text:
How to determine order of nucleophilicity among commonly used nucleophiles?
How to determine order of nucleophilicity among commonly used nucleophiles? For example, in Finklestein reaction, $\ce{NaI}$ (reagent) - taken in acetone (solvent) - displaces $\ce{SH-}$ ion from alkyl group via SN2 mechanism to form an alkyl iodide. But in a separate question we may be asked to write the product formed when $\ce{NaSH}$ reacts with an alkyl iodide in the presence of acetone. Since I do not know which of the two is a stronger nucleophile, I will displace $\ce{I-}$ and form $\ce{alkyl-SH}$. But definitely only one of the reactions shall occur.
So, my question is that is there an order of strength of nucleophilicity (like a metal reactivity series)? I am preparing for JEE Advanced 2015 (college entrance exam in India), so I am expecting answers within its scope.

"(...) in Finklestein reaction, NaI (reagent) taken in acetone(solvent) displaces SH− ion from alkyl group via SN2 mechanism to form alkyl iodide" this certainly won't happen. https://en.wikipedia.org/wiki/Finkelstein_reaction

IMO your problem is not that much order of nucleophilicity as difference betweennucleophile and leaving group

The relative strengths of nucleophiles can be correlated with three structural features:

A negatively charged nucleophile is always a more reactive nucleophile than its conjugate
acid. Thus $\ce{HO^-}$ is a better nucleophile than $\ce{H_2O}$ and $\ce{RO^-}$ is better than $\ce{ROH}$.
In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the following order of reactivity:
$$\ce{RO^- > HO^- >> RCO_2^- > ROH > H2O}$$
This is also their order of basicity. An alkoxide ion ($\ce{RO-}$) is a slightly stronger base than a hydroxide ion ($\ce{HO^-}$), a hydroxide ion is a much stronger base than a carboxylate ion ($\ce{RCO2^-}$), and so on.
When the nucleophilic atoms are different, nucleophilicities may not parallel
basicities. For example, in protic solvents $\ce{HS, CN}$, and $\ce{I}$ are all weaker bases than $\ce{HO^-}$, yet they are stronger nucleophiles than $\ce{HO^-}$.
$$\ce{HS^- > CN^- >  I^- > HO^-}$$

Nucleophilicity versus Basicity: While nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by $\mathrm{p}K_\mathrm{a}$, is measured by the position of an equilibrium involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by relative rates of reaction, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group.
For example, the hydroxide ion $\ce{OH^-}$ is a stronger base than a cyanide ion $\ce{CN^-}$; at equilibrium it has the greater affinity for a proton ($\mathrm{p}K_\mathrm{a}(\ce{H_2O})= 16$, while $\mathrm{p}K_\mathrm{a}(\ce{HCN})= 10$). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.

The answer is from Organic Chemistry by Solomons, Fryhle, Snyder.

Did you use an OCR or typed the whole thing from the book?

Hint: nucleophilicity is different from leaving group strength.
In fact, both are possible it depends what drives the reaction. Though it usually is not possible to displace $\ce{SH^-}$ ion, but precipitation drives it.
UCDavis article on nucleophilicity and solvent effects

Yes, I do know that precipitation of Na salt is the dr

This article most probably clear all your doubts on the topic: Types Of Reactions with Alkyl Halide
As a supplement :

This would clear your concept of strength of nucleophile- What Makes a good nucleophile

And this will teach you how to compare nucleophilicity Comparison of nucleophilicity

I would recommend you go through these links and clarify your doubts.


text:
How can meta-cresol be made, given that the hydroxyl group is ortho,para-directing?
How is it possible for m-cresol (3-methylphenol) to be synthesized?
The methylation of phenol would lead to the ortho- and para-isomers instead.

Here is a synthesis of m-cresol starting from benzene.  It achieves the desired meta orientation of the final substituents by introducing a meta-directing substituent (the aldehyde group) early-on in the synthetic scheme. In the nitration step, the meta isomer is by far the major product (o:m:p 19:72:9), but a distillation or recrystallization would be needed (or you could carry all the isomers along and separate later, but you'd probably have the widest range in boiling points at this stage with the nitro group present). The only novel step is the double Wolff-Kishner used to reduce both the aldehyde and nitro groups in a single reaction.

+1, thanks for the link to that article. Very interesting. I wasn't aware of that use for the Wolff-Kishner reduction. Previously I would've chosen Clemmensen reduction, which should also reduce both groups in one step, but Wolff-Kishner seems eminently preferable given the option.

From the comments above, its clear you are asking how m-cresol is prepared.
I know one method, which involves diazotization of m-toluidine.
In this way we should make m-toluidine first but it making m-toluidine is more difficult than m-cresol.
m-toluidine is prepared by m-nitrotoluene but unfortunately m-nitro toulene is minor isomer in nitration of toluene and this is really so difficult to separate it.
But you can prepare M-cresol thorugh this process, though the yield of the reaction will be small.
Though I will be looking for more answers on this question if someone knows another way of preparing this particular compound.
Hope this helps.

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text:
Stability of Chromic Acid
I am told that chromic acid is exceedingly unstable and that it cannot be purchased as a pure reagent.
Is that so?
I don't find any mentions of stability or lack thereof on Wikipedia, and hey, if I wanted to, I could purchase it here:
http://www.grainger.com/product/LABCHEM-Chromic-Acid-8FV43
So what's the deal with having to make chromic acid in situ? Does this have to do with chromic acid's stability? Or something else?
EDIT: oops I realized that the linked reagent for purchase is just chromic trioxide, which can be used to make chromic acid.

"chromic acid" is also a nickname for chromic trioxide.  It is stable enough to be sold commercially.  Chromic acid in the strict sense ($\ce{H2CrO4}$) is not sold commercially, I think.  For most purposes what you're looking for is Cr(VI) plus acidity, so the usual practice of adding concentrated sulfuric acid to sodium or potassium dichromate (which have multiple uses and can be found in every lab) is preferable to having to buy another reagent.

Why isn't the strict sense of chromic acid not sold commercially?

According to chemicalbook.com some Chinese companies do sell it, probably wholesale (http://www.chemicalbook.com/ChemicalProductProperty_EN_CB52130391.htm).  Lack of demand is probably the reason it's not more widely available; as I said, in practical applications you can start from CrO$_3$ or a dichromate + acid.

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text:
Is the E in Arrhenius equation the change of enthalpy?
I'd like to know whether the activation energy in Arrhenius equation is the enthalpy difference between the transition state and the reactant. If it's not enthalpy, is it Gibbs energy or something else?

The "E" is the height of the thermodynamic potential energy barrier. Which potential that is (enthalpy, Gibbs energy, Helmholtz, or internal energy) depends on the conditions under which the reaction takes place - which thermodynamic state variables are held constant.
For example, in a system where the number of moles, the pressure, and the temperature are held constant (called NPT), then the thermodynamic potential energy (and thus the activation energy) is measured using Gibbs free energy.

You can understand it as a difference in the potential energies of the reactants and the transition state on the potential energy surface of the reaction. Worth of note is the fact that the energy of activation should be treated more as an experimental parameter than a "real" value - for example you can write the Arrhenius equation for a complex reaction mechanistically composed of a few elementary ones - the E won't be a true barrier height then, more of a combination of the barriers for all elementary reactions.
Have a look at the Eyring equation as well, that one uses the Gibbs energy change.

I know that the gibbs energy is in the eyring equation. I just wonder what the E in Arrhenius really is.


text:
Relating Thermodynamic & Transport Properties to Electrochemical discharge curve
What are the fundamental thermodynamic & transport properties that the discharge cycle of a rechargeable battery with an intercalating cathode (like Li-ion) should depend upon? The few that I could think of are

transport properties of the ion through electrolye & cathode material.
stress/strain properties of the cathode material due to intercalation.
chemical interaction of the solvent & ions with the cathode material.
electrochemical potential difference between anode & cathode.

My question is what formalism relates all these properties systematically to the experimentally observed discharge curve such as voltage as a function of capacity, repeatability of discharge curves etc.


text:
Molecules that get charged in Water
I am a chemistry super noob and I was watching a MIT video where the professor states that:
"Certain molecules become charged when they come in contact with water in the natural state"
Can someone give me an example of one or more such molecules ?

It depends on the context how he meant it. I could imagine two scenarios:

Certain materials can produce ions when they dissolved in water. A typical example is table salt, NaCl. Table salt is solid form has a lattice that contains Na+ and Cl+ ions, and when you put it in water, the water molecules can help releasing these ions into the solution, where you now can find the ions. Note, that in this case the ions are atomic entities but molecules with charge are also ions, like $\ce{MnO4-}$ or $\ce{SO}_4^{2-}$.
Another scenario when the charged entities are formed as a reaction with water by e.g. giving an electron (oxidation) to water molecules or to other dissolved components e.g. oxygen in water. As you can imagine, it happens generally with very reactive materials (materials that are very sensitive to moisture), but essentially the rusting of iron is a similar reaction.

An addition: amino acids often gain or lose charge in water depending on the pH.

Terrible phrasing. Molecules by definition have no overall charge. Ions have overall charge. Not molecules. Molecules becoming charged become ions.
If you want to get into the nitty gritty you can try looking up more information about zwitterions, which are actually not ions in the strict sense as they possess no net charge.
This is the IUPAC definition of molecule:

An electrically neutral entity consisting of more than one atom (n > 1). Rigorously, a molecule, in which  n > 1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

This is the IUPAC definition of ion:

An atomic or molecular particle having a net electric charge.

This is the IUPAC definition of zwitterion:

Neutral compounds having formal unit electrical charges of opposite
  sign. Some chemists restrict the term to compounds with the charges on
  non-adjacent atoms. Sometimes referred to as inner salts, dipolar ions
  (a misnomer). E.g. H3N+CH2C(=O)O− ammonioacetate (glycine),
  (CH3)3N+–O− trimethylamine oxide.

http://goldbook.iupac.org/M04002.html

So, betaine is a molecule and not an ion?

Yes, it would. It is also a "zwitterion" since it has formal charges of opposite signs on its constituent ions. See here for the IUPAC definition of molecule - it does not contradict anything I said. http://goldbook.iupac.org/M04002.html

Good point, it is better be careful about this detail. Interesting that the adjective form "molecular" is still used in the other sense, irrespective of charge.

