text:
Do each and every resonant structures contribute to a resonance hybrid?

There can be many resonating forms for a compound. It is to be noted that they are not in any equilibrium to each other. Each and every structure will contribute to the ultimate hybrid, according to their stability. The most important point is that resonating hybrid is not a mixture of the resonating structures, but rather an intermediate of them.

This is what my book writes. But if it is intermediate, it must contain at least some properties of each and every structure. Now, one structure has no charge, while other structure has one positive charge and one negative charge. In the final structure, it still contains both the charges. So, how can it be an intermediate as it got no contribution from the first?

What molecule are you referring to?

Brief answer about charge - if the molecule has no net formal charge, then it will always have no net formal charge. Every resonance structure must always have no net formal charge. However, individual atoms may have partial formal charges, and the net partial charge on an atom will be the weighted average over the formal and partial charges on that atom in each resonance contributor.

@Marko: I asked the quo thinking about phenol. Phenol has 5 resonating str. 3 of them contain $+$ & $-$ charges;the other being neutral. In the hybrid also, charges remain present. So, if the hybrid is intermediate, what does it take from the two neutral strs.?

You can understand the relationship between the resonating forms for a compound and the ultimate hybrid by comparing it with the relationship between the different projections of a vector on the three coordinate axes and the vector itself. None of the projection can lonely represent the vector. But all of them can define the vector precisely in the space. Each projection contributes in defining the modulus of the vector according to its own value.

That's a great analogy. Take an example: phenol. It has 5 resonating structures. 3 of them are charged; the other 2 being neutral. Now, like the original vector, the hybrid must be contributed by the 5 strs. But the hybrid is also charged. So, how can it be an intermediate, since it has no contribution from the 2 neutral strs??

And neutral str. is more stable than dipolar str. So, why don't the more stable neutral structures contribute to the hybrid of phenol??

First of all the the neutral stru. is not more stable than dipolar str. (phenol is a slightly polar molecule). Secondly, you have three charged resonating structures counter two uncharged. So, it's normal that the charged str. weighed more in the hybrid of phenol. You should also remember, that I've given you an analogy to explain the mesmerism. So, it has its own limits. It cannot explain everything.

Sir, firstly, your analogy was great & has no flaw. Actually both the neutral ones contribute as the hybrid contains only partial charge,not whole charge. Now, regarding neutral being more stable, my  book...

...cites an example: there are two resonanting strs of 1,3 di pentene . First,neutral and second the polar. Then the book says, the neutral one is more stable.

Regarding 1,3 di pentene,  the neutral resonating str is,of course, more stable because this molecule is non-polar. This is not the case for phenol.


text:
Why are lactones more acidic than esters?
Lactones ($\mathrm pK_\mathrm a \sim 25$) tend to be more acidic than esters ($\mathrm pK_\mathrm a \sim 30$). I know that the lactone is fixed in a (E)-conformation, due to the ring:

and that in normal esters, the (Z)-conformer is favoured based on an anomeric effect, namely the donation of the $\mathrm{sp^2}$ lone pair on the ester oxygen into the C–O σ* orbital (image taken from Clayden, Greeves & Warren, Organic Chemistry, 2nd ed. (OUP), p 805):

Is this effect responsible for the increased acidity? If so, how?

What orbital is the one depicted as black, adjacent to the carbonyl carbon?

@ΚελεσίδηςΑντώνιος that is part of the C–O σ*

This article seems to address the question, albeit from a slightly different angle.

@orthocresol Yes, forgive me, stupid question I didn't read the text carefully enough.

Overview
The only interesting thing about lactones is that if the ring size is relatively small, they necessarily adopt the (E) conformation. Therefore, the question essentially boils down to: for a generic ester, why is the (E)-conformer 1 more acidic than the (Z)-conformer 2? This is a question that is easier to investigate, since we remove all other variables e.g. molecular structure.

Before diving into the details, I'll state the answer: 2 is more stable than 1. Likewise, if one compares the enolate anions, 2a is more stable than 1a, but the difference in stability between the anions is not as large as in the esters. Hence, $\Delta G^\circ$ for the deprotonation of 2 is larger than that for the deprotonation of 1, and 2 is less acidic. Schematically:

In physical organic chemistry the difference between the two values of $\Delta G$ is referred to as $\Delta \Delta G$. In this case, we can define $\Delta G_2 - \Delta G_1 \equiv \Delta \Delta G$, which we expect to be positive.

Neutral molecules
The two conformations of esters have been studied for quite a long time; as far as I can tell there seem to be two major reasons put forth as for why 2 is more stable than 1.
In the literature, the accepted argument seems to be based on a minimisation of dipole-dipole repulsions in 2.1–4 In particular, Roos et al.2 investigated the three possible contributing factors of (1) dipole-dipole repulsions, (2) repulsions between lone pairs on oxygens, and (3) steric repulsions between R and R'. Their conclusion was that the dipole-dipole repulsions were the most significant, and this analysis seems to have been accepted by later authors.

