?
I understand that that energy is required to overcome intermolecular forces holding solute particles together in the crystal, but I don't understand how this relates to lattice energies and respective solubilities in water.

When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy of solvation of the ions (gain) the compound won't dissociate, and it will be insoluble.
There is a large grey area where lattice energy and solvation energy are close, and for those compounds temperature has a big influence on solubility.

You can break the process of dissolution down into sub-processes and use Hess’ law to calculate the overall process’ enthalpy. Specifically, the processes you need to consider are:

destruction of the lattice, loss of $\Delta_\mathrm{L}H^0$
solvation of the cation, gain of $\Delta_\mathrm{solv,1}H^0$
solvation of the anion, gain of $\Delta_\mathrm{solv,2}H^0$

According to Hess’ law, the total enthalpy of dissolution $\Delta_\mathrm{D}H^0$ is thus:
$$\Delta_\mathrm{D}H^0 = \Delta_\mathrm{solv,1}H^0 + \Delta_\mathrm{solv,2}H^0 - \Delta_\mathrm{L}H^0$$
Thus, a higher lattice enthalpy will decrease the value of $\Delta_\mathrm{D}H^0$, i.e. decrease the enthalpy gained from dissolution, i.e. make dissolution ‘harder’.
There is also a kinetic effect associated: even if the overall process is favourable, it may just be a very slow reaction for the lattice to break down. This may initially be perceived as a lower solubility. However, given time (and potentially heat to increase molecular motion), you will arrive at a thermodynamic solubility which will be higher than the initially perceived one.


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Can water be produced by mixing hydrogen gas and oxygen?
Could there be some kind of process, a kind of "retrograde electrolysis" where hydrogen gas (or in some other form) and oxygen gas (or in some there form) are mixed with required energy to combine to form water as a byproduct?

While the combustion is a correct answer, if you are looking for "retrograde electrolysis" than you best check out hydrogen fuel cells where are real electrochemical reaction, opposite of water splitting, takes place and produce directly electricity:
http://en.wikipedia.org/wiki/Fuel_cell
There are different types, even different fuels, but the basic idea is the same: the oxidation and reduction process is separated by a membrane and the two reactions takes place on separate electrodes.

Indeed, it's called combustion! When you burn hydrogen and oxygen gas, the reaction is:
$$\ce{2H2_{(g)} + O2_{(g)} -> 2H2O_{(g)} + energy}$$
Basically, you could see the former U.S. space shuttle as a large machine that turns hydrogen and oxygen into water. (Note: the space shuttle used liquid hydrogen and oxygen, but otherwise the principle and formula is exactly the same)
Since the Gibbs free energy of the forward reaction is negative ($-242\textrm{ kJ/mol}$), you don't need to add energy to the system, the reaction is spontaneous (mind you, you generally need a spark to ignite the combustion because there's a kinetic barrier to the reaction called the energy of activation).
The reverse reaction, water to hydrogen and oxygen, has a positive Gibbs free energy, so you need to add energy in the form of electricity to the system to make the reaction happen.

Does a fuel cell have a  similar process?

The overall equation is the same but the mechanism is different. A platinum catalyst splits H2 into 2H+ (protons) and 2e- (electrons). A polymer membrane allows only the protons to pass through, forcing the electrons to take another route, creating an electric circuit. When the protons and electrons meet again they combine with oxygen from the air, giving water.

Could liquid hydrogen and liquid oxygen SOMEHOW be combined in some energetic reaction so that one byproduct is water? Or is this science fiction?

It sure can, the U.S. Space Shuttle's rockets used liquid hydrogen and liquid oxygen as fuel. I realize I implied the shuttle used gaseous fuel, I've corrected my answer above, sorry about that.


text:
How is the oxidation number related to the group number?

If an element $\ce{X}$ forms the highest oxide of the formula $\ce{XO3}$, then it belongs to group:
A) 14
B) 15
C) 16
D) 17

How is an oxidation number of an element related to its group?
I tried comparing $\ce{X}$ with $\ce{N}$-, $\ce{S}$-, $\ce{As}$-compounds, but all 3 come in the form $\ce{XO3}$ ($\ce{NO3}$, $\ce{SO3}$, $\ce{AsO3}$), so I can't guess the group number.

Read the text more carefully. It says that the "highest oxide" has the formula $\ce{XO_3}$, while the 3 possibilities you mention contain only one oxide and 2 oxyanions.

Isn't $\ce{SO3}$ the highest oxide of $\ce{S}$?

actually SO4- is the highest oxide. Oxide means that the oxygen is more electronegative, not necessarily that the compound is also neutral.

For main group elements (groups 1, 2, 13-18), the highest possible oxidation state $O_\text{max}$ is given by the group number $G$ (for groups 1 and 2) or by the group number - 10 (for the rest):
$$ O_\text{max} = \begin{cases} G &\text{if } G \in \{1,2\} \\
G - 10 &\text{if } 13 \leq G \leq 18 \end{cases} $$
This means that for your compound $\ce{\stackrel{VI}{X}O3}$, so with the oxidation state VI, the corresponding group would be group 16.


text:
Is there a means to measure pH without indicators?
I have created a solution that should contain hydroxide ions. Therefore, the pH > 7, but is there any way to roughly estimate the pH in absence of any indicator? Maybe using current, or checking whether it dissolves certain chemicals?
Please mind that I am limited in equipment and so complex science is not feasible for me.

You could make your own red cabbage indicator

I cannot make that,... I am not allowed to by the household ;p so in that respect I am restricted

Basic solutions are slippery (like soap).

How accurate do you need to be, and what kinds of materials are realistically available to you? Is there any particular reason why you can't purchase some inexpensive litmus paper or a universal indicator? If you expect the solution to be highly basic, then you may be able to get a very rough qualitative sense of its basicity by reacting it with an acid and measuring temperature change. Another possibility is reacting it with soluble salts of common metals (e.g., calcium or magnesium) and looking for a precipitate of the insoluble hydroxide. Use a control sample of plain water for comparison.

precipitation reactions are possible, ill take a look at that. Any suggestions of certain salts that are readily available?

Calcium chloride is the first one that comes to mind, since it's readily available for purchase and highly soluble, while the hydroxide is orders of magnitude less soluble.

If the household don't allow you to use red cabbage indicator I really don't know what else more easy and safety to do you can use :-)

Another possibility that comes to mind is zinc oxide. This can be purchased in bulk, and while it dissolves in both acidic and basic solutions (it is an amphoteric compound), it is very insoluble in neutral ones.

Zinc oxide sounds promising

I decided to check whether it dissolved organic materials like small leaves or so.. than I know that it is very strong. I also have acid that I can mix in miser quantities to check heat formation.

Have you tried with red or black tea? It turns lighter in acidic media and turns dark in basic media. I tried once adding the pulp juice of one lemon to a cup of tea and it lighted up (mainly for acetic acid), then I added NaOH-based salt and turned very black and opaque.
Maybe that suits for your solution.

The ideal solution for me! Thank you! Maybe you have a chemical explanation? That would be very nice.