Well that's an unfortunately loose application of terms.


text:
Why put zinc solution in a galvanic cell?
In a galvanic cell with Cu and Zn electrodes, I was taught that each electrode is immersed in a solution containing it's own positive ions. I understand why you need copper ions, so they can gain electrons, but why do you need zinc ions - the reaction is going to produce zinc ions anyway, and the negative ions from the salt bridge will migrate to neutralize them, so I don't see why you need zinc ions at the first place. Is there anything I'm missing in how the cell works?

You're right: you don't need a zinc solution to start with.  In fact the lemon battery experiment (http://en.wikipedia.org/wiki/Lemon_battery) shows you don't even need the copper solution to generate a current. (The reaction at the copper end in that case is 2H$^+$ + 2e$^-$ = H$_2$ and doesn't involve reduction of copper.)
The reason the textbook description is the way it is has to do with reproducibility and simplicity.  I'll address the question of reproducibility later on, but first consider:
What's the simplest thing the zinc could be immersed in? If you put the zinc electrode in the same copper solution as the other electrode, you'll get copper-plated zinc instead of a battery.  If you put it in distilled water you won't have a circuit.  There are other things you could put it into, such as salt water, but in any case two things are true: It needs to be a solution different from the one where the copper electrode is sitting; and it will have zinc ions as soon as you have a closed circuit, as the zinc electrode starts to dissolve and the copper electrode starts acquiring more copper.  So in a way the simplest answer is, stick it in a solution of zinc ions.
Now about reproducibility.
The potential of the cell depends on (among other things) the concentration of Zn$^{2+}$ in solution near the Zn electrode, and the concentration of Cu$^{2+}$ in solution near the Cu electrode.  Other things being equal, if you increase the concentration of Zn$^{2+}$ by a factor of 10, the potential will drop by about 0.118 volts (Nernst equation).   So in a way it's actually good to have a low concentration of Zn$^{2+}$; but that concentration will never be zero, at least after you close the circuit, and when the concentration is very very low it will be growing very fast, relatively speaking (the first micromole of Zn to dissolve might increase the concentration a million times, say), which means the potential will drop quickly in the initial moments.   You  might as well start with a concentration of  Zn$^{2+}$ that is not going to change so fast (in relative terms).

Thanks for the answer! What I missed is that pure water conducts very poorly so you can't leave the zinc without a solution. About this lemon experiment though: why do electrons travel in the wire, if the zinc electrode can react with the H+ straight inside the solution without the electrons traveling to the cathode to produce the reaction there?

My guess is that it is a matter of kinetics. In a weak acid, reduction of H$^+$ at the zinc electrode occurs so slowly that (so long as the wire has low resistance) it's just as easy for it to occur at the other electrode (that it, H$^+$ gets reduced at all metal surfaces joined by the wire and immersed in the juice).  When the pH is lower (stronger acid)  most of the reduction of H$^+$ does occur on the surface of the zinc itself.  See the Experimental section in the wikipedia article linked to in my answer.

It's tempting to look at the whole cell here and ask: hmm what's going on? But try taking a reductionist approach and look at each electrode system one at a time, in isolation.
Let's consider a basic model.If we immerse an electrode of $\ce{Cu}$ metal into a solution of $\ce{Cu^2+}$, the following can be thought to take place:
$$\ce{Cu(s) <=> Cu^2+(aq) + 2e-}$$
Where the ions in the solution are bombarding the electrode surface, and are being taken up by gaining two electrons from the metal, whilst the metal is giving up ions from its surface, keeping hold of two electrons. At equilibrium, this forward and backward bombardment occurs at the same rate.
You might say to me: okay but we can't actually probe this system with a voltmeter to say that this is going on, because this would require a circuit to be made with an external system, and hence would perturb any effect we wish to observe. I'd then say to you: you're correct but bear with my argument, just noting what I've said.
By a similar token, in our other isolated beaker we have:
$$\ce{Zn(s) <=> Zn^2+(aq) + 2e-}$$
Now, the whole idea behind an electrochemical cell is to extract chemical energy by systems which are not at equilibrium. If a system is at equilibrium then no energy will be available. The equation which gives the free energy available from an electrochemical cell is (derivation not given):
$$\Delta G=−RT\ln K_\mathrm{eq}=−nFE_\text{cell}$$
Where $R$ is the ideal gas constant, $T$ is the parameter of temperature, $F$ is the Faraday constant (the amount of charge carried by a mole of electrons) and $E_\text{cell}$ is your available electrochemical energy.
So for these two individual single-electrode systems at equilibrium, can we extract chemical energy? Well no, because Gibbs at equilibrium is zero, and hence so is $E_\text{cell}$.
What then do we do?
Complete the circuit between the two, and if the potential between two half-reactions is sufficient to warrant the moving of electrons, a dis-equilibrium will be created until the reaction has completed (which is indeed what we get for a system of zinc/copper). Once the electron transfer between zinc and copped has been achieved i.e. equilibrium has been achieved, then we can no longer extract any useful chemical energy. Our cell is dead!
So in summary: the dual metal electrode/same-metal ion solution is needed for your initial half-reactions, which you then perturb by completing your circuit. This perturbation of the individual half-reactions results in our being able to extract useful chemical energy by shifting the equilibrium until a new equilibrium is established.

Thanks for the answer, thought I can't see why your analysis makes justice with the real reaction: the reaction that's actually happening in the cell is different from the two reactions uncombined. Every metal in it's own solution, the reactions are as you noted, but in the combined cell the copper and zinc are exchanging electrons with each other, unlike when the reactions are seperated and every metal transfers electrons to/from it's own ions.


text:
Elementar cell of Y2O3
I think this picture from wikipedia is wrong because I count:
\begin{align}
\left.
\begin{array}{ccccr}
1&×&1 &= &\ce{1~Y} \\
6&×&\frac{1}{8} &= &\ce{\frac{3}{4}O}
\end{array}
\right\} \text{ gives }\ce{Y4O3}
\end{align}
How would be the correct elementary cell of $\ce{Y2O3}$?
(I didn't found any correct elementary cell using google)

The picture in Wikipedia doesn't give you the primitive cell. In Yttria, $\ce{Y2O3}$, 8 metal ions are in the special positions $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{4}$; the remaining 24 metals occupy the sites $u$, $0$, $\frac{1}{4}$. The 48 oxygen ions are in general positions $x$, $y$, $z$, and are arranged in distorted octahedra around the metal ions, the metal-oxygen bonding distances being unequal. So, the stoichiometry is $\ce{Y32O48}$ which gets simplified to $\ce{Y2O3}$. This structure is generally referred to as the C-type metal oxide structure. All the sesquioxides of the rare earths belong to this system, as do $\ce{Ga2O3}$, $\ce{In2O3}$, $\ce{Tl2O3}$ and the mineral bixbyite, $\ce{(Fe, Mn)2O3}$. The primitive cell looks like this (oxygen ions are red, metal ions are gray)

Here another view which might show more easily that indeed the positions of all the 48 oxygen atoms are different in such a way that no smaller primitive cell can be found.
Reference
M. G. Paton, E. N. Maslen, Acta Cryst. 1965, 19, 307-310


text:
What is the oxidation state of nitrogen in ammonium nitrate?

The equation :
$\ce{NH4Cl + AgNO3 -> NH4NO3 + AgCl}$

I know that oxidation number of $\ce{N}$ in $\ce{NH4Cl}$ is -3 and in $\ce{AgNO3}$ is +5.
But what I am confused about, is the oxidation number of $\ce{N}$ in the product $\ce{NH4NO3}$. Should I consider $\ce{N}$ as $\ce{N2}$ in this compound or should I get its oxidation number in $\ce{NH4+}$ and in $\ce{NO3-}$ and then get the sum of them?

First off, is this a redox reaction?

There are 2 different nitrogen atoms in $\ce{NH4NO3}$, one having oxidation number -3 and one with +5. They are different, having separate oxidation numbers independently of each other. It is perfectly fine to have several atoms of same element having different oxidation state in same compounds, for example in thiosulphate  $\ce{SO3S^{2-}}$ one (central) sulfur atom has oxidation number +4 and another has 0, and $\ce{Fe3O4}$ is actually $\ce{\overset{(II)}{Fe}\overset{(III)}{Fe2}O4}$ and many more.

A reasonable trick to get the oxidation state of both the $\ce{N}$ atoms would be to write down the ionic components of $\ce{NH4NO3}$. They are-

$\ce{NH4+}$
$\ce{NO3-}$

Now these ions can be considered separately to get the answer which is $-3$ and $+5$ respectively.

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text:
How can I identify whether a substance is KOH or NaOH?
I've bought $\ce{NaOH}$ or $\ce{KOH}$ (I don't remember which one!) for my electrolyzer and I've tested it. Now I want to improve its performance and I was searching for a better electrolyte when I realized I've forgotten what my current electrolyte is! It is white and has crystal shape, but which one is it, $\ce{KOH}$ or $\ce{NaOH}$? If they are both white how can I know?

I edited your title to make the question a little more specific. I also suggest that you remove the last three sentences and make them a separate question - it will be easier to get an answer that way.

I'm sorry to say, you won't improve your mileage by using your alternator to electrolyse water then burning the hydrogen produced. Given the efficiency of the engine/alternator/electrolyser, you'll be lucky to get 1 litre of electrolytically produced hydrogen for every 5 litres of hydrogen (or the gasoline equivalent thereof) burned in your engine.

Mind if I ask (out of curiosity), what did it end up being? ;)

A flame test should tell you pretty quickly. Make a small wire loop (preferably steel or nichrome), heat in a bunsen burner or propane torch flame until red hot and all the contaminants are burned off, then dip it in your solution and put it in the flame. If sodium is present, you should get a bright yellow flame, and if it is potassium, the flame will be a dim purple (lilac) color. But a small amount of sodium present can mask that color.

Back when we used a piece of cobalt glass to filter out the sodium line.

An alternative, non-destructive, physical method:
Potassium hydroxide (KOH) is radioactive due to the presence of naturally occurring 40K. With sufficient counting time, even a simple radiation detector can be used to distinguish the radiation of a nice big sample of potassium hydroxide from the natural background radiation in the room.


text:
What is meant by complete outer shell? Why do the noble gases have zero valency?
Does having 8 or 2 electron in the outmost shell mean its outmost shell is full and its valency is zero?
I know that the 3rd and 4th shell can contain 18 and 32 electrons.
Then how can Argon's (2,8,8) outmost shell be full, though it does not contain the highest number of electron (18) in it's 3rd shell?
It is the same in the cases of other noble gases too? Why?