In textbooks, however, the n-to-σ* anomeric donation seems to prevail.5–9 However, no literature references are given, and Deslongchamps6 admits that "there is no direct experimental evidence to show the importance of [these] secondary electronic effects" (p 57). This was written in 1983, and even computational studies since then have not mentioned the existence of any such effect.
What everybody can agree on, though, is that 2 is more stable than 1. The energy difference was determined by IR spectroscopy to be $\pu{4.75 kcal mol-1}$ for methyl formate, and $\pu{8.5 kcal mol-1}$ for methyl acetate (the latter being larger because of steric repulsions between methyl groups).10 These values are corroborated by multiple computational studies.2–4

Enolates
I actually get the feeling that some of the textbooks get sloppy, or gloss over this aspect. Reading Clayden5 for example, the impression I get is that because 1 is more unstable than 2, it reacts more readily (with bases or nucleophiles). However, in comparing $\Delta G$ for two different reactions, one must be careful to look at the relative stabilities of not just the reactants, but also the products. If 2a is also more stable than 1a, and to the same extent ($\Delta\Delta G = 0$), then there is no reason to expect any difference in acidity.
The question that we now turn to is therefore: why is the relative stabilisation of 2 partially lost upon deprotonation to form 2a? The answer will, of course, depend on which of the two above explanations you subscribe to.
First, the dipole minimisation argument. Upon deprotonation, negative charge develops on the α carbon; hence as a whole, the enolate moiety has a smaller dipole moment than an undeprotonated carbonyl moiety. (Since the carbonyl oxygen bears more negative charge than the α carbon, the direction of the dipole moment is roughly the same.) Houk3 has reported this in the form of Mulliken partial charges on atoms, and Wiberg4 in the form of electron populations, but the key point is the same: when the dipole of the carbonyl group is smaller, the dipole-dipole interactions are weaker. Consequently, while 2a is still more stable than 1a, the energy difference is not as large as it is in 2 versus 1. Houk's ab initio calculations (6-31+G*//3-21G) on methyl acetate3 gives the energy difference between 2a and 1a as $\pu{3.8 kcal mol-1}$, and that between 2 and 1 as $\pu{9.2 kcal mol-1}$. Therefore, the difference $\Delta\Delta G$ is roughly $\pu{5 kcal mol-1}$.
Extra: in a later article, Houk and Jorgensen11 revisited the topic to calculate the potential effects of solvation. A polar solvent would be expected to preferentially stabilise the more polar (E)-conformers 1 and 1a (which are more polar since the dipoles reinforce each other). It turns out that solvation in either water or acetonitrile only reduces $\Delta \Delta G$ by $\pu{1 kcal mol-1}$.
As for the stereoelectronic argument, only Alabugin7 offers any possible explanation:

A number of factors can contribute to this difference but, at least partially, this effect can be explained by the reduced acceptor ability of the C–O bond of the enolate where the oxygen atom has significant negative charge and by the subsequent decrease in the energy of the nO to σ*C–O stabilisation.

Further reading
Much of the research carried out has been targeted at understanding the anomalously high acidity of Meldrum's acid, and a literature search for this will yield many of the references cited here. The $\mathrm pK_\mathrm a$ values below (DMSO) are taken from Arnett and Harrelson12:

The dramatic increase in acidity upon "cyclisation" of a diester - to the point where the cyclic diester (Meldrum's acid) was more acidic than a cyclic diketone (dimedone) - was traditionally explained with the above information (i.e. it is more acidic because it has two (E)-esters).
However, newer research suggests that there is also an extra stereoelectronic factor that accounts for the unusually high acidity, namely donation from the ester oxygen lone pair into the σ* orbital of the adjacent $\ce{C(sp^3)–O}$ single bond. This stabilising effect is more pronounced in the conjugate base because the oxygen has greater electron density and the donor orbital has a higher energy.13

Notes and references

Huisgen, R. Neuere Beiträge zur Chemie mittlerer Ringe. Angew. Chem. 1957, 69 (11), 341–359. DOI: 10.1002/ange.19570691102. Disclaimer: I can't quite read German, but none of the figures indicate any sort of stereoelectronic effect.
Wennerstrom, H.; Forsen, S.; Roos, B. Ester group. I. Ab initio calculations on methyl formate. J. Phys. Chem. 1972, 76 (17), 2430–2436. DOI: 10.1021/j100661a015.
Wang, X.; Houk, K. N. Theoretical elucidation of the origin of the anomalously high acidity of Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1870–1872. DOI: 10.1021/ja00214a032.
Wiberg, K. B.; Laidig, K. E. Acidity of (Z)- and (E)-methyl acetates: relationship to Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1872–1874. DOI: 10.1021/ja00214a033.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012.
Deslongchamps, P. Stereoelectronic Effects in Organic Chemistry; Pergamon Press: Oxford, U.K., 1983.
Alabugin, I. V. Stereoelectronic Effects; Wiley: Chichester, U.K., 2016.
Kirby, A. J. Stereoelectronic Effects; Oxford UP: Oxford, U.K., 1996.
D. A. Evans' famous Chem 206 notes also only mention this stereoelectronic effect. The set of notes is downloadable here (the relevant part is in Lecture 2).
Blom, C. E.; Günthard, H. H. Rotational isomerism in methyl formate and methyl acetate; a low-temperature matrix infrared study using thermal molecular beams. Chem. Phys. Lett. 1981, 84 (2), 267–271. DOI: 10.1016/0009-2614(81)80342-9.
Evanseck, J. D.; Houk, K. N.; Briggs, J. M.; Jorgensen, W. L. Quantification of Solvent Effects on the Acidities of Z and E Esters from Fluid Simulations. J. Am. Chem. Soc. 1994, 116 (23), 10630–10638. DOI: 10.1021/ja00102a032.
Arnett, E. M.; Harrelson, J. A., Jr. Ion pairing and reactivity of enolate anions. 7. A spectacular example of the importance of rotational barriers: the ionization of Meldrum's acid. J. Am. Chem. Soc. 1987, 109 (3), 809–812. DOI: 10.1021/ja00237a028.
Byun, K.; Mo, Y.; Gao, J. New Insight on the Origin of the Unusual Acidity of Meldrum's Acid from ab Initio and Combined QM/MM Simulation Study. J. Am. Chem. Soc. 2001, 123 (17), 3974–3979. DOI: 10.1021/ja001369r.