Good and not a simple question, I believe oxidation - reduction of polyphenols play that role. You can look at this for approach: www.worldoftea.org

I just tested it with earl grey and it works!


text:
Why does fusing benzene rings not produce polycyclic alkynes?
It would make sense that when the benzene rings are fused they keep their pi bonds and form a polycyclic alkyne. However this is not the case.
My hypothesis is that because monocyclic alkynes at 8C or above are stable polycyclic alkynes would be even more stable and more stable with smaller rings than monocyclic alkynes.
However with 7 benzene rings fused in a hexagonal tiling type fashion the center ring has all single bonds like cyclohexane.
Why is it that fusing benzene or really any aromatic compound does not form polycyclic alkynes?

A resonance structure of this would be a polycyclic alkyne along with just the regular pi bond delocalization.
Bicyclo[4.4.0]deca-1,3,6,8,tetraene-4-yne is what I am thinking about. I am moving just the pi bonds in the resonance structure to get this. I am not moving actual carbon nuclei.
I can't find a picture of this but hopefully from the name you can figure it out.

To have an alkyne in a ring, we would need to have two adjacent carbons that don't have a C-H bond. In naphthalene, if you break the central 9, 10 bond and preserve the outer ring, you don't have two such adjacent carbons without C-H bonds.  Also, in resonance structures we can't move nuclei.  Could you include a picture of the alkyne you had in mind?

See my edit and tell me where to but the 4 double and 1 triple bond.

If a polycyclic aromatic compound were to ring open, an allene would be the initial, symmetry-allowed, ring-opened product instead of an alkyne.

I've built a Dreiding model of the bis-allene product derived from naphthalene and it does not appear to be particularly strained.  I suspect the reason we do not see this process occurring is because the starting polycyclic aromatic is highly (resonance) stabilized, whereas the bis-allene product, while not strained, has no resonance stabilization due to the lack of pi-overlap in the cyclic structure due to the orthogonality imposed by the allene units.  In other words, the activation energy would be too high and the reaction would not proceed.  However, if we were to pass naphthalene through a tube furnace at very high temperature (this has been done), it would be interesting to see if any of the resultant products could be explained on the basis of the bis-allene intermediate.
EDIT  Here's the bicyco[4.4.0] skeleton.  Using these numbers tell me where you want the 4 double and 1 triple bond.

Ah, so this (the structure on the right) is what you mean!  There are 2 issues with it

In this structure the bridgehead carbons each have 5 bonds, but
carbon can only have 4 bonds (at least in compounds such as these) making this an unrealistic structure.
This structure has 12 pi electrons, whereas the starting naphthalene only had 10.  Resonance structures involve the movement of electrons, there is no way to go from 10 pi electrons to 12 pi electrons while keeping all of the sigma bonds intact, so this is not a resonance structure for naphthalene.

but a possible resonance structure of this is a polycyclic alkyne which would result in a monocyclic alkyne in the ring opening. Isn't it that if something is stable with 1 ring that multiple makes it even more stable and so polycyclic alkynes are more stable than monocyclic ones?

If so than I still don't understand why aromaticity is more important in polycyclic compounds than stability of possible resonance structures(including ones with triple bonds)

It would help me understand if you could edit your question and include a picture.

I want the triple bond between the bridgehead carbons and the double bonds at C's 2,4,7,and 9. This is what I thought you could get as a resonance structure of naphthalene without moving any carbon nuclei.

The ideal bond angles for alkynes are $180^{\circ}$, which is necessary for optimal overlap of atomic orbitals combining to produce mutually perpendicular $\pi$-type molecular orbitals. Any significant deviation from this is normally inherently destabilizing. Rings, by the very nature of their geometry, often incur additional strain due to their conformational inflexibility and possible demands for non-ideal bond angles (relative to acyclic molecules, that is). Introducing $\pi$-bonds into rings can potentially compound this problem, since they force at least local planarity and shorter bond lengths. These factors all contribute to the immense instability of cyclic alkynes in general.
When you mention stable cycloalkynes, keep in mind that stability is a relative term in this context. While those with a sufficient number of carbons can be produced and isolated under specific experimental conditions, they are not thermodynamically stable relative to their open-chain equivalents, or to most of their various possible structural isomers. In some cases, they can, however, be made kinetically stable by the addition of flanking $\alpha$-substituents which inhibit reaction by their steric bulk1.
Syntheses of cycloalkynes below cyclooctyne are often elaborate, and frequently proceed through extremely reactive intermediates that decompose by specific reaction pathways so as to carefully avoid other products. Benzene (and the various polycyclic aromatic hydrocarbons, for that matter), on the other hand, is an extremely stable molecule which participates in quite a limited collection of reactions. Whatever particular reaction you're suggesting would almost certainly be hugely uphill in energy (i.e., yields a significantly less stable product), and hence very thermodynamically disfavored; it would also certainly have immense kinetic barriers as well (since any such reaction pathway implies at least temporary loss of aromaticity in the transition state[s] and intermediate[s]).
Relevant reading might include Dehydrobenzene and Cycloalkynes by Reinhard W. Hoffmann. There is extensive discussion of the reactivity of cycloalkynes, and numerous examples of spontaneous dimerizations, isomerizations, etc. Evidently, isomerizations to cycloallenes and cyclobutadiene dimers are common, which is a profound testament to the instability of cycloalkynes.
1. For example, 3,3,7,7-tetramethylcycloheptyne, a good description of which (and synthetic scheme for) is given in Strained Hydrocarbons by Dodziuk.


text:
How to dissolve grass/leaves without affecting enclosure?
I've got two questions:

What sort of compound can dissolve grass/leaves fastest without leaving "much" trace?
What sort of enclosure is required for this to be doen safely and without damaging the enclosure?

I am doing a small experiment and would like to try dissolving few leaves within a glass or something of the sort.

Plant matter is mostly cellulose which is insoluble in most solvents so it is unlikely you can "dissolve" the material.  You can react the material.  Two possibilities are nitric acid and sulfuric acid but there are a number of safety precautions you need to follow.

Ok, thank You very much. Is there something that is safer but yields similar results?


text:
What's the purpose of a Grignard reagent initiation?
Preface: I've read this question about water preventing an initiation from occurring.
I understand why it's important to slowly add the alkyl halide to the dry ether and magnesium for exotherm control. But why is the first addition called an initiation and what's its purpose?
Is it just to confirm that the conditions are ready to continue slowly adding the rest of the reagent? I know it's a free-radical reaction, maybe that's where the term "initiation" comes from. But I've seen the reaction described as "self-sustaining" after the initiation, but it's not a chain reaction. Perhaps that's referring to a predictable reaction proceeding forward instead of stalling then blowing up your glassware.

The first step in the Grignard reaction involves transfer of an electron from the surface of the metal to the alkyl halide to form a radical anion.  The radical anion then disproportionates into a free radical and halide anion.
\begin{aligned}
\ce{R−X + Mg &-> R−X^{•−} + Mg^{•+}}\\
\ce{R−X^{•−} &-> R^{•} + X^{−}}\\
\ce{R^{•} + Mg^{•+} &-> RMg^{+}}\\
\ce{RMg^{+} + X^{−} &-> RMgX}\\
\end{aligned}
When you take the magnesium out of the jar, the surface has a thin oxidation layer on it that prevents the initial electron transfer from occurring.  We used to take one piece of magnesium and scratch it good with a knife to expose a clean, fresh magnesium surface.  We would place this piece of magnesium in the flask (dry ether and the rest of the magnesium already added) and drip in a few drops of the alkyl halide.  Without fail, within a minute or two, mild bubbling would be noticed (local ether boiling), which became vigorous as the initiation phase of the reaction "cleaned" the surface of other pieces of oxidized magnesium.  After the vigorous bubbling started to subside we would drip in the remaining alkyl halide plus ether solution at a rate to maintain ether reflux.  So the reaction does involve a free radical but the initiation phase really has more to do with removing the oxidation layer from all of the magnesium and allowing that initial electron transfer step to occur.

