text:
How do we know that water has [H+] = 10^-7?
In the derivation of $K_\mathrm{w}$, the ionic product of water, it is stated that
$$[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$$
and substituting $[\ce{H+}] = [\ce{OH-}] = 10^{-7}$ we can find that $K_\mathrm{w} = 10^{-14}$.
How do we know, though, that $[\ce{H+}] = 10^{-7}$? And why is $[\ce{H+}] = [\ce{OH-}]$?

Why is $[\ce{H+}] = [\ce{OH-}]$?

Because the water is electrically neutral.
In general, for all solutions, the sum of the concentration of each ion, multiplied by the ion's respective charge, must be zero.

How do we know $[\ce{H+}] = 10^{-7}$?

This is found by experimentally by determining $K_\mathrm{w}$ and substituting, or by measuring the pH.
For details see: Bandura, A. V.; Lvov, S. N. The Ionization Constant of Water over Wide Ranges of Temperature and Density. J. Phys. Chem. Ref. Data 2006, 35 (1), 15–30. DOI: 10.1063/1.1928231.

Thanks ! But
How do we know concentration of H+ is 10^-7 while deriving Kw  ?

$K_w$ is measured. It does depend on temperature. pH of pure water is 7.47 at 0 degrees C and 6.14 at 100 degrees C. http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html


text:
Typesetting oxidation state and charge simultaneously
I'm currently trying to typeset the ion $\ce{Cr2O7^{2-}}$ with oxidation numbers on. My first thought was to put the oxidation state above the atom, but according to IUPAC, the oxidation state should be in the right superscript. Combined with charge, it looks ugly and wrong:
My current ways of doing this are below

The oxidation number in right superscript. Looks weird with both numbers so close.
The oxidation number above. Looks better, but is against IUPAC.

My question therefore is: Which is the preffered way to typeset this?

Also, I would like to excuse beforehand if this should rather be in tex.stackexchange.

My personal preference in this situation would be to put square brackets around the whole ion and put the charge outside.
$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$ gives:
$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$
This is sometimes done with polyatomic species anyway (particularly metal-ligand complexes) to emphasise that the charge is on the whole ion, not on any one particular atom.

Take a look at this answer, which was edited to match IUPAC convention.  Oxidation states in formulae are actually written as Arabic numerals, with either a + sign or a - sign before the number (where it is nonzero), on top of the element symbol.  Compare with ion charges written as ordinary superscript with the charge following the number:
$\ce{\overset{+4}{Si}\overset{-1}{H}_4}$
$\ce{\overset{+6}{Cr}_2\overset{-2}{O}_7^{2-}}$ or $\ce{[\overset{+6}{Cr}_2\overset{-2}{O}_7]^{2-}}$

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text:
Variation of angle in GeX2 molecules where X is an halogen
I'm wondering what is the exact reasoning to explain why in $\ce{GeX_2}$ the angle $\widehat{XGeX}$ is smaller when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$.
I ask this question because I had an exams some days ago and I failed in this question and for me my reasoning is not false.
What I said is that

From $\ce{F}$ to $\ce{I}$ the difference of
  electronegativity between the germanium and the halogen is
  decreasing. Then the electrons in the $\ce{Ge-X}$ bond are more near
  $\ce{X}$ when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$. Then the two
  bonds repulse themselves more for $\ce{GeI_2}$ than for $\ce{GeF_2}$ and so the angle increase from $\ce{F}$ to
  $\ce{I}$.

Why my reasoning is absolutely false, and what could be the perfect one?

http://chemistry.stackexchange.com/questions/15620/what-is-bents-rule

This is not a duplicate of the question linked by @Mithoron. Yes, the short version of the answer is "Bent's rule", but just because the answers are similar, it does not mean the questions are the same.

For me bent' s rules don't explain it and i want to know why my reasonning is false.

As for all atoms of period 3 and higher, we can consider germanium to be essentially unhybridised and the bonds being formed entirely by $\unicode[Times]{x3c3}_\mathrm{p-p}$ overlaps. Therefore, we expect a perfect $90^\circ$ angle for all three compounds.
However, while fluorine is small, the size of the halogens increases from fluorine to iodine. The larger the halogen the more space it requires. Since the central atom remains the same size (and, let’s approximate, is roughly the size of bromine) it cannot offer more space for a strict $90^\circ$ angle without lengthening (= weakening) the bonds which would be unfavourable. It is less unfavourable to widen the angle to generate more space. This requires the s-orbital to hybridise slightly (less favourable) but keeps the desired shorter $\ce{Ge-X}$ bond length (more favourable than anything else in this consideration). Not the bonding electrons repulse each other, but the steric stress of the halogen’s lone pairs causes them to move apart. The same effect can be observed in $\ce{NH3}$ versus $\ce{PH3}$ or $\ce{PH3}$ versus $\ce{PF3}$.
Note the following: If you have two bonds from a central atom which are each described by a bonding and an antibonding orbital, then these two sets of orbitals will always be orthogonal to each other (the integral of overlap is $0$). Thus they will be indifferent to each other, not repulsive.

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text:
Oxidation state of iron ions in a unit cell

Consider UC (unit cell) for a cubic ionic lattice containing 4 oxide ions - what is the avg oxidation state for Fe ions in this lattice?

The four oxide ions create a -8 charge that must be counteracted by the iron ions. In a FCC unit cell, there are 3 total Fe ions per UC; therefore each ion must have a +8/3rds oxidation state.
Is this correct?

It's mixed oxide - average ox state is 8/3, but one center has +2 and two have +3.

Fe oxides are, in fact, notorious (famous?) for having mixed oxidation states.


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How to determine the configuration of an propargylic alcohol?
I am trying to determine whether the stereocenter of 2-methyl-1-penten-4-yn-3-ol is R or S.

With the ghost atom approach I am determining S:

My supervisor said the stereocenter is actually R configuration. Could someone please explain this and tell me where I went wrong?

I think your working is correct, and the molecule should be assigned as S, with the alkyne taking priority over the disubstituted alkene such that the priority goes:

alcohol
alkyne
alkene
hydrogen

This leads to the S configuration, as you propose.
There is a chance that your supervisor is confused because in IUPAC naming nomenclature, the alkene often (and indeed, in this case) takes priority over the alkyne, which would of course lead to assignment of the molecule as R.

Chemdraw agrees with this analysis.

Chemdoodle disagrees and says the molecule is (3R)-2-Methyl-1-penten-4-yn-3-ol.

Chemspider agrees with S (CSID:61500095).

As always, it should be noted that many chemical software packages do not completely conform with IUPAC rules. The most reliable software by far is by ACD (e.g. online at http://ilab.psds.ac.uk/), which gives (S).

Spoiler alert: the answer is (S).
The complete rules for determining relative priorities of substituents is described fully in P-92 of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book).
As stated in P-92.1.4.2, double- and triple-bonded groups are expanded into versions where each multiply-bonded atom is duplicated as many times as is necessary. However,

only the doubly bonded atoms themselves are duplicated, and not the atoms or groups attached to them; the duplicated atoms may thus be considered as carrying three phantom atoms [...] of atomic number zero.

Phantom atoms have the lowest priority, below even hydrogen atoms, since their atomic number is zero. Using this, we can construct the full digraph. Here, (C) represents a duplicated carbon atom, which is a carbon atom carrying three phantom atoms (denoted as zeroes).

The atoms are colour-coded according to the sphere that they are in: essentially, this refers to the number of bonds from the stereogenic unit (which, in this case, is a chiral centre). The blue carbon is directly attached to the chiral centre, and so is in "sphere I". The green carbons, two bonds away from the chiral centre, are in "sphere II"; and the red atoms are in "sphere III".
P-92.1.5 describes the rules for precedence. The wording is quite difficult to understand, and I cannot yet convince myself to directly quote from the rules. However, having studied many of the examples given in the subsequent section, the analysis broadly proceeds as follows (I think):

Sphere I for both groups simply consists of 'C', and no comparison can be made.
Sphere II is 'C(C)(C)' for the alkyne, and 'CC(C)' for the alkene. No comparison can be made at this point (there is no rule stipulating that normal carbon atoms have greater priority over duplicated carbon atoms).
Moving on to Sphere III, for the alkyne, we have the three groups 'CCH > 000 > 000'; and for the alkene, we have 'CHH > HHH > 000'.
Comparing the first of each group of three, since CCH has a greater ranking than CHH, we can assign the greater priority to the alkyne group.

See also: Assign stereodiscriptor to 4th carbon of 3-ethyl-2,4-dimethly-5-(propan-2-yl)heptane-1,7-diol for a similar situation.

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text:
Is the triiodide ion polar?
Three professors argue it is non-polar.
My professor argues that it is a monopole, like most ions.
The structure of the triiodide ion places a negative formal charge on the central iodine atom. The molecular geometry is also linear (at least according to VSEPR), and its electronic geometry is trigonal bipyramidal. Given the high amount of symmetry, shouldn't triiodide be non-polar? Sure, it might have a charged central atom, but so does carbon dioxide.
What's the answer?

$D_{\infty\mathrm{h}}$ molecules like the triiodide ion cannot be polar. Polarity in the context of chemistry would mean having a dipole moment so imo the monopole is irrelevant

related http://chemistry.stackexchange.com/questions/7800/why-fluorine-doesnt-act-as-the-central-atom-in-inter-halogen-compounds/17074?s=12|0.0561#17074 - seems there's actually partial positive charge on cent. at.

@orthocresol You should know better than to describe polarity only with dipole of molecule.

also somewhat related http://chemistry.stackexchange.com/questions/27920/why-is-nacl3-possible

@Mithoron this is going into semantics already - if you consider triiodide polar because each individual I-I bond is polar, then you must also consider CO2, BF3 and CCl4 to be polar. In all cases there is a partial positive charge on the central atom.

Yes, polarity isn't well defined and I'm not even sure if it makes sense to ask about polarity of anion.

If one defines polarity as having a non-zero electric dipole, then in that strict sense, yes, there are non-polar anions and cations. Whether such a classification is useful is another matter. The thing is that cations and anions are almost always found together, and the smallest unit of the compound (a single ionic pair) does have an electric dipole, whether or not each of its ions has one too. See my comment in this previous related question.

http://goldbook.iupac.org/P04710.html - IUPAC definition says nothing about any dipoles, and it's rather macroscopic property not microscopic, seems "polar" was simply convenient adjective for describing molecules, but isn't necessarily proper for it.

I think you have to ask the professor to define 'polar' and 'non-polar'. This sounds like a miscommunication.

The structure of the triiodide ion places a negative formal charge on the central iodine atom.

No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).
That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.
The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.

Can you include the resonance structures you are talking about? I have a hard time seeing the formal charge being on the terminal iodine.