I’ve gone and read Huisgen’s essay up to the section on N-nitroso lactames for you. Indeed, he discusses polarity effects only and does not mention a stereoelectronic effect. When moving on to lactames, he writes that these do not show a noticeable difference in dipole moments between cis and trans forms so he concludes that lactames prefer the (Z) configuration for steric reasons.

Just to clarify, is this considered s-cis/s-trans because the C-O bond is a single bond?

@TanYongBoon Actually, no (my earlier comment was wrong)... s-cis and s-trans are descriptors used for a single bond between two conjugated double bonds, which isn't the case here. The 2013 IUPAC Blue Book (P-94.3.1.3, p 1290) says: "This descriptor should not be applied to other systems such as N-alkyl amides (use 'E/Z' or 'sp/ap')."

You speak of the n to $\sigma^$ donation in your answer as an explanation for the conformational preference. May I know which n and which $\sigma^$ exactly? I presume it is n of the carbonyl oxygen and $\sigma^*$ of the single C-O bond?

@TanYongBoon no, the other way round. It’s illustrated in the question.

source: Stack Exchange

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authors: Αντώνιος Κελεσίδης

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text:
Chemicals in a Rubber Compound - NSR 20, Accel CS, Accel DPG, Anitiox Q, Antioz PMB
I have a recipe for a rubber compound and am trying to understand some of the ingredients.
The ones I do not recognize are
Compound A

NSR 20
Accel CS
Accel DPG
Anitiox Q
Antioz PMB

Compound B

SMR 20
Accel NS
Accel DPG
PVI

Compound C

Orox HR
Orac CS
Orac TMTD

Is there a place to look these up on the web?

NSR 20 and SMR 20 are natural rubber (Nigerian Standard Rubber and Standard Malaysian Rubber)
"Accel" and "Orac" mean accelerator.
Accel DPG is diphenyl guanidine.
"Antiox" and "orox" mean antioxidant
"Antioz" means antiozonant
"Orac CS" is likely N-cyclohexyl-2-benzothiazole sulphenamide
"Orac TMTD" is likely Tetramethylthiuram disulfide
"PVI" may refer to pre-vulcanization inhibitor

Dave. This helps a lot. Do you know if "Orac" and "Orox" are industry terms? Could Antiox PMB be p-methoxybenzyl?

I don't think "Orac" and "Orox" are industry-wide terms.  Different companies use different names for the same compound.  If you looked through catalogs of chemical companies that see rubber antioxidants, antiozonant,  accelerators, etc. you might find more information.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid?
Why does the methyl group at the ortho-position stabilize the carboxylate ion? I recognize that there will be a steric clash between the carboxyl group and the methyl group but I don't understand why this would help increase acidity. If there was any resonance occurring then this steric clash would DESTABILIZE the anion and make it LESS acidic given that the carboxylate group would not be in the same plane as the ring and thus delocalization, which would stabilize negative charge, would be impossible. Clearly there is no conjugation with the aromatic ring in "benzoic acid based compounds" from the carboxylate group. So why does having a methyl group at the ortho-position increase acidity?

Having one methyl group ortho to the carboxylate group will not prevent the carboxylate group from becoming reasonably planar with the aromatic ring.  If you had a second methyl group in the other ortho position, then planarity would be a problem.
A methyl group attached to a benzene ring is electron releasing in both an inductive and a resonance sense.

Inductively, the electrons in an $\ce{sp^3}$ hybridized methyl group prefer to shift towards the lower energy (more s-character in the orbital) $\ce{sp^2}$ aromatic carbon.
The methyl group is also electron releasing in a resonance sense due to carbon-hydrogen hyperconjugation (see figure below).

The resonance structure pictured below on the left is one of many that we can draw for the o-toluic carboxylate anion.  Note that it places a positive charge on the methyl bearing ortho carbon. Both inductively and through resonance (see the hyperconjugated resonance structure below on the right) the methyl group will stabilize this resonance structure.  It is this stabilization provided by the methyl group that makes o-toluic acid a stronger acid (pKa=3.91) than benzoic acid (pKa=4.19).

Note: As pointed out by both @Mithoron and @GaurangTandon, this answer does not explain why p-toluic acid is a weaker acid than o-toluic acid or benzoic acid. The answers in this earlier question better answer this question.

o-toulic acid is stronger than benzoic acid because the conjugate base of o-toulic acid is stabilized by hydrogen bonding (C–H···O interaction).

Wow. That is a very untypical hydrogen bond.

I downvoted this answer because this is the same answer as Klaus' answer from more than two years earlier


text:
Why does hydrogen bonding in salicylic acid make it more acidic?
When salicylic acid deprotonates (losing the proton from the carboxyl group) it forms a hydrogen bond with between the oxygen in the carboxylate anion and the hydrogen in the alcohol group. However, surely there will be a steric clash between these groups which would surely be make more significant by the long hydrogen bond? Could the hydrogen bond still form with the carboxyl group twited out of the plane? Also, why doesn't it hydrogen bond intramolecularly before deprotonation which would disfavor the loss of a proton?

If the COO(-) group on salicylic acid's conjugate base didn't hydrogen bond, then wouldn't the COO(-) therefore be more inclined to pick up a proton and return the entire molecule to salicylic acid? In other words H-bonding is a stabilizing interaction. You can view H-bonding as diluting the negative charge density in the COO(-) group. This in turn makes proton abstraction less of a favorable move.
Structure of salicyclic acid's conjugate base:

Presumably it's not planar though?

Why wouldn't everything be planar? Each atom is sp2 hybridized.

There might be some steric hindrance to planarity though

Conformational isomers are possible with rotation about the single bonds between the carbon in the ring and to the hyroxyl group and the carboxylate group.