When we refer to "outer shells" we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the effective nuclear charge "felt" by the valence electrons changes as you move through successively higher energy configurations.
In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the next n fill before the current n (they have lower energy).
In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:
$$
\ce{Ar: \space [Ne]}3s^23p^6
$$
For the next noble gas, krypton, the condensed electron configuration is:
$$
\ce{Kr: \space [Ar]}4s^23d^{10}4p^6
$$
Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking  for the valence or "outermost" electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.


text:
Advanced Level Chemistry Book Recommendation
I have a test comming up on Chemistry in Advanced Level and I sincerly ask for a book recommendation for chemistry. I already got the book chemistry by Ramond Chang. And I am pretty near to the exam. What I really want is a book with Lesson and Questions related to that specific lesson book.

Can you clarify what you mean by "lesson book"?  Do you mean you want a workbook for the Chang text?

Yes. I want the work book and also a book that explains the lesson like Chang's.

Related: http://chemistry.stackexchange.com/q/37303/4945

Here are a few recommendations for A-level chemistry, made by fellow students such as yourself. TSR is a good student resource for A-level prep since there is are masses of students talking about the same topics.
(Zumdahl is a good general chemistry textbook as well, with quite the number of practice problems. However it is a full-fledged textbook you probably don't need a few days before an exam.)

Try 'Advanced Chemistry by Philip Mathews ' - a red large book.
That explain very much in chemistry for advanced level.


text:
Psychoactive ingredients in this soap?
Could any ingredients from this product be psychoactive in some people, if the are sensitive to them?
https://www.lush.co.uk/products/honey-i-washed-kids
I have found the product to be relaxing as advertised, but more so that I would expect from a nice scent alone.

None of the ingredients listed would be classified as a psychotropic substance.  However, many of the compounds, when studied in mice, do have effects on brain activity and behavior characteristics.  For example,

rapeseed oil (in a mixture with other vegetable oils): "the rapeseed oil group exhibited much higher locomotor activity than that expected from the alpha-linolenate/linoleate ratio. Additionally, the rapeseed oil group exhibited unusual behavior patterns, including higher ambulation and rearing activities, faster acquisition of the water maze task and slower habituation behavior as compared with the control group. Susceptibility to pentobarbital anesthesia tended to be higher in the rapeseed oil group." abstract link
bergamot: "produces dose-dependent increases in locomotor and exploratory activity that correlate with a predominant increase in the energy in the faster frequency bands of the EEG spectrum" abstract link
linalool: "Inhaled linalool showed anxiolytic properties in the light/dark test, increased social interaction and decreased aggressive behavior" abstract link

So I am not surprised to read that when used in combination they have a noticeable affect on you.


text:
Monobromination vs. Dibromination
My professor noted that bromination with light and methylene chloride gives you exclusively the monobromination product.
However, bromination without light will give you the multi-brominated product.
I have no found any reference to bromination in my textbook nor online. Is the second statement an accurate statement?

bromination of what?

Bromination of a hydrocarbon

Bromination of alkanes with Br2 will not occur in a dark environment. In the presence of light, multiple halogenations can occur given enough time.
For more information, including the full mechanism, check out: https://en.wikipedia.org/wiki/Free-radical_halogenation

So he's wrong about monobromination?

Let's just say my first guess is you misheard him.

My first guess is bromination using $\ce{Br2}$ in methylene chloride as a solvent.

The rate determining step (rds) in the free radical bromination of an alkane involves hydrogen abstraction to yield the corresponding alkyl radical.  Whether the chain initiating bromine radical is generated in the dark (thermally or by using N-bromosuccinamide for example) or light (where the Br-Br bond is homolytically cleaved using light energy), does not particularly matter; the same mechanism and (roughly the same) product distribution will result.
Selectivity (which hydrogen will be abstracted) will be determined by the relative strengths of the various C-H bonds in the molecule.  A tertiary C-H bond (tBu-H) has a bond dissociation energy (BDE) of around 93 kcal/mol, a secondary C-H bond (iPr-H) has a BDE around 95 kcal/mol, while a primary C-H bond (Et-H) has a BDE around 98 kcal/mol and methane has a BDE of 103 kcal/mol.  I wouldn't expect a bromine atom attached to carbon to increase the BDE for a hydrogen also attached to this carbon, so multibromination at that carbon should also occur.  Further, even in a simple molecule like ethane, once bromoethane is formed, a second bromination at the non-brominated carbon should occur as readily as bromination at the initially brominated carbon.  So both 1,1 and 1,2-dibromoethane should result as the secondary products.
In my opinion, both reactions will proceed through the same free radical - hydrogen abstraction mechanism, and, if 1) there is not a large excess of alkane to bromine and 2) the reaction is not terminated shortly after initiation, then both reactions will produce multi-brominated products

My first guess is that in light alkanes forms alkyl halides and in dark alkenes form dihalo-alkanes with bromine in non polar solvent as $CCl_4$; anyways:
Bromine is less reactive than other halogens toward alkanes in general but bromine is
more selective in the site of attack.
Bromine shows a much greater ability to discriminate among the different types of hydrogen
atoms. The reaction of 2-methylpropane and bromine, for example, gives almost exclusive
replacement of the tertiary hydrogen atom.
$$\small\text{2-methylpropane}\overbrace{\longrightarrow}^{\text{Br}_2,h\nu,127^oC}\text{tertiarybutylbromide}+(99\%)+\text{sec-butylbromide(trace)}$$
Abstraction of hydrogen by a bromine atom is endothermic in both cases. The transition states are more product-like and possess more radical character; therefore, the difference in radical stability is more strongly expressed, and $\Delta E_{act}$is larger. The larger $\Delta E_{act}$ is associated with greater product selectivity, since the tertiary bromide is obtained from the tertiary free radical.
In the laboratory it is more convenient to use light, either visible or ultraviolet, as
the source of energy to initiate the reaction. Reactions that occur when light energy is
absorbed by a molecule are called photochemical reactions. Photochemical techniques
permit the reaction of alkanes to be performed at room temperature.


text:
What are the parent compounds of the following?

I just need these two answered correctly so I have examples to work with, also if someone could label the longest continuous chain, that would be great too!
Thanks!

Find the longest chain by starting at one end of the molecule and counting the number of carbon atoms along each possible path.  I've attached a figure with numbering of the longest chain.  Hopefully it will answer your question and you will understand how to find and number the longest chain.  In the future, post your thoughts, so we can better understand how to help.


text:
Is the photoelectric effect possible for semiconductors?
Is the photoelectric effect possible for semiconductors?
I imagine that this might be, if possible, a two-photon process: excitation of an electron from a filled to conductance band, and then on to ejection. Is anyone familiar with such materials, or even whether this is possible?
I can't readily think of a theoretical reason against it. Two-photon events are often less likely than one-photon, however I could imagine irradiation by a laser of bandwith covering both the band-gap wavelength and that necessary for ejection would do it.
If it does indeed happen, I'd be glad to hear a short summary of how it differs from the photoelectric effect of metals.

Given that Jaques Pankove's excellent book on 'Optical Processes in Semiconductors' devotes chapter 13 to 'Photoelectric Emission', you can rest assured that the photoelectric effect is alive and well in semiconductors.  While somewhat more complex than emission from simple metals, there is no need for two-photon processes. The 'tricky' part is that you have to look at where the electron states are, and then how much energy is needed to take it out of the semiconductor.
The work function is the energy difference from the Fermi energy to the vacuum level.  There may not be any states at the Fermi energy though.
The electron affinity is the energy difference from the bottom of the conduction band to the vacuum level.
One has to take the band structure into account, since direct and indirect transitions will likely have different energies required.
Also, surface band bending may make interpretation of experimental results more difficult. Again, perusing Pankove may prove useful.

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


text:
Aerobic respiration to produce water
Could the aerobic respiration reaction occur outside the body to produce water? If it is possible how would it occur

If you look at the net reaction, it is just combustion of glucose:
$$
\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}
$$
So you could just burn a sample of glucose, and it would produce water.


text:
In what ways does the transfer of heat affect bond stability?
It is to my understanding that certain amount of joules may change an element's/compound's physical state given a certain temperature. How does a certain added number of joules affect the orbital of a bonded electron ( I realize my question's ambiguity by not stating a specific type of bond).

How does a certain added number of joules affect the orbital of a
  bonded electron

If you pump enough energy into an element\molecule you can promote an electron from one orbital to the next highest.  For example, the hydrogen atom has a single 1s electron, put enough energy in and you can promote that electron to the 2s orbital.  That means that, on average, the electron would spend more time further away from the nucleus.  In molecules, there are numerous "sub-levels" between each (molecular) orbital.  These "sub-levels" relate to rotational and vibrational energy.  So as you pump energy into the system the molecule will rotate and vibrate more violently.  If you put enough energy in, the bond will break and the atoms connected by that bond will fly apart.


text:
Subtle implications of quantum numbers
Question:

Given the subshells $\ce{1s, 2s, 2p, 3s, 3p}$ and $\ce{3d}$, identify those
  that meet the following descriptions:
  a) has $l=2$
  b) Can contain two electrons with spin $m_{s}=\pm\frac{1}{2}$.

For a): from my understanding $l=n-1$. Therefore $\ce{3s, 3p}$ and $\ce{3d}$ should be correct but according to my solutions manual, the answer is only $\ce{3d}$. Why is this so?
For b) The answer is $\ce{2p, 3p}$ and $\ce{3d}$. Why are these the answers?

a) You are wrong. $l=2$ simply means $\ce{d}$-orbitals, thus, only $\ce{3d}$ is the right answer.
b) Assuming that the meaning of the question is to specify subshells that can contain two electrons in the same spin state, i.e. both with $m_s = +1/2$ or both with $m_s = -1/2$, $\ce{s}$-orbitals should indeed be excluded.

Can you explain why s-orbitals should be excluded?

@user3814584 each $s$-subshell consists of just one orbital, an orbital holds at most 2 electrons, but they should have the opposite spin if placed on the same orbital. Thus, $s$-subshell could contain 2 electrons, but not with the same spin. Starting from $p$-subshell, you have more then one orbital in a subshell, and thus, the subshell could contain 2 electrons even with the same spin.


text:
What is the meaning of n in n-octane and similar compounds?
Straight-chain octane ($\ce{CH3(CH2)6CH3}$) is also called n-octane. What does the n stand for, and where did the term originate?
For reference, see the PubChem entry for octane (beside "Chemical Names:")

In this case, n means "normal", i.e. straight chain. This is somewhat of a holdover from the olden days, but in the case of octane in particular, it makes sense to specify n-octane if that is what you have. For example, the "octane" used as a standard gasoline engine fuel against which gasolines are rated (by definition, 100 octane) is actually not n-octane, but rather 2,2,4-trimethylpentane - a branched isomer of octane - in contrast to the straight chain version.