@Martin-マーチン I can’t draw it atm, but it is: $\ce{I-I}~\ce{I-}$

I see. I did not think about this. So the resonance you are talking about would be $$\ce{I-I\bond{<-}I- <=> {}^{-}I\bond{->}I-I}?$$ But I think we shall not forget about other resonance structures, too. $$\ce{{}^{-}I\bond{->}\overset{+}{I}\bond{<-}I- <=> I-{I- }-I}$$

@Martin-マーチン And I disagree with the last one because it creates a dectet on the central iodione.

You realize that all those resonance contributors violate the octet rule, at least when you want to describe their bonding as covalent.

@Martin-マーチン If I see that arrow as a bond, yes.

I would argue it is a polar molecular ion.‡ In the comments to this question there has been already pointed out that the term "polarity" is quite ill-defined.
Many chemists understand that if a molecule has a vanishing dipole moment, the molecule itself is non-polar; at least this is true since molecules are neutral by definition. If you actually can extend that definition to charged species is open for debate. I personally dislike this definition - molecule or not.
In this framework carbon dioxide is non-polar, but it reacts with polar substrates, like Grignard agents. The reason for this is its quadrupole moment.
A much more general definition of the term polarity hence should include any molecule that has a multipole of any kind. We can then even include ions.

A polar molecular entity is a molecule that possesses a multipole.

Since the triiodide anion has a non-zero quadrupole moment, I would consider it polar. This point has been and will always  be open for debate.
A quick DF-BP86/def2-SVP calculation determines the natural atomic charges as $q(\ce{I_{\mu}})=-0.11$ (the atom in the middle) and $q(\ce{I_{term}})=-0.44$ (the terminal two; deviations from $1.00$ due to rounding). The same calculation finds:
 Traceless Quadrupole moment (field-independent basis, Debye-Ang):
   XX=             10.9525   YY=             10.9525   ZZ=            -21.9050
   XY=              0.0000   XZ=              0.0000   YZ=              0.0000

‡: It is not a molecule, since it is not neutral, which is a prerequisite from the IUPAC goldbook definition.

♦ - just to make sure, your calculations show that the central iodide ion has a -0.11 charge, correct?

@Dissenter Yes, that's what the calculations provide. But since I did not do any calculation, take it with a grain of salt.

@Martin if you define "polar" as having any multipole, you have to consider H-H, N-N, and every molecule as polar.  "It is true, that for
H2,  N2 etc., wave mechanics, too, gives at least quadrupoles." Fritz London http://pubs.rsc.org/en/content/articlepdf/1937/tf/tf937330008b

@DavePhD I would not dismiss that thought. But what does that add to the specification of the molecule any way.

@Martin-マーチン Well, if you define polar as having a permanent electric dipole moment, it specifies some subset of molecules.  On another note, is triiodide really linear? "Isolated Triiodide and Pentaiodide Anions..." is saying bent 174.8 degrees http://onlinelibrary.wiley.com/doi/10.1002/zaac.200800115/abstract

@DavePhD All I am trying to say is that looking at the dipole moment for clarification is not enough. And I am saying that that classification does not add anything to the discussion anyway. It's too simple. Also, we might not want to have this discussion in the comment section. Maybe chat, but I am on the road tonight....

Just to clarify, what is meant by a "vanishing dipole moment"?

@TanYongBoon a vanishing dipole moment is used to describe a molecular dipole that is zero due to some symmetry constraint.

If you accept the Atkins Physical Chemistry definition of polar:

A polar molecule has a permanent electric dipole moment arising from the partial charges on its atoms

Then the answer turns upon whether triiodide is actually linear.
Isolated Triiodide and Pentaiodide Anions in the Crystal Structure of [Rb(Dibenzopyridino-18-Crown-6)2]2(I3)(I5) Zeitschrift für anorganische und allgemeine Chemie Volume 634, pages 1247–1248, says it is bent with an angle of 174.8 degrees.
Theoretical Study of the Solvent Effect on Triiodide Ion in Solutions J. Phys. Chem. A 1998, 102, 2065-2071, says that triiodide is linear in the gas phase, but "The free-energy surface of the same ion in
aqueous solution is distinctly different and indicates a drastically
enhanced probability for structures with lower symmetry."
So in the gas phase it would be non-polar by the Atkins definition of polar, while in other situations it can be slightly bent and therefore slightly polar.

According to the IUPAC: A molecule is an electrically neutral entity, which would exclude the $\ce{I3-}$, even in Atkin's definition. I made the same mistake though. For the rest of your statements I am happy to give you a thumbs up.

@Martin-マーチン I wouldn't say Atkins is limiting the word "polar" to molecules.  Instead, he is only defining the term "polar molecular" rather than "polar".  The journal Inorganic Chemistry seems to think it's ok to refer to polar ions, "The Polar [WO2F4]2- Anion in the Solid State", http://chemgroups.northwestern.edu/poeppelmeier/pubs/HT/polar_-WO2F4-2-.pdf .


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What is the functional group derived from benzene called?
If you were asked in an exam to name this functional group, would you call it benzene or an aromatic hydrocarbon?

It's not a functional group, it's a molecule. If it were to be bonded something else within a molecule, then it would be called a phenyl group.

I would call it an arene, but if you're just worried about your exam, why ask us? You should probably ask your teacher because they are who will be marking your script.

I know it is a molecule, what functional group would it be in?

And I'm studying by myself at home, I don't have a teacher with me at this very moment

Functional groups derived from benzene are various aryl groups like phenyl, three phenylene groups etc.

"Benzene" characterises a compound and not a functional group, "Aromatic Hydrocarbon" characterises a class of compounds. I would call your structure as a functional group, "phenyl".

Is Benzene a Phenyl then?

Benzene is the name of an actual compound. Phenyl is the name of a functional group. The structure you drew can be seen to represent either a "real" compound and then you call it "benzene" or it can represent the abstract concept of a functional group and then you call it "phenyl". You might say for example that benzene is an aromatic hydrocarbon and that phenyl is an aromatic functional group


text:
Dissociation of silver chloride in a solution
When silver chloride is exposed to light it can dissociate to silver metal and atomic chlorine. It might be a silly question, but could the chlorine go back into the solution as chloride ions ? if so is this likely to occur? does the outcome change if the solution is being boiled?
Thanks!

The answer is yes and no. Silver chloride doesn't "dissociate", it decomposes into silver metal and dissolved chlorine.  The chloride reacts with the water to form hydrochloric acid and hypochlorous acid as shown below:
$$\ce{2AgCl_{(s)} +H_2O_{(l)} ->[h\nu] 2Ag_{(s)} + Cl_2_{(aq)} +H_2O_{(l)}->2Ag_{(s)} + HCl_{(aq)} + HClO_{(aq)}}$$

Temperature makes a big difference in solubility of AgCl: it increases by over an order of magnitude from room temperature to boiling. See http://www.saltlakemetals.com/Solubility_Of_Silver_Chloride.htm

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

It really seems to me that for clarity there should be two separate reaction schemes. Making the whole thing one long line is just confusing.


text:
In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital?
Suppose we have the complex $\ce{[Fe(H2O)6]}$. Iron has 26 electrons, and its electronic configuration is therefore $[\ce{Ar}]\,\mathrm{4s^2 3d^6}$, but in crystal-field theory, we say that the $\mathrm{3d}$ orbitals have 8 electrons instead of 6. Why?

$\ce{[Fe(H2O)6]^{\pm 0}}$? I’m pretty sure that doesn’t exist; at least due to the 18-electron rule.

Your example is a little questionable. I would think you are talking about $\ce{[Fe(H2O)6]^2+}$, in which case you are actually supposed to have an $\ce{Fe^2+}$ ion with a $[\ce{Ar}]\mathrm{(3d)^6}$ configuration. There are no $\mathrm{4s}$ electrons in the $\ce{Fe^2+}$ ion.
Nevertheless, it is true that in crystal field theory, electrons are assumed to be entirely in the d-orbitals, even for low-oxidation state complexes. For example, nickel(0) in $\ce{[Ni(CO)4]}$ is assumed to have a "$\mathrm{d^{10}}$" configuration. There is a reason for this, the explanation of which requires a molecular orbital treatment. Consider a generic octahedral complex, $\ce{[MX6]^{n-6}}$, which is formed from a $\ce{M^n+}$ ion and six $\ce{X-}$ ligands. We could construct an MO diagram:

(Source: Miessler & Tarr, Inorganic Chemistry 3rd ed., p 347)
Each $\ce{X-}$ ligand contributes one pair of electrons, which gives rise to a total of six pairs; these six pairs of electrons enter the low-energy bonding MOs $\mathrm{a_{1g} + e_g + t_{1u}}$. These MOs are similar in energy to the ligand AOs, and therefore are likely to be mostly centred on the ligand, which means we can essentially ignore these MOs when we are describing the properties of the complex. (Of course, these MOs are important in the sense that they are responsible for the bonding between metal and ligand. If these bonding MOs did not exist, the complex would not form. However, we are usually much more interested in the higher-energy orbitals.)
The electrons in the $\ce{M^n+}$ ion can be in any of the AOs ($\mathrm{3d}$, $\mathrm{4s}$, or even $\mathrm{4p}$) - it does not matter. The only important thing is that, once the complex is formed, these electrons will enter the lowest-energy MOs according to the aufbau principle. Since those low-energy bonding MOs are already occupied by ligand electrons, the metal electrons will have to go into the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs, which in the simplistic treatment of crystal-field theory are called the "d-orbitals".
The truth is that those two MOs have quite a lot of metal d-orbital character (as you can see from the dotted lines), so crystal-field theory is not entirely wrong. Nevertheless, you can see from the MO diagram why it would be wrong to assume that the metal $\mathrm{4s}$ orbital is filled: what CFT calls the $\mathrm{4s}$ orbital is actually the $\mathrm{a_{1g}^*}$ orbital in the MO scheme. This orbital can only be filled if the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs are already filled, which is not the case unless we are talking about some kind of zinc(0) complex (very unlikely!).
Similar diagrams can be constructed for other geometries (tetrahedral, square planar etc.) and the same point will hold. Therefore, even though $\ce{Ni}$ itself has a $[\ce{Ar}]\mathrm{(3d)^8(4s)^2}$ configuration, the tetrahedral $\ce{[Ni(CO)4]}$ complex will have a $\mathrm{(e)^4(t_2)^6}$ configuration (as you are probably familiar with, in tetrahedral complexes, the d-orbitals are split into $\mathrm{e + t_2}$). Again, in the language of crystal-field theory, both the $\mathrm{e}$ and $\mathrm{t_2}$ orbitals are simply called "d-orbitals", so this is equivalent to having a "$\mathrm{d^{10}}$" configuration.

After realizing the confusion, I've tried to stick to "valence electrons" for the metal.