@Dissenter but H-bonding is present even before deprotonation why does it not make the acid weaker?


text:
Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution?
If I were to attempt to nitrate Pyridine (essentially benzene but with one carbon replaced with Nitrogen) would I get substitution at the Ortho-, Meta- or para position? Also why is this? I can only think of an inductive effect that the Nitrogen might have but I can't think of more factors and I also can't think whether it will be directing towards o-,m- or p- nor why. Can you help?

Sorry, but this is rather common knowledge. The Wikipedia article on pyridine alone would have answered most of your questions and in nearly any general textbook on organic chemistry you would have found it as well. Your question triggered a nice answer by @ron but in general it would be nice if you'd do some research before asking a question here. We don't want to repeat what you can find in most textbooks or easily via google. Rather, we want to help with problems that can't be easily solved this way.

@Philipp I don't agree with you. Virtually all questions on this website can be solved by searching in textbooks/internet/scientific articles. If the question is not extremely trivial and not asked before and shows at least some kind of though process and think it is suitable and on topic here.

@Jori Certainly, all questions on this website can be solved by searching in textbooks/internet/scientific article. But that was not my point. My point was that this question is concerned about the fact that a simple Wikipedia search would have solved the problem or a look in any general textbook on organic chemistry. I think this kind of research effort prior to asking a question here should be required by any OP because that's what we require for homework questions anyway.

@Philipp I agree that the OP should have put in more effort in this case. I only meant to say that the question is in principle not off topic or bad.

@Jori I would agree with you that it is not off-topic, but it borders on a homework question. Also I would concede that it is not a particularly bad question but still I don't like such questions. It would be a perfectly acceptable question if the OP says that he looked it up at Wikipedia or some textbook and didn't understand this or that point about it.

The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.

Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons.  Below are the resonance structures we can draw for pyridine.  They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects.

This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).
If we look at the resonance structures that we can draw for electrophilic attack at the ortho

meta

and para positions,

we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen.  Therefor, meta-substitution is preferred in electrophilic attack on pyridine

So electrophilic attack on pyridine will produce meta-substituted pyridines.  Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation.  The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred.  In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.

Why are these resonance structures in pyridine even possible? The negative charge is surely occupying a p orbital on the Nitrogen but that is part of the delocalised system already

@user3764899 What do you mean? Why would these resonance structures not be possible?

@ron Any particular reason for using sodium nitrate instead of nitric acid? I also wondered: this meta substitution is obviously kinetically controlled, but is it also thermodynamically more favorable?

@user3764899  Any resonance structure you can draw is a "possible" contributor to the total picture of a molecule.  The ones drawn above with a negative charge on nitrogen are significant contributors to the overall description of pyridine because they are placing the negative charge on a more electronegative element.  These resonance structures tell us that in pyridine, the pi electron density is more concentrated in the region around the nitrogen - unlike benzene where the pi density is uniformly distributed.

@Jori  Yields with nitric acid are very low.  There are many "tricks" used in pyridine chemistry in order to get good yields (like the use of pyridine N-oxide).  Sorry, don't know the relative thermodynamic stabilities of the products.

My question about whether the resonance structures arose because the molecule is similar to benzene and the resonance structures of benzene show now charge separation.


text:
What's the meaning of the $-\equiv-$-like bond symbol?
In the following image, $\mathrm{CH}_2$ is connected to benzene rings with a $-\equiv-$ like symbol. What's the meaning of this symbol?

This is a triple bond between two $\ce{C}$ atoms, i.e. $\ce{-C#C -}$. It is drawn that way because the carbon atoms in a triple bond are $\mathrm{sp}$-hybridised and thus exhibit a linear arrangement.

If this is a triple bond, then why here the benzene rings are with a single bond?

@Sparkler Sorry, I don't understand your question. Benzene rings can only directly be linked by a single bond, because you can't have a carbon atom with 5 bonds (notwithstanding very rare exceptions). If you link them via a carbon chain then this chain can contain single, double or triple bonds.


text:
Antacid tablet is acid?
Doing a grade 9 experiment comparing the alkalinity of antacid tablets.
Using bromothymol blue indicator.
Dissolved tablets in both tap water and distilled water and in all situations antacid came out  yellow in with bromothymol blue indicator ... so the antacid is an acid?
I was intending to have students compare brands by titrating with HCl and comparing the volume of HCl used, but have nowhere to go as the antacid is already an acid?
Can anyone explain what is wrong?

I would definitely check the pH of your water ... tap water is usually not neutral and distilled water may or may not be neutral depending on how well controlled the distillation was.

Which antacid brand are you using? You could use methyl orange (pH red 3.1-4.4 yellow) or bromophenol blue (pH yellow 3.0-4.6 blue) as an indicator instead.

You can do your experiment. The active ingredient in most antacids is bicarbonate, and bicarbonates are amphoteric: they can be both acids and bases. In most cases, bicarbonate is a weak base, so despite your bromothymol blue outcome, bicarbonate is a much weaker acid than $\ce{HCl}$, and you should be able to do your titration.


text:
What determines how favorable (exothermic) the electron affinity of a certain element is?
(Favorable means how exothermic a reaction is, more favorable= more exothermic)
There are some abnormalities in the trends of how favorable the electron affinity is.
From Al to Cl, the
Electronegativity is increasing, but the energy releasing from the electron affinity of P is smaller than that of Si. Why? Is it because of how the electrons are arranged?
There are many determinants of how favorable electron affinity is. For example, oxygen and sulfur. Generally, we will think oxygen to have a more favorable electron affinity, due to its stronger electronegativity. Both have the same electrons config, except for the number of shells. But sulfur appears to have a more favorable electron affinity because of its larger size. The smaller size of oxygen atom produces more electron-electron repulsion when one or more electrons are added.
All I ask for is a list of determinants. I tried to find one but couldn't.