Okay I found a source for this information (under "Etymology"): http://en.wiktionary.org/wiki/n-octane

Amusingly, 2,2,4-trimethylpentane is colloquially called iso-octane, which totally flies in the face of the n/sec/tert/iso convention.

@Lighthart Iso-octane would mean an isomer of octane, which 2,2,4-trimethylpentane qualifies. So why do you think it is wrong?

Iso typically refers to a 1-methylethyl motif.

Please note that the completely correct, exhaustive, preferred (and only?) IUPAC name of ‘n-octane’ is octane. Current IUPAC Nomenclature of Organic Compounds does not even mention the ‘n-' prefix anywhere (if I did not miss something).


text:
Boiling solutions with two salts
If you dissolve two different ionic compounds (that don't make a precipitate) in water and then boil the solution so all the water evaporates, the salts will re-crystallize. Is it possible that you will get two compounds that are different from the ones you orignially dissolved? Is there any way to predict whether you will get the original compounds or new ones?

Mixing and matching different anions and cations can easily produce different compounds.  For instance, sodium and potassium chloride, and sodium and potassium nitrate, are all soluble in water and do not react to form a precipitate. When the aqueous solution dries, all four compounds are possible.  So, mixing sodium chloride and potassium nitrate will result in 2 anions and 2 cations in solution, allowing all four pairs to possible precipitate.
Now, will the precipitate as pure compounds or not is another question.  That requires looking at the relevant phase diagrams and seeing what is thermodynamically favored (and consideration of kinetic factors like how fast you evaporate the liquid).
As an example, the phase diagram for KCl and NaCl exhibits a miscibility gap starting at 505C.  This means that below that temperature they will phase separate into a Na-rich chloride and a K-rich chloride.  At 100C the concentration of the 'impurity' phase is less than 1%.
The nitrate diagram also shows a two phase region below 100C.
So, mixing (Na,K) (Cl,NO3) will result in all four precipitating out.


text:
Two-electron Integrals over Gaussian Plane Waves
Is there an efficient method to compute the two-electron integrals over the basis set represented by a product of a Gaussian function $g(r)$ and plane wave:
$\psi(r)=g(r)e^{ikr}$
where $\lambda=1/k$ is approximately one fifth of the width of gaussians.

This question appears to be off-topic because it is about physics mathematics.

where could i ask it?

You could try the Physics Stackexchange

you might be interested in helping us start a brand new stack exchange just for this type of question: https://area51.stackexchange.com/proposals/122958/materials-modeling

I'm not sure if anyone has done it, but the closest I can think of is the GPW (mixed Gaussian Plane Wave) method. I think that involves a mapping between Gaussian and PW basis. It's DFT, but just look for the Coulomb integrals.
General idea:
Lippert, Gerald, et al. "A hybrid Gaussian and plane wave density functional scheme." Molecular Physics 92.3 (1997): 477-488.
The implementation can be found here:
VandeVondele, Joost, et al. "Quickstep: Fast and accurate density functional calculations using a mixed Gaussian and plane waves approach." Computer Physics Communications 167.2 (2005): 103-128.


text:
Order of second ionization energies of lithium, beryllium and boron?
The correct arrange of them is like this:
Be < B < Li ... But why?
Li will have the greatest IE2 (second ionization energy) because that will involve removing a core electron, but what I am confused about is about Be and B.
I think Be will have greater IE2 than B, because B has 2 electrons in the same orbital, so the repulsive force between them will make it easier to remove one of them, but in Be there is only one electron in the orbital and there is no repulsive force in the same orbital, therefore
$$\ce{ B < Be}$$
What do you think?

The key factors are the effective nuclear charge and the principal quantum number n Read my answer below

To reach the second ionization potential we remove a second electron from an already ionized element
$$\ce{M^{+1} -> M^{+2}}$$
Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol)
\begin{aligned}
\ce{& Li^{+1} & 1s^2~~ &~~ 7297}\\
\ce{& Be^{+1} & 1s^2~ 2s^1~~&~~ 1757}\\
\ce{& B^{+1} & 1s^2~ 2s^2~~&~~ 2426}\\
\end{aligned}
We see that both $\ce{Li^{+1}}$ and $\ce{B^{+1}}$ have completely filled shells, the stable "inert gas" configuration.  We would expect it to be difficult to remove one electron from these two ions and destroy this stable configuration.  Further, we would expect removal of one electron from $\ce{Li^{+1}}$ to be more difficult than from $\ce{B^{+1}}$ because in the former case we are removing a 1s electron which is closer to the nucleus than the 2s electron in the case of $\ce{B^{+1}}$.  On the other hand, if $\ce{Be^{+1}}$ loses one electron it will have a $\ce{1s^2}$ electron configuration and achieve a stable "inert gas" configuration.  This should make it much easier to remove one electron from $\ce{Be^{+1}}$ than from the other two ions.

Li+ is fairly obvious, the 1s electrons do not shield each other well so each is attracted to an effective nuclear charge approaching 3 resulting in a very high IP. Be+ is different, the 1s2 2s1 configuration, isoelectronic with Li atom, has a well shielded 2s1 electron. The effective nuclear charge approaches 2 and the distance is greater. The result is a large second IP but less that of Li+ [and much greater than Li atom]. B+ is similar to Li+ in that the 2s2 electrons do not effectively shield each other, but they are shielded strongly by the inner 1s2 electrons resulting in an effective nuclear charge approaching 3 at a greater distance than Li+. This results in an IP less than Li+ but greater than Be+.
The reason that inert gas configurations or filled orbitals are of lower energy is not magic. Electrons added to the same orbital or to orbitals of the same [or very close] energy do not effectively shield each other from the nucleus. The increased nuclear charge means each electron is more strongly attracted to the nucleus, is of lower energy. These effects are very evident in the first and second periods and become more subtle at higher atomic numbers.


text:
How are chemistry and materials science related?
I'm currently a chemistry major and now I am considering doing a minor in materials science.
Before making a decision, I would like to know how the two disciplines are related and what are their differences such as techniques, approaches, emphasis, etc.
How will my knowledge in chemistry be useful in my understanding of materials science, and how useful is a knowledge of materials science to a chemist?

An undergraduate chemistry course here in the UK typically has a materials component to it in the form of a module or two, so you bet they're related!
It's sometimes not that fruitful to view materials science and chemistry as a different subject—much like physic vs chemistry—however we can associate certain motivations to either topic as we see fit. Please note that this is quite subjective; this view does indeed vary from person to person.
First of all, materials is more interested in the solid state than chemistry in general. We might like to look at crystal structures via x-ray crystallography (see Bravais lattices, space groups etc.) as well as the rather intriguing quasi-crystals (for which Dan Shectman won the Nobel in Chemistry in 2011). The skills gained in a materials crystallography course would no doubt help with chemistry. The converse is also true since many substances in chemistry are amenable to study via crystallography, such as proteins. You therefore may also have a crystallography module in your chemistry course—looking at molecules in the solid state not typically viewed as a 'material'. If it has appreciable electron density, you can study it by x-ray crystallography!
Another aspect of materials science looks at properties of materials, such as stress/strain curves etc.I've had such a lecture series specifically dealing with polymers in a chemistry course; one of the challenges of modern day chemistry is to invent and modify polymer architectures for specific industrial applications. It's clear then that chemistry and materials science complement each other here. Whilst materials looks at interesting properties of given materials, chemistry constructs those polymers.
Materials science can differ from chemistry in that it focuses (or at least did in my case) on the band-structure of solids more than chemistry does. So it's possible that a materials course would give you a more nuanced appreciation of semi-conductors and the like. Again, this would help overwhelmingly if you decided to go into the field of conducting polymers—a currently very active field in R&D, especially if it involves replacing rare earths.
To summarise: materials science and chemistry go hand in hand in my quick overview, however some aspects of chemistry (such as organic) view the solid-state as the 'graveyard of chemistry'—you'd be hard pressed to find reactions which occur in the solid state! If you want to find out a little more before delving into materials, then finding a quick Youtube course might help to sway you either way. In my opinion there are far more ridiculous pairings in the world than chemistry and materials!


text:
tap water and distilled water for electrolysis
my water electrolyzer which I've built using this pdf, produces less gas than I expected. In that file the author insists on using distilled water instead of tap water and I have used tap water. would it be the reason?
Why is it so important to use distilled water since it has high resistance for electricity?

This article tells you to use distilled water because the salt in tap water will damage the booster. The author also tell you to add NaOH or KOH to the distilled water. Did you add these salts?

Yes I did add NaOH. So distilled water is only for the protection of booster and has nothing to do with performance?

In this article, the author only mentions the damage to the booster. Theoretically, tap water contains metal ions which are more prone to get electrons than hydrogen ions are, like Fe3+ and Cu2+, so they will hinder the process of producing hydrogen gas from water, while Na+ and K+ will not. It seems that the author considers tap water's damage to the booster more important than its ability to decrease the speed of making hydrogen gas, which should be the actual situation.


text:
Why do unbonded electrons exist in pairs?
Basically the term which we use to refer them is lone pair. In Lewis structure why we represent those unbonded electron in pairs. Like here (structure of SO2)

Here if we assume both the unbonded electrons to be existing in pairs then it's geometry will be different from what we would observe in case when both electrons exists freely (without pair).
So why we basically consider those electrons to be existing in pair? Is it any kind of observation or any theory or both?

Maybe you have better, more explanatory answers, if you briefly comment how far you are familiar with quantum chemistry, especially VB.

I've studied almost complete basic of chemical bonding (and that includes Valance Bond Theory). However I am in search of a descent explanation.

Why unbonded electrons exits in pair

Unbonded, non-bonded or lone pair electrons are terms used to describe electrons that surround an atom, but don't play a direct role in bond formation.  Importantly, these electrons do not always exist as lone pairs where the electrons reside in a common orbital.  Sometimes they exist as single non-bonding electrons residing in separate orbitals.
Let's compare the non-bonding electrons in water to those in molecular oxygen.  Here is an illustration of the molecular orbital arrangement for water.  As we fill the orbitals with electrons according to the Aufbau Principle we arrive at the point where we have two electrons remaining and the next available molecular orbital is the $\ce{1b_1}$ orbital.  According to Hund's Rule, we must place these two electrons into this orbital with their spins paired.  The result is that we have two non-bonding electrons paired up in the same orbital - a lone pair of electrons.

