text:
Why is isoindole unstable?
According to this paper, isoindole is unstable because the 1-carbon in the isoindolidine tautomer is very electron-deficient.  I figured that the electron-deficiency of this carbon must be because it is bonded to an aromatic carbon and double bonded to a nitrogen, both of which atoms are electron-withdrawing.
By this logic, I would expect other carbons in this situation (C(aryl)-C=N) to also be very electron-deficient.  However, aromatic Schiff bases seem pretty common, for example salen.  What am I missing?

Of course, the carbon in Schiff bases is still very electron deficient as is evident by its NMR signal. Note that $\text{electron deficient}\ne\text{unstable}$. However, I think that an elaborate answer which I cannot satisfactorily provide at the moment would be very interesting.

Kopecký et.al. indeed utter the word unstable, but not in the sense of its recommended definitions, which were established later.

Although the parent compound, isoindole (I), has resisted isolation, the existence of this unstable species has been shown by trapping with dienophiles (i).

The problem here is that stability always requires a point of reference, which is not given in the quote above.
I would assume that the authors use unstable (or instability) as a way to say the compound cannot be isolated. They go on and justify why this compound behaves in this way, describing what I would say is its reactivity. While their argument is sound and I believe correct, the choice of their wording is a little bit ambiguous (in today's terms).
We clearly need to distinguish a compounds reactivity (towards a certain reaction) from its stability within a common set of parameters. As such the following statements are correct, but neither of them is a justification for the other.

2H-Isoindole is a less stable $\ce{C8NH7}$ isomer than 1H-indole.
2H-Isoindole is more reactive towards dienophiles than 1H-indole.

While statement one is easy enough to prove computationally, it is a lot harder to explain or analyse. Quick calculations on the DF-M06L/def2-TZVPP level of theory gives the following hypothetical reaction energy:
$$\begin{align}
\ce{1$H$-indole &-> 2$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +38.6~\pu{kJ mol^-1}
\end{align}$$
Unfortunately Kopecký et.al. offer no explanation for this, because they are more or less focused on explaining why it is "unstable", by which they most likely mean highly reactive or not isolatable. They are concerned with the reaction paths that lead away from this species and find that 1H-isoindole plays a big (probably most important) role in this.

The [...] electron deficiency for position 1 of the isoindolenine [...] leads us to conclude leads us to conclude that the reactive species responsible for the instability of this series of compounds probably is the isoindolenine [...].
(Note that the numbering differs from the IUPAC name and the position referred to as 1 is actually 3.)

I think this analysis is spot on a holds up to modern standards of computational methodology. The 1H-tautomer indeed opens up a very big variety of decomposition and reaction pathways, and it is readily accessible at room temperature (DF-M06L/def2-TZVPP):
$$\begin{align}
\ce{2$H$-isoindole &-> 1$H$-isoindole} & \Delta_\mathrm{r}G^\circ &= +3.4~\pu{kJ mol^-1}
\end{align}$$
The reason for this is - and this is the point where Joule and Mills get it completely wrong - the retention (not the creation) of the aromatic system in the benzene moiety. The chemical behaviour (reactivity towards cycloadditions) is not determined by the lack of a 'complete' benzene ring. But that is us slipping into statement 2, which we are not quite ready yet.
Let us first dissect the latter statement in the same book (p.447), which is equally as wrong:

Isoindole, benzo[c]thiophene and isobenzofuran are much less stable than their isomers, indole, and benzo[b]thiophene and benzo[b]furan.

Although they offer no appreciable effort to prove this statement they are probably correct. At least the calculations for the $\ce{C8NH7}$ system confirm that. They even use the correct (or recommended) definition of stable in this case.

This is undoubtedly associated with their lower aromaticity, which can be appreciated qualitatively by noting that in these [c]-systems, the six-membered ring is not a complete benzenoid unit.

Again, they offer no proof for that other than a hand-wavy argument based on a single Lewis-structure. If we include resonance into the mix then we have an at least as reasonable argument as this to disprove it.

If we have a look at the molecular orbitals of the compounds, we will see that there is more to this than just one resonance description. I have pulled out two characteristic molecular orbitals for example, but the complete set can be found here. On the left side (top) we have 1H-indole and on the right side (bottom) we have 2H-isoindole. Apart from a light shift of the electron density due to the loosened symmetry in indole, the orbitals are pretty much the same.

When we take the 1H-isoinole isomer into the mix, we see that the aromatic system of the benzene ring is retained completely.

Aromaticity is still not a fully understood concept and as a result there is not really a rigorous concept that can be applied (see the definitions below). There will probably be debate about it for the next decades. (Hopefully my bounty will pay off and we can find out more in this question.)
As a result we don't really have a handle on judging whether or not 2H-isoindole has a lower aromaticity than 1H-indole. And according to Kopecký's paper, with available methods at this time, this cannot be backed up with theoretical results.
While I hoped to find some experimental evidence for the existence of isoindole due to the general progress within the last fifty years, I was still unable to find a single source.
I'll skip over the rest of the quote because the stated facts are certainly accurate, just the reasoning with the 'incomplete' benzene ring is false.
Let's look at statement two for a while.
The reactivity of 2H-isoindole towards dienophiles is very easily explained looking at the HOMO of the compound; for comparison the HOMO of butadiene on the right (bottom).

The very easy rationale is that the HOMO is of the right symmetry for a Diels-Alder reaction, which additionally retains the aromatic character of the benzene moiety.
Now a look at the HOMO (left/ top) of 1H-indole will tell us that the only viable attack can be carried out on the 4,7-positions, leaving a much smaller aromatic moiety, i.e. pyrrole, than isoindole. The only way for an inverse electron demand  Diels-Alder reaction would also be in the benzene moiety as the LUMO (right/ bottom) shows.

This is why one molecule reacts readily, while another has to be forced (really, really hard) or reacts as the dienophile via 'the double bond in the pyrrole system'.

After all that information the question still remains, or rather the question has become
"Why is isoindole less stable than indole?"
As I said before, this is really not easy to interpret. In general, lifting symmetry restrictions results in a more stable configuration, as fewer boundary conditions equal more flexibility and that usually gives a lower energy.
One aspect is certainly also that the lone pair of the nitrogen will be better delocalised in indole as there are more carbon atoms in proximity. This is actually something visible in the molecular orbitals.
Apart from that, neither geometries differ much from idealised benzene ($\mathbf{d}(\ce{C-C})=138.7~\pu{pm}$) or pyrrole($\mathbf{d}(\ce{C-N})=136.7~\pu{pm}$, $\mathbf{d}(\ce{C-C'})=137.1~\pu{pm}$, $\mathbf{d}(\ce{C'-C'})=141.4~\pu{pm}$, picture), so that a significant distortion energy is probably not a large contributing factor.

Other than that, I don't know. I also cannot explain, why the other aromatic $\ce{C8NH7}$ isomers are even less stable than isoindole. That would probably require more extensive calculations and analyses.
\begin{array}{lr}\hline
\text{Isomer}& \Delta E_\mathrm{el}/\pu{kJ mol^-1} & \Delta G/\pu{kJ mol^-1}\\\hline
\text{1$H$-indole} & 0.0 & 0.0\\
\text{2$H$-isoindole} & 37.3 & 38.6\\
\text{indolizine} & 52.0 & 50.6\\
\text{1$H$-cyclopenta[$b$]pyridine} & 83.4 & 82.4\\
\text{2$H$-cyclopenta[$c$]pyridine} & 79.7 & 79.1\\\hline
\end{array}

By this logic, I would expect other carbons in this situation, $\ce{C_{ar}-\color{\red}{C}=N}$, to also be very electron-deficient. However, aromatic Schiff bases seem pretty common, for example salen. What am I missing?

It is true that the carbon atoms in question will be very electron deficient, and they will probably be a good attach point for nucleophiles. In the salen ligand the nitrogens are probably stabilised by intramolecular hydrogen bonds, and while in complexes the entropy factor should make is very stable, too.
On another note, the isoindole moiety is also  very, very common. Just think heme ligands and complexes.
One final thought: Carbonic acid decomposes rapidly in water, however, it is seldom questioned, that the molecule itself is stable under certain conditions.

My personal preferred definition for stable:  An certain arrangement of atoms that forms a local minimum on a potential energy hypersurface and therefore could theoretically probed is stable.

Helpful definitions from the IUPAC gold book

unstable
  The opposite of stable, i.e. the chemical species concerned has a higher molar Gibbs energy than some assumed standard. The term should not be used in place of reactive or transient, although more reactive or transient species are frequently also more unstable. (Very unstable chemical species tend to undergo exothermic unimolecular decompositions. Variations in the structure of the related chemical species of this kind generally affect the energy of the transition states for these decompositions less than they affect the stability of the decomposing chemical species. Low stability may therefore parallel a relatively high rate of unimolecular decomposition.)

stable
  As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species A is more stable than its isomer B if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B},$$
  under standard conditions. If for the two reactions:
  \begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G_1^\circ)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G_1^\circ)\\
\end{align}
  $\Delta_\mathrm{r}G_1^\circ > \Delta_\mathrm{r}G_2^\circ$, P is more stable relative to the product Y than is Q relative to Z. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

aromatic

In the traditional sense, 'having a chemistry typified by benzene'.
A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.

Good grief, Mart, you're never going to get your research work done if you keep putting together such enormous answers!

Thanks a lot, this is really helpful!  The distinction between instability and reactivity is something I have been unknowingly confused about for a while.    You say that "the 1H-tautomer indeed opens up a very big variety of decomposition and reaction pathways" -- do you know what these are?   And I'm not convinced that heme is an example of isoindole.  Phthalocyanine, for example, is constructed from diiminoisoindoline, which has no resonance structure that is a substituted isoindole, so it seems something very different is going on here.

@JoshuaMeyers heme was a misfire, I'm not sure what was going on in my head. Maybe to many structures after a long answer. I'm already in the weekend, but I try to look into the other structures next week. About the pathways, I'll try to get to that also. Once the aromatic system is destroyed, you can pretty much do imine chemistry. The electron deficient carbon is a good access point for nucleophiles, there could be more. We'll see. We could talk in a chat room some more. It's an interesting topic that requires some more research...

@Martin-マーチン Sure, we could chat sometime.  It seems that you aren't the only one who's wrong about heme.  Strangely enough, Joule and Mills also seem to list phthalocyanine as a isoindole derivative.  Wikipedia misleadingly has hemes in the isoindole page but lists them as isoindole-1,3-dione derivatives.

Great, another answer I can bounty soon =D

@Martin-マーチン Is there any way to understand why these molecules have the molecular orbitals that they do, and not just take the program that generated them as a black box?