Good answer, I'd like to add a point to be considered: The initial electron transfer is usually slightly endothermic. In the initiation process you will usually need to add some energy - the warmth of the hand might do (depending on the used halid). After the reaction started, the formation of the Grignard is exotherm, hence providing enough energy to overcome the activation energy of the electron transfer. It will then proceed until the limiting reactant is used up - hence it is called self-sustained.

@ron That makes a lot of sense. My group has been performing Grignard reagent prep and reactions on and industrial scale, and your answer gives much needed perspective to why we have the initiation step, and how to troubleshoot it if it doesn't go right away.

@Martin That's an interesting point. Thanks for explaining the self-sustaining aspect of the initation.

First, you are mixing magnesium metal with an alkyl halide. The product of this reaction is $\ce{R-Mg-I}$, where $\ce{R}$ is the alkyl group and $\ce{I}$ is the halide. It is called an initiation reaction because you first need to make the reactive compound.
Next, you add the alcohol to produce the product. Mixing an alkyl halide and an alcohol together will not yield the product you want.


text:
Can electrolyzing ice cause an explosion?
I know electrolyzing water separates it into hydrogen and oxygen.  However, I am curious, if ice is electrolyzed, could it explode?  My thought process is that oxygen and hydrogen bubbles would build up in the ice, and continued electrolysis would ignite the gasses (forming water again).  However, since it is contained within the ice, it could ignite explosively.  Is this what would indeed happen?  One way to try this might be using a tazer or bug zapper on ice.  I believe this could be dangerous, and appropriate precautions would need to be taken.

Bulk electrolysis of ice would likely not get very far. Electrolysis works in liquids because the liquid is able to transport an ion to complete the circuit. In the case of water electrolysis, water is needed to transport protons (or hydroxide, but we'll stick to just one thing).
The half reactions for electrolysis of water are
Oxidation at anode:
$$\ce{2 H2O (l) -> O2(g) + 2H+(aq) + 2e-}$$
The electrons travel through the wires and the protons through solution to the cathode where the following happens:
Reduction at cathode:
$$\ce{2H+(aq) + 2e- -> H2(g)}$$
Electrolysis requires ion mobility. In solid water this mobility is severely decreased. All water molecules are locked in a rigid hydrogen bond network. A proton can propagate slowly through this network by proton exchange, but it will be slow. Thus electrolysis will be slowed.
Also, electrolysis only occurs at the surface of the electrodes, which must provide a means for the gas to escape (whether they are on the surface of the ice or driven into the ice, gaps will exist for gas to escape).

You're also forgetting the very low autoionisation of water 10^-7 meaning pure water electrolyses slowly and since freezing eliminates impurities from the lattice ice tends to be very pure indeed


text:
What is the difference between protein domains, motifs, and supersecondary structures?
I was just wondering if there is any difference between supersecondary structure and motifs. As well, is a protein domain an independently folded region of a protein that has a particular function associated with it - that may or may not contain one or multiple motifs.

All supersecondary structures are motifs, but not all motifs are supersecondary structures.
Motifs, in a biological sense, are very similar to fashion motifs. They are patterns that repeat in a lot of different places. Supersecondary structures are motifs that are made of several secondary structures. Some motifs are much smaller (pockets for holding ions are 3-4 amino acids long), or involve no secondary structure interactions (see SLMs).
Regarding the second half of your question, a protein domain (depending on who you ask) is any region or regions of a protein that do something, regardless of how it is folded. Very often, domains can fold correctly on their own. Generally, the required protease/functional assay to determine whether a possible domain is really technically a domain is not usually done. For example, the kinase insert domain found in some tyrosine kinase receptors is considered a domain, despite being a) quite small b) unable to perform its function in isolation (not entirely fair, as its function is to phosphorylate the rest of the receptor).
In sum: protein domains have a couple of competing definitions, so you cannot assume that something referred to as a 'protein domain' is capable of performing its function when separated from the rest of the protein.
Domains can be of nearly any size (due to their functional and not structural classification), but very small and very large ones are rare. They can be made of any number of motifs, or none at all.

A few well-defined searches can quickly resolve questions like these:
1: is any difference between supersecondary structure and motifs?
Motifs and supersecondary structures are synonymous. src
2: is a protein domain an independently folded region of a protein that has a particular function associated with it? (that may or may not contain one or multiple motifs?)
"Each domain forms a compact three-dimensional structure and often can be independently stable and folded. Many proteins consist of several structural domains." Wiki: Structural Domain
"Several motifs pack together to form compact, local, semi-independent units called domains." Wiki: Protein Domain


text:
Determine the ΔH for the reaction in kJ/mol of ethanol?

A $1.000~\mathrm{g}$ sample of ethanol, $\ce{C2H5OH}$, was burned in a bomb calorimeter whose heat capacity had been determined to be $2.71~\mathrm{kJ/^\circ C}$. The temperature of $3.000~\mathrm{kg}$ of water rose from $24.284~^\circ\mathrm{C}$ to $26.225~^\circ\mathrm{C}$. Determine the $\Delta H$ for the reaction in $\mathrm{kJ/mol}$ of
  ethanol?

I said :
\begin{aligned}
q_\mathrm{cal} &= C_\mathrm{cal}\cdot T\\
        &= (2.71\ \mathrm{kJ/^\circ C})(1.94\ \mathrm{^\circ C})\\
        &= 5.26~\mathrm{kJ}\\
\end{aligned}
Then the $q_\mathrm{r}=-5.26\ \mathrm{kJ}$
Then calculated the number of mol by dividing the mass g over molar mass and I got $0.02171~\mathrm{mol}$.
Finally I divided $-5.26\ \mathrm{kJ}$ by $0.02171\ \mathrm{mol}$ and I got $= -242.3~\mathrm{kJ/mol}$.
But the correct value is $-1365~\mathrm{kJ/mol}$. What's wrong?

What about the 3kg of water?  The equation you chose only works if you factor the water's heat capacity into the Ccal, which did not happen here.

So,I should get the q of water by the formula s.m.T and then take the negative value of it .. and then divide it by the number of mol. Right?

@brinnb I don't get the point of Ccal ??

The calorimeter is made of matter so it absorbs heat too.  Ccal tells you how much.

$$|Q|=|(ms+C)\Delta T|=(3\times4.18+2.71)\times1.941\;\mathrm{kJ}=29.6\;\mathrm J$$
$$\Delta H_n=\Delta H/n=-29.6/(1/46)\;\mathrm{kJ}=-1361\;\mathrm{kJ}$$

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


text:
What is the difference between a bond, force, and interaction?
I am trying to explain why some solids are more soluble in water than others.
So the energy released from the formation of bonds between the solid solute molecules and water molecules compensate the energy needed to break the intermolecular bonds between water molecules and the intermolecular bonds between the solute molecules.
The problem is that I'm not sure if I have used the correct terminology, for example I have never used intermolecular 'bonds', I've only used intermolecular forces before.

Here is how I think of the situation.