Electron affinity is a number, it is not favorable, nor does it release energy. Although the question is probably a valid one, the unclear formulation makes it very difficult to decipher. Please reformulate.

The electron affinity (EA) of an atom (A) is defined as the minimal energy that is required to free an electron (e$^-$) from the associated anion (A$^-$)
$$
\text{EA: }\qquad\text{A}^-\text{(g)}\rightarrow \text{A(g)}+\text{e}^-,
$$
in other words, the electron affinity is the ionization energy of the associated anion.
A large positive EA means that the anion A$^-$ is stable, while a negative EA indicates that the anion ${\text{A}^-}$ is unstable (such as He$^-$). Halogens have the largest electronegativity as the associated anion obtains a completely filled shell.
In general, if an atom A obtains a full/half full shell/sub shell by addition of an additional electron, then A will have a large EA, while A will have a small EA if it already has a full/half full shell/sub shell.
In addition there are some competing effects along a group in the periodic table. Consider the EA of the halogens for example:
Atom:      F       Cl      Br      I
EA /eV     3.40    3.61    3.36    3.06

(i) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the attraction between the nucleus and electrons and the larger is EA. This would predict the trend to be EA(F) > EA(Cl) > EA(Br) > EA(I).
(ii) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the electron repulsion the smaller is EA. The expected trend is EA(F) < EA(Cl) < EA(Br) < EA(I).
As these effects are in competition with each other, there is a maximum in the observed trend for the halogens around chlorine. Where the maximum occurs, is unfortunately difficult to predict.


text:
What is the specific heat of copper through its phases?
Specific heat is the amount of heat required to change the temperature of 1 gram of a substance by 1 Celsius.
For instance, the specific heat of water in the solid phase is 2.059 J/gC and 4.184 J/gC in the liquid phase.
That being said, what is the specific heat of copper in all three of its phases (liquid, solid and gas) in J/gC st standard pressure? How would one figure this out?

The NIST Webbook is generally a good place to look for thermochemistry data. You can see here that there are data for all three phases. It looks a bit daunting, but because heat capacity is not constant with temperature, it's expressed as a Shomate equation. Basically, you find the equation that starts with $C^°_p$ and substitute the parameters from the table underneath along with the desired temperature and you can calculate the heat capacity for any temperature within the range shown at the top of the table. (note that these are molar heat capacities and you'll have to convert if you want it in terms of mass)
If you prefer, you can just click View table to get computed values at certain temperatures, which will save you the calculation.

The heat capacity of a material is a measure of the heat energy (in J) that the material (of mass M in g or moles) absorbs or releases for every unit rise or fall in temperature (in °C or K).

Whilst the heat capacity of most materials will generally be different at different temperatures and pressures, as they undergo internal structural changes with temperature, the heat capacity is generally quoted at a given temperature and pressure. These are typically quoted as 'standard conditions' (as defined by the various standard authorities around the world). For example, the National Institute of Standards and Technology (NIST) defines 'standard conditions' as 20°C and 1atm.
Now, copper is a solid below the temperature of 1358 K (1085°C) and has a specific heat capacity of 0.386 J/g.K or 24.5 J/mol.K (at 20°C and 1 atm).
Various empirical formulae exist which allow one to calculate the specific heat capacity of copper at other temperatures, based on measurements taken in various experiments and fitting the data to various curves such as cubic splines.
Refer to http://www.nist.gov/data/PDFfiles/jpcrd263.pdf for more information.
Copper is a liquid between the temperature of 1358K and 2835 K (2562°C) with true specific heat capacity of 0.572 J/g/K or 36.33 J/mol.K at a temperature of 1400K.
More accurate values of specific heat are given by suitable curve-fitting formulae, based on experimental data. (See: http://link.springer.com/article/10.1007%2FBF02755998)
Above it's boiling point of 2835 K (2562°C) copper vapourises. Copper gas has specific heat capacity of 25.14 (at 3000K) or at other temperatures, given by the Shomate Equation: $c_p(t)=A+Bt+Ct^2+Dt^3+Et^{-2}$
where
$t$ is the  temperature in Kelvin / 1000
$A = -80.48635$
$B= 49.35865$
$C=-7.578061$
$D=0.404960$
$E=133.3382$
Source: http://webbook.nist.gov/cgi/cbook.cgi?ID=C7440508&Type=JANAFG&Table=on#JANAFG
References:
http://www.nist.gov/data/PDFfiles/jpcrd263.pdf


text:
Write the combustion equation of an organic compound

Write the combustion equation of the organic compound $\ce{C5H10O}$

I have no idea at all how to write the equation. Should I add $\ce{O2}$ to make it $$\ce{C5H10O + O2 → ... ?}$$

combustion is a reaction using O2 to burn the other reactant and will result in CO2 and H2O
$\ce {\_\_\ C5H10O + \_\_\ O2 -> \_\_\ CO2 + \_\_\ H2O}$
then just balance
so..
$\ce {C5H10O + 7O2 -> 5CO2 + 5H2O}$


text:
Why is bipolar resonating structure less stable than monopolar structure?
In my book, it is written:

Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures.

The book cites example:

Carboxylate ion which contains negative monopolar charge is in fact more stable than the corresponding carboxylic acid which contains both negative and positive charge.

But what's the reason?? What is going on?? Please help me with a math-free explanation.

This sounds weird to me, and googling for "monopolar resonance structures" gives only this page (besides antenna construction).

@user36790 Could you cite the source of the quote, too, please.