@Joshua I recommend learning molecular orbital theory and basic quantum chemistry. It's more or less the understanding you need how the program 'generates' the solutions to quite complex mathematical problems. Once you have a deeper understanding of that, the program is no longer a black box. It basically becomes a very advanced calculator.

@Martin-マーチン What book would you recommend?

Perhaps that paper is wrong, and instead isoindole is unstable because it doesn't have a complete benzene ring.  This quote from Heterocyclic Chemistry, by Joule and Mills is relevant (p. 372):

Image text: The chemical behaviour of isoindole, benzo[c]furan and benzo[c]thiophene is dominated by their lack
  of a ‘ complete ’ benzene ring: these three heterocycles undergo cycloaddition processes across the 1- and
  3-positions with great facility, because the products do now have a regular benzene ring. Often, no attempt
  is made to isolate examples of these heterocycles, but they are simply generated in the presence of the
  dienophile with which it is desired that they react. As a result of this strong tendency, few of the classical
  electrophilic and nucleophilic processes have been much studied.

Can someone tell me if this is correct?
Edit: Later (p.447) the book explicitly states that this is why isoindole is unstable:

Image text:  Isoindole1, benzo[c]thiophene2 and isobenzofuran3 are much less stable than their isomers, indole, and
  benzo[b]thiophene and benzo[b]furan. This is undoubtedly associated with their lower aromaticity, which
  can be appreciated qualitatively by noting that in these [c]-systems, the six-membered ring is not a complete
  benzenoid unit. Of the three unsubstituted heterocycles, benzo[c]thiophene is the most stable – it survives
  as a solid for a few days at −30 °C – but most chemistry has been carried out with substituted derivatives.
  The instability manifests itself in a strong tendency to add reagents so as to generate products that do have
  a complete benzene ring, in particular these heterocycles are susceptible to cycloaddition of dienophiles.
  In this context, then, it is not surprising that for isoindole, for which an alternative tautomer (1H-isoindole,
  sometimes called ‘ isoindolenine ’ ) is possible, which does have a complete benzenoid unit, an appreciable
  percentage of that alternative exists in equilibrium.4  Indeed, some isoindoles exist largely as the tautomer
  with a C–N double bond – 1,3,4,7-tetramethylisoindole is an example5 - but 1-phenylisoindole favours the
  2H-tautomer to the extent of 91%.6 6
   Substituents on the benzenoid ring can also influence both the stability
  of the isoindole and the position of tautomeric equilibrium, for example 4,5,6,7-tetrabromoisoindole is a
  stable crystalline solid which exists wholly as the 2H-tautomer;7
   a pivaloyl substituent at C-5, though remote
  from the sensitive heterocyclic ring, can also stabilise an isoindole.8 The position of such tautomeric equilibria can be altered by changing solvent - solvents such as dimethylsulfoxide tend to favour the N-hydrogen
  tautomer, where protic solvents like alcohols favour the imine tautomer.

This argumentation is flawed, if not completely wrong. The idea that either of these compounds exists in a tautomeric equilibrium with a C=N bond is ridiculous; this is trying to interpret resonance structures as actual structures, which is obviously wrong. In any of these cases the aromatic system (10 pi e) is as delocalised as in the respective diastereomer. Simply for symmetry reasons indole cannot act as a dien, because that would completely annihilate the aromaticity while in isoindole part of it can be retained.

@Martin-マーチン The idea of a tautomeric equilibrium is certainly not ‘ridiculous’: see Z. Naturforschg. 1965, 20b, 75.

@Jan oh well, I misread that there is a proton shift involved. The rest of the argument is still shaky at best; no, I'm sure it's wrong. In either compound the benzene ring is already complete. But I don't want to go deeper into this, especially not in the comments. I doubt my analysis would be popular, because it wouldn't be simple.

@Martin I'm curious about your analysis, please share it as an answer if you have the time.

@Martin I think you meant "isomer" instead of "diastereomer".

@Martin Why isn't it possible for there to be two tautomers, one of which has a 6pi aromatic system and the other of which has a 10pi aromatic system?  It is possible to make compounds that are locked in either 10pi or 6pi+4pi configurations: http://pubs.acs.org/doi/abs/10.1021/ed063p306


text:
Why is HCN not a Base and why does HNC does not dimerise?
We know that species that have a lone pair of electrons are Lewis bases (like ammonia).
The structue of $\ce{HCN}$ is $\ce{H-C#{N:}}$; nitrogen has a lone pair of electrons and it is still acidic. Why?
We know that species with odd electron counts such as nitrogen dioxide (having odd electrons) dimerise to acheive greater stability.
The structue of $\ce{HNC}$ is $\ce{H-N#C}$.  So why it does not dimerise?

How is HNC an odd-electron species?

Nitrogen has 5 electrons in its outermost orbit. Out of 5 electrons, 4 electrons are bonded. Hence N has 1 electrons left.

You heard about dipolar bonds and compounds? https://en.wikipedia.org/wiki/Dipolar_compound

The usual rule is one question per post unless strongly related; while your questions are not really related I’ll still answer both of them since it’s only two.
Why is $\ce{HCN}$ not a base?
Well — it is a base. You seem to be thinking that being an acid and being a base is a dichotomy for some reason; rest assured that aside from the very simple Arrhenius theory no acid-base theory predicts that bases cannot be acids and vice-versa.
You correctly point out that there is a lone pair on nitrogen, and that lone pair can indeed act as a (Brønsted-Lowry) base. The only ‘difficulty’ is that the carbon side of the $\ce{^-C#N}$ entity is more basic than the nitrogen side; thus protonation will happen on carbon first.
$$\ce{H-C#N^+-H <=> H-C#N + H+ <=> ^-C#N + 2 H+}\tag{1}$$
Equation $(1)$ shows the complete acid-base equilibrium.
Why is $\ce{NCH}$ not an odd-electron compound?
I took the liberty of rephrasing your second question. If you count electrons as is done in equation $(2)$, you will realise that $\ce{CNH}$ comes in at ten electrons — an even number — just like $\ce{HCN}$.
$$\sum n_{\ce{e-}} = n_{\ce{e-}}(\ce{C}) + n_{\ce{e-}}(\ce{N}) + n_{\ce{e-}}(\ce{H})
= 4 + 5 + 1 = 10\tag{2}$$
When writing the Lewis structure, you should remember formal charges. Indeed, your description of $\ce{CNH}$ is bad; instead you should write its Lewis structure as $\ce{^-C#N^+-H}$. All valences are satisfied, there is no reason for the compound to dimerise. Assuming five electrons to be on nitrogen (and thus four to be on carbon) would generate a very unstable diradical which would violate the octet rule in a bad way (nine electrons on nitrogen). Lower electron count violations of the octet rule are fine but going over the electron count for a main group element does not happen.
However, there is a great tendency for the compound to de- and reprotonate as written in equation $(3)$:
$$\ce{CNH <=>> CN- + H+ <=>> HCN}\tag{3}$$

The structure of hydrogen cyanide is $\ce{H-C#N}$ not $\ce{H-N#C}$ so there is no lone electron on the carbon atom to dimerise as $\ce{H-N#C-C#N-H}$.
Hydrogen cyanide is an acid ionizing as $\ce{H^+}$ and $\ce{^-C#N}$ in aqueous solution.

Nitrogen has 5 electrons in its outermost orbit but in cyanide and isocyanide only 4 electrona are bonded . 1 electron is left

This does point out some inconsistencies but does not answer the question.,

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text:
References for studying Winkler's method and CO2 estimation by APHA method
What are the best sources to study and understand

Process (reactions involved)
Calculation

of Winkler's method and APHA method for CO2 estimation?

Related (with source): http://chemistry.stackexchange.com/questions/27450/finding-out-the-concentration-of-oxygen-in-water/

@KlausWarzecha The cited document is not in English.

Ludwig (actually Lajos) Winkler, a chemist of hungarian heritage, published his method in 1888 in a German Journal: Ber. Dtsch. Chem. Ges., 1888, 21, 2843-2854.
Given that and the fact that Hungary was part of the Austro-Hungarian Empire at that time, it isn't overly astonishing that the publication is in German, rather than in English.
A nice description in English, with a discussion of technical obstacles, but without the math or equations, was published by Patrick G. McCormick in J. Chem. Educ., 1972, 49, 839-841 under the title The determination of dissolved oxygen by the Winkler method. A student laboratory experiment.

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Clarification needed regarding description of molecules without hybridization using pure p orbitals
I need some clarification regarding this answer given by the user @Jan.

@Jan says:
Therefore, the a priori predicted most stable configurations of
  certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 p orbitals to form bonds; keep one lone pair in an s type orbital. (Predicted bond angle:
  $90^\circ$)

I could not understand the meaning of the first bullet point clearly.
I cannot agree that in mono, di and tri coordination pure $p$ orbitals are always used. Pure $p$ orbitals have bond angles of $90^{0}$. In that case, how can we explain the structure of compounds like $\ce{BF3}$ or ethene where $\mathrm{sp^2}$ hybridization is said to exist and bond angles are $120^{0}$ ?
Or did @Jan imply that we need to use pure $p$ orbitals in mono, di and tri coordination in case no hybridization takes place ? If so, how can we understand when hybridization will take place and when it will not? Please give some examples.

In brief: the hybridization model fails badly for hypervalent species. All have large contributions from charge-shift bonding, which leads to the actual valence electron population in the region around the central atom being $\leq 8$ in all cases. I'll post an answer with more detail when I can.

Yes, those bullet points are only very brief xD They can fail quickly; $\ce{BF3}$ does indeed fail the first bullet point because I did not consider phrasing it in a way to include boron. By the way, I’m leaning towards voting to close this as too broad since the two concerns you raise should be answered in two very different ways (and they are not really conceptionally related, except for both being simplified). This is coming at the danger that the pentacoordination part may be closed as a duplicate of one of the many questions here dealing with the topic.

@Jan Umm, before you vote to close I would ask you whether the question can be edited. Should I divide the question into two or more parts ? BTW your answer to my previous question was not really clear and the bullet points are really too brief.