The term "bond" is generally reserved to describe an attractive, stabilizing interaction between atoms or groups in the same molecule or in different molecules, that has a magnitude of at least several kcal/mol.  Everything from hydrogen bonds at the low end up to full chemical bonds at the high magnitude end are included.
The terms "force" and "interaction" can be used to describe any attractive or repulsive situation, of any magnitude, between atoms or molecules. So these terms would be used to describe the van der Waals "interaction", where the "forces" involved are much less than those involved in bonding.  However, the terms "force" and "bonding" can also be used to describe situations involving larger magnitude interactions such as, "bringing the atoms together led to a bonding interaction."

"Force" and "interaction" can be used to describe any situation - attractive, repulsive, high energy, low energy; whereas "bonding" is typically reserved to describe stabilizing, higher energy interactions.

While this answer is nicely written and very clear in the general sense, it does not really address the specifics of the question.  Please consider explicitly detailing how these terms are related to water solubility.

As always, my understanding may be limited.  Please feel free to post a more complete answer.


text:
Wavelength of X-rays
What will be the wavelength of an X-ray, if we apply a potential difference of, say $\pu{20 kV}$ across the ends of the X-ray tube?
Alright, I started with the modified de-Broglie equation-
$$\Lambda=\frac{h}{\sqrt{meV}}$$
I pinned the values of Planck constant, electronic charge and potential difference but can't guess what would be the mass $m$ here.
The answer comes $\pu{0.62*10^-10m}$ if solved perfectly. Can anybody confirm?

check  Planck–Einstein equation.

But actually, the question provides insufficient information. X-ray from X-ray tubes is part of hi-energy electronic spectra of the anode material, so in reality the wavelength of most photons would be defined by anode material, and the only requirement for the potential difference is to be enough to cause excitation of the anode material. (It is somewhat oversimplified description, but reasonably adequate)

$$\lambda=\frac h{\sqrt{meV}}$$
$$E=mc^2$$
Also $$E=\frac{hc}\lambda$$
Therefore,
$$mc^2=\frac{hc}\lambda$$
$$mc=\frac h\lambda$$
Hence $$m=\frac h{\lambda c}$$
Plugging the value of $m$ from here we get an equation with all known constants. Solving for $\lambda$, we can get the result as $0.62\times10^{-10}\ \mathrm m$.

This gives the shortest (highest energy) x-ray. The overall spectra from the x-ray tube would be Bremsstrahlung radiation (sort of like blackbody radiation) with line spectra from the element of the anode material of the x-ray tube. The electron beam of an x-ray tube is in a vacuum. The x-rays are generally emitted through a thin beryllium window which absorbs lower energy (long wavelength) x-rays.

Firstly as @MaxW pointed out, using the given information, it is possible to find the shortest wavelength (or maximum frequency) X-ray only.

In an X-ray tube, electrons are accelerated in a vacuum by an electric field and shot into a piece of heavy metal (e.g., $\ce{W, Rh, Mo, Cu, Ag}$) plate. X-rays are emitted as the electrons decelerate in the metal. The output spectrum consists of a continuous spectrum of X-rays, with sharp peaks at certain energies as in the graph. The continuous spectrum is due to bremsstrahlung (German for "deceleration radiation"), while the sharp peaks are characteristic X-rays associated with the atoms in the target.
The spectrum has a sharp cutoff at low wavelength (high frequency), which is due to the limited energy of the incoming electrons (which is equal to the voltage on the tube times the electron charge). This cutoff applies to both the continuous (bremsstrahlung) spectrum and the characteristic sharp peaks, i.e. there is no X-ray of any kind beyond the cutoff. However, the cutoff is most obvious for the continuous spectrum.
Hence, the Duane-Hunt law comes handy in here, which gives the maximum frequency of X-rays that can be emitted by Bremsstrahlung in an X-ray tube by accelerating electrons through an excitation voltage $V$ into a metal target.
The maximum frequency $\nu_\text{max}$ is given by-
${\displaystyle \nu _{\text {max}}={\frac {eV}{h}},}$
which corresponds to a minimum wavelength
${\displaystyle \lambda _{\text {min}}={\frac {hc}{eV}},}$
where $h$ is Planck's constant, $e$ is the charge of the electron, and $c$ is the speed of light.
Thus putting the values given-
$$\lambda _{\text {min}}={\frac {hc}{eV}}={\frac {\pu{1240 eV nm}}{\pu{20*1000 eV}}}= \pu{0.62*10^-10 m}$$
For further references check the Wikipedia page.

I speculate what reads like a (of course incomplete) list of (solid) targets required the correction Mb -> Mo, and benefits the addition of Cu and Ag so often used for the characterization of organic molecules (proteins), regardless if «classical» tube, rotating anode (especially in biochem), or microfocus X-ray tube.

@Buttonwood Ah...sure! I was in a hurry, and I simply omitted that part while doing the corrections. Thanks! I wasn't aware of the usage of $\ce{Ag}$ though.

Some of them are listed in IUCr's teaching, and of course by the manufactures of X-ray tubes. Beside micro focus tubes (e.g. on diffractometers with both a Cu and a Mo source), be aware of the brighter liquid targets, e.g. background for small crystals.


text:
Why is oxygen absorbed in a colorless alkaline solution of pyrogallol and why does it turn dark brown?
My book says:

Oxygen is absorbed in colorless alkaline solution of pyrogallol and turns dark brown?

I'd like to know why and how.

Pyrogallol is the common name for 1,2,3-trihydroxybenzene.  Hydroxy groups are strong activavtors of aromatic systems.  By the time 3 hydroxyl groups have been added to a benzene nucleus you have a very reactive molecule, reactive enough to react with oxygen in the air.  The reaction is very complex and free radicals are thought to be involved.  Free radical reactions involving activated aromatic rings often lead to colored, high molecular weight products that are not easy to characterize.  Here's a link to a paper that suggests the initial products are probably the ortho-quinone and 2-pyrone pictured below.  Certainly the quinone will be colored, and as it reacts further to produce the higher molecular weight products, they will be colored too.  Here's another link to a very old paper suggesting that some of the higher molecular weight products contain at least 20 carbon atoms.

The balanced equation describing the initial radical oxidation of pyrogallol to the quinone can be written as
$$\ce{2~ pyrogallol + O_2 -> 2 ~o-quinone + 2H2O}$$


text:
Recipe for paper glue
I need to produce paper bags. Is there any simple receipt for making paper glue so that brown (packaging) paper will stick to itself?

...do you have access to polyvinyl alcohol and other chemical ingredients, or you just want to use around-the-house stuff?

both recipes would be appreciated

This looks promising for homemade

There are a few different ways of making your own glue, here are four different versions you can try :
Simple Glue Ingredients:
1/2 cup flour
1/3 cup water
Paper Mache Glue Ingredients:
1 cup flour
1/3 cup sugar
1-1/2 cup water with 1 tsp white vinegar mixed in
Cornstarch Paste Ingredients:
1-1/2 cups cold water
2 tablespoons cornstarch
2 tablespoons corn syrup
1 teaspoon white vinegar
Simple No-Cook Flour Paste:
1 cup flour
1/2 cup water
1/8 teaspoon salt
Taken from http://www.wikihow.com/Make-Glue


text:
Acid-Catalyzed Hydration
Is there any particular reason why this mechanism is used in the solution manual rather than an acid-catalyzed hydration?

I was thinking of having the double bond attack a hydronium ion, which would create a carbocation. This would then create a place for water to attack. Finally, the water would be attacked by water to reform hydronium ion.
Why wouldn't this work?
Does this have anything to do with the fused ring?