Let's next examine the situation in $\ce{O2}$.  Here is the molecular orbital diagram for $\ce{O2}$.  As we again fill the orbitals with electrons according to the Aufbau Principle we again arrive at the point where we have two electrons remaining and the next available molecular orbital is the $\ce{\pi^{x}_{g}}$.  But we see that there is a second orbital at exactly the same energy (degenerate) and it is the $\ce{\pi^{y}_{g}}$ orbital.  According to Hund's Rule, rather than put both electrons into one of these orbitals, we place just one electron in each of these two degenerate orbitals.  So while
$\ce{O2}$ does have two non-bonding electrons, they are not paired up - there is no lone pair, these electrons exist in separate orbitals and $\ce{O2}$ consequently has a triplet ground state resulting from these two unpaired non-bonding electrons.

So sometimes a non-bonding pair of electrons will exist as a lone pair (as in the case of water), sometimes they won't exist as a lone pair as in the case of oxygen.  To solve these kinds of problems one must know (or guess) the structure of the molecule and then determine the molecular orbital arrangement.  Then, when we fill these orbitals with electrons we will see if any non-bonding electrons will exist as a lone pair in a single orbital, or if the two non-bonding electrons will exist in separate orbitals and remain unpaired.

A single unpaired electron is called a radical, and usually aren't stable. Each electron orbital can hold two electrons, one spin up, other spin down. So a lone pair is just a full electron orbital, and cannot accept another electron. An orbital with 1 electron will readily react to get another electron and form a pair.
Check this: Free Radical
The first reaction they show is UV light being used to break $\mathrm{Cl}_2$ gas into $2\mathrm{Cl}$ radicals, each with a single lone electron, forming free radicals. These radicals could be used as free radical polymerization initiators, or for other reactions.
In the structure you show, the molecule forms a trigonal planar configuration, with the lone pair forming the third arm. If you could stably achieve two free radicals instead of the lone pair, it would have tetrahedral configuration, with the radicals forming 2 of the 4 arms.
The theory behind it is more quantum mechanics than I can remember, but I know it involves the Pauli exclusion principle and Hund's rule.

Thank You very much but I think this answer is rather explaining about the bond formation of free electrons (free radical), not the logic behind the existence of lone pairs. Don't you think?

Well, adding in some information from the other answer, all atoms and molecules have electron orbitals, and each orbital can hold up to 2 electrons. Also, each orbital has an energy level. Sometimes orbitals have the same energy level. When this happens, electrons will fill the orbitals one at a time, only adding a second electron if no other empty orbitals exist at that energy level. Look at the O2 MO diagram above. Lone pairs are formed because two electrons had to fit into 1 orbital at that energy level. This also explains magnetism, unpaired electrons will respond to a magnet.


text:
Crystal structure in function of pressure
If you were to investigate crystal structure changes as a function of pressure, which metals would you study first? Last? Why?

Seems like homework, so I'll give some hints:
What do you expect to happen to a crystal under pressure?
What does the crystal structure have to do with the response to pressure?
Why would the crystal structure want to change under pressure?
What crystal structures are common for metals? (there aren't many)
Since there aren't many, what distinguishes their responses to pressure that might make the metal want to change crystal structure (a phase change).
That should get you going.


text:
Are protons or electrons responsible for the chemistry of an atom?
Is the question in a correct form? I can't add anything as a comment.

Sirs, I'm not sure that the "closed" tag is fair.

The electrons are responsible for the actual chemistry happening around an atom, however the number of electrons in an atom is determined by the charge of the nucleus, i.e. by the protons. The protons are not really participating in the reactions, but they are the ones who compensate the charge of the electrons, therefore they are important to determine the energy levels of the free atom.
As an illustration, you can take helium $(\ce{He})$ and a hydride ion $(\ce{H-}).$ Both have 2 electrons on an 1s, however the $\ce{He}$ is neutral, while the nucleus of $\ce{H}$ is compensating only for one charge. This result in totally different energy levels and stability, different affinity for oxidation or other chemical reactions.

Of course then there is the whole 'nuclear chemistry' thing, but I always found that label to cause more confusion than insight!


text:
What makes sulfur-containing compounds such good antioxidants?
Many sulfur-containing compounds (like glutathione and cysteine groups) have high antioxidant activity. Why?

There are several factors that combine to make sulfur compounds such effective antioxidants.  First, sulfur compounds often contain S-H bonds (mercaptans) and are excellent radical scavengers because sulfur radicals are thermodynamically more stable than the analogous oxygen radical.  That is to say that the following equilibrium lies far to the right
$$\ce{ROS^{.} + RS-H <=> ROS-H + RS^{.}}$$
where ROS represents a "reactive oxygen species" such as hydroxyl radical ($\ce{HO^{.}}$), superoxide radical ($\ce{O_2^{-~.}}$, etc.  Sulfur compounds (sulfides and disulfides, for example) are also very easily oxidized to their corresponding sulfoxides.
$$\ce{R-S-R +ROS -> R-SO-R + deactivated ~ROS}$$
If multiple mercaptan groups exist within a molecule (for example multiple cysteine residues in an amino acid) they can be oxidized to the corresponding disulfide while reducing an oxidizer.
$$\ce{2~ RS-H + [oxidized~molecule] -> RS-SR + [H-reduced~molecule-H]}$$
This oxidative process provides sulfur compounds with a second pathway to trap or transform, and consequently inactivate, ROS.
In addition to being very effective ROS scavengers, sulfur compounds also bind strongly to many metal ions.  These same metal ions often play a role in the generation of ROS.  By binding to them, sulfur compounds effectively remove them from the ROS generating process thereby further lowering the concentration of ROS in the cell.


text:
What exactly does it mean for the reaction of superoxide with non-radicals to be spin-forbidden?
From Wikipedia:

The reaction of superoxide with non-radicals is spin forbidden. In biological systems,   this means its main reactions are with itself (dismutation) or with another biological radical such as nitric oxide (NO) or with a transition-series metal.

I know that one of the configuration bands is paired, and one is unpaired. Does the unpaired one have to bond with an unpaired one from another radical? What does this also say about reactions of the excited state of singlet oxygen (shown in the middle below) with non-radicals?

Also, is superoxide a doublet?

Yes, superoxide is a doublet. You have a similar electron configuration as $\ce{O2}$ but not 2 electrons in the $\pi^{*}$ orbitals but 3 (so two are paired and one is unpaired).

What exactly does it mean for the reaction of superoxide with
  non-radicals to be spin-forbidden?

For clarity, let's start with a simpler case.  If we take a ground state singlet molecule (singlet means all electrons are paired) and we irradiate it with photons, photon absorption will excite an electron to an excited state.  Will the excited state be a singlet or triplet immediately after it is produced?  For the transition to the excited state to occur readily it must be spin-allowed.  That means that spin must be conserved.  If we started with a singlet, then our excited state must still be a singlet.  In that excited state the excited electron will still have its spin paired with the electron that was its partner that remains in a lower energy orbital.  Creation of a triplet state, where the excited electron and its ground state partner no longer have their spins paired is spin-forbidden and is therefor much less likely to occur.  Spin must be conserved.
Well rules are made to be broken.  What can we do to increase the probability of a singlet -> triplet conversion that is formally spin-forbidden. Spin-Orbit coupling (the coupling of the electron spin with the orbital angular momentum of non-circular orbits) is our way out. We need to find a spot where the Morse potential curves overlap for both the singlet and triplet. Then, if there is a large spin-orbit coupling interaction we can readily cross over from singlet to triplet, or back.

Large atoms such as bromine and iodine, as well as transition metals with unpaired electrons have large spin-orbit couplings.  If such groups are attached to our molecule of interest, then spin need not be conserved and formally "spin-forbidden" reactions become more likely.

Does the unpaired one have to bond with an unpaired one from another
  radical?

Yes, if spin is to be conserved.

What does this also say about reactions of the excited state
  of singlet oxygen (shown in the middle below) with non-radicals?

The singlet oxygen shown in the middle of your figure ia a singlet species and will react readily with other singlet (non-radical) molecules.

is superoxide a doublet
  Yes superoxide is a doublet.  Ground state $\ce{O2}$ is a triplet - it has two unpaired electrons in energetically degenerate orbitals, if we add one electron to form superoxide radical anion, that electron will pair with one of those two electrons leaving only one unpaired electron.


text:
UV-VIS spectra of n-decanes
I was wondering if anyone knows where I can find UV-VIS spectra of the n-decanes.  Specifically tetradecane and dodecane. I can't seem to find such information using my regular search methods, so I was wondering if there was a common database that contains such things that I am simply unaware of (likely).
All I really need to know is if these chemicals will absorb radiation with wavelengths between 355-375 nm. A source to back this up will only help.

They are colourless liquids ....

To the best of my knowledge, ordinary alkanes do not typically absorb anywhere in the UV/Vis region. The organic molecules that do typically contain electrons in $\pi$- or non-bonding orbitals, which have energy level transitions corresponding to UV/Vis wavelengths. That said, I did find one paper1 describing absorptions around the far UV ambit, where $\lambda < 200\mathrm{nm}$.

DOI: 10.1021/jp903435c

Thank you for this information.  In addition to the paper you cited there, I found a book - "Spectrometric identification of organic compounds" - that confirms this information.


text:
Balancing ionic equations and determining number of ions
There were two questions that I've listed below that I've had troubles on. I got an answer in both, but both of them were incorrect.
For the first question, they list the answer to be D and then for the second they list the answer as A.