@Doraemonドラえもん Yeah sure you can edit! In fact, the intended way to address closed (or: ‘on hold’) questions is to edit them to make them fit again. In this case, I would suggest asking about each of the bullet points in a separate question, i.e. editing this one down to one of the two and putting the other into another question ^^

@Jan Done. Please check if my questions are making sense, individually. http://chemistry.stackexchange.com/questions/64673/how-to-explain-shape-of-molecules-if-hybridization-involving-d-orbitals-in-main

Yes, what I wrote on the topic in the previous question was very, very brief and while writing it I was not thinking of boron — an element that usually calls for exceptions due to it only having three valence electrons.
However nonetheless, the first thing you should do for any compound is to consider all atoms in ground state configuration and see whether you can form the required bonds from occupied orbitals in that state. Let’s take the example compounds $\ce{BF3, NF3, AsH3, C2H4}$ (tricoordinate) $\ce{CO2, H2O, H2Se, BeCl2, CH2}$ (dicoordinate) and $\ce{HCl, N2, CO}$ (monocoordinate) for the discussion. Any structural data referenced herein is taken from the corresponding Wikipedia articles.
Ideal cases
I consider the examples $\ce{AsH3, H2Se}$ and $\ce{HCl}$ to be ideal cases. Taking each atom in its ground state ($\ce{As}:{ [\ce{Ar}]}\,\mathrm{4s^2\,4p^3}; \ce{Se}{: [\ce{Ar}]}\,\mathrm{4s^2\,4p^4}; \ce{Cl}{: [\ce{Ne}]}\,\mathrm{3s^2\,3p^5}; \ce{H}{: \mathrm{1s^1}}$), arsenic has three half-filled p orbitas, selenium has two and chlorine has one. Each of these elements (except hydrogen) has a single lone pair in an s orbital and if there are further lone pairs, they are in p orbitals (arsenic has no more). Thus, we can construct $\sigma(\ce{s-p})$ bonds until all valences are satisfied.
Moving on to bond angles: well, $\ce{HCl}$ has no problems as there are only two atoms. But $\ce{H2Se}$ and $\ce{AsH3}$ also have hardly any problems since the central atoms are so large that the comparatively smaller ligands can be attached at almost perfect $90^\circ$ angles without significant steric repulsion. The bond angles repored ($\ce{AsH3}{: 91.8^\circ}; \ce{H2Se}{: 91^\circ}$) are in adequate agreement.
Ideal cases with steric distortion
The next set of examples along the route you should be going are $\ce{NF3}$ and $\ce{H2O}$. These compounds fulfil the general criteria outlined in the previous section. However, their central atoms are decidedly smaller. Thus, perfect $90^\circ$ angles are unobtainable as there would be a steric clash of the ligand atoms. The ligand atoms move apart to prevent steric repulsion; this is achieved by mixing in s-contribution into the bonding orbitals. $25~\%$ s-contribution would correspond to a tetrahedral bond angle of $109.5^\circ$ and $\mathrm{sp^3}$ hybrid orbials. The table below compares the bond angle and length data of the lighter homologues discussed here and the heavier ones discussed above (consider the fluorines in $\ce{NF3}$ to be pseudohydrogens).
\begin{array}{lcccc}\hline
\text{element} & \ce{N} & \ce{As} & \ce{O} & \ce{Se} \\ \hline
d(\ce{X-H})/\mathrm{pm} & 137 & 151.9 & 95.84 & 146 \\
\angle(\ce{H-X-H})/^\circ & 102.5 & 91.8 & 104.5 & 91 \\ \hline\end{array}
$\ce{CH2}$, carbene, is a dangerous case. As far as I know, no definite structures are reported (although Wikipedia mentions bond angles determined by EPR) since it is too reactive and since it is susceptible to dimerisation to form ethene. It also comes in three distinct forms: a singlet form, a linear triplet and a bent triplet. The most ideal structure of these three is the singlet carbene, which also fits nicely into this subsection.
For a singlet carbene, you would also start off from carbon’s ground state $[\ce{He}]\,\mathrm{2s^2\,2p^2}$. The two p orbitals each form a $\sigma(\ce{s-p})$ bond to the hydrogen atoms. The two hydrogens, however, need to move away from each other to ease steric repulsion; thus a little bit of s-contribution is mixed in for a final bond angle of $102^\circ$. A third p orbial remains unoccupied — note the similariy to $\ce{H2O}$ which has everything carbene has plus two electrons in a p-type lone pair.
Practically ideal cases which naturally lead to multiple bond
Herein, we shall discuss $\ce{N2}$ and $\ce{CO}$. For both, we can apply the method outlined above; I’ll start with the simpler $\ce{N2}$ case. We initially have one s lone pair and three, singly occupied p orbitals on each nitrogen. One of these forms a standard $\sigma(\ce{p-p})$ bond with the other nitrogen. Now, thankfully, the other p orbitals are nicely aligned and fit perfectly for generating two $\pi(\ce{p-p})$ bonds. Together, that gives us three bonds or a triple bond, composed of one σ and two π bonds.
The carbon monoxide case is essentially the same except that carbon would have one empty p orbital and oxygen would have one lone pair in a p orbital; only one π bond can be formed in the way outlined above. However, the unoccupied carbon orbital and the fully populated oxygen orbital can align likewise and form what could formally be described as a second, dative π bond. That renders nitrogen and carbon monoxide isoelectronic. Note, however, that the simplification introduced herein is not able to explain the big differences between $\ce{CO}$ and $\ce{N2}$ that arise from their actually rather different molecular orbital schemes.
Cases, in which an s type lone pair is impossible
I’ve cleared away all the simple cases. If you go through the list, the only ones that remain are $\ce{BF3, C2H4}$ and $\ce{BeCl2}$ — although we might as well include linear and bent triplet $\ce{CH2}$ in here. These all have in common that assuming an s type lone pair would not allow for sufficient electrons available for bonding.
If that is the case, the simple answer always suggests hybridisation (athough the MO description does not require it and nonlocalised molecular orbitals would not support it). At least one p orbital must be mixed in to allow for the generation of enough σ bonds; yet you should always try and make the p contribution of the hybrid orbitals as small as possible (in simplified terms) as that means the s lone pair ‘loses’ less energy en route. (Beware, the terms and images I used are inaccurate!) Any orbitals, populated or not, that are not required for σ bonds remain as p type ones. For $\ce{CO2}$ and linear triplet $\ce{CH2}$, that gives us an $\mathrm{sp}$ configuration of the central carbon with two unused p orbitals; in $\ce{C2H4}$, bent triplet $\ce{CH2}$ and $\ce{BF3}$, the central atoms are $\mathrm{sp^2}$ configured with one additional p orbital.
If possible — i.e. in $\ce{CO2}$ and $\ce{C2H4}$ — π bonds should then be formed. Each of the oxygens in carbon dioxide has one p-type semipopulated p orbital available for a π bond as does the second carbon of $\ce{C2H4}$. The carbenes cannot form π bonds due to the lack of partners, and thus they remain as diradicals. The linear one has two degenerate p-type radical orbitals, the bent one almost has an $\mathrm{sp^2}$ type radical orbital and one p type. (In fact, bent triplet carbene has a larger bond angle than the $\mathrm{sp^2}$ predicted $120^\circ$. This is in line with removing part of the s contribution of the radicals and supplying that to the bonding orbitals. Nobody expects you to correctly analyse the structures of carbenes, though.)
Beryllium
You may have noticed that that element had gone missing at the end of the previous section. This is because it is the most special of all cases. All other alkaline earth metals behave as metals (form dications and practically ionic structures) while beryllium still features notable non-metal properties.
I think that the best description of $\ce{BeCl2}$ is via a four-electron-three-centre bond akin to $\ce{XeF2}$. That again requires an unhybridised central atom. The lowest of the three relevant MOs of the 4e3c bond would be formed of the fully bonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ combination; the second bonding MO would then be $\mathrm{p}(\ce{Cl}) + \mathrm{p}(\ce{Cl})$ where the nodal plane traverses $\ce{Be}$ — mixing in its 2p orbital probably increases the bonding character — and a final antibonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ orbital wherein the phase of beryllium’s orbital is of opposite sign with respect to the chlorine orbitals’. If that is the case, it should possibly be discussed with the ‘hypervalent’ cases of the other question.
I could be terribly wrong, though. Thankfully, nobody will request you to predict the MO scheme of $\ce{BeCl2}$ until at least the final bachelor’s examination.


text:
Bonding of the complex when both weak and strong field ligand are present

In the complex $\ce{[Ni(H2O)2(NH3)4]^2+}$ the magnetic moment ($\mu$) of $\ce{Ni}$ is –

Zero

$2.83~\mathrm{BM}$

$1.73~\mathrm{BM}$

$3.87~\mathrm{BM}$

It is easy to find the unpaired electrons and hence magnetic moment if we know the bonding strcture.
$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series. I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.
Nickel has $\ce{d^8}$ configuration here. I dont understand whether I should pair this electron or not. If I pair this electron, I get this:

If I pair the electron then I have to use one $\ce{3d}$ orbital and one $\ce{4d}$ orbital. I have not used both orbitals for overlap in any other question before so I don't know if that's right. Well, I think it should be because anyway these orbitals will make hybridized orbitals of equal energy for bonding. In that case, we get $\ce{d^2sp^3}$ (octahedral geometry).
If I do not pair the electron, I get $\ce{sp^3d^2}$ hybridization.
I am not sure but is this logic correct that since there are more number of strong field ligands than weak field ligands, pairing will occur?
Should I pair the electrons or not and why?

I don't think that the electrons would pair up. This pairing even does not happen in Ni(NH_3)_6 . Anyways, I am unsure why dsp3d is not generally observed.

$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series.

No, this is not correct. I keep saying this time and time again. Ammonia is a medium-field ligand at best, and water is the defined ‘zero point’ below which weak field ligands begin.

I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.

This is simplified at best. The transition from low spin to high spin depends on both the metal and the ligands (and also the geometry). In fact, some complexes are known both in a high-spin and in a low-spin state with identical central metal and ligands. You always need to consider the big picture to determine whether a high spin or low spin complex will result.
I am also going to completely ignore the stuff you wrote about hybridisation. You should not consider hybridisation in transition metal complexes as I have posted numerous times.

To answer this question you should ask yourself how many ligands you have around the nickel centre and which geometry that implies. The complex being tetraamminediaquanickel(II) results in six ligands and thus an octahedral configuration. The next step should be to consider the molecular orbital scheme of an octahedral complex (although the d-orbitals alone carry all the necessary information so even the crystal field model can answer it).