I think ron's answer is correct. Incidentally, another synthetic route avoiding a carbocation might be epoxidation by a peroxyacid (e.g., mCPBA) followed by treatment with a hydride donor (e.g., LAH). Personally, I would rather deal with a pyrophoric substance like LAH over a mercury compound, but you pick your poison either way.

Acid catalyzed hydration probably would work, but might well be accompanied by other materials derived from the carbocation intermediate, such as rearrangement and elimination products, which would make this route less desirable.  The mercuric acetate route does not involve free carbocations, so such side products will not contaminate the desired product.

+1, this was my thought as well. Isomerization to spiro[4.5]decan-6-ol seems possible as far as rearrangements go, but I see no reason to suppose it'd be favorable. Elimination to give the tetrasubstituted double bond seems plausible, though.

Yes, the 9,10-octalin would be the concern.


text:
Is Chemical Engineering a good pathway to Biomedical Engineering for graduate school?
I have been thinking about my future goals and I have recently decided to go for Biomedical Engineering. sites recommend to start in Chemical Engineering because its a good pathway for biomedical engineering. is this true? please share what you think and from your perspective, thanks

May be more relevant to se.academia? I am doing something along those lines, because of the flexibility of the chemE undergrad degree (solid) vs the BME undergrad degree (meh) if your school offers a biochemical emphasis,  do that. Or take side courses in bio courses in your interests, do research in biochem/bme labs. These will strengthen your graduate apps.

Sweet! Thank you for your advice it is really helpful getting feedback from other people

Are you taking about doing an undergraduate degree in Chem E, leading to a graduate degree in biomed?

Yes!! That's exactly what I am thinking of doing?? And for my undergrad focusing a bit more in the biochemistry side

If you want to get into a biomedical engineering program in graduate school, chemical engineering is not a bad option. You would of course probably have better luck if you majored in biomedical as an undergrad as well, since your coursework would more directly prepare you for those programs.
A couple of things to note though:
There is a (big) difference between bio-medical engineering and bio-chemical engineering. Biomedical engineering is more closely related to pre-med and materials science than to biochemistry. That said, bio chem courses probably wouldn't hurt you, but if you have the option, taking real biomedical engineering courses as electives (or as a minor) would be better. For bio-chemical engineering, these days regular chem E programs almost always have at least an elective component in biochem engineering, and often times departments are called "Chemical and Biological Engineering" due to the similarities. In other words, chemical engineering undergrad degrees are very well suited (and possibly over-prepare) you for biochemical engineering graduate work.
The other thing to know is: the real secret to getting into a good graduate program is to get research experience as an undergrad in the field that you want to work in. This demonstrates to the acceptance committee that you meet the two main requirements PIs have for grad students:

You are interested in the subject.
You have the aptitude to actually accomplish something.

A chemical engineering degree demonstrates aptitude on its own, and as @rch mentioned, is a pretty good "fall back" plan in case you change your mind later (although chem E is a tough major if it's just a fall back plan - I would not recommend it unless you really like chemistry and physics and could imagine working in a chemical plant, since that is likely what you will do if you "fall back" on it).
The problem is going to be that lots of other people will be able to demonstrate aptitude, with degrees in biomedical engineering, materials engineering, chemical engineering (lots of them get into biomedical later), pre-med, and other stuff that is just as well suited. In addition to aptitude, a lot of them (if not most of them) will also have research experience through undergrad projects or internships. To be competitive with them, you will want some as well.
The best easiest way to get research experience as an undergrad is to find a professor who is working on something you are interested in, and volunteer. Many times they will have paid positions specifically for undergrads, but if you can put in volunteer hours and are reasonably good at communicating, managing your time, and getting things done, most professors would be happy to put you to work for free. If you can manage to get a publication before you finish your undergrad degree, you will have an excellent shot at getting into the school of your choice, provided your GRE scores and grades are good (another reason to carefully consider chem E - not an easy major).
Another option is internships, although a research internship carries more weight in grad school circles than a regular old "gets the coffee and fills out spreadsheets" internship. They are much harder to get, but usually pay pretty well, so it is definitely worth trying for one.
The main point I am trying to make here is - major in (almost) whatever you want, but if you want to go to grad school in a particular field, get research experience in that field as an undergrad. Go for paid work first, but be prepared to volunteer if you have to, at least for a semester. You can quickly turn a volunteer position into a paid one if you do good work. Work experience also goes a long way toward getting you a job if you don't end up going to grad school. I'd put your odds of getting a good job with a biomedical engineering undergrad degree with experience at higher than getting a job with a chemical engineering degree without experience, especially these days.
Also, just so you don't think I'm trying to talk you out of a Chem E degree, Chem E was my major (BS, MS, and PhD) and I really liked it. If I didn't like chemistry and physics so much though, I probably wouldn't have made it through.

Thank you so much for your feedback it's making my decisions much clearer in what to do in my undergrad


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What is the mechanism of monofunctional glycosylase catalysis?
I am learning about base excision repair mechanisms. I understand the main idea how glycosylases are working, but I am trying to understand this in much more detail. I found a paper where they have written:

Monofunctional glycosylases catalyze the single-step displacement of damaged base via attack of an activated water molecule on c1' of the substrate

I am not sure what an "activated water molecule" is, how, for example, an aspartate residue is making a water molecule 'activated', and how this "activated water molecule" is removing the damaged base (hydrolysis N-glycosyl bond).
Source:
J Christopher Fromme, Anirban Banerjee, Gregory L Verdine: DNA glycosylase recognition and catalysis.

This is a duplicate of a question on biology SE: http://biology.stackexchange.com/questions/19667/question-about-monofunctional-glycosylases/19703#19703

Cross-Posting is strongly discouraged! @canadianer I flagged it for the moderators - but you can also do this yourself, the next time you see something like this.

I was going to flag it as a duplicate but it said that the duplicate post must be on chemistry SE so I wasn't sure if it was allowed or not.

This already has an excellent answer on Bio.SE.  As noted above, please do not cross-post because it results in the duplication of efforts and is unfair to those who have put in a lot of time to answer on the other site.


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Lewis Structures - Peripheral Octets
1) Which Lewis structure is preferential of the two below? EN considerations suggest the first structure. Size/charge density suggests the second one; the second structure would likely be more stable because if we think of the halide series of binary acids - the bromide anion is more stable than the chloride anion - and hence hydrobromic acid is a stronger acid than hydrochloric acid - the sheer size of the bromide ion is able to effectively dilute its charge and overcome its lack in electronegativity.

2) I am told that peripheral atoms should have octets. Why is this so? What is the physical reasoning behind this? Does this have anything to do with minimizing the overall potential energy of the molecule - i.e. the peripheral atoms are perhaps more subject to attack than internal atoms (due to steric factors) - so molecules try to minimize such a possibility by completing the octet for peripheral atoms? On the other hand, the octet rule is more of a guideline rather than a rule.
3) What is the bond-angle of the Cl-Br-Cl bond? I am told that the bond angle is a perfect 180 degrees. What about the first structure? If we consider the first structure a resonance contributor, wouldn't that lead to a distorted/non-linear bond angle? Or would that still lead to a perfect 180 degree bond angle, because to me, it seems that all the lone pairs on the bromine can participate in resonance (if not all at once).

Aren't the 2 resonance structures you've drawn the same?