6.Determine the net ionic equation for the reaction between $\ce{HNO3}$ and $\ce{Mg(OH)2}$.
a. $\ce{2H+(aq) + (OH)2^{2-}(aq) -> H2(OH)2(aq) }$
b. $\ce{2HNO3(aq) + Mg(OH)2(aq) -> Mg(NO3)2(aq) + 2H2O(l) }$
c. $\ce{Mg2+(aq) + 2NO3 (aq) -> Mg(NO3)2(s) }$
d. $\ce{H+(aq) + OH(aq) -> H2O(l) }$
e. $\ce{Mg+(aq) + NO3 (aq) -> MgNO3(s) }$
8.Determine the number of ions that are present in $38.8~\mathrm{mL}$ of $0.113~\mathrm{M}$ $\ce{(NH4)2CO3. }$
a. $7.92\cdot 10^{21}~\text{ions }$
b. $2.64\cdot 10^{21}~\text{ions }$
c. $2.07\cdot 10^{23}~\text{ions }$
d. $6.20\cdot 10^{23}~\text{ions }$
e. $1.32\cdot 10^{23}~\text{ions }$

For the first question, I did this: $$\ce{2 H+ + 2NO3- +Mg(OH)2 -> 2H2O + Mg+2 + 2NO3-}$$ The question I have is why $\ce{Mg(OH)2}$ splits up. I don't get it because things you split up are things that are aqueous or soluble. $\ce{Mg(OH)2}$ isn't soluble so I listed it as a solid and thought it wouldn't break down. All I need to know is why I have to break down $\ce{Mg}$ and the $\ce{OH2}$, because they aren't soluble.
For the second problem I got an answer of $2.64\cdot 10^{21}~\text{ions}$ but the review said it was A. I set it up as $.113~\mathrm{M} = \text{moles} / 0.0388~\mathrm{L}$ and then multiplied $.113$ by $0.0388$ and then multiplied that amount by $6.022\cdot 10^{23}~\text{ions}$ which got me the answer B.

You are correct on the net ionic equation.  For the other, think about how many particles each ammonium carbonate generates when it dissolves.

Actually $Mg(OH)_2$ has (low) solubility ($K_{sp}=1.5\cdot 10^{-11}$). Since it is asking for the net ionic equation, the only ions reacting (since $NO_3^-$ stays in solution) are $H^+$ and $OH^-$.
You forgot that each molecule of ammonium carbonate produces 3 ions, so multiply your answer by 3.


text:
Finding orbit radius using the Bohr model and Rydberg equation
To start with its a homework problem, quite lengthy.

A particle of mass equal to 208 times the mass of an electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr model of the atom is applicable to this system,

Derive an expression for radius of $n$th Bohr orbit.
Find the value of $n$ for which the radius is equal to the radii of first orbit of hydrogen.
Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.

Now, I did the first part and got the answer correct. Here's what I did.
Suppose the mass of the particle revolving is $M$, its speed is $v$, and $M = 208 m_{e}$. Electrostatic force is the centripetal force. Therefore
$$
\begin{align}
\frac{Mv^2}{r} &= \frac{(ke)(3e)}{r^2} \\
v^2 &= \frac{3ke^2}{208m_{e}r}
\end{align}
$$
From the Bohr model,
$$
m_{e}vr=\frac{nh}{2\pi}
$$
where $h$ is Planck's constant. Therefore,
$$
v = \frac{nh}{2\pi m_{e}r}
$$
Squaring it,
$$
v^2 = \frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}
$$
Equating the two equations which have $v^2$ in them,
$$
\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2} = \frac{3ke^2}{208m_{e}r}
$$
After solving for $r$, we get something like this,
$$
r = \frac{n^2h^2}{4(\pi)^{2}3ke^{2}208m_e}
$$
All of the above is correct. The problem is in the second and third parts; when I put $r =\pu{0.53*10^{-10} m}$ I do NOT get the required answer. To approach the third part, I started with the standard Rydberg equation,
$$
\frac{1}{\lambda} = \mathcal{R}Z^2 \left( \frac{1}{n_f^2}-\frac{1}{n_i^2} \right)
$$
I plugged in each value, $n_i = 3, n_f = 1, Z = 3$; but again didn't get the answer correct.

The answer to the second part is 25 $(n=25)$; and to the third is 55.2 picometers.

To answer the second part:
We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$.
Part one has a mistake, as it is
$$
\begin{align}
&&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\
&&Mvr&=n\hbar\\
\implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2}
\end{align}
$$
We also know the Bohr radius:
$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$
Therefore we can write and cancel:
$$
\begin{align}
&& r&=a_0\\
&& \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\
\therefore&& Z\frac{M}{m_e}&=n^2\\
\therefore&& n&=\sqrt{Z\cdot208}\approx25
\end{align}
$$
The third part:
The Rydberg Formula is given as
$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
with the Rydberg $\mathcal{R}$ constant defined for a photon emitted by an electron. We'll assume that the mass of the nucleus is 7 atomic units (three protons + four neutrons). Taking into account that $m_p\approx 1836m_e$, we arrive at
$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{7\cdot1836m_e}}$$
Now the Rydberg constant has to be modified to include the mass of the particle:
$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$
With $\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$ (wikipedia), I got to $\lambda_\mathrm{vac}=55.6~\mathrm{pm}$.
Without taking the reduced mass into account, i.e. $\mathcal{R}\approx\mathcal{R}_\infty$ I arrived at $\lambda_\mathrm{vac}=54.8~\mathrm{pm}$.
Both values are reasonably close to the solution given.
(If the question was really about the muon, the more accurate weight ratio is 206.77 and the corresponding wavelengths 55.1 pm and 56.0 pm.)


text:
What is the result when the polarity of the solvent is increased in nucleophilic substitution?
I read in the Guidebook to Mechanism in Organic Chemistry that increasing the polarity of the solvent slightly reduces the rate of an $S_N2$ reaction. Why does this happen?
Another question:
We know that in $S_N2$, using a polar protic solvent would reduce the rate as the nucleophile gets solvated by hydrogen bonding, but I found in Solomon's Organic Chemistry that the $S_N1$ reaction rate increases in a polar protic solvent. Why does this occur? The nucleophile still gets solvated right?

I read in a (book) that increasing the polarity of the solvent slightly reduces the rate of SN2 reaction. Why does this happen?

Increasing the polarity of the solvent increases the solvation of the nucleophile. There will be increased dipole-dipole interactions between a solvent and a nucleophile as polarity of the solvent increase.
This will reduce the mobility of the nucleophile and in turn reduce the rate of reaction in an SN2 reaction. The SN2 mechanism's rate is determined by the rate at which the nucleophile and the substrate can randomly collide into each other (in the right way - i.e. you want the nucleophile to attack the backside of the leaving group).

But I found in another book that the SN1 reaction rate increases in polar protic solvent. Why does this occur? The nucleophile still gets solvated right?

Yes, the nucleophile is still solvated, but in an SN1 reaction, the rate-determining step is the rate at which the leaving group leaves. That is by far the slowest step in an SN1 reaction because of the high activation energy involved in creating a carbocation, which is highly unstable. So it doesn't really matter if the nucleophile is heavily solvated or not so heavily solvated - the slowest step determines the rate.
Also, regarding why polar protic solvent increase the rate of SN1 reactions - think about the carbocation. There is a high activation energy barrier in forming this carbocation - it's extremely unstable. Protic solvents can stabilize this carbocation through hydrogen bond formation. This lowers the activation energy barrier.

The standard textbook explanation is based on a comparison of charge separation in the ground state and the transition state (TS).  If there is more charge separation in the TS than the ground state, then polar solvents stabilize the TS more than the ground state and the reaction will be accelerated. On the other hand, if there is more charge in the ground state than in the TS, then polar solvents will stabilize the ground state more than the TS and consequently the reaction will slow down.
Let's look further at the effect of charge separation and solvent on these two specific reactions and see how they are affected.  Let's start with the $\ce{S_{N}1}$ reaction.

I found in another book that the SN1 reaction rate increases in polar
  protic solvent. Why does this occur?

In the $\ce{S_{N}1}$ reaction, the rate determining step (rds) is formation of the carbocation intermediate. Let's say we start with a charged nucleophile and a neutral compound.  Moving along the reaction coordinate, by the time we reach the carbocation intermediate, the starting materials (charged nucleophile plus neutral compound) have transformed into a charged carbocation plus a charged leaving group plus the charged nucleophile.  Moving from starting materials to the intermediate has created additional charged species (carbocation plus leaving group).
$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [R^+ +X^{-} + Nu^- ]~~ (the~ TS)}$$
Since the transition state leading to these intermediate species will strongly resemble the intermediate, it will also have more charge separation then the starting materials and therefor, will be more stabilized by polar solvents then the starting materials.  This stabilization of the transition state will lower its energy and speed up the rds.  In other words a polar solvent has sped up the rate of the $\ce{S_{N}1}$ because more charge exists in the transition state than in the ground state.

I read in a (book) that increasing the polarity of the solvent
  slightly reduces the rate of SN2 reaction. Why does this happen?

Now, lets perform the same analysis on the $\ce{S_{N}2}$ reaction.  Again we start out with a charged nucleophile and a neutral compound.  The rds in this case is the formation of a trigonal bipyramid species.
$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [\delta^{-} ~~ X...R...Nu ~~\delta^{-} ]~~ (the~ TS)}$$
In the ground state there was one unit of negative charge residing fully on the nucleophile. In the TS that unit of charge is now spread out over both the nuceophile and the leaving group; the charge has become much more diffuse, the charge density per unit volume has decreased in the TS compared to the ground state.  In this case, we would expect a polar solvent to stabilize the ground state (starting materials) slightly more than the still charged - but more diffusely charged - TS.  Stabilizing the starting materials slightly more than the TS would slightly increase the energy required to pass over the TS.  Hence we would expect polar solvents to slightly slow down the rate of an $\ce{S_{N}2}$ reaction.


text:
Why does bond angle decrease in the order H2O, H2S, H2Se?
I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how?

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:
\begin{array}{lcc}
\text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\
\hline
\ce{H2O} & 104.5 &  96 \\
\ce{H2S} &  92.3 & 134 \\
\ce{H2Se}&  91.0 & 146 \\
\hline
\end{array}
The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$.  On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization.  That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively.  Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons.  If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$.  We see from the above table that we are very close to the measured values.  We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider.  This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.
The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum.  The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

Neat answer Ron. The H-S-H bond can open up a bit because the other side of the p-orbital is more empty as a result of the S-H bond, but not too much of course because there is still electron density there. Is that the force stopping it from going all the way to a 180 degree bond? Or are there other forces involved? (hope this was a bit clear, just curious)

Thanks Jori.  Each S-H bond uses one p orbital and each p orbital is oriented roughly 90 degrees from the other.  Px and Py, or PX and Pz, or Py and Pz - pick which two you'd like to use to make the two S-H bonds, but they are all 90 degrees from one another.  There's no way to get bonds 180 degrees apart using pure p orbitals.