Figure 1: simplified MO scheme of an octahedral $\ce{[ML6]}$ complex ignoring π $\ce{L\bond{->}M}$ interactions. Originally taken from Professor Klüfers’ web scriptum of his coordination chemistry course at the LMU Munich.
The important bit to note is that the d orbitals split into a $\mathrm{t_{2g}}$ and an $\mathrm{e_g}$ part. $\mathrm{e_g}$ corresponds to the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2- y^2}$ orbitals. Thus, a perfect octahedral geometry will generate two degenerate orbitals which should be filled with parallel spins by Hund’s rule. Even if you start to assume Jahn-Teller distortions here (which are not likely, since a high-spin $\mathrm{d^8}$ octahedral configuration is not susceptible to Jahn Teller because there are no unevenly populated degenerate orbitals), that would not lead to an energy difference large enough to induce spin pairing. Remember that a typical 3d-metal octahedral complex is high-spin even in $\mathrm{d^6}$ cases; the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ is generally much larger than what can be achieved by Jahn-Teller distortion.
Now you may decide to argue using square planar complexes. However, the complex in question cannot be square planar since there are six ligands.
Therefore, a magnetic moment of zero is not an option in the case presented; $\ce{[Ni(H2O)2(NH3)4]^2+}$ cannot reasonably adopt a low-spin configuration. The high-spin complex is paramagnetic and in a triplet state ($2S+1 = 3$).

Can you simplify this answer considering that students are not taught jahn teller distortion, molecular orbital theory or crystal field theory in detail in high school? I first will have to learn these topics to be able to comprehend your answer.

@Blue No, that is not possible. You cannot comprehend coordination complexes without crystal field theory. Any school teaching otherwise is teaching inconsistently and inconsequentially.

Well i said we are not taught these in detail. I know about eg and t2g orbital s but i have no idea about a1g, t1u....its not there in the school textbooks as well,...we do most of the questions using hybridization of the central atom and use crystal field theory to explain some exceptional cases, d-d transition, etc...we are only taught molecular orbital theory for homodiatomic molecules..nothing has been said about jahn teller distortion in my textbook ...

@Arishta I presume that you had been reading NCERT Textbooks which of course don't have all this information. For class 12, you should consider reading Inorganic Chemistry by JD Lee or atleast the concise Indian version by Wiley India. All doubts regarding Jahn Teller effect and rest of the chapter will be cleared.


text:
Estimation of CO2 by titrimetric method using phenolphthalein indicator
We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association (APHA) method. It was a titrimetric method using phenolphthalein indicator.
Titrant used was $\ce{NaOH}$ and analyte was sample water. Reactions involved:
$$\ce{CO2 (g) ⇌ CO2 (aq)}$$
$$\ce{CO2 aq + H2O ⇌ H2CO3 ⇌ H+ + HCO3^2-} \label{rxn:1}\tag{1}$$
(Only $10\%$ of dissolved $\ce{CO2}$ get converted into $\ce{H2CO3}$.)
On addition of $\ce{NaOH}$:
\begin{align}
\ce{2NaOH + CO2 &-> Na2CO3 + H2O} \label{rxn:2}\tag{2} \\
\ce{Na2CO3 + H2O + CO2 &-> 2NaHCO3} \label{rxn:3}\tag{3}\\
\ce{NaHCO3 &-> Na+ + HCO3-} \label{rxn:4}\tag{4} \\
\ce{\underset{from \eqref{rxn:1} and \eqref{rxn:4}}{HCO3-} + H2O &-> H2CO3 + OH-} \\
\ce{OH- + H+ &-> H2O} \\
\ce{HIn &⇌ H+ + In-}
\end{align}
(The equilibrium shift to the right and when $[\ce{HIn}]:[\ce{In^-}]=1:10$ pink colour appears.)
But when we calculate the concentration of $\ce{CO2}$ we consider only reaction $\eqref{rxn:2}$, which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in step 3 and the $\ce{CO2}$ (the $10\%$) that was directly converted  into $\ce{H2CO3}$ (Rxn. $\eqref{rxn:1}$) are not taken into account.
So it means we are calculating only a part of the free $\ce{CO2}$ ($45\%$) and not all.
From this can we conclude that APHA's method is not very accurate at estimating the free $\ce{CO2}$ in water?
P. S. I read the procedure and also the calculations in a practical book.

The equations are all a bit convoluted, but remember that CO2 + H2O and hydrogen carbonate + hydronium are also in equillibrium.

In equation 1, you have to check the pH of your water sample and check the pKa of carbonic acid, bicarbonate and carbonate equilibrium.

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction 2 which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in Step 3 and the CO2 (the $\pu{10\%}$) that was directly converted into $\ce{H2CO3}$ (eq 1) are not taken into account.

Almost $\ce{H2CO3}$ in the sample will be converted to $\ce{HCO3-}$ as you proceed with the phenolpthalein endpoint. Search for "Speciation of Carbonate as function of pH" (Sorry I do want to immediately link the graph but it is only a http and stackexchange won't accept it).
At ~8.3 or so, all carbonate species is at the form of bicarbonate. All carbonic acid in the solution will be transformed as bicarbonate. The amount of carbonate ion is almost nonexistent. The $\ce{H+}$ released will be neutralized by NaOH, so aside from reacting directly with $\ce{CO2}$, it will also react with the $\ce{H+}$ released by $\ce{H2CO3}$, so (1) is accounted on the amount of NaOH that will be used.
Also, the sum of all equations (2) and (3) is just
$$\ce{2 NaOH + 2CO2 -> 2NaHCO3}$$
So in essence, for 1 mole of NaOH you react with one mole of $\ce{CO2}$ to produce one mole sodium bicarbonate.
In totality, for every one mole $\ce{H2CO3}$, you react with one mole NaOH. For every direct reaction of NaOH with $\ce{CO2}$, the ratio is $1:1$.
So I guess that's all reactions accounted for.

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text:
Why does insufficient data lead to random errors?
I've read up on the causes of random error and noticed insufficient data to be one of the causes of random error. However,  I'm confused as to why insufficient data could possibly lead to random errors. Could someone provide some context for why this might be?

Please define "random error" and "insufficient data". By the first, you probably mean "statistical error" (as opposed to "systematic"), but the latter can be many things.

Random Error - error in which there's equal likelihood of data value being too high or too low.

Off topic , recommended to move

More datapoints (if everything is done correctly) lead to more trustworthy results. With less datapoints, the possible (statistical) error grows, until at one point, when you only have as many datapoints as you have parameters, it looks like you have no error at all. (That is of course stupid, you should always have derived the error range of the individual data points.)
Now what is "insufficient"? Is that when the statistical error grows too large for your feeling? Or when you only have two points to draw a straight line through? Or when you only have two points to draw a parabola? That's your prerogative.
(note that I have also above used the word "error" in the same sloppy way your books seem to do. The "errors" are in your data, and they have a systematic component (hopefully small) and a distribution. The derived values (e.g. the slope of a linear regression) have a confidence interval.)

When you discussed the idea that too few data points are likely to exacerbate the size of random error, I think that's what my textbook was getting at when it mentioned insufficient data. Thanks!

When making a series of measurements we implicitly assume that there is a an 'error', or better, an uncertainty associated with the measurement and that this uncertainty is random and not so influenced by that of any of the others, i.e. the errors and random and independent of one another.
In virtually every case, no matter what the noise on a measurement is, if  a sufficiently large number are made then the histogram of the measurements will follow a gaussian or normal distribution, which is a 'bell shaped' curve. The peak of the curve is then taken as the average or mean value of the measurement. Depending upon how wide the bell shaped curve is allows us to put an error limit on the measurement. (The theorem supporting this is called the Central Limit Theorem)
However, we cannot always make a huge number of measurements and so if we make only a few then they could fall anywhere on this distribution; they could be exactly on the mean value or way off this value, one just does not know. In this case we have to use make estimate of the true value by averaging the measurements  and by assuming that the noise is random and gaussian (normally) distributed we can estimate the 'error', called the standard deviation.
Usually the error is stated as a mean value plus or minus the standard deviation ($\pm 1 \sigma$) for example $23.4 \pm 0.3$ which means that $\approx 68 $ % of the time the measurement falls within the limits $23.7$ and $23.1 $ but by random chance has a $32 $ % of being outside these.  This is not very convincing so it is better to try to obtain $\pm 2 \sigma$ which is $\approx 94 $ % chance of the measurement value being within the range, and by random chance $6$% of being outside this.  Of course it is up to you to decide what is appropriate.
An additional point to bear in mind is that the signal to noise ratio is reduced as $\sqrt n$ for n measurements. Thus to improve the signal to noise ratio of a measurement can involve a lot of extra work.
So to answer your question directly; the errors are there all the time, insufficient data does not make them greater or smaller but instead does not average them away to a smaller value.


text:
Saturated vapor pressure
From the table of the saturated vapor pressure of the water, at 20 degrees C, the saturated vapor pressure is 2.3388 kPa.
My question is: If I put a water in a closed container, and I pressurize the container up 2.3388 kPa (above the atmosphere pressure) by nitrogen or any other inert gas, will the water evaporate, or it will never evaporate because we are already above its saturated pressure?
In other words, is the saturated vapor pressure of a substance affected only by the partial pressure of the substance itself, or by the amount of pressure above its surface regardless of the substances cause that pressure?

Yasser, if you would like to thank somebody for a really helpful answer, there is a much better way than writing a comment (bad) or using the answer field (very bad): Just accept the answer.  Have a good time at Chemistry.SE!

This is purely an equilibrium phenomenon. The vapor pressure of a fluid is generally just a function of the temperature. This pressure is just the partial pressure of the fluid, not anything else. To rationalize why this is, consider what the reaction looks like:
$$\ce{A(l) <=> A(g)}$$
$$\mathrm{K}_{eq} = \frac{P_{\ce{A(g)}}}{\ce{[A(l)]}}$$
The partial pressure is only dependent on factors that affect the activity of the pure substance. Since the liquid is essentially incompressible, the total external pressure does not have an effect.


text:
Should β-mercaptoethanol and anything that touches it be handled only under fume hoods for safety aspects?
Background: I recently started in a lab that regularly uses β-mercaptoethanol (βME) for protein purification. I know that the chemical is an irritant for the lungs if inhaled (mucus membranes), damaging if in contact with eyes, toxic if in contact with skin. However, I have had projects in different labs that handle this chemical differently. Some strictly only open it in the fume-hood, have designed waste in the fume-hood for any pipette tips that touch βME and always discard any protective gloves that they used while handling this chemical directly after use – so not to spread residual βME all over the lab. This also goes for any buffers that have βME in it.
In the lab that I currently work in, they open the concentrated bottle of βME in the fume hood, but they don't care too much about discarding pipette tips in designated waste nor change gloves after use (and I know that the gloves smell like βME after opening the bottle). In addition, they do not work with lammeli buffer in the fume hood (they use 4 × lammeli for SDS-PAGE). However, from the safety manual of lammeli-buffers — which also contains βME (and you can immidiately smell that) it seems to be just as hazardous as the concentrated βME.
I don't have hypochondria when it comes to lab safety, but I would like to develop a rational understanding of how dangerous this is common chemical is. In addition I believe that the fridge which we store βME in smells quite strongly of βME. Should the bottle be stored in a secondary container (plastic bag) in addition?
Question: How do people who commonly use this chemical treat it during their everyday lab routines? How hazards is it to regularly inhale trances of βME from e.g. the fridge or when people work with lammeli buffers on the lab bench? What do people think about discarding their gloves after working with it, before they start touches other lab equipment?