Yes, they are the same.

Since they are the same, what do you mean by, "Which Lewis structure is preferential of the two"?

No, the pictures in the top row have a double-bonded chlorine, while the picture in the bottom row has no double-bonded chlorine.

Quoting Greenwood & Earnshaw's Chemistry of the Elements, discussing the triatomic polyhalide anions:

[I]nteratomic distances [...] are individually always substantially greater than for the corresponding diatomic interhalogen.

Unfortunately, the following data table does not contain an actual value for the bond length, though it does indicate that the two $\ce{Br-Cl}$ bonds are of equal length (which we probably could've safely assumed from symmetry considerations and the energetic degeneracy of the orbitals of the two chlorine atoms, if nothing else), and that the $\ce{Cl-Br-Cl}$ bond angle is (approximately) $180^{\circ}$. Bond length can often be correlated to both bond strength and bond multiplicity, especially when comparing compounds of the same atoms. Since the bonds in $\ce{BrCl2-}$ are longer than those in $\ce{BrCl}$, it's reasonable to infer that they are also weaker, and do not have greater multiplicity (i.e., there is no additional $\pi$-bonding). I would further reason that any $\pi$-orbital overlap would have to be very poor (judging by the mismatch in both size and energy of the $3p$ and $4p$ orbitals of chlorine and bromine, respectively), which makes $\pi$-bonding unlikely. (For the sake of completeness, I would note that I could be mistaken here, since $4p$ orbitals are somewhat smaller and lower in energy than expected, given the poor shielding of the nucleus by $3d$ electrons. That said, I don't think the magnitude of this effect is large enough to make a major difference here.)
A full quantitative molecular orbital theory treatment of the molecule would require computation, and I wasn't able to find many useful literature references. However, as the molecule is relatively simple, we can probably make reasonable progress with even a very naive analysis. On the valence level(s), we have a total of $22$ electrons in $12$ twelve atomic orbitals, all $s$ or $p$. These will combine to form twelve molecular orbitals, eleven of which are going to be doubly occupied. Note that for any bonding MO generated, an anti-bonding MO will also be generated. Hence, even if we had six bonding MOs (and, in reality, some would almost certainly be non-bonding), we would have five anti-bonding MOs occupied as well. This would yield an overall bond order of $1$, indicating a bond order of $\frac{1}{2}$ for each individual bond. By analogy to similar (but more common and hence extensively studied) polyhalogen anions (e.g., $\ce{I3-}$), this conclusion seems justified. $\ce{I3-}$, and numerous similar isoelectronic molecules, are often described using the three-center four-electron (3c4e) bond, and I think this model is probably applicable to $\ce{BrCl2-}$ as well.
The upshot of all this is that, while it's true that you can draw certain valid resonance contributors, they may not accurately reflect the structure and electron density of the actual molecule. The structure with the negative formal charge on the bromine is best, which agrees with your intuition regarding its size and polarizability being able to stabilize the charge. Due to the disparity in electronegativity, however, the bonds are nevertheless polarized towards the chlorine atoms, so the formal charge alone doesn't fully describe the electron density correctly.

Adding to your answer, it is obvious, that a good, representative Lewis structure for these kinds of molecules is impossible (And automagic detection by an NBO analysis will fail). Considering its structure, it is fairly obvious, that there has to be $\pi$ bonding (and antibonding), making these 3c4e bonds. I loved your poor-mans MO approach, which is not necessary as poor - pardon naïve - as you described it.

@Martin, thanks for the feedback. I can't take full credit for that MO analysis; I was borrowing the methodology one of my early professors applied in lecture (some years ago) to explain another 3c4e molecule, $\ce{XeF2}$. I think there's an elegant simplicity to that approach, which agrees well with more sophisticated results, and I felt confident that it should apply to this case as well.


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Is there an easy way to remember charges on ions?
I have a chemistry test coming up and I might need to know the charge that goes with the different ions like $\ce{SO4}$ has $-2$, $\ce{NO2}$ is $-1$ and  $\ce{PO4}$ is $-3$.  Is there an easy way to remember this by looking at the periodic table or something like remembering that the second row elements are all going to be $\ce{something-O3}$  and the next row is all $\ce{something-O4}$ for the -ate endings (except $\ce{Cl}$ which is the odd man out). It will make writing out the different acid combos easier, if I can remember the charge difference that needs offsetting by the hydrogen.
I'm just trying to lay down some associative triggers for myself.

I don't believe there is any simple way to determine what they are except by just plain memorization. I have given some tricks below that have helped me.
When I write down my poly-atomic ions, I generally put them in similar groups in a certain order. From least amount of oxygen to most amount oxygen.

hypo (name of ion) ite -> least oxygen
(name of ion) ite -> one more oxygen than above
(name of ion) ate -> one more oxygen than 'ite'
per (name of ion) ate -> one more oxygen than 'ate'

Understanding the above helps most with 'ite' and 'ate'. 'Ate' generally being 1 more than 'ite' for oxygen.
Then I list them from least charge to most charge. I order them this way:

Nitr ite $\ce{NO2-}$
Nitr ate $\ce{NO3-}$
Sulf ite $\ce{SO3^2-}$
Sulf ate $\ce{SO4^2-}$

You can see that 'ate' has one more oxygen, and that I have listed them in charge order. Sometimes looking at the trends in oxygen and charge will help you memorize. I have had to remember them by writing them down several times and now I am comfortable.
Here is another example: Hypochlorite is
$\ce{ClO^-}$
So, chlorite should be?

 $\ce{ClO2-}$, because it has one more oxygen, and

Chlorate should be?

 $\ce{ClO3-}$, because it has one more oxygen than the above, and

$\ce{ClO4-}$, will be called?

 Per chlor ate, because it has the maximum amount of oxygen.

The above is not a hard and fast rule. One exception I can think of is manganate and permanganate.
Manganate is $\ce{MnO4^2-}$ but permanganate is $\ce{MnO4^-}$; same amount of oxygen, just a different charge.
Trying to figure out any of the above from a periodic table didn't make a difference to me because my professor tests us without it. Yikes!

There is a very useful trick that I was taught to remember the charges and formulas:

Nick the Camel ate an Icky Clam Booger for Supper in Phoenix.

This trick helps with most of the polyatomics. The first letter of each word stands for the first letter of there compound.

Nick Nitrate. The number of consonants is the number of oxygen so. Nick = 3 O's.  The number of vowels stand for the charge. One vowel in Nick = -1 charge. So, $\ce{NO3-}$
Camel Carbonate; 3 consonants= 3 O's; 2 vowels = 2-; $\ce{CO3^2-}$
Icky Iodate; 3 consonants = 3 O's; 1 vowel = 1-; $\ce{IO3^-}$
Clam Chlorate; $\ce{ClO3^-}$
Booger Borate; $\ce{BO3^3-}$
Supper Sulfate; $\ce{SO4^2-}$
Phoenix Phosphate; $\ce{PO4^3-}$

Hopefully this helped, you can always add word to the sentence to help remember other polyatomic ions.

Eww. I'm glad I memorized this the old-fashioned way.