Well I imagined that they could turn a little bit (like you said in the answer) as the direction the orbital turns has a lower electron density (it's a bit hard to explain for me without making some graphs) because the S-H Se-H or w/e bond is on the other side. In an extreme case where the bond takes all the electron density away from the other lobe of the p orbital I could imagine two adjacent p orbitals overlapping (i.e. the bonding lobe part and a non-bonding lobe part of the other p orbital). But like I said that probably isn't the case, because there will be still some electron (cont.)

density on the other lobe of the p-orbital. I just wondered if that was the correct force in the process or if there is some other force restricting the bending of the H-S-H (resp. H-Se-H) bond. (I could add a drawing to show more clearly what I mean, but I'm no good at drawing haha, but if necessary to get my idea across I'll try it)

Yes the bonds can bend a little, but two p orbitals (or more accurately, their wavefunctions) cannot interact since they are orthogonal to each other.  Also, yes, there will always be electron density throughout the entire p orbital.  Bonding will shift the density somewhat, but it will still exist throughout the entire orbital.  A picture would probably help a lot.

Very well spoken! You can make a 180 deg bond with only p orbitals, but this will be a three centre bond, just like in the xenon fluorides. In terms of MO theory, the bending is described by mixing in a little s- and a little d- character (the latter ones are polarisation functions). You can also apply Coulson's theorem and see, that there is tiny bit hybridisation, but of course you knew that already. @Jori I think I get what you want to say, and it might be worth asking this as a new question.

Hybridisations are a self-convinient model to explain certain facts. Why have we ignored the hybridisations in H2S H2Se? Was it just to support our experimental observations or does it have a concrete reason?

In addition to "the sp3 explanation does not fit with all of the experimentally observed data," it is also inconsistent with our theory of molecular orbitals, from which the diagrams in porphyrin's answer come. To quote Albright's "Orbital interactions in chemistry", the idea that S uses pure p orbitals for bonding in SH2 "is far removed from reality."

The question asks why water has a larger angle than other hydrides of the form $\ce{XH2}$ in particular $\ce{H2S}$ and $\ce{H2Se}$. There have been other similar questions, so an attempt at a general answer is given below.
There are, of course, many other triatomic hydrides, $\ce{LiH2}$, $\ce{BeH2}$, $\ce{BeH2}$, $\ce{NH2}$, etc.. It turns out that some are linear and some are V shaped, but with different bond angles, and that the same general explanation can be used for each of these cases.
It is clear that as the bond angle for water is neither $109.4^\circ$, $120^\circ$, nor $180^\circ$ that $\ce{sp^3}$, $\ce{sp^2}$ or $\ce{sp}$ hybridisation will not explain the bond angles. Furthermore, the UV photoelectron spectrum of water, which measures orbital energies, has to be explained as does the UV absorption spectra.
The way out of this problem is to appeal to molecular orbital theory and to construct orbitals based upon $\ce{s}$ and $\ce{p}$ orbitals and their overlap as bond angle changes. The orbital diagram was worked out a long time ago is now called a Walsh diagram (A. D. Walsh J. Chem. Soc. 1953, 2262; DOI: 10.1039/JR9530002260). The figure below sketches such a diagram, and the next few paragraphs explain the figure.

The shading indicates the sign (phase) of the orbital, 'like to like' being bonding otherwise not bonding. The energies are relative as are the shape of the curves. On the left are the orbitals arranged in order of increasing energy for a linear molecule; on the right those for a bent molecule. The orbitals labelled $\Pi_\mathrm{u}$ are degenerate in the linear molecule but not so in the bent ones. The labels $\sigma_\mathrm{u}$, $\sigma_\mathrm{g}$ refer to sigma bonds, the $\mathrm{g}$ and $\mathrm{u}$ subscripts refer to whether the combined MO has a centre of inversion $\mathrm{g}$ (gerade) or not $\mathrm{u}$ (ungerade) and derive from the irreducible representations in the $D_\mathrm{\infty h}$ point group. The labels on the right-hand side refer to representations in the $C_\mathrm{2v}$ point group.
Of the three $\Pi_\mathrm{u}$ orbitals one forms the $\sigma_\mathrm{u}$, the other two are degenerate and non-bonding.
One of the $\ce{p}$ orbitals lies in the plane of the diagram, the other out of the plane, towards the reader.
When the molecule is bent this orbital remains non-bonding, the other becomes the $\ce{3a_1}$ orbital (red line) whose energy is significantly lowered as overlap with the H atom's s orbital increases.
To work out whether a molecule is linear or bent all that is necessary is to put electrons into the orbitals. Thus, the next thing is to make a list of the number of possible electrons and see what diagram predicts.
\begin{array}{rcll}
  \text{Nr.} &
  \text{Shape} &
  \text{molecule(s)} &
  \text{(angle, configuration)} \\
  \hline
  2  & \text{bent}   & \ce{LiH2+} & (72,~\text{calculated})\\
  3  & \text{linear} & \ce{LiH2}, \ce{BeH2+} &\\
  4  & \text{linear} & \ce{BeH2}, \ce{BH2+}  &\\
  5  & \text{bent}   & \ce{BH2}   & (131, \ce{[2a_1^2 1b_2^2 3a_1^1]})\\
  6  & \text{bent}   & \ce{^1CH2} & (110, \ce{[1b_2^2 3a_1^2]})\\
     &               & \ce{^3CH2} & (136, \ce{[1b_2^2 3a_1 1b_1^1]})\\
     &               & \ce{BH2^-} & (102)\\
     &               & \ce{NH2+}  & (115, \ce{[3a_1^2])}\\
  7  & \text{bent}   & \ce{NH2}   & (103.4, \ce{[3a_1^2 1b_1^1]})\\
  8  & \text{bent}   & \ce{OH2}   & (104.31, \ce{[3a_2^2 1b_1^2]})\\
     &               & \ce{NH2^-} & (104)\\
     &               & \ce{FH2^+} &\\
  \hline
\end{array}
Other hydrides show similar effects depending on the number of electrons in $\ce{b2}$, $\ce{a1}$ and $\ce{b1}$ orbitals; for example:
\begin{array}{ll}
  \ce{AlH2}  & (119, \ce{[b_2^2 a1^1]}) \\
  \ce{PH2}   & (91.5, \ce{[b_2^2 a_1^2 b_1^1]}) \\
  \ce{SH2}   & (92)\\
  \ce{SeH2}  & (91)\\
  \ce{TeH2}  & (90.2)\\
  \ce{SiH2}  & (93)\\
\end{array}
The agreement with experiment is qualitatively good, but, of course the bond angles cannot be accurately determined with such a basic model only general trends.
The photoelectron spectrum (PES) of water shows signals from $\ce{2a1}$, $\ce{1b2}$, $\ce{3a1}$, $\ce{1b1}$ orbitals, ($21.2$, $18.7$, $14.23$, and $\pu{12.6 eV}$ respectively) the last being non-bonding as shown by the lack of structure. The signals from $\ce{3b2}$ and $\ce{3a1}$ orbitals show vibrational structure indicating that these are bonding orbitals.
The range of UV and visible absorption by $\ce{BH2}$, $\ce{NH2}$, $\ce{OH2}$ are $600 - 900$, $450 - 740$, and $150 - \pu{200 nm}$ respectively. $\ce{BH2}$ has a small HOMO-LUMO energy gap between $\ce{3a1}$ and $\ce{1b1}$ as the ground state is slightly bent. The first excited state, is predicted to be linear as its configuration is $\ce{1b_2^2 1b_1^1}$ and this is observed experimentally.
$\ce{NH2}$ has a HOMO-LUMO energy gap from $\ce{3a_1^2 1b_1^1}$ to $\ce{3a_1^1 1b_1^2 }$, so both ground and excited states should be bent, the excited state angle is approx $144^\circ$. Compared to $\ce{BH2}$, $\ce{NH2}$ is more bent so the HOMO-LUMO energy gap should be larger as observed.
$\ce{OH2}$ has a HOMO-LUMO energy gap from $\ce{3a_1^2 1b_1^2}$ to $\ce{3a_1^2 1b_1^1 4a_1^1 }$, i.e. an electron promoted from the non-bonding orbital to the first anti-bonding orbital. The excited molecule remains bent largely due to the strong effect of two electrons in $\ce{3a1}$ counteracting the single electron in $\ce{4a1}$. The bond angle is almost unchanged at $107^\circ$, but the energy gap will be larger than in $\ce{BH2}$ or $\ce{NH2}$, again as observed.
The bond angles of $\ce{NH2}$, $\ce{NH2-}$ and $\ce{NH2+}$ are all very similar, $103^\circ$, $104^\circ$, and $115^\circ$ respectively. $\ce{NH2}$ has the configuration $\ce{3a_1^2 1b_1^1}$  where the $\ce{b1}$ is a non bonding orbital, thus adding one electron makes little difference, removing one means that the $\ce{3a_1}$ orbital is not stabilised as much and so the bond angle is opened a little.
The singlet and triplet state $\ce{CH2}$ molecules show that the singlet has two electrons in the $\ce{3a1}$ orbital and has a smaller angle than the triplet state with just one electron here and one in the non-bonding $\ce{b1}$, thus the triplet ground state bond angle is expected to be larger than the singlet.
As the size of the central atom increases, its nucleus becomes more shielded by core electrons and it becomes less electronegative. Thus going down the periodic table the $\ce{X-H}$ bond becomes less ionic, more electron density is around the $\ce{H}$ atom thus the $\ce{H}$ nucleus is better shielded, and thus the $\ce{X-H}$ bond is longer and weaker. Thus, as usual with trends within the same family in the periodic table, the effect is, basically, one of atomic size.
Molecules with heavier central atom, $\ce{SH2}$, $\ce{PH2}$, etc. all have bond angles around $90^\circ$. The decrease in electronegativity destabilises the $\Pi_\mathrm{u}$ orbital raising its energy. The $\ce{s}$ orbitals of the heavier central atoms are larger and lower in energy than those of oxygen, hence these orbitals overlap with the $\ce{H}$ atom's $\ce{s}$ orbital more weakly. Both these factors help to stabilise the linear $3\sigma_\mathrm{g}$ orbital and hence the $\ce{4a1}$ in the bent configuration. This orbital belongs to the same symmetry species as $\ce{3a1}$ and thus they can interact by a second order Jahn-Teller interaction. This is proportional to $1/\Delta E$ where $\Delta E$ is the energy gap between the two orbitals mentioned. The effect of this interaction is to raise the $\ce{4a1}$ and decrease the $\ce{3a1}$ in energy. Thus in going down the series $\ce{OH2}$, $\ce{SH2}$, $\ce{SeH2}$, etc. the bond angle should decrease which is what is observed.
Example have been given for $\ce{XH2}$ molecules, but this method has also been used to understand triatomic and tetra-atomic molecules in general, such as $\ce{NO2}$, $\ce{SO2}$, $\ce{NH3}$, etc..