Good thinking, reasonable lab safety: People who overdo it tend to forget where the real, specific dangers are, and around them also nobody learns about them.

β-Mercaptoethanol is not that dangerous a chemical. Yes, you should not drink it, you shouldn’t dip your bare fingers in it, you shouldn’t inhale tons of it and you shouldn’t use it to rub yourself down. But other than that it is hardly more dangerous than most other thiols. Simple methanol is probably more toxic and is also routinely handled.
The thing about β-mercaptoethanol is that it  stinks like hell.  And it has the tendency to stick around stinking for quite some time. Like with most thio-compounds, you can smell it long before it becomes seriously dangerous; your sense of smell is a warning against being too risky. You know yourself that the gloves stink if you used the bottle, that the pipette tips stink after pipetting it and that any flasks that contained it tend to stink for quite a while afterwards.
And that is the reason why some labs rigorously segregate it from other chemicals; have dedicated wastes for pipette tips and gloves all kept under a fume it. It is all simply meant to stop the stench as early as possible, to keep it confined in a single fume hood and keep most of the lab air smelling better. The safety aspect is secondary. If you develop a routine of changing equipment every time after handling β-mercaptoethanol, you are less likely to forget to take off your glove and the lab has less smell to deal with.
Therefore, you would be doing your lab colleagues a favour by keeping the bottle confined within its own small plastic bag; but it’s only an olfactory favour, not that much a health favour.

to add to the answer; thiols and other sulfur compounds can be especially corrosive to metals and help them oxidize more rapidly.

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text:
Identifying the C3, C4, S4, and S6 symmetry operations in the Oh point group
The Oh character table has ten symmetry classes. I am trying to identify them, but my stumbling block is in identifying the order of each class.
I will use this diagram as a reference, from Sydney Kettle's Symmetry and Structure: Readable Group Theory for Chemists pg 191 (Wiley, 3rd edition).

The classes (with their respective orders) for the Oh table are:
$$\begin{array}{cccccccccc}E & 8C_3 & 6C'_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d}\end{array}$$

$E$ and $i$ are obvious.
$6C'_2$ and $3C_2$ are shown in the diagram.
$8C_3$ is shown in the diagram above; however, it only displays 4 axes. The character table indicates there are 8 equivalent rotations for this symmetry operation. Where do the other 4 axes for the $C_3$ class come from?
$6C_4$ is the same issue. It's only shown as 3 axes. Where are the other 3 $C_4$ axes from?

The improper rotations are given in another diagram, this time on page 194 from the same book.

The mirror and dihedral planes, $\sigma_\mathrm{h}$ and $\sigma_\mathrm{d}$, are both shown to match the order in the character table.
Similar to above for $C_3$ and $C_4$, the $S_4$ and $S_6$ rotations are only half the order supplied in the character table.

In summary, from 10 total classes in the octahedral table, I can identify all the corresponding axes/planes for 6 of those 10 classes. For the remaining 4 classes, $C_3$, $C_4$, $S_4$, and $S_6$, I can only identify half the axes. At first I thought the character table was counting both positive and negative rotations, but then the order for the $C'_2$ and $C_2$ rotations should be doubled if that were the case ($12C'_2$ and $6C_2$).
Can anyone please advise on exactly what the $C_3$, $C_4$, $S_4$, and $S_6$ operations correspond to in an octahedral complex?

Many rotatable 3D complexes are shown with their symmetry operations in the web site molecule-viewer.com . You can add/remove each of the elements at will.

As porphyrin mentioned, there are pages that show the operations, for example: http://symmetry.otterbein.edu/gallery/index.html

At first I thought the character table was counting both positive and negative rotations, but then the order for the $C_2'$ and $C_2$ rotations should be doubled if that were the case ($12C_2'$ and $6C_2$).

You were very close. The missing symmetry operations are indeed ‘positive and negative’ but only those that lead to non-identical transformations. In a $C_2$ axis, it does not matter whether you turn clockwise or counter-clockwise. Consider the following mock-up of the central planar square of an octahedron with the ligands labelled A, B, C and D:
 A
BMD                      (1)
 C

If you apply $C_2$, you will arrive at the same orientation (2), regardless of which direction you rotate in:
 C
DMB                      (2)
 A

However, when rotating by $C_4$ clockwise and counterclockwise actually lead to two different results (3) and (4):
 B
CMA                      (3)
 D

 D
AMC                      (4)
 B

These two are not identical transformations. In fact, they have distinct symbols; depending on your direction of rotation one is $C_4$ and the other is $C_4^3$. You can also interpret them as rotation by $90^\circ$ and rotation by $3 \times 90^\circ = 270^\circ$. $C_2$ and $C_4^2$ correspond to the same transformation as well.
The same case can be made for $C_3$, the reverse rotation is $C_3^2$.
The $S_n$ cases are not quite as simple. Well, actually $S_4$ is ($S_4$ and $S_4^3$) but that’s a coincidence. $S_6$ shows this more clearly. Remember that $$S_6 = C_6 \oplus \sigma_\mathrm{h}$$
Therefore, performing $S_6$ twice is equivalent to $C_3$:
$$S_6 \oplus S_6 = S_6^2 = C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_3 \oplus E = C_3$$
Thus, $S_6$ applied twice and $S_6^5$ applied twice give $C_3$ and $C_3^2$, respectively, so $S_6^2$ and $S_6^4$ are already taken care of. A last case needs to be made for $S_6^3$: This is identical to $i$ as can be seen below:
$$S_6^3 = S_6 + S_6 + S_6 = C_6 \oplus C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_2 \oplus \sigma_\mathrm{h} = i$$
Thus, while you could imagine up to six different $S_6$ transformations, only two of those ($S_6$ and $S_6^5$) give new transformations not covered elsewhere.

Keep in mind that a class doesn't indicate equivalent symmetry elements, merely equivalent operations.
For example, given $\ce{O_h}$, the ${C_3}$ and ${C_3^2}$ operations are class-equivalent. You're correct that there are only 4 axes, so the other 4 operations are ${C_3^2 = C_3^{-1}}$. Similarly, the 4 ${C_4}$ axes generate the ${C_4}$ operations and their inverse - which are in the same class: ${C_4^3 = C_4^{-1}}$.
It's similar with ${S_4}$ and ${S_6}$.


text:
How to explain shape of molecules in penta and hexa coordination if hybridization involving d-orbitals (in main block) is considered incorrect?
In this answer @Jan says:

penta- and hexacoordination; tetracoordination with additional lone
  pairs and related: attempt to form as many normal bonds with p
  orbitals as possible; keep one lone pair in an s orbital if possible.
  Use remaining p orbitals to construct four-electron-3-centre bonds to
  the remaining atoms. (Predicted bond angles: diverse. $90^{∘}$ going
  from 2e-2c bonds to 4e-3c bonds; $180^{∘}$ between a pair of coordinating
  atoms contributing to the same 4e-3c bond.)

Let us consider $\ce{PCl5}$. According to several books the hybridization in this compound is $\mathrm{sp^3d}$. But @Jan says that pure $d$ orbitals never take part in hybridization in case of main group elements. Okay. So, if we assume that there is no hybridization in $\ce{PCl5}$ then as Jan says we form "as many normal bonds" with $\ce{Cl}$ as possible using $p$ orbitals. Since, $\ce{P}$ has 3 p orbitals we use them to form mutually perpendicular bonds. Suppose they are the top, bottom and leftmost bonds as in the following diagram:

After this, we are still left with two $\ce{Cl}$ atoms and only one $s$ orbital. How can we place two $\ce{Cl}$ atoms in only one pure $s$ orbital ? This part makes no sense to me.
Now, let us consider the $\ce{ClF3}$ example that @Jan gives. If we follow Jan's method we keep one lone pair in a pure $s$ orbital , then the shape can never be trigonal bipyramidal like this:

This is because $s$ orbital is spherical in nature and does not look like a dumbbell shaped $p$ orbital.
Also, could someone please give me some examples in cases of "penta- and hexacoordination; tetracoordination with additional lone pairs", where $3c-4e$ bonds are formed using remaining $p$ orbitals when no hybridization occurs (as mentioned by @Jan) ?

I think they’re looking good =)

This is very broad as written, and it also reads more like a critique of another person's answer to a question than as a question itself.

@ToddMinehardt Well, Jan himself/herself told me that this question is fine. And it simply asks for clarification. No one is criticizing anyone or anyone's answer here.

@Jan Do you also think that is very broad as Todd says? Should I edit it further ? (In my opinion if I trim this question further it will lose its essence )

@ToddMinehardt Imho, the question is fine. The critique is on scientific level imho; they question the (very simplified) concept I gave in another answer asking for expansion. It was too much to include in the other answer but I think it’s good here.

In fact, I even think this can serve as a canonical dupe target for ‘My book says compound X uses d orbitals; here I read it doesn’t. Why/how?’

One possibility, PCl3^2+ Cl- Cl-, P uses sp2 hybridization.

Sounds like it's overflow from the other question - I trust your assessment, withdraw close vote. Thanks for clarifying.