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How can a chemical structure be stored in a matrix?
How do I represent a chemical structure in a matrix format in Matlab? Can I input a chemical structure into Matlab using the S.M.I.L.E.S notation?
I have to specify the connectivity of different atoms with others and type of bonds between them. I have to write the structure in matrix form, so that it can be utilized in some linear programming. For example, if one atom is connected to another atom I have to store 1 in matrix otherwise 0 (binary variables).
I need to develop back end code for my purpose to design some molecules using this approach. As of now I am dealing with 2D structures.

what do you need it for? Depending on the need, different representations may be adequate, so, please, provide more info.

utilize that in some linear programming | it still not enough. There are tens of ways to store molecular structure in matrix format. The problem arises from the fact that molecules are 3-d flexible objects, so you have to specify explicitly what molecular traits are critical for you. AFAIK Matlab does not have tools explicitly designed to read molecular data in any format (though it is not personal experience but second-hand knowledge). However openbabel allows conversions between tens of formats and some are either very simple plain text or xml-based, that's the only advice I can get now.

Ok, I got your case, here is the deal. There is no good solution for your problem (and people worked on it since start of computer era) The best approach I know is to develop library of 3d fragments and combine them using genetic algorithm approach, building molecular surface with desired electrostatic field piece by piece. The details may vary, for example the surface may be provided implicitly in form of target protein. 2d approach has shown .... inadequacy. In case of integral properties like melting points extrapolation from known structures still works better then computers.\

@permeakra I would be delighted to upvote your comments, if you converted them into an answer (and then the comments can be removed).

@permeakra i would be doing the things suggested by you like GA approach and so later on..but the thing is even before to proceed to that step i need to input the structure into a databse(or an arrray ),which i am unable to do..

@Martin any help from your side how to proceed with the problem i asked??

I'm quite confused. The question appears to answer itself in the second paragraph, but that edit wasn't made by the OP, and the original problem (still) isn't that well defined.

@chipbuster These are the comments of the OP, I edited them into the question and then flagged them for removal. I agree, that the problem is still quite fuzzy...

@Martin Ahh. Thanks for taking care of that, though it's a pity that it makes this conversation quite difficult to read. Kishore, We're really going to need more details than what you have. One of the basic rules of computing is that there is no "best" way for general problems like these, there's just a way that's good for what what you want to do. Without knowing what you want to do, I'm just spinning my wheels.

Are the use of MATLAB (or OCTAVE) and a self-designed molecular representation mandatory for you project?
Established 2D- and 3D representations for molecules beyond a connectivity matrix, i.e. with coordinates, (partial) charges, annotation of isotopes, etc. exist, such as in the form of the Chemical Markup Language (CML).
In addition, well-documented toolkits, frameworks and modules are available to operate on these data are available for C++, Java, Perl, Python, and Ruby.

i have developed some part of code in MATLAB for my project...it would be better if i proceed with matlab itself..

how can we proceed with coding in python?? i never used python before..

@Kishore for specific programming questions I recommend StackOverflow, however make sure to read their Help Center so your question isn't closed


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Van der Waals Radius clarifications
I just read about van der Waals radius and had some questions about it.

Would it be correct to say that in general, van der Waals radius decreases as we move from left to right in the periodic table?
The van der Waals radius of the noble gas in a period is larger than the radius of halogen that precedes it. Why is it so?
What is the difference between van der Waals radius for monoatomic (say neon) and non-monoatomic (say fluorine) gases?

Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions,

Yes, that is a general trend, but as you can see there are many exceptions.  The trend exists for the same reason that covalent radii generally shrink as we move from left to right across the periodic table, every time we add another proton and electron to an atom, the electrons do not effectively screen the outermost electrons from the positive nucleus so they become drawn in closer to the nucleus.
& 3. There are a lot of different ways to measure the van der
Waals radius of an atom, but in recent times the preferred method is
to examine the crystal structure of a molecule.  Using $\ce{F2}$ as
an example, the crystal would be composed of fluorine molecules.  if
we measure the crystallographic distance between two adjacent, but
non-bonded fluorine atoms and divide by two, we would have a single
value that could be used to estimate the van der Waals radius for
fluorine (we would probably want to measure a number of similar
values for fluorine from other molecules, maybe using $\ce{CH3F,
NF3, ClF,}$ etc., for example, and average them together). Our
crystal might look something like the drawing below.

Because diatomic and polyatomic molecules are polarizable, the bonded fluorine atoms will likely be slightly polarized when packed in the crystal in order to stabilize (lower the energy) the overall system.  This polarization will cause non-bonded fluorines to draw a bit closer to each other then they would if there were no polarization.  A crystal of a noble gas (let's say neon) will just be a collection of monatomic neon atoms, no bonds, no polarization.  Therefor there is no "attractive" force to draw adjacent neon atoms closer together like there was in the case of fluorine.  So even before we make our measurements we might expect the van der Waals radius for neon to be larger than that for fluorine, which turns out to be the case.

Let me just point out, that there are intermolecular forces between noble gasses, usually the quite weak London dispersion interaction. Noble gasses are known to form clusters at low temperature. The energy to break apart these "bonds" (they are not really bonds in the goldbook definition) will increase in the group, as the polarisability will also increase. We had already a small discussion about this in the comments section to this question/answer.


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What is the plastic material of the CD cases?
What is the material of which the transparent part of CD cases are made of? It is nicely rigid and really lightweight.

These are called slimline jewel cases.
From the Wikipedia article on Optical Disc Packaging:

The bulk of slimline cases are made with translucent or transparent polystyrene, and are available in multiple colors. A stronger alternative is made from semi-opaque, semi-flexible polypropylene which is strong enough to protect the disc, but flexible enough not to break easily.

Here is an image of a polystyrene case to compare to yours:

The back part of the case in this picture is HIPS, PS-PB copolymer, as the description on wikipedia says, the transparent front cover plain PS. The case in the OPs picture is likely all standard PS.

CD jewel cases are made from injection molded polystyrene.  Clam-shell alternatives use polypropylene.


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Why doesn't the viscosity of water change much with temperature like it does for other substances?
I would like to know why water doesn't ever become like syrup before it freezes when many other liquids, such as hydrocarbons, seem to have a more gradual change from becoming more viscous and then turning solid.
As I understand it, the more viscous a material is, the stronger the inter-molecular forces are. When the liquid is heated up, the molecules have more kinetic energy and so they can more easily break free of these forces--the liquid will become less viscous.
Why doesn't water have the same spectrum of viscosity as other materials? It seems like it is either solid, or it has very low viscosity.

This is something I spent a lot of time thinking about during my PhD, so let me see if I can explain it.
What you are talking about is the relationship between macroscopically observable properties (like viscosity), the relaxation time spectrum (which is a way of looking at the time scales involved in molecular movements), and the structure of the molecules themselves. It's kind of complicated, but I think I can explain it in way that will make sense.
You know that as we increase the temperature, molecules have more kinetic energy and can more easily break free of the forces holding them together. I have some other answers here and here that explain this in more detail. With something like viscosity, however, there is more to the picture.
The first thing you need to know is:

Viscosity is a measure of momentum transfer in a fluid

In other words, it is a way of quantifying the effect that movement in one part of a fluid has on the parts immediately adjacent to it. If the movements are highly coupled (meaning that if one layer moves, the layers next to it move almost as much), the viscosity is high. If not, the viscosity is low. Wikipedia has a pretty good explanation of this on the macroscopic scale.
So what causes a fluid to have a viscosity? Since you know about intermolecular attractive forces, this is an obvious candidate. And indeed, the stronger the intermolecular forces are in a fluid, the higher the viscosity tends to be. However, there is another factor involved - intermolecular repulsive forces (sometimes called steric effects or excluded volume interactions). The idea is that if one molecule wants to move in a dense fluid, the other molecules have to get out of the way first.