@orthocresol thanks for this answer. However, using just CFT arguments, do you think this could just be explained by arguing that s orbitals are more diffuse and, therefore, experience more repulsion from the ligands and are destabilised more? Moreover, I still don't see why LFT (using MO), which assumes pure covalency, is necessarily better than CFT, especially for hard-hard complexes?


text:
A Tool to atomize liquid in the milliliter range
for some graduate project I'm trying to create a mist from some water based solution to apply it to a surface, basically the surface is solid but very gentle and I need to spray it with this solution as evenly as possible, I cannot apply the solution with a syringe and then spread it around since this will pattern the surface, also the volume applied to the surface is important, I need to spray 2ml+-0.5ml, spraying 4ml for example would be too much
I've looked into an air brush but I couldn't find a way to control the volume and I've looked into a spray bottle but the volume there is too big.
any advice or suggestion would be welcome for a tool that can do this would be great!

You might want to look into using an inkjet (salvaged from an old printer). They can produce a regular, directed stream of small droplets and have a capacity of at least several mL. You can power them and control their behavior electronically. There are probably inkjet printers available that print onto a fixed platen, so that would take care of the production aspects.

You might use a perfume atomizer, if the solution is compatible with the materials of the sprayer. Measure the volume supplied with each squirt; it should be quite consistent, due to the constant volume of the plunger.
You could also construct an atomizer from glass tubing, as is done with soda straws in this Do-Science article. Again, I would suggest measuring the efficiency with which the spray is delivered... there will be some over-spray and drifting with any atomizer.


text:
What is the relationship between configuration & optical rotation?
Is there a clear cut rule between the two? For example, is R always (+), or is S always (-), or is there no relationship? And if so, why? Wouldn't it make sense that plane polarized light would rotate in the direction ascribed to chiral molecules?

You have to remember that (R) and (S) are essentially arbitrary rules that are created by humans. There is no way for nature to assign the Cahn-Ingold-Prelog priority of substituents because nature doesn't know what the priority rules even are and even if she did, she wouldn't have a way to evaluate them. IUPAC could create an entirely new arbitrary naming system under which we would label chiral molecules as (A) and (B) and that would make absolutely no difference whatsoever to the (+)/(-) optical rotation of the molecules.

There is no general correlation between the the rectus (R) and sinister (S) prefixes and dextro/levorotation. In some cases the (R)-enantiomer is the dextrorotary enantiomer (as in the case of D-alanine), and in other cases the (R)-enantiomer is the levorotary enantiomer (as in the case of L-cysteine). The relationship can only be determined on a case-by-case basis with experimental measurements or detailed computer modeling.
For a really detailed explanation of exactly how chiral molecules rotate planes of polarized light, I recommend you check out this physics post, but to answer your other question, no it doesn't make sense for planes of polarized light to rotate according the Cahn–Ingold–Prelog Priority Rules. This system only takes into consideration the atomic number and number of bonds to atoms of a higher atomic number. Even though $\ce{-Cl}$ might rank higher than $\ce{-C(CH)3}$, wouldn't it make sense for the two groups to have different effects on rotation, in the cases of both direction and magnitude? This is why dextro/levorotation must be determined experimentally.


text:
At what temperature (in kelvin) are most of the elements on the periodic table liquids?
This question is out of pure curiosity.
At what temperature are a majority of the elements on the periodic table in a liquid state/phase of matter? For the purpose of this question, assume the standard pressure (100,000 Pa). Is there some sort of formula or theorem that explains this? Please give a detailed explanation on your conclusion.

The Dynamic Periodic Table is a great site for questions about the properties of different elements. For your purposes, it even has a slider for the phases of different elements at different temperatures. I will say though that carbon is the last element to melt, and at atmospheric pressures it doesn't even do that-- it actually sublimes at $3,900:\mathrm{K}$.

Now that's a great question indeed! Evidently, at 0K all elements except helium are solids, at 10000K they are all gases, so someplace in between the number of liquids must reach a maximum; what and where might that be?
Well, there is no formula or theorem that says liquid hydrogen must boil at 20K, nor is there such a thing for any other element, so this line of thinking hardly provides any insight. Let's try another way.
Let's take a look at all elements' melting and boiling points. (I used the great site referenced in a comment by ringo, where all data are conveniently stored on line 47 in the file http://www.ptable.com/Static/interactivity-2fcd37b.js). Let's simply make a joint list of all melting and boiling points, sort it (keeping track of which is which), and then walk through it from bottom to top, calculating the number of liquids on each interval. If you know some kind of programming, you can do it with a script.
It turns out that the maximum number of simultaneously liquid elements is 49, and it is reached several times between 2183K (melting of vanadium) and 2792K (boiling of aluminium). By dragging the slider you may actually see which elements are included in these 49. Now, the data are not terribly reliable (for example, in higher actinides they claim some melting points obtained by God-knows-what extrapolation, but no boiling points at all), yet still this is something to think about.

The boiling point of hydrogen is $-252.87~^{\circ}\rm C$ or $20.28\rm\,K$, which is the least boiling point so you can expect that all other elements would liquidify or solidify at or below this temperature.
But the concept of phases is much more complicated than that. Aside from three states (namely solid, liquid and gas) there are other states such as plasma, fermionic condensate, Bose–Einstein condensate and top-quark condensate.
At Bose–einstein condensate, at 170 nano-kelvin rubidium atoms coalesce to form one particle that acts like a wave forming Bose-gas.
So, from this example you can perceive that on super-cooling you can form gas. Hence, it is very difficult to specifically answer your question as technology improves nature tend to show its diverse beauty.


text:
What is the limiting value of band gap that makes a material semiconductor?
What is the limiting value of band gap (in eV) that makes a material semiconductor? Or what is the value of band gap that separates insulator and semiconductor?

There is no commonly accepted definition. If you can make a standard semiconducting device in the material, it is a semiconductor.

The transition of electron from the valence band to the conduction band is a probabilistic event.  Electrons get this excess energy equivalent to the band gap ($\Delta E$) from the thermal energy ($kT$, where $k$ is the Boltzmann constant and $T$, temperature in Kelvin). Hence, the probability of such an event will be proportional to $e^{-\Delta E/kT}$.
Answering your question, although there is no standard defined value of band gap, you can see since it's an exponentially falling smooth curve that one can define a cutoff $\Delta E$, saying that for band gap $> \Delta E$ there are practically no electrons transiting to the conduction band, qualifying the material as an insulator (i.e., exhibiting insulator-like properties).

However, this completely ignores the dopants, which are close to the band edges and thus can get carriers in to the conduction (electrons) and valence (holes) bands at reasonable temperatures. Silicon carbide has a band gap in excess of 3eV, more than twice that of GaAs. Aluminium nitride, a so-called ultra-wide bandgap material, has a band gap of 6 eV, which is getting pretty close to silicon dioxide at ~8.9eV. Yet, aluminum nitride can be used to make devices because good dopants exist. So, the ability to dope is more important than a specific band gap.


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When fugacity and z-compressability equations instead of Gibbs free energy equations?
In very high pressure environments such as mud circulating systems, conditions for polymers contain fugacity/z-compressability criteria instead of Gibbs free energy, more in the answer. So
When fugacity and z-compressability equations instead of Gibbs free energy equations?
and helper questions to the latter

when are fugacity and z-compressability equations used?
When are Gibbs free energy equations used?
By which criteria can you justify your decision to choose between Gibbs free energy criterion and fugacity/compressability criterion?

Are you really referring to liquid systems (that are nearly incompressible), or are you referring more to gaseous systems beyond the ideal gas region?

Elastic solids (like gelled drilling muds and elastomers) can not be treated in the same way as liquids and gases, using the z factor and fugacity (unless the loading on the solid is purely isotropic).  This is because, at equilibrium, a solid is capable of storing elastic energy (involving anisotropic loading and strains) after it is deformed.  As a result, the equilibrium state of an elastic solid cannot be described in terms of only two intensive variables (like it can for a liquid or gas).  In the case of an isotropic elastic solid, you need to specify either (a) temperature and three principal stresses or (b) temperature and 3 principal strains).  So, rather than 2 intensive variables, you need to specify 4 intensive variables.  All the thermodynamic functions, U, H, S, A, and G are functions of these 4 intensive variables.


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Are non-exothermic explosions possible?
Explosives like picric acid and trinitrotoluene are unstable due to the close packing of so many $\ce{-NO2}$ groups onto a benzene ring. These highly strained bonds require only a relatively small amount of energy to cause the molecule to rip itself apart, forming a lot of more stable gas molecules, and releasing a lot of heat in the process.
My question is: is it thermodynamically feasible to have a reaction that rapidly evolves a lot of gas without being extremely exothermic? Would this still be considered an explosion?
It doesn't seem to me that this could be possible, but I never cease to be amazed at how little I know about chemistry. If it is feasible, and the gas molecules are more stable, then where does the energy go? If the gas molecules aren't more stable, then what's the driving force?

There is a precise term for this: entropic explosions. I don't know to what extent they are "low temperature", though.

Sodium azide evolving nitrogen "explosively" (not a detonation) in an automotive air bag might be an example of a not highly exothermic but highly gas-producing reaction.

@NicolauSakerNeto - I think I'd post your comment as an answer.

If you pour liquid nitrogen into a "weak" container and seal it, or pack dry ice into a "weak" container and seal it, then you're going to get an "explosion." In such a case the physical change is purely endothermic.

Ultimately, explosions are just a type of spontaneous reaction, and even endothermic reactions can be spontaneous, so long as entropy increases enough over the course of the reaction to compensate it. Near room temperature, however, the entropic factor is usually small compared to the enthalpic factor. This is why there are comparatively few examples of reactions which produce strong cooling upon mixing reagents.
That said, none of this strictly precludes slightly-exothermic or endothermic explosions, to the point that the concept has been legitimately investigated. As I mentioned in the comments, there is even a specific term for these types of explosions; entropic explosions.
For a spontaneous reaction to be explosive, there is one property it must have above all others: the reaction must happen very quickly. Now we are dealing with kinetics, which is much harder to analyse theoretically. It is possible, however, to point out another factor that makes endothermic explosions difficult.
By far, most reactions speed up when temperature is increased. In a conventional, exothermic explosive, this establishes a positive feedback effect, where heat from the reaction causes it to react faster, which produces more heat, which causes even faster reaction, and so on, facilitating a runaway reaction and subsequent detonation. If the exotherm is weak, then there is a larger chance that the explosion will fizzle out before enough heat is produce to get things going, or the reaction might be too slow to explode and instead steadily combusts. For an endothermic explosive, very much the opposite happens, and a negative feedback is established; once started, the reaction would decrease the temperature of the explosive, hindering propagation.
So, while there are no arguments which completely forbid slightly-exothermic or endothermic explosions, it seems to be a rather unlikely scenario in practice. Wikipedia only lists one possible contender for an entropic explosion, which is triacetone triperoxide. The original article making the claim is mostly theoretical, and there seem to be some rebuttals to it, so even that is not certain. Of course, we may be in the middle of a giant selection bias effect, because almost all research into explosives is focused on getting more energetic materials, not less. Maybe there are examples waiting to be found, but no one has had the interest in doing so!
As a sidenote, explosions actually don't necessarily produce gas. There are some rare examples, such as cuprous acetylide ($\ce{Cu2C2}$) and silver acetylide ($\ce{Ag2C2}$) where all the products are solids. Also, there are examples of non-chemical explosions which are only slightly exothermic or even endothermic. As MaxW mentions, a pressure explosion from a liquefied gas will be endothermic, as energy is consumed to evaporate the liquid. Another interesting example of an explosion which is probably only slightly exothermic is a Prince Rupert's drop being set off.