This has been asked (and explained) in one form or the other multiple times: http://chemistry.stackexchange.com/q/29142/4945 http://chemistry.stackexchange.com/q/18427/4945 http://chemistry.stackexchange.com/q/28692/4945 http://chemistry.stackexchange.com/q/6785/4945 Simply let go of the hybridisation concept, forget the VSEPR model and try to form a MO scheme with ligand group orbitals, the lone pairs will fall in place. An while I disagree with the name 'hypervalent' the wikipedia article is quite well explained. (cc @jan )


text:
Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry
Contextual background: I am trying to reproduce the symmetries in the graph on this page (MO diagram for $\ce{ML6}$ complex). I have reproduced the metal s and p orbital representations as well as the linear combination of representations for the $\ce{L6}$ fragment. All that remain are the metal d orbitals.
So I know the five d-orbitals collectively reduce to the $\mathrm{e_g}$ + $\mathrm{t_{2g}}$ representation. I put my five d-orbitals at the origin and went through the operations for the reducible representation. Everything else adds up, except I can't find that $-1$ in the $C_3$ and $S_6$ operations.
Character table
$$\small\begin{array}{c|cccccccccc|cc}\hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & & x^2+y^2+z^2 \\
\mathrm{A_{2g}} & 1  & 1  & -1 & -1 & 1  & 1  & -1 & 1  & 1  & -1 & & \\
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0  & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T_{1g}} & 3  & 0  & -1 & 1  & -1 & 3  & 1  & 0  & -1 & -1 & (R_x,R_y,R_z) & \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  & & (xy,xz,yz) \\
\mathrm{A_{1u}} & 1  & 1  & 1  & 1  & 1  & -1 & -1 & -1 & -1 & -1  & & \\
\mathrm{A_{2u}} & 1  & 1  & -1 & -1 & 1  & -1 & 1  & -1 & -1 & 1  & & \\
\mathrm{E_u}    & 2  & -1 & 0  & 0  & 2  & -2 & 0  & 1  & -2 & 0  & & \\
\mathrm{T_{1u}} & 3  & 0  & -1 & 1  & -1 & -3 & -1 & 0  & 1  & 1  & (x,y,z) & \\
\mathrm{T_{2u}} & 3  & 0  & 1  & -1 & -1 & -3 & 1  & 0  & 1  & -1 & & \\ \hline
\end{array}$$
Work
$$\begin{array}{c|cccccccccc} \hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0 \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  \\ \hline
\Gamma_{\text{d-orbitals}} & 5 & ? & 1 & -1 & 1 & 5 & -1 & ? & 1 & 1 \\ \hline
\end{array}$$
As can be seen, everything matches up but the $C_3$ and $S_6$ operations. I tried using Molecule Viewer and rotating the d-orbitals, but I still don't see the $-1$. They all look like zero to me.
I would expect the $\mathrm d_{x^2-y^2}$ orbital to correspond to $E_g$ from the character table, but I only see it rotate into a "$\mathrm d_{y^2-z^2}$" orbital, with the orbitals along the y and z axes instead of x and y axes. The $\mathrm d_{z^2}$ orbital rotates into a "$\mathrm d_{x^2}$" orbital, lying along the x axis instead of the z axis.
Anyone know which d-orbital rotates into $-1$ for a $C_3$ operation? I assume a $-1$ for $S_6$ is the same orbital, but if not then does anyone which that one is as well?
Thank you

Please don't use MathJax in titles. And if you can, please use the native image uploader, that helps to prevent issues with dead image links.

Thank you for the corrections. I found both yours and Jacob's answers helpful for different reasons. Is it only possible to award a check to one response?

Yes, you can only accept one answer; just accept whichever one you like, don't worry about it.

The $C_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$.
Hence,
$$\phi_1=2z^2-(x^2+y^2)$$
rotates into
$$-(1/2)\phi_1 +(3/2)\phi_2$$ and
$$\phi_2=x^2-y^2$$
rotates into
$$-(1/2)\phi_1 -(1/2)\phi_2$$
so, in the basis
 $( \phi_1, \phi_2 )$ the matrix that represents $C_3$ is
$$\left [\begin{matrix} -1/2 & -1/2  \\ 3/2 & -1/2  \end{matrix}\right]$$
Denoting this 2 by 2 matrix by M what we have is
$C_3(\phi_1  \phi_2) = (\phi_1 \phi_2)M$.
The trace of this matrix is $-1$.
This is what you find in the character table.
Since $C_3^3=I$ (identity) its eigenvalues are the cubic roots of unity:
$1, \lambda$ and $ \lambda^*$
where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$,
and $\lambda^*$ is the complex conjugate of $\lambda $.
It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].
You can find a linear combination of $\phi_1$ and $\phi_2$ that will be an eigenvector of $C_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$)
and another linear combination that will have eigenvalue $\lambda^*$. Now, the matrix representing $C_3$ will be diagonal with $ \lambda$ and $\lambda^*$ along the diagonal. The trace remains $-1$.
Something similar should work for $S_6$.

From reading what you wrote I think you don't entirely understand how degenerate irreps work. Firstly, it's incorrect to say "$\mathrm{d}_{x^2-y^2}$ corresponds to $\mathrm{E_g}$". The $\mathrm{d}_{x^2-y^2}$ orbital by itself does not transform as anything. It only makes sense to say that the two orbitals transform together as $\mathrm{E_g}$. But anyway, let's go from the start.

For non-degenerate irreps, the behaviour of an orbital is that the symmetry operation transforms it into some multiple of itself:
$$\hat{R}\psi = \chi^R\psi \tag{1}$$
where $\chi^R$ is the character under the symmetry operation $\hat{R}$. For example if $\psi$ is an s-orbital, then all the symmetry operations leave it untouched: $\hat{R}\psi = \psi$. Therefore $\chi^R = 1$ for all symmetry operations, and $\psi$ transforms as the totally symmetric irrep ($\mathrm{A_{1g}}$ in this case).

Taking this logic and applying it to degenerate irreps is very dangerous. When two orbitals $(\psi_1, \psi_2)$ transform together as a degenerate irrep, what happens is that
$$\begin{align}
\hat{R}\pmatrix{\psi_1 & \psi_2} &= \pmatrix{\psi_1 & \psi_2}\pmatrix{\Gamma_{11} & \Gamma_{12} \\ \Gamma_{21} & \Gamma_{22}} \\[8pt]
&= \pmatrix{\Gamma_{11}\psi_1 + \Gamma_{21}\psi_2 & \Gamma_{12}\psi_1 + \Gamma_{22}\psi_2} \tag{2}
\end{align}$$
where $\Gamma = \pmatrix{\Gamma_{11} & \Gamma_{12} \\ \Gamma_{21} & \Gamma_{22}}$ is some matrix that you have to find out by investigating the effect of $\hat{R}$ on the two functions.
$\Gamma$ is called a "matrix representation" of the symmetry operation. The character under this symmetry operation (which is what is listed in the character table) is the trace of this matrix $\Gamma$, i.e.
$$\chi^R = \Gamma_{11} + \Gamma_{22}. \tag{3}$$
(Notice that in the one-dimensional case the matrix $\Gamma$ simply becomes a number and the character is simply that number itself, which regenerates equation $(1)$.)
Now let's say $\psi_1 = \mathrm{d}_{x^2-y^2}$, $\psi_2 = \mathrm{d}_{z^2}$, and $\hat{R} = \hat{C}_3$. The issue is that you seem to expect that the rotation $C_3$ to have a nicely well-defined effect on $\psi_1$:
$$\hat{R}\psi_1 = k_1\psi_1 \tag{4}$$
But for a degenerate irrep this is simply not true. From equation $(2)$ one can see that
$$\hat{R}\psi_1 = \Gamma_{11}\psi_1 + \Gamma_{21}\psi_2 \tag{5}$$
Furthermore there is no guarantee that any of these two numbers $\Gamma_{11}$ or $\Gamma_{21}$ are related to the number in the character table. The only requirement is that $\Gamma_{11} + \Gamma_{22}$ (the latter of which we haven't even started talking about!) is equal to the character under the $C_3$ operation, i.e. $-1$. So now you see why asking "which d-orbital rotates into $−1$ for a $C_3$ operation?" doesn't make much sense. Neither of those two d-orbitals do.
Hopefully with this description you are in a better position to understand the other answer, which is essentially telling you that $\Gamma_{11} = -1/2$ and $\Gamma_{22} = -1/2$.

This makes a lot of sense. In basic chemistry they gloss over this theory, and we (or at least in my case) are taught a shortcut: performing the operation for all relevant orbitals, then add up which ones transform into themselves. I repeated this for t2g and noticed that "transforming into themselves" corresponds to the trace. Hence it works most of the time. Aiming deliberately for the trace component, it seems like "eyeballing" it may still be possible for eg. Incidentally, is $d_{z^2}$ always shorthand for $d_{2z^2-x^2-y^2}$?  I always treated it as $z^2$, which fails here for the matrix.

Yes, it is indeed shorthand.


text:
Why do atoms get smaller moving left to right on a periodic table?
I get that it's because of the increase number of protons, but I still don't understand the concept. When you move left to right on a periodic table, the protons increase at the same rate as the electrons. In fact,  the number of electrons end up being the same as the number of protons, no matter how far you are across a period (unless it's an ion).
So shouldn't the attractive force of the nucleus and the forces of electron-electron repulsion be cancelling out due to the equal number of protons and electrons in an atom? Or is it because protons have a greater force of attraction compared to the repulsion force between electrons?

@Mithoron Certainly a duplicate, but the answers there are rather sloppy.

Atomic size across a period, in fact, has everything to do with electron shielding and the effective nuclear charge (charge felt by the electrons).
Across the period, electrons are added to the same outer level, meaning that shielding by inner electrons does not change, however, the numbers of proton increases. This results in the effective nuclear charge rising significantly (as no. protons increases while added electrons are shielding inefficiently), which of course, means that the atom has a smaller size.

Through the period, the core electron shell shrinks, because the charge of the nucleus increases. The outer electrons have no influence on the core electrons: In an electrically charged sphere, there is no electric field in the inside. (Actually the valence electrons do have a finite propability density in the core, so this is not 100% correct, but the conclusion holds.)
The core electrons shield the valence shell from the core, so the effective charge of the nucleus is always identical to the number of valence electrons, but also the repulsion from the outermost electrons of the core only kicks in at a smaller radius.
Also the electrons tend to avoid each other pair-wise (Pauli principle), so the electrostatic repulsion between them is weaker than expected. This makes for the funny sawtooth line of atomic diameters vs. group number.
And, for heavy elements, the innermost core electrons become (relativistically) heavier, because they "move" very fast, so their orbital radius shrinks even more, letting the rest of the atom shrink as well (like above).


text:
Trying to understand why Sn2 is not possible in this reaction
I am trying to understand why the red Br doesn't react? Is it because there can't be a flip of groups? Or steric hindrance?
Thank you!

Related: http://chemistry.stackexchange.com/questions/6404/will-bridged-compounds-undergo-sn1-reaction

Backside attack on the red bromide is not possible. The rest of the norbornyl skeleton completely blocks the trajectory required for $\mathrm{S}_{N}2$ reaction.

Thank you for your answer! If possible, I would like to ask you this question: would the reaction if there were no bridge in the middle? Thank you!

Yes. Because then, it would just be cyclohexane. As long as you could access the conformation where the red bromide group is axial, you could have a productive substitution reaction.