As an analogy, imagine moving through the crowd at a packed concert or club. In order for you to move, someone else needs to move, and for them to move, someone else has to move, and so on. You can be polite and wait patiently for them to move, and eventually someone will. Or you could be assertive and ask, and move faster. If you are really impatient, you could barrel through people, knocking them out of the way (and hope you don't get beat up). Your impatience in a crowd is analogous to shear stress in a fluid. At lower temperatures, the molecules are closer together (since their kinetic energy isn't high enough to move very far out of the potential energy wells of their neighbors), and they move more slowly in general. This means that the time it takes for neighbors to move goes up, and as a result, viscosity goes up too - if you force them to move, they will knock others out of the way, transferring momentum through the fluid. At higher temperatures molecules move much more frequently, and so it is relatively easy to "drag" some through the fluid - they don't have to knock other molecules out of the way, because by the time they get there, they have moved on their own. The viscosity is lower.
Now, for simple molecules, the line between "can't move" and "moves easily" is pretty distinct. Once the molecules are mobile enough that excluded volume interactions are overcome, the material melts, and further changes to viscosity are small and mostly due to intermolecular attractive forces. For example, here is a plot of the viscosity of water vs. temperature

As you can see, it decreases smoothly as temperature increases. Below the freezing point, the viscosity is harder to define (although large masses of ice do flow), but the curve is discontinuous and it "jumps up" to some higher value.
But what about complex molecules? Let's link the monomers together:

Now it becomes much harder for the molecules to move. At a given temperature, you have to wait about the same amount of time for a single monomer to move, but it takes much longer for a pair to move together. Going back to the crowded room analogy - imagine trying to move through a crowd in which everybody - including you - was handcuffed in pairs to someone else. It would be a lot more difficult! In general, it turns out that the relaxation time for linear polymers increases as (approximately) the square of the chain length (for short lengths below the entanglement length). This means that if you double the number of monomers in a polymer chain, the average time it takes for them to move out of the way goes up 4 times. This is true up to a certain length of chain called the entanglement length (which depends on the polymer), at which point the likelihood of coordinated movement of the whole chain becomes so low that each polymer is sort of "trapped" in a tube made of other polymers. The analogy often used to explain this is a nest of snakes - they have to slither around each other. As a result, this type of motion was named reptation - after reptiles.
The interesting thing is that for shorter lengths (below the entanglement length) each length scale of the chain moves essentially independently of the rest. This means that in a longer chain, on short time scales the shorter lengths act as though they were in a low-viscosity fluid - they move easily and motion is not correlated with their neighbors. For longer length scales, the chain is essentially "frozen" - it can't move until its neighbors move - and so it acts more like a solid.

The result is that short lengths of chain can "wiggle" on short time scales, but longer lengths move more slowly. This gives us a relaxation time spectrum - it means that if you apply very fast changes (high shear rates) the molecules don't have time to get out of the way, and so the viscosity is high. If you increase the temperature, things move faster, and so at the same shear rate, the viscosity will be lower (more of the chain can move out of the way). The same is true at lower shear rates for the same temperature - the shorter length scales can move out of the way, and so it becomes more liquid - the viscosity decreases. On the longest time scales (lowest shear rates, coldest temperatures) the entire length of the chain can move, and so you see the lowest viscosity.
This is called shear-thickening, and is commonly seen in linear polymer melts or concentrated solutions. It is an example of viscoelastic behavior, which means that depending on the conditions (temperature or time scale of interaction) the material will exhibit a range of liquid-like (viscous) or solid-like (elastic) behavior. An example commonly used to demonstrate this is silly putty (although silly putty is not a pure linear polymer.) On short time scales, the polymers don't have time to relax completely and so it behaves like an elastic solid. This is why you can bounce a ball of silly putty on a table. At longer time scales, the polymers can fully relax, and so the silly putty flows. This is why if you let a lump of it sit out for a while, it will form a puddle. Silly putty is unique in that the relaxation time spectrum ranges from elastic to viscous over time scales that are commonly observed at room temperature, so you can see these effects in high contrast very easily.
There are many other types of non-Newtonian behavior (which just means viscosity is not constant), but they are all fundamentally connected to the relaxation time spectrum of the fluid, which is directly related to the molecular structure.
To sum up:

Water has a narrow viscosity spectrum (as a function of temperature) because it has a narrow stress relaxation time spectrum.
It has a narrow stress relaxation time spectrum because it is a relatively simple, small molecule.

Brilliant answer; I'm very fortunate. Because you're so knowledgeable I would ask a follow up question. Does water jump from its viscosity level to a solid when it freezes or is there a point (even on the miniscule scale) when it becomes more viscous? Also, from reading your explanation, can I assume that the behavior of silly putty is a consequence of the relaxation time spectrum?

I would also like to say thank you for such a wonderful answer. It is a very good thing to see someone put so much effort into an answer and share the knowledge.

You are welcome. I edited my answer to address your questions, but the short version is:
Yes, water jumps in viscosity as it freezes - but it also becomes more viscous as temperature decreases.
Yes, silly putty is a great example (probably the most common example) of a viscoelastic material whose behavior is related to its relaxation time spectrum.

At higher temperatures molecules move much more frequently, and so it is relatively easy to "drag" some through the fluid - they don't have to knock other molecules out of the way, because by the time they get there, they have moved on their own. The viscosity is lower.

I think this explanation by Thomij is a little misleading as the increase in temperature also causes increase in the rate of collision among molecules. So the molecules that are subjected to shear stress will encounter more collisions, hence more momentum transfer, which results in higher viscosity. But in reality, the main dominating factor in liquid viscosity is the intermolecular force rather than collisional or knock of energy transfer. But when it comes to gases, the collisional energy transfer becomes more dominant. That's why in gases, the viscosity increases with temperature.


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Atmopsheric Gas IR-Spectra
I am looking for normal resolution IR-spectra of: $\ce{O2}$ and $\ce{N2}$ at high concentrations and; common minor gas in atmosphere: $\ce{H2O}$, $\ce{CO}$, and $\ce{CO2}$ at environmental concentrations. I have found some reference on internet (mostly from Green House Effect articles), but I cannot found it on a free, accurate and reliable source even on the free part of NIST databases.
Where can I find good references for such spectra?
Thank you.

Completely symmetrical diatomic molecules like $\ce{O2}$ and $\ce{N2}$ are not IR-active because they have a zero dipole moment, even when oscillating within their one vibrational mode. Hence, there is no change in dipole moment possible, so these molecules will not be detectable by IR. The other molecules you mention, however, are IR-active, and I was able to find spectra of all three in the NIST database.
There are other spectroscopic methods that can detect IR-inactive molecules, however. Raman spectroscopy may be applicable, and I'm dimly aware of sophisticated methods based on absorptions in electronically excited states, but I don't know nearly enough about the subject to offer any concrete information on those techniques.

+1, marked as answer. I have missed the IR-Spectra check box.

@jlandercy, ah, that's an easy mistake to make. Personally, I don't like the NIST search form very much. I often just use Google instead, along with a site:nist.gov query modifier.

@ Greg E: Thanks for the advice. By the way, do you know a good article which address IR measurment technique non-linearity?

@jlandercy, no, I'm sorry, I don't know of any off-hand.