Baking soda and vinegar while not necessarily explosive is endothermic and releases a fairly significant amount of gas.


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In electrolysis, why is hydroxide oxidised in preference to anything(?) but halides?
In electrolysis of aqueous solutions, I can loosely explain what happens at the cathode in terms of reactivity - any metal more reactive than hydrogen stays in solution, effectively because they're more attracted to water molecules. I could also explain this in terms of redox potential if I had to (I'm teaching this at GCSE level right now - to 16-year-olds - and we don't bring in redox potential or the electrochemical series until later).
But what happens at the anode is currently puzzling to me. We're told that the only anions which get oxidised in preference to hydroxide are the halides, but I can't explain that in terms of reactivity, and so far I'm not sure how to explain it in terms of standard reduction potential, either.
So - why the halides? And are they really the only anions oxidised in preference to hydroxide?

http://chemistry.stackexchange.com/questions/15435/why-is-chloride-oxidised-instead-of-water-in-copper-chloride-electrolysis

Thanks @Mithoron, I did see that, but if it answers my question, I'm not entirely following it. The fact that (e.g.) chlorine is oxidised comes down to the overpotential of oxygen, right? How do we generalise this observation to account for the fact that halides are the only ions oxidised in preference to oxygen (if indeed they are, and my GCSE course isn't lying to me)?

I think the statement is too general. Certainly because of the overwhelming concentration of water the redox reactions of water must be considered in any aqueous electrochemical cell.

I did wonder about that. But again I lack the in-depth chemistry knowledge to assess how much too general...


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Why is OH group activating towards electrophilic aromatic substitution?
In phenol, the OH group has -I effect due to electronegative oxygen. So, OH group should deactivate the benzene ring towards electrophilic substitution.
[P.S.: I am not talking about the o,p-directing ability of OH, which appears due to lone pair. Also please do not suggest me the question at the link Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?, because that is not what I am looking for.]

The oxygen in the $\ce{-OH}$ group in phenol has unpaired electrons that are capable of being donated to the ring. Although oxygen is electronegative and does exhibit some I- effects, the electron sharing that oxygen is capable of (shown by when drawing the resonance structures) has a stronger effect and results in $\ce{-OH}$ being a good ring activator. I've included the resonance structures below.

in such case, the halogens should also work as activating groups, but they are deactivating. How do you explain it then???

Halogens are weekly deactivating, there are two competing mechanisms: inductive (deactivating), resonance (activating), overall = weakly deactivating

Halogens are deactivating groups because they are much so electronegative (even more than oxygen). Because of this, electrophilic addition reactions with halobenzenes will be considerably slower than with phenols. They are, however, ortho/para directions like oxygen because, like oxygen, they have unbonded pairs of electrons that get donated to the ring, forming similar resonance structures.

You could even improve this answer with orbitals; the hydroxy group introduces a seventh p-orbital into the pi system and two more electrons; the latter make for the stronger activation.


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Does an increase of area lead to more collisons in a confined jar of gas?
I'm reading a lecture by Richard Feynman in which he is talking about the behaviour of gas molecules/atoms within a confined vessel.
The confined vessel is a cylinder with a piston to stop the water molecules from flying out.
However this follow sentence has me stumped
"Clearly, the force is proportional to the area, for if we increase the area but keep the number of molecules per cubic centimeter the same, we increase the number of collisions with the piston in the same proportion as the area was increased."
If the area of the piston is increased, but number molecules remain the same, would there not be less force as the collisons are spread out?
I'm thinking of this in the same vain as shoes exerting less pressure than high-heels due to the increased area.

Pressure is force over an area, or as an equation you might state that as:
$$P=\frac{F}{A}$$
Which can be rearranged as
$$F=P \cdot A$$
You are correct in saying that if the number of molecules are held constant, that the force would be over a larger area and therefore decreased. As you increase the area but keep the number of molecules constant, the pressure decreases proportionally, and no change in force is seen, but what Feynman was talking about was keeping the amount of molecules per cubic centimeter constant, or in other words keeping pressure constant. This way as area is increased, you see a proportional increase in force.

Not quite. The pressure of a gas on an area is given by:
$$ P =\frac{F}{A} $$
and the force of gas on an area is given by:
$$ F = P*A $$
The question is then is the force constant or the pressure constant?  If the force is constant then Pressure will have an inversly proportional relationship to area.
$$P \propto \frac{1}{A} $$
If the pressure is constant then Force will have a directly proportional relationship.
$$ F \propto A $$
If you have the same number of molecules but a larger area it just means that more molecules are pushing on the piston at any given time.  Thus for a piston acting on a gas of constant pressure as area increases force will increase proportionally.  For a human foot of constant force (i.e. your weight) as the area increases of the foot the prssure exerted will decrease inversly proportional.


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Is this a feasible synthetic route for 1-(phenylethynyl)cyclopentan-1-ol?

From my notes $\ce{NaNH_2}$ is a good base for the deprotonation of alkynes. After the deprotonation I thought the alkyne would attack the carbon on the cyclopentanone because of the $\delta^+$ charge on the carbonyl carbon due to the EWG effects of the oxygen. Should the protonation of the $\ce{O^-}$ be carried out with conditions specified or can I just leave it as the addition of $\ce{H^+}$?

There is no such compound as $\ce{NaNH3}$. It is easy to mix up sodium in liquid ammonia $\ce{Na}/\ce{NH3}$ with sodium amide $\ce{NaNH2}$. Make sure you know what you are writing. As for the final workup, you definitely need some kind of proton source.

Thank you i must have taken that down incorrectly, ok so would $H_3O^+$ suffice? @orthocresol

Again, please make sure you know what you are writing. If you treat the alkyne with $\ce{Na}/\ce{NH3}$, you are going to get a trans alkene, not the acetylide ion. For the final workup, it depends on how strict your marking scheme is, but I would recommend you write an actual compound like water. You can't go out and buy a bottle of "$\ce{H3O+}$" from Sigma-Aldrich.

Ok so would the conjugate acid of the sodium amide protonate the $O^-$? @orthocresol

Or would that be too weak an acid to do so?

What do you think? You could look up some typical pKa values of an alcohol.

The route definitely looks feasible and it is more or less what I would have done to arrive at your desired product. A few notes:

Make sure you actually have sodium amide $\ce{NaNH2}$ and not sodium in ammonia $\ce{Na/NH3}$. As pointed out in the question comments, they react very differently; the latter generates a trans-alkene.
I’m not sure I would have chosen the same base. But the choice of base is likely arbitrary anyway. n-Butyllithium would definitely also do the trick but might be overkill.
Typically in a lab setting, reactions with strong bases (as is the case here) are carefully quenched with something aquaeous. So you can be sure that there will be an aquaeous proton source ($\ce{H2O}$ or $\ce{H+/H2O}$, depending on your markers’ preferences) in the end. There should be no need to additionally specify conditions. It would make no sense to attempt to isolate the alcoholate anyway.

The reaction looks OK to me, but I would like to point a few things out. When I was in Aberdeen I used to do a lot of alkylation chemistry in liquid ammonia.
I have always found that the reaction of sodium metal with ammonia to make sodium amide is a mighty pain. I have always found that the reaction is very sluggish at first and then it suddenly takes off and generates a lot of heat which then greatly increases the boiling rate of the ammonia.
I found in Scotland that potassium metal reacts with ammonia nice and gently if you put a little of an iron salt into the liquid ammonia. The first time I did it the reaction put the fear of God in me. But I found that it is easier and safer to use potassium to make potassium amide than to make sodium amide.
I would suggest that you should make sure that all of the solvated electrons have reacted before you add the alkyne (acetylene). The solvated electrons are an inky blue. It is interesting that solvated electrons form if you either dissolve sodium in water or bombard it with high energy electrons or gamma rays. In water the rate of the reaction of solvated electrons with solvated protons to form hydrogen atoms and water is far faster than the rate of reaction of solvated electrons in liquid ammonia with the ammonia to form hydrogen atoms.
Maybe after some years of doing things with atoms you will start to see everything through the lens of nuclear chemistry.
But back to the organic chemistry, I would make the slurry of potassium amide in liquid ammonia and then cool the mixture to -78 oC with a dry ice / acetone bath. Then I would slowly drip in an ether solution of the phenyl acetylene. I would allow it to react for a while and then slowly add the ketone. I would leave it stirring at -78 oC for a while and then the quench I would use would be ammonium chloride powder. You just add the NH4Cl and then allow the mixture to slowly warm up.
The next day I would add ether to the reaction pot, if you then filter off the solids (KCl + NH4Cl) then you will have a solution of the product. I would be careful with the distillation as I recall that this sort of compound can be a bit acid sensitive. I think that vacuum distillation will be the best way to make it nice and pure.


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Why is o-halophenol more acidic than m-halophenol?
Why is o-halophenol more acidic than m-halophenol?
My thoughts: m-halophenol has only the -I effect which favors acidity while o-halophenol (which has greater -I) but also has +r which should reduce the effect caused by -I.

Related: Acidity of substituted phenols

It's because the halogen is highly electronegative. Therefore, the closer the halogen is to the alkoxide, the more it helps in stabilizing its negative charge after it's been deprotonated. Conversely, the further away the halogen is, the less it is able to help stabilize the negative charge. The more stable the conjugate base, the stronger the acid.
This question is basically testing induction, which is covered pretty well here:
http://orgomadesimple.com/organic-chemistry-acids-and-bases/


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Valency and chemical formula
If an element valency corresponds to the group number in the periodic table, oxygen would be 6 right?
The question states; Species: Li, O
I assume Li is 1 and O is 6, so my chemical formula is $\ce{Li6O}$ but the correct answer is $\ce{Li2O}$
Why is the valency of Oxygen 2?
similarly Pb(II) and Iodine, which iodine is in group 7 but just like the answer it said the valency is 1.
 If I drew the electron shells for oxygen I realize there are 2 electron spaces, if I did the same for Iodine there are 7 spaces. so this did not answer my question
then I saw number of bonds = (full valence shell) - (n. of valence electron)
for oxygen it returned 2, through: $8 - 6$
then it was the same for Iodine: $8- 7 $
This did not answer my question either, as Lithium and Sodium correspond with the group number. Am I missing something OR doing something wrong OR is there some sort of principle among this periodic table?