Excellent answer by Zhe. I'd also like to point out that the red Br is on a tertiary carbon anyway - makes SN2 nearly impossible.


text:
What does the molecular orbital scheme of beryllium chloride and hydride look like?
Beryllium is a somewhat fascinating element since it is the only member of the second group that behaves somewhat non-metalish and, e.g. forms a somewhat covalent chloride and a covalent hydride.
In an answer to a different and unrelated question, I tried guessing what the bonding picture of $\ce{BeCl2}$ would be. Given the monomer’s linear shape, I was inclined to assume something that introduces a four-electron-three-centre bond, which has the nice side-advantage of allowing beryllium to contribute to bonding orbitals using its s orbital only. However, I would guess that the more traditional picture would include two $\mathrm{sp^2}$ type bonds to the neighbouring chlorines and then a nonzero amount of π backbonding making the MO similar to that of $\ce{CO2}$.
What does the actual, calculated MO scheme of $\ce{BeCl2}$ look like; what does $\ce{BeH2}$’s look like, how do they compare and what can we learn from them concerning the bonding properties of beryllium?

Given that you're the one asking this, I assume you are looking for a quantitative (computational) MO diagram as opposed to the simple qualitative one, which isn't really difficult to derive.

Related: http://chemistry.stackexchange.com/questions/21764/which-lewis-structure-for-becl2-is-more-commonly-seen/21786#21786

http://chemistry.stackexchange.com/questions/22088/dimers-and-polymers-of-electron-deficient-compounds

I calculated molecular orbitals for $\ce{BeCl2}$ at MP2/jun-cc-pVDZ level of theory; given below are the doubly occupied MOs at this level of theory along with their symmetry designations and energy in a.u. For some reason my Avogadro kept crashing when I tried to select an iso value I thought appropriate for the last two MOs; anyway these are completely localised $\ce{Be}$ atoms. If we look at the 8 valence orbitals, there is indeed delocalised $\pi$ bonding as you suggest, and do bear a striking resemblance to MOs for $\ce{CO2}$.
For $\ce{BeH2}$ I again ran into trouble generating nice visuals using the cube file in Avogadro. I will update this post with visuals as soon as possible (It would be great if someone can help out).
I got the following doubly occupied MOs at MP2/jun-cc-pVDZ level of theory:      -4.644660 ($\ce{a_g}$) ;     -0.504366  ($\ce{a_g}$) ;  -0.469857  ($\ce{b_{3u}}$). However, it is evident from the symmetry designations that $\pi$-type orbitals aren't involved in bonding.
P.S: If you are interested, I can provide the input, output, and cube files on request.


text:
Device to Save Fuel
If I have an electromagnetic device connected to an internal combustion engine, such that the fuel particles passing through the device are "sorted", allowing the fuel to burn better and thus generating fuel economy.
This is the typical device that is offered on thousands of websites, but the truth is that this device does not work, but:

what is the reason it does not work?

i.e. if the device is able to rearrange the particles should be affected the entropy, but I do not think that the device change the temperature of the system ... If someone can help me and tell me why this device does not work, I would be very grateful
Pd: Maybe is enough, compared to the efficiency of the engine with and without the device, but How I should compared this? .. .Thanks

Ever heard of energy conservation law?

@IvanNeretin Yes, but I don't see how explain this with this laws, can help me pls! Thanks!

Please don't make destructive edits to your post. This is not nice to future visitors, who won't understand what's going on when they see half a post - and the Terms of Service explicitly specify that content that you post on Stack Exchange no longer belongs to you.

I'll set aside for the moment the question of whether an "electromagnetic device" of any kind that is suitable for aftermarket installation on a consumer vehicle has any capability to "sort molecules," which is terrifically suspect in and of itself.
For the sake of argument, let's say such a device exists, and that it is able to rearrange the molecules in the fuel flow in a way that is better suited for combustion. Let's further say it were technologically possible to put the "active portion" of the device, the part that "sorts" the molecules, right at the outlet of the fuel injectors.  (This is of course impossible to achieve with an aftermarket device, but let's go with it for the sake of argument.)  This would achieve "sorting" of the molecules right at that outlet orifice.
Molecules vibrate and rotate very quickly. For example, Wikipedia puts the temporal frequency for vibrations at somewhere between $10^{12}~\mathrm{Hz}$ and $10^{14}~\mathrm{Hz}$.  At typical freeway speeds, automobile engines operate at around $3000~\mathrm{RPM}$, tops, or $50~\mathrm{Hz}$.  This corresponds to a time-per-cycle of ${1\over 50}~\mathrm s$, or $20~\mathrm{ms}$.
Let's say that only a tiny fraction of the cycle elapses between when the fuel is injected and combustion occurs, something like $0.01\%$ of the cycle, which translates to $0.002~\mathrm{ms}$, or $2~\mu\mathrm s$.  In this time interval, the molecules of the fuel will have vibrated literally millions of times:
$$
\left(2\times 10^{-6}~\mathrm s\right) *
  \left({10^{12}~\mathrm{vibrations}\over \mathrm s}\right) =
  2\times 10^6 ~\mathrm{vibrations}
$$
Similarly, the molecules will have rotated, translated relative to one another, and collided with one another many, many times in that (unrealistically short) $2~\mu\mathrm s$ span of time before the spark plug triggers combustion. No matter how you look at it, the fuel molecules would have long "un-sorted themselves" by the time ignition occurs in the cylinder.
(Not to mention, all of those assumptions in the first two paragraphs are also totally bogus.)


text:
Why is the pKa of a ketone higher than of water?
The pKa of water is 16, while the pKa of a ketone, HC3(CO)CH3, is closer to 20.
The lower the pKa the stronger the acid. When a ketone loses a hydrogen, the NBE are resonance stabilized into the carbonyl group, lowering their energy, increasing the acidity. When a proton is removed from water, the NBE have no resonance structures. Yet it has a lower pKa, so it is a stronger acid.
Why is this?

dont forget that bond polarity also plays a role

As I'm sure you know, (p)Ka is a measure of the equilibrium between an acid and its conjugate base (what you're calling NBE?). Both the conjugate bases of water (hydroxide) and a ketone (enolate) have a negative charge on the oxygen. I'm summarizing your question as: If both conjugate bases have a negative charge on the same atom type, but one base (the enolate) is stabilized by resonance, shouldn't that corresponding acid (ketone) be more acidic? (Correct me if I'm misreading you.)
I think the key here is what (p)Ka really represents. Fot water, with room temp pKa 15.7, the equilibrium favors water but some dissociation is present. For acetone, pKa 19.2, the equilibrium lies further on the non-dissociated side than for water. We take this to mean that acetone is a weaker acid that water. If we only looked at the stability of the conjugate bases, we might conclude the opposite (previous paragraph). But as pKa measures a balance between a conjugate pair, we must consider the acid forms as well.
The ketone contains the very "stable" (energetically favorable) C=O carbonyl bond. When a ketone donates an alpha proton to become the enolate conjugate base, the carbonyl bond is disrupted. This is an extra energetic cost that is not present in the water equilibrium, and I propose it is the reason why the ketones tend to dissociate less than water.
By this logic, water is a better acid than a ketone because acting as an acid requires a ketone to disrupt an energetically preferable C=O bond. Thus the balance lies closer to the undissociated side for a ketone (lower Ka, higher pKa) than for water (lower pKa), so we say that water is a stronger acid and a ketone is weaker.
(One might argue that the carbonyl need not be disrupted, the lone pair could localize on the alpha carbon, but I hope we can agree such a carbanion is quite unstable, and a carbanion not preferable to breaking the carbonyl and localizing on the more electronegative oxygen.)

ok thanks that makes sense. I agree with you on the breaking of the favorable C=O double bond is not something the ketone is looking to do. Thanks for the explination.


text:
Finding the value of pressure equilibrium constant
Question:

A rigid vessel containing only $\ce{NO2(g)}$ is heated to $\pu{337^{ \circ}C}$  and allowed to come to equilibrium according to the following reaction:
  $$\ce{2NO2 <=> 2NO + O2}$$
The density of the resulting mixture is measured to be $\pu{0.52 g/L}$ at a total pressure of $\pu{ 0.75 atm}$. What is the value of $K_\mathrm p$?

The answer key says it should be $\pu{0.65 atm}$, but I can't get that number. I first tried to find mols using the given density and molar masses, and then finding the partial pressure of each component using the mol fractions, but it doesn't work out.

The density is given.From here you will get a relation between initial concentration and volume.From $K_c$ you will get relation between volume,extent of dissociation and initial concentration.From $K_p$ you will get relation between pressure,extent of dissociation and initial concentration.After that use the formulas that Klaus Warzecha gave.I think that will be enough to solve this.

Just a hint to get you started

From the balanced reaction equation , it seems that

$$K_p = \frac{p_{\ce{NO}}^2 \cdot p_{\ce{O2}}} {p_{\ce{NO2}}^2}$$

The relation between $K_p$ and $K_c$ is given as
$K_p = K_c\cdot(\mathrm{R}T)^{\Delta n}$, where $\Delta n$ is the difference between the number of moles of product and reactant gases.


text:
What is the IUPAC name of the following compound?
What is the IUPAC name of this molecule with a dihalide alkyl chain?

Welcome to Chem.SE! We ask that you share some of your work in solving your homework problems, so we can best help you learn.

You also might want to consider reading our guidelines on how to ask nomenclature questions.

Related: IUPAC Naming: 2-phenyloctane or 2-octanylbenzene

When listing preferred IUPAC names, the ring prioritized over the chain (making it a substituent), especially when the chain is shorter than the ring.
First we label the carbon at the end of the chain as '1' and continue across the chain. The bromines are at positions 2 & 3, and it is a 5 carbon chain. So, we label it as pentyl chain and it therefore follows the IUPAC name is
(2,3-dibromopentyl)benzene

When i tried to the nomenclature, i was getting the iupac name as 2,3-dibromo-1-phenylpentane. This is because i think that the benzene ring should be considered a substituent as it contains the functional group (halides).

What about 2,3-dibromo-1-phenylpentane???

@sankalprajabhoj 2,3-dibromo-1-phenylpentane is a correct systematic name in accordance with current IUPAC recommendations; however, it is not the preferred IUPAC name.


text:
How do I sterilize an obsidian blade without damaging it?
I recently came across this article on the use of obsidian scalpels in surgery: How Stone Age blades are still cutting it in modern surgery

Now these scalpels are pretty fragile, with some of them having a thickness of the order of 0.1 mm.
I don't actually plan on operating on someone with this, but all the same, I'd like to identify potential methods to sterilize such a blade.
Knowing how delicate this thing is, I doubt conventional methods (such as autoclaving it) used to sterilize surgical tools could be safely applied in this case.
As the use of obsidian blades in surgery is relatively new, I'm not surprised that I can't find any literature dealing with the sterilization of obsidian blades online.
I did try looking up the structure that obsidian assumes at the molecular level, with the hope that the knowledge would help me narrow down possible sterilization methods I could use here, but I drew a blank there as well.
I have considered subjecting the scalpel to strong dose of UV radiation or perhaps 'cleaning' the scalpel with Povidone-Iodine (Betadine). However, I fear that these methods could possibly weaken the (already delicate) structure of the obsidian.
Would anyone be able to suggest a safe way of sterilizing obsidian scalpels?