In addition to Max's good answer, it is a complete logical fallacy to go into "this resonance contributor is better than that resonance contributor". Neither contributor is a real thing. However, you can talk about "which contributor more closely resembles the true structure of the hybrid".

In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right).

The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard.
The "resonance stabilization" refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.
The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable.
Diagram soure: http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids


text:
What is the mathematical foundation for the probabilistic nature of chemistry?
Is this best answered through quantum theory, Boltzmann statistics, or what?
For example, in a reaction energy diagram, only some of the reactants go over to the side of the products, and then get stuck there. So the progression of the reaction can be thought of in probabilities - what are the chances that something will go to products, and what are the chances the products will revert back. So you'd use the Arrhenius equation here. Could you dig deeper and find a quantum mechanical reason for this?

You might want to check the Eyring equation. That equation is derived from general principles, while the Arhenius equation (powerful though it may be) is empirical.

Actually, a huge amount of the probablistic nature of chemistry is simply the Law of Large Numbers (LLN). A mole of any substance is approximately $6\times10^{23}$ molecules, which is 600 sextillion molecules! The LLN kicks in at quantities much smaller than the sextillion level. Put simply, there are so many interactions occurring in any macro-level reaction that these are inevitably going to converge to the theoretical mean by the LLN.
Now your question adds in the issue of activation energy, but I still think it fits this model, because activation energy tends to produce a sharp difference: too low, and little reaction happens; but high enough, and we mostly go to completion. Once activation energy is high enough, the LLN applies.

If you want to address only chemical reactions, than best answer is the statistical mechanics, which exercises Maxwell-Boltzmann statistics, exactly as you suggested.
In any reasonable chemical experiments, you are dealing with enormously large ensembles, Avogadro number is $6 \times 10^{23}$ and extremely long times (compared to all motions accessible to molecules).
Therefore, you can follow quasi-equilibrium reasoning, leading to transition state theory.
Sometimes, this is not enough and you want to go further and include coupling between the motions, which leads you to Reaction Dynamics (no wiki article yet, but this is an excellent book).
This approach was needed to describe the crossed molecular beam experiment (Nobel prize 1986).


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Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place
Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place.
I don't understand what I'm meant to put for this.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

Carbonate contamination in oxides compounds often happens after a certain time and air exposure because of the $\ce {CO2}$ in the air.
For your magnesium oxide it follows the reaction
$\ce {MgO + CO2 -> MgCO3}$
with $\ce {MgCO3}$ being the magnesium carbonate.
A solution to prevent this kind of polution is to keep the oxide in a dessicator or to treat them in a furnace at a temperature around to $500 °C$ wich is the decomposition temperature of most of the carbonate compounds.

This is the most likely culprit. It also has nice applications for carbon storage and sequestration.


text:
Quick Stoich & Gasses
I have a quick question in Stoich and Gasses, I however could not get any far, any answer will be appreciated. Thank You.

A sample of $\ce{Pb(NO3)2}$ is heated, yielding $\ce{O2, NO2\ gas,\ and\ solid\ PbO}$. A
  $\ce{293 \ ml}$ gas sample, measured at $\ce{200C}$ and $\ce{1\ atm}$, is collected. What
  was the mass of the $\ce{Pb(NO3)2}$ sample?

I feel like in this situation I need to have only one gas so I can apply the Ideal Gas Law to solve and find the mass of $\ce{Pb(NO3)2}$. What should I do there are two gasses and I don't know the volumes or the masses?
$$\ce{2Pb(NO3)2 -> O2 + 4NO2 + 4PbO}$$

You can still assume ideal gas conditions even if it's a mixture of several gasses. You also need a reaction equation.

@tschoppi I don't quite grasp that.

You don't need a molar mass. If you're looking for one, you're completely on the wrong track.

The equation of the reaction provides you with the composition of the gas liberated.
You assume ideal gas, which means that the gas particles are indistinguishable. This means the reaction equation could look like
$$ \ce{2Pb(NO3)2 -> 5X + 4PbO} $$
where 5 moles of gas are liberated for every 2 moles of lead nitrate.
Reforming the ideal gas equation to $n=pV/(RT)$ will lead you to the numerical value. The amount of lead nitrate will then be $$n_{\ce{Pb(NO3)2}} = \frac{2n}{5}$$
From that calculating the mass is a trivial task.

Thank you for the answer. So you can add the gasses together? That's a new concept, thanks for that.


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Bond Potential Energy Lowering
"Bonds will form only if the potential energy will be lower when the
atoms are bonded than when they are separate."
Is this statement true? If so, how do endothermic reactions exist? Can't you add energy to a system to make the bonds form albeit unfavorably?

Two different factors play a role when considering a reaction. The energy of the bonds etc. (enthalpy) and the entropy. Together, these two components make up what is called the 'Gibbs Free Energy'. (Wikipedia)
The change in Gibbs free energy ($\Delta{}G$) of a system must be $< 0$ for a reaction to be spontaneous. From this it is possible to see that a reaction where the entropy increases drastically, but the enthalpy increases. In this case a reaction can be endothermic.
A case of terminology: The 'more correct' term (my teacher always said) is exergonic or endergonic. What this means is that the reaction has a $\Delta{}G < 0$ or $\Delta{}G > 0$ respectively.
As @Cohorent mentioned before, it is also possible to have a system that is larger than just the chemical reaction (adding a battery for example). Meaning that the reaction is maybe endergonic, but a different energy source is used to drive the reaction. Note then, that the $\Delta{}G$ for the entire system is still below zero.

They only form of the overall energy of the process is lower. You can form a less stable bond if you put in energy either through the formation of another bond which is more stable or through energy in form of light or electricity.
Example: you can electrochemical my form aluminum from aluminum oxide.