Then it asked, $\ce{Ag}$, $\ce{OH-}$. because Ag is a transition metal, so how do I find its valency? OH is a polyatomic ion, is there a way to find out the valency for polyatomic ion or am I suppose to memorize it for exams?
 The question in my book mostly gives 2 species then asked for the chemical formulas. I'm just very confused at the valency part.

In a chemical bond some elements like Li are electron donors, and some like oxygen are electron receptors.

Max's comment sums it up. Beware, I'm gonna have to use many simplifications in order to explain this.
A bit of introduction
The species you mentioned are ionic compounds. For your level it's useful to take ionic bonds as bonds between atoms that have such different electronegativities (ChemGuide typically provides useful introductions, I recommend clicking on that link and reading it carefully) that we can imagine them as different ions. Hence, this bond is the attraction between charged particles. What does that mean? It means that one of the species will have a positive charge and the other a negative charge.
Electronegativity increases in diagonally, as we move from bottom left to top right.
$\hspace{5ex}$
$\hspace{30ex}$ Electronegativity (Periodic Trend) – Source
For now just don't apply the electronegativity concept to noble gases. If you really want to know how their electronegativities are determined, take a look at interesting posts like this.
The more electronegative the atom, the less the chance that it'll bond by losing its electrons. Thus, we can conclude two important rules that occasionally become useful, while they're not highly accurate at higher levels:

Metals in general, and especially alkali and alkali earth metals, will bond ionically by losing their valence electrons.
Non-metals in general, and especially halogens, oxygen and nitrogen, will bond ionically by gaining electrons.

I repeat: The two bullet points above are just rules of thumb. Don't expect them to work all the time. But they're good starting points, so I included them in the answer. Don't hesitate to resort back to the first sentence whenever you ran into cases where you don't get a definitive answer by them.

Lithium oxide's case
So let's apply what we learnt to the $\ce{Li2O}$'s case. Lithium will lose its electrons in order to gain the configuration of a noble gases, in which case it would be helium.
$$\ce{Li(s)->Li(g)}^{~1}$$
$$\ce{Li(g)->Li+(g)}$$
Oxygen, however, doesn't favor losing its electrons at all. It will, instead, gain electrons and its electronic configuration would arbitrarily look like that of neon, a noble gas.
$$\ce{O2(g)->2O^2- (g)}$$
Finally, the two ions will form a lattice. Billions of billions of these ions form $\ce{Li2O}$.
$$\ce{Li+ (g) + O^2- (g) -> Li2O(s)}$$
Hence the overall reaction is
$$\ce{2Li (s) + O2 (g) -> 2Li2O(s)}$$
So let's review what we understood: In lithium oxide, oxygen will gain electrons while lithium will lose electrons, since their electronegativities differ greatly. So we should see how many electrons it will lose in order to get to the configuration of the previous noble gas in the periodic table, while oxygen will gain electrons to reach the configuration of the next noble gas in the periodic table. See this Q/A to gain more insight on how to figure this out using their electronic configuration or group number if you don't already know how.
Note that the formula for ionic compounds is the smallest relation between its ions. By $\ce{Li2O}$ we indicate that for every two lithium ions, one oxygen ion is present in the lattice, crudely, and assuming the crystal is without defect. Think of it like this: There was a need for two lithium ions to provide a single oxygen ion with the electrons it needed to fulfill its stable electronic configuration.
Silver hydroxide's case
Let's solve the case of silver and hydroxide.
The hydroxide ion has a charge of $-1$, hence the trouble of determining the charge it will have while bonded ionicallly with silver is gone! Unfortunately, for now, the best way to find the charge out is memorizing it. You may get some fuzzy speculations by drawing Lewis structures, but poly-atomic ions are sometimes exotic chemical species and unpredictable, since many effects are at play to determine the charge of the ion. You can find a list of them here, or more just with a simple Google search.
You have no choice but to memorize the charge of transition metals though, since they tend to assume abnormal electronic configurations. Over time, you'll face it so much you'll learn by heart that $\ce{Ag+}$ is silver's most stable ion.
Now try to apply the logic and reasoning we used back in $\ce{Li2O}$'s case. What would the formula be?

 $\ce{AgOH}$

Useful tips:

Don't forget, that if the poly-atomic ions are more than one in the ionic formula, you should indicate their number with them in parenthesis.
$\Longrightarrow \rm\color{red}{Wrong:} \quad \ce{(NH4)F,~Ca3PO42}$
$\Longrightarrow \rm\color{green}{Correct:} \quad \ce{NH4F,~Ca3(PO4)2}$
Beware of poly-atomic ions that only consist of one atom. These make for the biggest traps and confusions of the students. Good examples are peroxide, superoxide, and azide. Hence $\ce{Li2O2}$ and $\ce{LiO2}$ are correct! Simply because, the ions in them are not the oxide ion, $\ce{O^2-}$, but $\ce{O2^2-}$ and $\ce{O2^-}$, respectively. 2
The oxidation states of the elements, if mentioned, give clue as to what charge their corresponding ion will have. $\ce{Pb(II) \Longrightarrow Pb^2+ ,~Fe(II) \Longrightarrow Fe^2+ ,~Fe(III) \Longrightarrow Fe^3+}$ and so on.

1: It's simplified if I recall correctly, but we take the lattice as formed by ions in the gas state. It helps in defining lattice enthalpy, and I haven't read anywhere that it's not what's really happening.
2: Actually, you can't necessarily make any ionic compounds with any ions you want. Some of them are too unstable to reasonably exist, like lithium superoxide. But it seems the examiners want you to be a magician anyway.

Actually, even silver(I) hydroxide is not particularly stable; it loses water to giver silver(I) oxide. But that certainly does not detract from the answer at all. The only thing that makes me sad is the question, which is confusing at best...


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Does the hydrogen ion actually exist?
A chemistry test I did was an IGCSE test. One of the questions was

Is Hydrogen Chloride ionic or covalent?

Now, on a sheet I have been given, it states the hydrogen ion is $\ce{H+}$. So, it forms a positive ion, like metals. I'm fine with that. I know that hydrogen is strange and I also know that the (I)GCSE chemistry curriculum simplifies/lies a lot. The bonding in Hydrogen Chloride is covalent. Now, since a 'Hydrogen ion' is just a proton, I wanted to know whether in reality it actually exists or not.

Yes free $\ce{H+}$ ions, protons, really exist.
Protons are constantly emanating from the sun and reaching Earth.
The proton flux is continuously monitored by satellite.
However, in a solution such as water, instead of bare $\ce{H+}$ ions, they are $\ce{H3O+}$ or larger ions such as $\ce{H5O2+}$ or $\ce{H9O4+}$.
When $\ce{HCl}$ dissolves, the proton is transferred to a water molecule to form $\ce{H3O+}$.
$\ce{HCl}$ only has 17% ionic character according to Pauling's The Nature of the Chemical Bond, so it is covalent, but reacts with water to form separate ions in solution.
For more information see Eigen or Zundel Ion: News from Calculated and
Experimental Photoelectron Spectroscopy:

The appearance
  of the hydronium ion was and is
  currently debated as several possible
  structures, three of them being $\ce{H3O+}$
  (the Eigen cation), $\ce{H5O2+}$ (the Zundel
  cation) and $\ce{H9O4+}$ (the strongly solvated
  Eigen cation)

and
Both Zundel and Eigen Isomers Contribute to the IR Spectrum of the Gas-Phase $\ce{H9O4+}$ Cluster:

The “Eigen cation”, $\ce{H3O+(H2O)3}$, is the most prevalent protonated water structure in the liquid phase

Wouldn't $\ce{HCl}$ dissociation also yield $\ce{H9O4+}$ and not only  $\ce{H3O+}$?

@mannaia yes, see here for details : http://www1.lsbu.ac.uk/water/hydrogen_ions.html


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Why does the PH3 geometry deviate more from the trigonal planar one, than does NH3?
$\ce{PH3}$ has a more bent structure than $\ce{NH3}$. The HOMO-LUMO gap for $\ce{PH3}$ is smaller than for $\ce{NH3}$, and so the distortion from the trigonal planar geometry is said to be larger. However, why is the HOMO-LUMO gap smaller for $\ce{PH3}$?
It should be possible, I suppose, to argue for why this is so by using qualitative molecular orbital theory, and from looking at the construction of the molecular orbital diagrams, but I am not sure how to construct accurate MO diagrams for $\ce{PH3}$ and $\ce{NH3}$.
In the answer to the possible duplicate question, it seems to be taken for granted that some distortion from the planar structure is stabilizing. In my question, the starting point is the planar structure, and then looking at why $\ce{PH3}$ is in a more distorted geometry than $\ce{NH3}$, relative to the planar structures.
This distortion is often explained in terms of the "HOMO-LUMO gap", but I fail to see how the HOMO-LUMO gap can be assessed qualitatively in order to predict which of $\ce{NH3}$ and $\ce{PH3}$ is in a more distorted geometry. So what factor ultimately leads to the smaller HOMO-LUMO gap in $\ce{PH3}$? $\ce{BH3}$ exists in a trigonal planar geometry - why is not this molecule pyramidalized?

also http://chemistry.stackexchange.com/questions/33818/bond-angles-for-the-hydrides?lq=1

Why should either $\ce{PH3}$ or $\ce{NH3}$ ever assume a planar geometry? Phosphorus and nitrogen have a lone pair that boron does not have.

I see two (or three) question here now. 1.) Why is the HOMO-LUMO gap smaller in PH3 than it is in NH3? 2.) Why is BH3 not pyramidalised? I am not really sure if I understand the connection between those. 3.) How is the distortion explained by the HOMO-LUMO gap? If it "... is often explained ...", then please add at least one reference so that we can see it in context. I don't understand in which way the HOMO-LUMO gap should relate to pyramidalisation. Additionally, I don't think a qualitative MO picture gives the answer for that, i.e. it has to be at least semi-quantitative.

A smaller HOMO-LUMO gap favours pyramidalisation via second order Jahn-Teller: see my answer on this question (larger barrier to inversion is the same as greater pyramidalisation).

Apparantly it was community consensus that this question be clear enough to stay open.
Neither $\ce{PH3}$ nor $\ce{NH3}$ have any incentive whatsoever to assume a trigonal planar geometry. The most stable ground state of both nitrogen and phosphorus atoms is $n\mathrm{s^2},\ n\mathrm{p^3}$ where $n = 2$ for nitrogen and $n=3$ for phosphorus. Since the p-orbitals are higher in energy per se, it makes sense to use them for bonding and lower their energy — which, as a side note leads to much more efficient overlap than using s-orbitals for bonding, since p-orbitals have a direction in space.
Therefore, the most stable geometry for both would be trigonal pyramidal with $90^\circ$ angles between the substituents. This would allow all-p bonding orbitals and an all-s lone pair on the central atom. If you check out $\ce{PH3}$, you will find almost exactly that geometry save a little distortion.
The distortion you see in $\ce{PH3}$ is of the same type as the one which moves the hydrogen atoms much further apart in $\ce{NH3}$: Steric stress. The hydrogens themselves repel each other; they require a certain space. Around a large phosphorus atom there is enough space for them to crowd at the ideal bonding angle while the small nitrogen does not offer that space. The final bond angle of $\approx 107^\circ$ is a fine-tuned result of trying to keep the hydrogen atoms as far away from each other as necessary and keeping the bonding angle as close to $90^\circ$ as possible.
$\ce{BH3}$ is an odd one out in this comparison. Boron does not have a lone pair and thus we expect isolated $\ce{BH3}$ to assume trigonal planar geometry. However, we also expect isolated $\ce{BH3}$ to have a low stability since boron does not have an octet. Hence it dimerises to $\ce{B2H6}$ with two $\ce{B-H-B}$ banana bonds of the 2-electron-3-centre type. The result can be described as a distorted tetrahedron around boron.