According to the Guideline for Disinfection and Sterilization in Healthcare Facilities, isopropanol does not seem to be an option.

Herr @Klaus ,I did toy with the idea of using alcohol earlier, since my primary concern was to maintain the integrity of the structure of obsidian, but the link you provided mentions a point I overlooked (alcohol does not have sporocidal action). So Danke for that. Also, I've wanted to ask you this for a while: Why don't you come over to The Periodic Table or the g-block. People would find it easier to reach you that way ;)

I don't see why Obsidian shouldn't survive a common autoclave? And here https://www.ncbi.nlm.nih.gov/pubmed/8415970 is a paper from 1993, not exactly new. I'm sure you'll find the sterilisation technique in there.


text:
Is there a least reactive chemical species?
As a total chemistry layman I enjoyed reading "Why doesn't $\ce{H2O}$ burn?", but it prompted another question in my mind. One of the answers there was that $\ce{H2O}$ can burn in the presence of a stronger oxidizer like fluorine, so burning stable compounds is just a question of using a stronger oxidizers.
Or is it? Is there a chemical species, or a "least reactive compound" that, once made, is difficult or impossible to chemically transform into something else? A chemical black hole, if you will.
Does the above answer change if we allow high temperatures and pressures versus restricting ourselves to roughly standard temperature and pressure?

No. You need to specify reactivity towards what.

There is neither least reactive compound, nor the most reactive compound, and indeed the very word "reactive" is not well-defined to the point that you'd better not use it at all.

Perhaps the question should be rephrased to use "most inert" rather than "least reactive." There is no single compound that is always the most inert, but there are definitely examples of compounds that are unusually inert with regard to other chemicals. Fluorocarbons like teflon and CFCs are good examples of "inert" compounds.

@IvanNeretin Could you suggest a better term? I haven't had any chemistry since high school (and about 12 years ago).

@barbecue What does it take to break apart fluorocarbons?

@david a lot of fluorocarbons are highly resistant to being broken down by chemicals alone. That's why PTFE is often used for storage containers for strong acids and bases, for example. And yes, changing temperature and pressure definitely changes the whole equation. They can easily be broken down by high heat.

I think a good argument can be made for either helium or neon, the most noble of the noble gasses. Those are the two prototypical unreactive elements. They are the only two stable elements for which no more complex compounds (i.e., other than the single atoms themselves) have yet been isolated, at any temperature. The slightly more reactive element argon will admit formation of compounds such as argon hydrofluoride ($\ce{HArF}$). This compound is only stable up to $\mathrm{17\ K}$, because any hotter and the frail bonds are overcome by random thermal collisions which break the compound apart into $\ce{Ar}$ and $\ce{HF}$.
Picking which of helium or neon is less reactive is a bit more difficult. A naive analysis of periodic trends would point to helium as the most inert, but more detailed computational studies suggest that at least in some cases neon may be less reactive. For example, the extremely Lewis acidic compound beryllium monoxide ($\ce{BeO}$) may potentially form an isolable, if very weakly bound, compound with helium, $\ce{HeBeO}$, but neon is not thought to form the analogous $\ce{NeBeO}$. $\ce{HHeF}$ may also be just barely stable, whereas $\ce{HNeF}$ is not thought to form. None of these have yet been observed in the laboratory, but there is definitely ongoing research into coaxing helium and neon to make isolable compounds.
All that said, we can get helium and neon to react, if we drop the requirement that the product must be isolated (that is, "put into a bottle"). Chemical species such as $\ce{He2^+}$, $\ce{Ne2^+}$ and $\ce{HeNe^+}$ have long been known from mass spectrometry experiments, they just can't be isolated because that would require the presence of a counterbalancing negative ion, which would immediately proceed to react with the positive ion and cause decomposition with release of the noble gas. You can also bring what is arguably the most reactive species in chemistry, the hydrogen cation ($\ce{H^+}$) into the fray. Helium and neon will both easily react with $\ce{H^+}$ to form $\ce{HeH+}$ and $\ce{NeH+}$ , as shown by the exothermic proton affinities of $\ce{He}$ and $\ce{Ne}$. Again, these composite particles cannot be isolated, as they are amongst the strongest Brønsted-Lowry acids in existence, and will protonate anything they come into contact with in order to release the neutral noble gas atom. Some other possible non-isolable relevant noble gas ions are $\ce{FHeO^{-}}$, $\ce{HeCCH+}$ and $\ce{PbHe15^{2+}}$ (!), among others.
Edit: I forgot to mention that there are a few other cases of helium or neon binding to other atoms. By exciting one of the electrons in the atom, it is possible to coax it to bond with another one, forming an excimer or exciplex. See for example, the dihelium excimer $\mathrm{He_2^{*}}$. This is a short-lived species that still can't be isolated, however, because in a matter of microseconds it releases a photon and de-excites, promptly separating into the two free atoms.

"isolable compounds" ? Not really. Matrix isolation is ... fine, but not really "put in a bottle", right? ;-)

The question specified compounds, which implies two or more elements together, rather than molecules of a single element.

Good answer. I suppose the intent behind my original question was whether or not it was possible to create a molecule so inert that it wasn't really possible to recover the constituent components without extraordinary measures. This is probably why using the term "reactivity" was a bad idea.

@Karl The bottle is part of the Matrix! ;)

If you want to limit the answer to "compounds", I would suggest sulfur hexafluoride and nitrogen gas as two of the most unreactive small molecules. Some high temperature ceramics might also qualify, although they are not really small molecules. Interesting question! - Which begs another question: how do you quantify how unreactive a compound is?


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Resonance structures
I'm working with a certain molecule, and could see that it was only on the ether group that it was possible to make these resonance structures. First, I'd like to know if the structures are correct and second how does it effect the molecule that this particular group can make these resonance structure. Does it contribute to its stability?

Some arrows are missing and others are drawn poorly but otherwise they are correct.

It's just a sketch -I'll draw a better one . Can you point out which arrows are missing?

I'm not sure what "It's just a sketch" means as a response here. If you're drawing resonance structures, you should either have no arrows because you think it's obvious, or you should have all arrows. The intermediate case isn't so good. The third structure is missing one arrow. In the second structure, the arrow at the 5 o'clock position should point to the carbon atom, not to the bond.

You should use mesomery arrows ($\ce{<->}$) and not equilibrium arrows ($\ce{<=>}$) for resonance structures.

Personally, I would recommend drawing the resonance process this way (first line):

Please note the difference in arrows between forms; you have used the reversible equilibrium arrows instead of the double-headed resonance arrows. This may seem like merely a grammatical concern, but the processes of equilibrium and resonance are fundamentally different and indicating the wrong process is incorrect.
Assuming the substituent is indeed a methoxy group as you have drawn it, this compound is known as anisole (not phenol as mentioned elsewhere).
These resonance structures have significance in rationalizing and predicting experimental observations. They demonstrate the electron-donating nature of the methoxy group, and the electron-rich nature of anisole's aromatic ring (relative to unsubstituted benzene) (see resonance hybrid).
These structures explain the experimental observation that, when anisole reacts with electrophiles, the electrophiles bond at the ortho (C2,6) and para (C4) positions, and not at the meta positions (C3,5). (See third line.) We say this occurs because the bond to an electrophile will be made by an aromatic ring donating electrons from the carbon(s) with the highest electron density (in the absence of steric factors), which are the three carbons featuring lone pairs in our resonance structures (line 2).

Well the following resonance structures are of the phenolate ion, which are correct. The existence of this resonance in phenolate ion provides it quite some stability, hence phenol is acidic. Phenolate ion, phenol's conjugate base, being stable by this resonance, is a weak base, so phenol becomes acidic.

Phenol is not a strong acid.

Depends upon with what it is being compared to. I just added the word strong to make it compatible for the conjugate base argument. Of course it is not a strong acid compared to HCl and other "strong" acids.

Phenol (pKa~10) is a stronger acid than most other alcohols, but weaker than a carboxylic acid. But that is NOT phenol, it is anisole.

Damn! Didn't look at the methyl attached to the oxygen!!


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How to make a placebo drink taste convincing?
I'm trying to make a placebo alcoholic drink - something strong (cognac, whiskey or vodka). Is it possible to make a drink containing absolutely no alcohol taste like it's got 30 or 40% alcohol? What can I add to give it that "kick"?

Genuinely interested if there might be a combination of chemicals out there safe for human consumption that would mimic the taste of spirits. I doubt I'll get my answer from cooking websites. My intention is not finding the recipie for the perfect mocktail; I'm conducting a small experiment. I have limited knowledge on chemistry which is why I'm here - please help! It's for a good cause.

I'm voting to reopen the question. The perception of what makes a typical cognac, brandy or whisky definitely has a scientific background. The fingerprint of  volatile compounds has been examined by GC/MS for brandy, cognac (brandy) and whisky

This is a tough one. Most of the things I can think of would have some combination of still being intoxicating, tasting bad, and giving you serious liver damage.


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Why does carbon dioxide not form addition compounds?
Although $\ce{CO_2}$ has four lone pairs on the two oxygen atoms, but it still does not form addition compounds by donating lone pairs. Why?
$\ce{CO}$ on the other hand, does form addition compounds by donating the lone pair from C atom.

A fairly simplistic argument is that in CO, the lone pair on oxygen is of a much lower energy than the lone pair on carbon (because oxygen has a greater effective nuclear charge and therefore pulls electrons closer to itself). Therefore it is quite rare (very rare indeed) to find any compound that has the oxygen acting as a donor atom; CO only really reacts via its lone pair on C.
Carbon dioxide only has low-energy lone pairs on O, and is consequently a lousy Lewis base.
One could easily refine this by bringing in MO schemes, but at 3 a.m. in the morning, this is good enough to convince me.

Wik sez CO2 complexes do form in certain circumstances. Your argument about the oxygens not being the electron donors seems to hold for CO2 too, though -- looks like it's pretty much only ever one of the C-O bonds that actually does the electron donating.

Actually, protonated CO2 is known, too. I think my argument is a little bit shaky with respect to CO. The better explanation is probably simply that the HOMO has larger coefficients on C (which can be linked to the relative energies of the AOs - but the link isn't quite as direct as I wrote above). But it'll do. I don't pretend that this is a rigorous answer.

@orthocresol A rigorous answer would be better.