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Can I break starch down into glucose units?
I was wondering if starch can be broken up into individual glucose units. Because on a site I saw, it said that through acid hydrolysis starch can be broken up into amylose and amylopectin. However, as I researched amylopectin I found out it is branched and had a different bond than amylose so just to simplify everything:
Can starch be broken up into individual glucose units and through what processes?

Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in survival.
There are different methods how to break it down. Plants, of course, use enzymes. Humans can also break down starch with the help of enzymes (which is why potatoes give us energy), much like most other animals, fungi and many single-cell organisms.
The first amylose-breaking enzyme is present in saliva, so it’s that early that the breakdown begins. However, distinct glucose usually isn’t liberated until the gut.
Chemically, the easiest method would be acidic hydrolysis, i.e. use hydrochloric acid and heat. The downside is, that this method might destroy the individual glucose units, too; so using enzymes is preferred if you want glucose.

Yes, you can. In fact you do it every time you eat a piece of bread or a potato; of course, when I say you, I mean the enzymes in your body.
But I am  guessing you wish to break it down in vitro instead of in vivo. Well you can do that too.
Here's a simple experiment: just filter an collect some saliva from your mouth, and get a hold of some starch solution. Add a few drops of Iodine to the starch solution, and it will indicate the presence of starch by turning bluish-black. Now add some of your saliva to it, and wait for a bit and the colour begins to fade away.
What's happening here is an enzyme called $\alpha$-amylase in you saliva is hard at working breaking the bonds in the starch chain. It acts at random locations along the starch chain, $\alpha$-amylase breaks down long-chain carbohydrates, ultimately yielding maltotriose and maltose from amylose, or maltose, glucose and "limit dextrin" from amylopectin.
Do note that salivary amylase doesn't bring about complete breakdown to glucose, instead you have some glucose but mostly di- and trisaccharides which are further worked upon by more enzymes from the pancreas further down your digestive tract.
There are different kinds of amylase proteins out there, but all of them  are glycoside hydrolases and act on $\alpha$-1,4-glycosidic bonds.
Now, you're not wrong to think acid hydrolsis would work, but that's a more inelegant solution in my opinion. Moreover, as @Jan pointed out, it's also likely to cause breakdown of individual glucose units.

so how would you go about using enzymes to create chemically pure glucose then?


text:
Carbocation rearrangement in alkene addition reaction
This is a question from my textbook:

Determine a step-wise $S_N1$ mechanism for the overall reaction:

The solution shows the following carbocation rearrangement after Bromine leaves in the first step:
Why does the carbocation rearrange like this, going from a tertiary allylic carbocation to a secondary allylic carbocation? I thought tertiary carbocations are more stable and thus favored to be the major product? I considered Zaitsev's Rule in which the more substituted alkene is the major product, but I thought that only applies in elimination reactions.
Any clarification is appreciated!

The mechanism shown is not a carbocation rearrangement. Those two structures are resonance structures, which means that the chloride anion can attack at either of the carbons that bear a positive charge. Because there are two different alkene products that are possible, the more substituted alkene will be favored. Strictly speaking, you're right, Zaitsev's Rule doesn't apply here, because this isn't an elimination reaction. However, the reasoning is the same: the most substituted, most stable alkene is the favored product.

A contributing factor might also be the steric effect, the middle carbon is secondary, whereas the 'right'-carbon is tertiary.


text:
Does baked bread have the same amount of calories as its ingredients?
I bake a loaf of bread containing:

500 g strong flour (1700 kcal)
40 g butter (300 kcal)
24 g yeast (80 kcal)

For a total of 2080 kcal.
How many calories does my finished loaf contain?
My assumption is that at most, some of the energy in the yeast will be used during the proving process but I realised I know very little about the actual processes that take place during proving and baking.

Yes, the yeast does consume sugar to produce the bubbles in the dough, but I don't think it would be more than a few grams. During the baking of the bread, some water and alcohol evaporates and proteins are denatured.
Denaturing proteins does not change the calories. Since the alcohol that was produced is lost, then the calories are reduced. Water doesn't have any calories.
So, to answer to the question in the title: there are fewer calories in the baked bread than in the ingredients. Unfortunately, it cannot be predicted, the bread needs to be tested.

The odd heat-induced decarboxylation or other reactions should also occur; they would further reduce the calories.

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text:
Does the hydrogen taken in an E1 reaction have to be antiperiplanar?
I'm wondering if the hydrogen stolen during an $E_1$ reaction has to be antiperiplanar/anticoplanar like the hydrogen in an $E_2$ reaction.
Intuitively I'd say no, because the carbocation is flat so there's less steric hindrance than an E2 reaction, but I want to check.
As an example, we could use 3-methyl-2-bromobutane. Once that carbocation forms and shifts, does it matter which hydrogen is stolen?

In an E2 reaction, there is a strong preference for the hydrogen to be coplanar with the leaving group and pointed in an anti-direction to the leaving group. In an E1 reaction, the hydrogen does not have to be anti-periplanar to the leaving group because in the E1 reaction the deprotonation event happens in a separate mechanistic step, after the leaving group has left. So there's no need to define the relative orientation of the hydrogen and leaving group in the E1 reaction.
However, for the E2 reaction, remember that the anti-periplanar arrangement is preferred because it places the C-H bonding orbital anti to the C-LG antibonding orbital. In the transition state, those are the orbitals that reorganize to create the C-C pi bond.
With respect to the E1 reaction, there is no C-LG antibonding orbital, but there is an empty p-orbital on the carbocation. The C-H bond should be parallel to the p-orbital to create a similar (but strictly speaking, not anti-periplanar) alignment of orbitals.