Why is the HOMO-LUMO gap in ammonia different than in phosphine?

@Goat'sMilk Even with the same geometry it would be because phosphorus is a different atom with different energy levels of its orbitals so the mixing in molecular orbitals creates different final levels. But apart from that, I wouldn’t know why to use the LUMO in this question at all.


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Initial guess for Unrestricted Hartree-Fock calculation
I am implementing an Unrestricted Hartree-Fock (UHF) calculation, following Ref. [1]. It works fine for a odd number of electrons ($N_\alpha\neq N_\beta$) since, despite the same initial guess for $\alpha$ and $\beta$ spins, the density matrices at the second step will be different and I get different orbital energies for different spin orbitals. The problem arises when I have an even number of electrons ($N_\alpha=N_\beta$) . Authors of Ref. [1] state that (p. 216)

When $N_\alpha=N_\beta$, a restricted solution of Roothaan equations is a solution to the unrestricted Pople-Nesbet equations. This restricted solution always exists and necessarily results if an initial guess $P_\alpha=P_\beta$ is used. [...] In seeking this second solution [the unrestricted one], it is imperative that an initial guess $P_\alpha \neq P_\beta$ be used or the equations will necessarily yield the restricted solution.

Unfortunately there is no clue on which initial guess should be taken. In fact in this book, for the restricted case, the Fock matrix is initially approximated by the core Hamiltonian (Guess=Core in Gaussian09), which consists in taking $P_\alpha=P_\beta=0$.
Since the core Hamiltonian guess cannot be differentiated for $\alpha$ and $\beta$ spins, I implemented Huckel guess (Guess=Huckel in Gaussian09) following Ref. [2] which consists in approximating the initial Fock matrix with the following:
$$
F_{ij} = KS_{ij}\frac{H_{ii}+H_{jj}}{2}
$$
$H$ is the core Hamiltonian, $S$ is the overlap matrix and $K$ is a constant. I tried to slightly change the coefficient $K$ (the suggested value is 1.75) for different spins in order to get a different initial guess. However, the SCF procedure converges towards a value that is higher in energy than the restricted solution. This should not be the case, in particular for a dissociating $\ce{H2}$ molecule (i.e. distance $R=4.4$ Bohr).
How I can properly differentiate the initial guess for $\alpha$ and $\beta$ spins?

EDIT
The reference of Gaussian09 states the following:

Requests that the HOMO and LUMO be mixed so as to destroy α-β and spatial symmetries. This is useful in producing UHF wavefunctions for singlet states. [...]

I belive that this is the solution to my problem, but what does actually means HOMO and LUMO be mixed in practice (i.e. before starting the calculation)?

[1] Szabo and Ostlund, Modern Quantum Chemistry, Dover, 1996.
[2] M. Lee et al., Extended Hückel and Slater’s rule initial guess for real space grid-based density functional theory, Computational and Theoretical Chemistry, 2015.

It only means spatial distribution of alpha and beta can be different, which is easiest to set up by mixing homo lumo levels

@Greg I tried to differentiate the spatial distribution by choosing a different coefficient for the Hückel guess, but apparently it does not work. In order to mix the orbitals I need to perform at least one SCF step (i.e. I need to compute HOMO and LUMO orbitals for my initial guess of Fock matrix). But I didn't find anywhere how to mix these orbitals in practice afterwards.

Generally you use a new HOMO = constant (old HOMO + old LUMO); new LUMO = constant (old HOMO - old LUMO). The trick is that you actually should know what you are doing, and choose a related pair of orbitals (i.e. sometimes you have to mix HOMO-1 or HOMO-2 .. instead of HOMO...). Note, it does not necessarily work out of the box. Present toolbox of  QChem is more about nice behaving closed shell molecules, open shell situations are often neglected as unimportant / difficult.

When $N_{\alpha} = N_{\beta}$, a restricted solution of Roothaan equations is a solution to the unrestricted Pople-Nesbet equations. This restricted solution always exists and necessarily results if an initial guess $P_{\alpha} = P_{\beta}$ is used. [...] In seeking this second solution [the unrestricted one], it is imperative that an initial guess $P_{\alpha} ≠ P_{\beta}$ be used or the equations will necessarily yield the restricted solution.

The quote you state from Szabo and Ostlund is not wrong, but isn't a practical quote in the sense that it helps to think about the different spin cases that one would want to handle. You already know how RHF/closed-shell calculations work in theory and in practice. I won't talk about restricted open-shell Hartree-Fock (ROHF), because it isn't relevant to this problem.
There are three UHF/open-shell scenarios, which will be considered in order:

Odd number of electrons
Even number of electrons, high spin
Even number of electrons, low spin

Odd number of electrons
For this case, take the copper(II) atom, which has a 3d$^{9}$ valence electron configuration; it is a doublet, so it has 1 more $\alpha$ electron than $\beta$ electron. There's no trouble here, because the indices for occupied orbitals run over different ranges; in the simplest case, the range for $\alpha$ spin will be 1 larger than for $\beta$ spin.
Even number of electrons, high spin
For this case, let's take triplet oxygen. In practice, this is actually identical to the above case, except we have 2 more $\alpha$ electrons than $\beta$ electrons. Again, we have differently-sized occupied blocks for each set of MO coefficients which will lead to different densities.
Even number of electrons, low spin
For this case, let's take singlet oxygen.
This is the difficult case, because the orbital degeneracy means for a proper qualitative description of the molecular orbitals, two determinants must be used, requiring a CAS(2,2) (at a minimum). Because this valence electron degeneracy isn't possible in a single determinant picture, the degenerate MOs will "split" into one lower in energy, the other higher. Requesting a singlet spin multiplicity means the two electrons will now occupy the same orbital; there's no reason for them to occupy different orbitals, at least for the initial guess. Hopefully it's (visually) clear that if the spatial components of $\alpha$ and $\beta$ spins are the same in this case, our guess is equivalent to the RHF guess. This means that for an open-shell singlet, one must break spatial symmetry in the initial guess, just as Szabo and Ostlund states. Since the RHF solution always exists as a local minimum on the "determinental" potential energy surface, whatever you choose as your SCF converger is most likely going to get stuck in this minimum.
This is a little hand-wavy, since we know that UHF calculations approximate two real determinants at the cost of spin contamination, but that isn't our goal.
Breaking spatial symmetry
How do we break spatial symmetry? Consider the initial steps of a typical RHF calculation that diagonalizes the one-electron (core) Hamiltonian:
$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{0'} &= \mathbf{\epsilon}^{0'} \mathbf{C}^{0'} \\
\mathbf{C}^{0} &= \mathbf{S}^{-1/2} \mathbf{C}^{0'} \\
P_{\mu\nu} &= \sum_{i}^{\text{occ}} C_{\mu i} C_{\nu i}
\end{align}$$
where the tick mark signifies a quantity is in the orthogonal AO basis, as opposed to the original (non-orthogonal) AO basis. Now assume we're doing a UHF calculation:
$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{\alpha, 0'} &= \mathbf{\epsilon}^{\alpha, 0'} \mathbf{C}^{\alpha, 0'} \\
\mathbf{F}^{0'} \mathbf{C}^{\beta, 0'} &= \mathbf{\epsilon}^{\beta, 0'} \mathbf{C}^{\beta, 0'} \\
\mathbf{C}^{\alpha, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\alpha, 0'} \\
\mathbf{C}^{\beta, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\beta, 0'} \\
P_{\mu\nu}^{\alpha} &= \sum_{i}^{\text{occ }\alpha} C_{\mu i}^{\alpha} C_{\nu i}^{\alpha} \\
P_{\mu\nu}^{\beta} &= \sum_{i}^{\text{occ }\beta} C_{\mu i}^{\beta} C_{\nu i}^{\beta}
\end{align}$$
This is misleading! Because we only have one initial Fock matrix, it will only have one set of eigenvectors, so $\mathbf{C}^{\alpha, 0'} \equiv \mathbf{C}^{\beta, 0'}$! If the ranges for forming the two densities are different (as it would be when $N_{\alpha} \neq N_{\beta}$), then having an initial pair of identical MO coefficient matrices is not an issue. For your Huckel guess (called the Generalized Wolfsberg-Helmholtz or GWH guess in Q-Chem, scf_guess = gwh), the initial MO coefficient matrices would still be identical, because you modify a quantity they have in common (the core Hamiltonian).
So, we must modify the MO coefficients (coefficient matrices) themselves. There are three options that I know of:

mixing
rotation
switching

These are variants of the same thing; rotation is the most general, since you can define switching or mixing of two orbitals via rotations.
Mixing
The most straightforward approach of the three that requires no prior knowledge of your system is mixing. The usual choice is to "mix" some amount of the HOMO with the LUMO of the same spin:
$$\begin{align}
C_{\mu,\text{HOMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( C_{\mu,\text{HOMO}}^{\text{old}} + k C_{\mu,\text{LUMO}}^{\text{old}} \right) \\
C_{\mu,\text{LUMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( -k C_{\mu,\text{HOMO}}^{\text{old}} + C_{\mu,\text{LUMO}}^{\text{old}} \right)
\end{align}$$
where $k = [0, 1]$. If $k = 0$, nothing happens; if $k = 1$, the HOMO and LUMO become the symmetric and antisymmetric linear combinations of the two, respectively. In theory, this doesn't need to be the HOMO and LUMO; any two MO indices could be chosen, as long as they will result in different densities. This is also only for a single $\mu$; because we want to work with matrices, we would be working with whole rows of the MO coefficient matrix. For Q-Chem, this is $\alpha$ spin only (scf_guess_mix = k); the Gaussian manual doesn't say which spin it takes.
Rotation
The most general form for operating on pairs of MOs is rotation. Recall the rotation matrix in two dimensions:
$$
R(\theta) = \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
$$
This can be used to "rotate" two arbitrary molecular orbitals $i,j$ by an arbitrary angle:
$$
\begin{bmatrix}
C_{i}^{'} \\
C_{j}^{'}
\end{bmatrix}
= \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
C_{i} \\
C_{j}
\end{bmatrix}$$
where again each $C$ is a row vector. This can be expended into a set of 2 equations:
$$\begin{align}
C_{i}^{'} &= C_{i} \cos(\theta) - C_{j} \sin(\theta) \\
C_{j}^{'} &= C_{i} \sin(\theta) + C_{j} \cos(\theta)
\end{align}$$
There are a few interesting cases.
180 degree rotation: change phase
$$
R(180^{\circ}) = \begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
$$
leads to
$$\begin{align}
C_{i}^{'} &= -C_{i} \\
C_{j}^{'} &= -C_{j}
\end{align}$$
This is kind of pointless, since the energy should be invariant to a phase swap.
90 degree rotation: interchange
$$
R(90^{\circ}) = \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
$$
leads to
$$\begin{align}
C_{i}^{'} &= - C_{j} \\
C_{j}^{'} &= C_{i}
\end{align}$$
This is the usual case when "rotate orbitals" is seen. This is equivalent to switching orbitals. I don't know about other programs, but if you do this in ORCA, the orbital energies will be swapped as well.
45 degree rotation: symmetric and antisymmetric linear combinations
$$
R(45^{\circ}) = \begin{bmatrix}
1/\sqrt{2} & -1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix}
$$
leads to
$$\begin{align}
C_{i}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} - C_{j} \right) \\
C_{j}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} + C_{j} \right)
\end{align}$$
This last example is identical to the $k = 1$ (100%) case from the mixing example shown earlier.
Switching
We could also alter occupation numbers, such as by placing the $\alpha$ (or $\beta$) HOMO electron into the same-spin LUMO, or some other occupied orbital, but this isn't an automated procedure like orbital rotation, because you need to know your occupied/virtual orbital indices prior to starting the calculation. This is equivalent to switching orbitals.
A short summary of program capabilities (to the best of my knowledge):

Q-Chem: mix, switch
Gaussian: mix, switch
GAMESS: mix, switch
DALTON: switch
MOLPRO: rotate
ORCA: rotate

An important note about mixing: it doesn't matter that choosing the HOMO and LUMO to permute may be incorrect. Programs that have baked-in support for symmetry (DALTON, MOLPRO, unsure about GAMESS) will only mix/switch/rotate orbitals within a symmetry unless symmetry is disabled, but the point is only to break symmetry in the initial guess so that the local RHF minimum is escaped. If you know beforehand that your system has particular symmetries that need to be preserved, then you would probably be doing something more complex than just mixing the HOMO and LUMO anyway.

So you start with $C^{\alpha}$ and $C^{\beta}$ as zeros, like in RHF calculation, then you apply mixing/rotation/switching during the SCF loop on one of the two density matrices?

@mle No, you typically only apply switching or mixing to the guess and not to further SCF iterations, since then you will almost certainly not obtain the minimized wavefunction according to the variational principle; you'll have some amount of "something else" added to your final MOs. Don't confuse this with damping or smearing, which you tend to apply in early iterations to overcome a bad guess, then turn off once you have more confidence SCF will converge to the right solution.


text:
Why is water not part of the equilibrium constant?
In the acid base reaction
$$\ce{NH4+ + H2O <-> NH3 + H3O+}$$
the acidity constant, which is a relation between concentrations is given by
$$K_{\mathrm{a}}=\frac {\ce{[NH3]} \cdot \ce{[H3O+]}}{\ce{[NH4+]} \cdot \ce{[H2O]}}$$
Since $\ce{[H2O]}$ is constant, it can be included inside the constant to obtain a new $K_\mathrm{a}$ which is $6.3 \times 10^{-10}$.
(Being a constant allows us to relate the molar concentration to the density of water.)
And so, even though some quantity of water reacts with $\ce{NH4+}$, but the volume of water varies in a way that $\ce{[H2O]}$ would still be constant.
What I don't understand is why the molar concentration of water is constant. Why isn't it handled the same way as the other reactant?
I understand that if $\ce{H2O}$ was the only liquid and the rest of the reactants and products were of different state (gaseous or solid), then water would exist alone in the beaker and its concentration would therefore remain constant, but this is not the case.
That was about heterogeneousness. Trying another guess; in the reaction of esterification $$\ce{{alcohol} + {carboxylic acid} <-> {ester} + H2O}$$
$\ce{H2O}$ is considered as a product which concentration is not constant, so it is counted explicitly in the equilibrium constant.
Could the answer be that $\ce{NH4+}$, $\ce{NH3}$ and $\ce{H3O+}$ are dissolved in water, while the alcohol, the carboxylic acid, and the ester are not?
I'm looking for an answer that clearly shows which volume is considered (water or the solution), because I'm confused as to why something would or wouldn't be included in the equilibrium constant.

To get your chemistry to appear pretty, you need to enclose it in $_$. So $\ce{NH4+}$ becomes $\ce{NH4+}$. Do not use MathJax/LaTeX in question titles as it messes with searching.

In general the concentration of water is assumed to be a constant in aqueous solution. That isn't absolutely true of course since for example the density of water changes with temperature.

related, but not duplicate: http://chemistry.stackexchange.com/questions/9847/what-is-the-pka-of-the-oxonium-hydronium-h%e2%82%83o%e2%81%ba-ion/27278#27278

There is no IUPAC definition of $K_a$, however, many reputable texts like Levine's Physical Chemistry define $K_a$ as:
$$\frac {a(\ce{A-})a(\ce{H3O+})}{ a(\ce{HA})a(\ce{H2O})}$$
where "a(X)" is the activity of species "X".
Then various approximations can be made, such as approximating $a(\ce{H2O}) = 1$ and approximating the activity of the solutes as the concentration of the solutes.
These approximations are only reasonable in dilute solutions.

The concentration of water in water (sounds weird but this is just liquid water) is found as follows:$$[\ce{H2O}]= \mathrm{\frac{1.0\frac{g}{cm^3}}{18.0148\frac{g}{mol}}}=55.5\mathrm{\frac{mol}{L}}$$
Compare this with the usual concentrations of reactants and products which is much smaller than that. Thus, the change in concentration of water throughout reactions which take place in water is considered to be constant.
The above is a very good approximation. While it is true that water's density varies some with temperature, at $98~\mathrm{^\circ C}$ the density of water is still $0.96\mathrm{\frac{g}{cm^3}}$, so the concentration of water will still be much much more than that of any of the reactants.
So, as you speculated, whether or not water is considered in the equilibrium constant entirely depends on whether or not water is a product and if water is the solvent.
If water is the solvent, it would be very normal to ignore the water which is produced or the water which might act as a reactant.
Similarly, any reaction which takes place by solvolysis will exclude the solvent from the equilibrium constant.
Also, you say that if water were alone in a beaker, then obviously all you have is water so the concentration is constant. Consider that water auto-ionizes by:$$\ce{2H2O <=> H3O+ + OH-}$$In this case, even though we know that we don't have just water, we know that the concentration of the other things is tiny compared to the concentration of water, so we just ignore the other stuff and drink the water.

Pro tip: Type entire equations with a single \ce{...}. $\ce{H2 + Cl2 -> 2 HCl}$ and $\ce{AB + C <=> A + B + C <=>> A + BC}$ give $\ce{H2 + Cl2 -> 2 HCl}$ and $\ce{AB + C <=> A + B + C <=>> A + BC}$, respectively ;)

So not only water must be a solvent, but also its quantity must be relatively large comparing to other soluble substances of the reaction in order to consider its concentration as constant and attaining the above value of 55.5 mol per liter. Agree?

Or I guess the solubility (which is the maximum amount of dissolved matter in a given temperature) which would be small will keep the concentration of other substances small relative to water, that's why we won't bother by comparing quantities put into reaction.


text:
Can 2 pieces of metals combine after long time contact?
Does 2 pieces of metals combine automatically after long time contact because of atom vibration and collision and finally form metallic bonds between the contact surface?

Well, they kinda can, but the surface must be really exceptionally clean.

See, this has been answered before: http://physics.stackexchange.com/questions/87107/why-dont-metals-bond-when-touched-together

Interestingly, in space if two metals make contact, they will automatically combine to form one metal

Gold will bond to gold  in a short time, and only a bit longer is needed when in contact with silver or copper. Indium also can bond quickly. BTW, I've had mercury wet a gold ring, leaving a silver-colored stain, that disappeared in a few days as the mercury flowed deeper into the ring.


text:
Reference value for two-electron repulsion integral over GTO's
I am currently trying to implement a Full CI program from scratch. The energies I get are a bit too high, so I'm looking for the mistake.
One possibility is my implementation of the two-electron repulsion integrals,
$$\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{e^{-\alpha_1 \boldsymbol{r}_1^2} e^{-\alpha_2 \boldsymbol{r}_1^2} e^{-\beta_1 \boldsymbol{r}_2^2} e^{-\beta_2 \boldsymbol{r}_2^2}} {\left| \boldsymbol{r}_1-\boldsymbol{r}_2 \right|\ } \mathrm{d}\boldsymbol{r}_1^3 \mathrm{d}\boldsymbol{r}_2^3 $$
with $\boldsymbol{r}_1=(x_1,y_1,z_1),  \boldsymbol{r}_2=(x_2,y_2,z_2).$
However, I cannot find any values I could compare my results to. I tried (numerical) integration in Maple, but that's way too slow and/or numerically unstable because of the singularity at  $\boldsymbol{r}_1=\boldsymbol{r}_2$. I tried installing the libraries libint and Libcint, and the python module pyscf, which all should be able to do this kind of computations, but I horribly fail at installing things not made for Windows (MINGW is only half working, and I don't have a proper Linux installation available right now...).
So, could someone who has such a program installed please give me a number this integral evaluates to, for whatever numbers $\alpha_1 \ldots \beta_2$ ?

I successfully installed libint some years ago using cygwin (free posix emulation layer for windows.)

Thank you, but the thing is, I already wasted about three days on trying to get some library to work (and still didn't, I know, that's pathetic ^^), and I'm exhausted. All I really need is a single number to compare... Here I think it's more likely someone has a quantum chemistry program installed that's willing to easily output such a number. It would really help a lot.
Well, might try the python thing again tomorrow on a friends laptop, if I can borrow it...

Try a live DVD/USB if you want to use Linux.