Adding a bit to the answers above, one factor that isn't shown in the Walsh diagram is that as the angle decreases, there is increased mixing between the central atom valence s and p orbitals, such that the 2a$_1$ orbital has increased p contribution and the 3a$_1$ has increased s. This is where one gets the result that Ron mentioned at the end of his answer that the lone pairs on water reside in a pure p (1b$_1$) and an sp (3a$_1$) orbital. That means the bonding orbitals shift from one pure s (2a$_1$) and one pure p (1b$_2$) to one sp (2a$_1$) and one p (1b$_2$) (ignoring the extreme case where 3a$_1$ actually gets lower in energy than 1b$_2$, which isn't really relevant). Mixing occurs to a greater extent in $\ce{SH2}$ relative to $\ce{OH2}$ because the 3s and 3p orbitals of S are closer in energy to each other than 2s and 2p on O.
If we hybridize the two bonding orbitals so that they are equivalent and do the same for the two nonbonding orbitals, we find that they start as bonding = 50% s/50% p (ie $sp$ hybrid) and nonbonding = 100% p and shift towards an endpoint of bonding and nonbonding both being 25% s/75% p (ie $sp^3$ hybrid).
Thus, the common introductory chemistry explanation that "bonding in $\ce{SH2}$ is pure p" is not supported by the MO analysis. Instead, $\ce{SH2}$ is closer to $sp^3$ than $\ce{H2O}$ is. The bonding orbitals in $\ce{H2O}$ are somewhere between $sp^2$ and $sp^3$. So it is correct to say that "the bonds in $\ce{SH2}$ have less s character than those in $\ce{OH2}$", but not to say that they are "pure p".
The fact that the $\ce{SH2}$ bond angle is around 90 degrees is not because its bonds are made from p orbitals only. That coincidence is a red herring. Instead, the fact that the bond angle is smaller than the canonical $sp^3$ is because the bonding and nonbonding orbitals are not equivalent. That means that the particular p orbitals involved in each $sp^3$ group do not have to have the same symmetry as in, for example, a tetrahedral molecule like CH4.

I will try to give u a most appropriate and short answer that u can understand easily
See  h20 has 104.5 degrees bond angle , h2s has 92degrees , h2se has 91degrees and h2te has 90degrees bond angles
Draw diagrams of these u will find all of them have tetrahedral shape with 2 lone pairs , assume that no hybridization occurs and all these central atoms are using pure p orbitals for bonding then because of repulsions by lone pairs the bond angle should be 90degrees between 2 surrounding atoms , now according to dragos rules when central atom belongs to 3rd period or higher and electro negativity of surrounding atoms is 2.5 or less then central atom uses almost pure p orbitals .
So the final answer is the extenend of hybridization decreases In this case which leads to decrease in bond angle . note that only in h2te no hybridization is observed.

We know that as the electronegativity of central atom increases, the bond angles also increase. The relevant electronegative order is $$\ce{O > S > Se}\,,$$ hence the bond angle order of $$\ce{H2O>H2S>H2Se}\,.$$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


text:
What sort of lab can determine PAH concentration in a soil or charcoal sample? How difficult is it to do?
Non-chemist here.  I'm curious how difficult it is to determine concentration of PAH in a sample of soil or powdered charcoal, or a sample of the two mixed together.
What would be involved in doing so?
Is it particularly specialized, or the sort of thing that any trained chemist could do?
Is it particularly difficult?  Costly?  Specialized?  Or would most any chemistry lab have the tools to do it?
Any idea if there are any commercial labs where samples can be shipped for analysis?  Anywhere in the world, or in the US in particular?

It is most likely done using chromatomass-spectrometry of toluene extract. Extract itself is pretty easy to make, but finding chromatomass-spectrometer tuned for specific needs is probably hard. Large chemical universities, oil and coal chemical plant should have proper facilities.

@permeakra I'd say if you can ballpark a cost (reagents, time, etc.) that's a great answer.

@jonsca The time is AFAIK reasonable, no more then a day for the analysis itself, no idea for other procedures. I'm not aware of the exact cost, this is why I posted a comment =). While materials used (liquid nitrogen for the mass-spectrometer, carrier for chromatographer) cannot cost much, the cost of the hardware, software and mass-spectrum library to identify components is likely to be high, so it must cost something. I guess it should be ~100-1000 dollars, but this is only a guess, and an uninformed one.

@permeakra Gotcha.  Well, we'll see if someone else has a better idea of those costs.  It might even work to say that it varies based on the volume of tests that the institution performs or something (since it's something that's going to vary by region, anyway).

Thanks folks for the ideas.  Any idea how common a procedure it is?  Are there any sorts of labs for which it'd be routine?


text:
Meaning of Cu1.5Al in Cu-Al phase diagram (I understand all other indexes incl. fractions like Cu0.75Al0.25)
I understand Cu0.75Al0.25 - that's just 75% Cu, 25% Al. I understand integer indexes - that's a normal chemical formula after all.
However, I have no idea how to interpret Cu1.5Al, I've never seen a fraction in a chemical formula, and if it was a composition-percentage it should be like in the 0.75-0.25 example?
Look at between 50% and 60%, below 630°C in the diagram below.
I read this page but still don't understand it.

As the page you referenced mentioned, things can get a bit tricky.
I will start with your example of $\ce{Cu_{0.75}Al_{0.25}}$ - that is not just $75%$ Cu and $25%$ Al - that phase has a particular crystal structure, with the prototype (basically the first identified alloy with that crystal structure) of W (tungsten). So, it has the arrangement of atoms in the unit cell in the same places as pure tungsten, but $1/4$ of them are Al, and $3/4$ of them are Cu. There is no ordering of these atoms on the W lattice sites, or else the symmetry would be different, and the prototype might just be $\ce{Cu3Al}$.
For $\ce{Cu_{1.5}Al}$ it is similar, but perhaps even more confusing.  The 'Published phase label' is $\ce{Cu_{1.5}Al}$, so that is what it is usually called in the literature.  The 'formula' is $\ce{Cu_{1.7}Al}$, which just means when people started referring to the phase they didn't have an accurate composition measurement.  Then, the 'Prototype' is $\ce{Co_{1.75}Ge}$.  That means that the first similar crystal structure was measured in the Co-Ge system.  The non-integer subscript(s) indicate that the unit cell has vacancies in the ideal structure (without looking it up, I would speculate that the Co sites are $7/8$'s occupied).
To further complicate matters, if you look at the Co-Ge phase diagram you will find that the phase label is now $\ce{Co_{1.67}Ge}$, Formula is $\ce{Co_{1.5}Ge}$, and the prototype is still called $\ce{Co_{1.75}Ge}$.
So: crystal structures can be complicated, identification of composition has been getting better over time (folks have been looking at the Al-Cu system for a long time!), and if you really want to know what is going on you probably have to dive into the x-ray diffraction literature and look at the published atom positions in a given phase.


text:
How to draw resonance structures for CH3COCN
What are the resonance structures of $\ce{CH3COCN}$ molecule?

You mean Pyruvonitrile?

Here are the significant resonance structures for pyruvonitrile.

just curios, how do you draw these structures?

I use ChemSketch, it's free.

what is ChemSketch?

How to show lone pairs electrons of atoms of these structures?

ChemSketch is a free software package that let's you draw chemical structures.  It's free, so it doesn't do things like show lone pairs very easily, but it can be done in a convoluted sort of way.  http://www.acdlabs.com/resources/freeware/


text:
Writing Topology file for Ligand for Gromacs
Is there any tutorial for how to manually write the topology file for any Ligand file(PDB for mol2)? There are so many sites which provides it like ProDrug or ATB but no where the actual process is written. Can anyone explain the process?
PS: I am actually not sure this question belong to which SO.

First off, you should not be manually writing topology files whenever you can avoid it - it's far too prone to making silly errors. That's why there are so many programs out there that attempt to automate the process.
If however, you just want to understand HOW the topology file is structured, I recommend reading the GROMACS manual (chapter 5). These tutorials also do a good job of describing some of the key aspects of the topology files.

Thanks for the answer but I know the tutorial and I have mentioned both ProDrug and ATB. But every time get errors so here in this post they ask to write it manually. Can you suggest any offline tool?

From the tutorials I linked, here are a few tools used for different forcefields. Not all of them are actively maintained so you do need to actually understand how the topology SHOULD look and double check the output. You should really take some time to read the chapter in the manual so you understand the structure of the files.
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin/gmx-tutorials/complex/02_topology.html


text:
How is ammonia used in production or processing of steel?
There are some large tanks of ammonia in nearby iron works.

I wonder, what do they need it for? What is the application of ammonia in steel industry - what processes of making or processing steel require this much ammonia?

Ammonia is used for production of Nitrogen and Hydrogen using Ammonia Cracking.

Likely, for nitriding http://en.wikipedia.org/wiki/Nitriding . Essentially, ammonia is used as nitrogen donor. Nitrogen atoms diffuse into steel, producing hardened, corrosion-resistant surface layer.

Indeed - N2 will not nitride steel in a thermal process.  NH3 breaks up much more easily, so that is used.


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What are the properties of alloys?
You spend a lot of time in chemistry talking about molecular amd ionic compounds, but I'm interested in what you can do with metals only. What is the difference between the properties (physical & chemical) of pure metals and those of alloys? Is it possible to predict the properties of alloys before making them? Can you make alloys with mercury?

What seems to be a set of simple questions leads one into the field of metallurgy, computational thermodynamics, mechanical properties, and goodness knows what else.
For example, steel has very different mechanical properties from pure iron.  There are about a bazillion different kinds of steel, each with different impurities mixed in to give specific mechanical properties, optimizing strength, fatigue life, corrosion resistance, etc. Much of the development has been through insight gained by experience.  More recently, the CALPHAD approach is giving more fundamental insight through thermodynamic modeling, but predicting the mechanical properties $\it a\ priori$ is still difficult to impossible.
As for mercury, there are a large number of binary phases of Hg with other elements.  These are readily found in binary alloy references, or online at the ASM Alloy Phase Diagram Database (assuming your library has access).  ASM has info on >60 Hg binary diagrams, and a number of ternary systems as well.