You don't state which method you implemented in order to compute two-electron integrals, therefore I will list all the main references at first.
Cook's book [1] contains analytical formulas for overlap, kinetic, electron-nucleus attraction and electron-electron repulsion integrals. The analytical formula for electron-electron repulsion integrals is wrong in the book, but you can look at this discussion for the errata.
However, a Full CI calculation is computationally costly. For this reason I believe a better approach would be to use a more efficient scheme to compute integrals. You can look at the following:

Obara-Saika scheme
McMurchie-Davidson scheme
Rys quadrature

You can find a good explanation of these methods in Ref. [2].
Now, if you want to check what you already implemented, you can find a list of the two-electron integrals in the STO-3G basis set for $\ce{HeH+}$ in Szabo's book appendix [3]. Here I can give you a set of two-electron integrals I obtain for $\ce{H2}$ (always in the STO-3G basis set) for a bond length of $1.4$ Bohr:
 (           1           1           1           1 )   0.77460834925515787
 (           1           1           1           2 )   0.44410904384277344
 (           1           1           2           1 )   0.44410904384277350
 (           1           1           2           2 )   0.56967771733030592
 (           1           2           1           1 )   0.44410904384277361
 (           1           2           1           2 )   0.29702946944511982
 (           1           2           2           1 )   0.29702946944511982
 (           1           2           2           2 )   0.44410904384277333
 (           2           1           1           1 )   0.44410904384277333
 (           2           1           1           2 )   0.29702946944511982
 (           2           1           2           1 )   0.29702946944511982
 (           2           1           2           2 )   0.44410904384277361
 (           2           2           1           1 )   0.56967771733030592
 (           2           2           1           2 )   0.44410904384277350
 (           2           2           2           1 )   0.44410904384277344
 (           2           2           2           2 )   0.77460834925515787

Note that these results came from a program I wrote myself, but it usually match very well Szabo's [3] and Gaussian09 values for the total energy. If these results match your calculations, then the problem might be in integrals with higher angular momentum.
[1] Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.
[2] T. Helgaker, P. Jørgensen and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000.
[3]  A. Szabo and N. Ostlund, Modern Quantum Chemistry, Dover, 1996.

Awesome, thank you very much! My integrals are indeed all about 0.02 too high, enough to explain the difference in my final result. I implemented the analytical solution, and you are right, the computational cost is incredible. I will look into the integration schemes you mentioned!

@Giogina You can stick with the analytical implementation in order to have a working program at first. Make sure you implemented well your calculation (Cook's correct analytical formula is quite straightforward) and make sure your problem does not comes from Boys function!

Hm, on the first glimpse my re-derivation of the formula does look very similar to Cook's one, but with 15 summations it's so easy to have a small mistake somewhere, I'll probably just have to rewrite the whole thing. I trust my boys function, since I'm still in Maple, it should do that well.


text:
Does Dean Kamen's Slingshot really filter virtually everything?
According to Dean Kamen (in the movie http://www.imdb.com/title/tt3242934/) and on the Net (http://www.popsci.com/article/science/pure-genius-how-dean-kamens-invention-could-bring-clean-water-millions) his device that uses pressure-Distillation to purify water, it would work with virtually any water and render medical-grade pure water.
I have a hard time to believe this as:
-chemicals could evaporate before water and so get into the distilled water
-I have no idea how he keeps the distillation chamber clean (prefiltering?) so it does not clog.
So I would like to know what are the limitations of distilling water to purify it?
I know that distilled water cant be drunk as it is without adding salts. I like the project of Mr. Kamen very much, I just want to know the boundaries, as the promises ("Fill in anything that is wet") seem way to optimistic.
Thx

All it does is distill water while keeping heat within the system.  There isn't much detail provided about it but heres what I can say.

I have a hard time to believe this as: -chemicals could evaporate before water and so get into the distilled water -I have no idea how he keeps the distillation chamber clean (prefiltering?) so it does not clog.

While you are right that chemicals can evaporate with the water this is not usually of concern if the water source is not severly contaminated with volatile chemicals (as opposed to biological and soluble toxins).  If you were to have chemical contamination you could use reverse osmosis or activated charcoal to remove residual contamination.  typically this is not an issue though.

I know that distilled water cant be drunk as it is without adding salts. I like the project of Mr. Kamen very much, I just want to know the boundaries, as the promises ("Fill in anything that is wet") seem way to optimistic.

Distilled water can be drank.  It may not be the most tasteful water but it does taste better than cholera.  If you want more information his TEDMED Talk is very informative.


text:
Peroxide and 2 extra electrons?
Trying to figure out where peroxide gains 2 electrons and why it has a 2- charge? On the Lewis dot structure it shows each O has 6 electrons and that it has a single bond between the two Oxygen atoms.
To me, each O atom would have 4 electrons and have a double bond between them?

Well, I might take a different approach at an answer, and try to get to a reasonable Lewis structure with data we already should have had from drawing structural diagrams! It's kinda upside down, since none of what's written below is happening in real-life chemistry.
The structure you have in mind is the diatomic oxygen molecule, with 12 electrons in its system overall.
$\hspace{33ex}$
$\hspace{45ex}$Source
While the peroxide ion overall has 14 electrons in its system, since each oxygen has 6 valence electrons and the system also has two extra electrons indicated by the $2-$ charge of the ion. Now, let's oversimplify this a bit: Imagine that in peroxide also, the oxygen atoms need to fulfill their octet configuration, hence each of them would have 8 electrons in its outer shell.
Let $N_b$ be the number of bonding electrons and $N_{nb}$ the number of non-bonding electrons of each oxygen atom. We know that

A single oxygen atom would have non-bonding electrons and bonding electrons that fill the octet.
All of the electrons in the system, which are the non-bonding electrons for each of the two oxygens, and the bonding electrons, make 14 electrons.

Hence we solve:
$$\Bigg\lbrace\begin{array}\\ N_b + N_{nb} = 8 \\ N_b + 2N_{nb} = 14 \end{array}\Bigg\rbrace \Rightarrow \boxed{N_b = 2},~\boxed{N_{nb} = 6} $$
Hence we could figure out how many bonding and non-bonding electrons there are in compounds we can draw Lewis structures for, by finding out the overall number of electrons (taking into account the number of electrons in the valence of each atom and the number of electrons lost or gained due to the charge of the ion) with a rudimentary approach.
Try to find out how many bonding and non-bonding electrons there are for the azide, permanganate, manganate, sulfate, oxalate, salicylate, and superoxide anions and thionyl, sulfuryl, nitrosyl and methanium cations.

Thank you so much. I still dont see where the ion picks up the additional two electrons however.. Are the two oxygen atoms that come together to form peroxide are they already ions ? I understand how its 14 and how to calculate the formal charge but how exactly do they pick up the additional 2? U

@Atticus283blink The same question applies to any other ion. $\ce{Na+}$, for example - where did it lose that missing electron? Or $\ce{Cl-}$ - where did it get that extra electron? Chemistry is full of mysteries, you know.

@ivan Neretin @M A R How is peroxide actually formed? Do two oxygen atoms that have already lost 1 electron each bind together or?

@Atticus283blink Like any other ion or molecule, it can be actually formed in a variety of ways. Losing electrons, however, is definitely not one of them. Losing electrons would get you a particle with positive charge.

I meant gain electrons. It says each oxygen in peroxide has an oxidation state of -1 what does this entail?

I guess what im asking is how does the peroxide form? Is it just two oxygen atoms that come together which both have a charge of 1- (each have picked up an extra electron) before bonding together and when they come together this forms a peroxide?

Like I said, it may have formed in many different ways. Maybe an oxygen molecule $\ce{O2}$ picked up two extra electrons. Maybe a hydrogen peroxide molecule $\ce{H2O2}$ lost two protons. It does not matter; also, it has nothing to do with oxidation states whatsoever.


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Do really the electrons could only orbit in specific orbits with a fixed radius?
According to Bohr's model of atom

The electrons could only orbit the nucleus in specific orbits or shells with a fixed radius. Only shells with a radius given by the equation below would be allowed, and the electron could not exist in between these shells.
  r(n)=n2
   ⋅r(1)

But when I learnt about different orbitals inside the atom in my textbook, it was said that you cannot precisely tell the location of an electron, as you know the velocity, according to Heisenberg's uncertainty principle. Besides in a video of khanacademy.org Sal said that orbitals means that if we took a number of snapshots of atoms, we would see that the electrons are denser in those orbitals but that does not necessarily mean that we wouldn't find it outside of the orbitals. The probability of finding electrons is higher in the orbitals but not 100%. But Bohr said electrons are quantized. It must be in a fixed orbital with a fixed radius. Isn't that contradictory? Which one is true? Or what is wrong in my understanding?

Forget about Bohr's model. It is but of historical importance, much like phlogiston theory.

http://chemistry.stackexchange.com/questions/32860/question-about-quantum-mechanical-model-of-atom-and-the-quantum-numbers

"According to Bohr's model of atom" -- If the model was accurate, it wouldn't have been replaced, no?

Bohr said electrons are quantized. It must be in a fixed Orbital with a fixed radius. Isn't that contradictory?

No it is not contradictory.
"Quantized" means that a continuous range of a variable, such as energy, is not allowed.  Only discrete values of the variable are allowed.
In Bohr's model, the radii and energies were quantized.

Which one is true?

It is not true that electrons orbit in fixed circular orbits.  Subsequent to the Bohr theory, there was Sommerfeld theory, which involved elliptical orbits and was a relativistic theory that correctly predicted the fine structure of the hydrogen atom (which Bohr's theory did not). Then there was Schrodinger's theory, which did away with the fixed orbits, but was non-relativistic and gave the same energy levels as Bohr's theory.  This was followed by Dirac's theory which was a relativistic theory and gave the same energy levels as Sommerfeld's theory.  However, the hyperfine structure of the hydrogen atom, such as the Lamb shift, was still not explained, until quantum electrodynamics was developed. Further developments included the recognition that the proton is not a point particle, but is composite.  So there has been no finding the truth, just more accurate, but more complex, approximations of the truth.

Sal said that orbitals means that if we took a number of snapshots of atoms, we would see that the electrons are denser in those orbitals but that does not necessarily mean that we wouldn't find it outside of the orbitals. The probability of finding electrons is higher in the orbitals but not 100%

Orbitals have an infinite spatial extent.  There are nodes where the wavefunction is zero for the Schrodinger equation solutions of the hydrogen atom, but nodes are specific planes, conical surfaces or spherical surfaces, not 3-dimensional regions.  People depict the orbitals as having finite volumes, but these depictions are the high probability regions not the entire orbital.  So if Sal said that, he should have said that the electrons can be outside of the high-probability regions of orbitals that people draw, rather than saying outside the orbitals.

I believe Schrodinger's theory is definitely a lot more realistic than Bohr's model. Bohr was not wrong about electrons being quantized. It is definitely true that electrons can only be in one orbital at a time. However, where Bohr was wrong, was to assume that electrons orbit around the nucleus in a circular motion. If this was true, then according to the laws of classical physics, the electrons would continuously lose energy and spiral into the nucleus. Obviously, this does not occur, so we can assume that electrons do not orbit the nucleus. Rather, they exist in these orbitals. Orbitals are more like an electron cloud that surrounds the nucleus. Therefore, it isn't contradictory, it's just that Bohr had some flaws in his model that were later 'fixed' so to speak, in Schrodinger's model. This is not to say though, that Schrodinger's model was perfect. As above comments have stated, there are theories that were developed after Schrodinger's that were definitely even more clear and advanced.
I hope this helped!