It actually does not matter whether or not it is antiperiplanar, because once the carbocation is formed, there is no sense of antiperiplanar. The two carbon atoms are in the same plane now. For a planar carbocation, all arrangements are equivalent(i.e. no sense of stereoisomerism). It does not matter which side the base attacks on. Also, since the bond angle in an $sp^2$ system is greater than that of an $sp^3$ system, the steric resistance to the base is less, which is the point you mentioned.
However, steric factors are not much significant in eliminations where protons are involved, because only a proton is required to be abstracted, and the incoming nucleophile(acting as a base ) is not required to attack or attach to a site.

Three or less carbon atoms are always in the same plane, therefore your last statement is quite confusing.

Yes, you're right. I should not have written $\alpha-\beta$


text:
Why is the halogenation of an alcohol a multistep process?
Refer to the mechanism posted on this site, if you will: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-3.html
When the alcohol attacks the sulfur center, the pi electrons in the oxygen sulfur bond are taken by the oxygen. Then pyridine sucks up the proton. Afterwards the oxygen donates those pi electrons back and the chlorine anion leaves.
Why doesn't the chlorine anion just leave? What's up with the oxygen shuffling electrons around?

The mechanism in the referred website should not be considered a $\ce{S_{N}2}$ mechanism, as it has a tetrahedral intermediate and therefore is more likely to be an addition-elimination reaction. A $\ce{S_{N}2}$ mechanism is concerted, where nucleophile attaches, while simultaneously the leaving group detaches.

If the chlorine left straight away I.e after step 1 in the mechanism you linked to then you would have three charges on the structure - charges on two oxygens and a charge on sulfur. Doesn't seem too stable to me.
Instead what happens after step 1 is a proton transfer. Proton transfers are generally rapid. This proton transfer is also thermodynamically favored as it relieves the oxygen of its positive formal charge.

If would still be 1 charge wouldn't it? The alcohol oxygen is +1, and the chloride is -1. The O double bonded to S is neutral.

As a general rule $S_{N}^{2}$ reaction do not happen at at an '$sp^{2}$' center (check here for help on hybridization: wikipedia). Unlike in an '$sp^{3}$' carbon, there is no likely transition state to accommodate the 'nucleophile' and 'nucleofuge'. Instead of this 'concerted' reaction, an addition-elimination takes place. Like @Dissenter already stated, the proton transfer reaction is favourable and fast, driving the reaction to completion.


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Is Sweetwater, Missouri, named due to water contamination with lead compounds?
The town of Sweetwater, Missouri is known for its mineral resources, particularly rich lead sulfide (galena) deposits. I am curious to find the origin of the name "Sweetwater" and if it has any connection with lead compounds that at one time or another leached into the local water resources (I believe lead acetate tastes sweet). In any event, I can't seem to find any history or such a connection through internet research.

Can you be more specific about where this town is? Is it in Newington County, MO?

'Sweetwater' is a pretty common name west of the Mississippi, seemingly used to lure the unsuspecting to some god-forsaken part of the desert where a trickle of water ran...  I'm not sure a connection to lead poisoning is warranted in all cases.  Often 'Sweetwater' is in comparison to horribly hard water (lots of calcite) encountered in much of the plains area.

According to The State Historical Society of Missouri:

Sweetwater is the name of a little trading-point established on Sweetwater Branch just after the Civil War. It was named for the branch.

and

Sweetwater Branch was named by the pioneers for the unusually good water in the stream.

The place (and branch) are both in Newton County, Missouri. The same reference linked above indicates that what was once called Sweetwater is now referred to as Boulder City (because of the boulders nearby).

Thanks for the answer. "unusually good water" seems to support that perhaps at one time there may have been natural contamination of the water by lead. That's if one believes "good" might imply sweet.

For me "good" implies clear, clean and fresh. The german word for fresh water is Süßwasser, which is translated word by word to "sweet water". I guess that this is due to the big difference to saline water and not a possible sweet taste of lead salts ...

@PH13 Thanks - so now I'm wondering about the dates/hertitage of the 'pioneers' that Todd Minehardt is referencing. I wonder if they were of German heritage. That could connect your note with his answer, and a possible explanation of the name.


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What would be SMILES notation for a compound with delocalized bonding?
For example, the cyclopentadienyl rings in ferrocene $\ce{Fe(}\eta^5 \ce{-C5H5)2}$ are often drawn with delocalized electron circles in them instead of explicit pi bonds.

The Wikipedia article for ferrocene provides the following SMILES string: [cH-]1cccc1.[cH-]1cccc1.[Fe+2]. In this SMILES string, the charge on the cyclopentadienyl ligands are explicitly attached to carbon atoms in the ring.

Can explicit or implicit delocalization be stored in a SMILES string? For example, would c1cccc1.c1cccc1.Fe also be an acceptable SMILES string for ferrocene that can be parsed and understood?
Does the localized version of the SMILES string provide enough information for the delocalization to be reconstructed when the string is parsed back into a structure? If not, how can this be done?

If you mean the aromaticity of the cyclopentadienyl rings, I think SMILES doesn't capture such information, and in general it can only be noticed after rebuilding the structure.

As an example, the wikipedia article for ferrocene $\ce{Fe(}\eta^5 \ce{-C5H5)2}$ has the SMILES as [cH-]1cccc1.[cH-]1cccc1.[Fe+2], which as Nicolau mentions contains no such delocalization information. The charges on the cyclopentadienyl ligands are explicitly attached to carbon atoms.

There is an excellent general question on what SMILES strings can and cannot convey buried in this question. I provided some improvements so that the question becomes answerable and the OP can still get the desired outcome.

Thanks Ben. I could not have put it in a better way. It is not a homework question. I am coming to it from statistics/maths side and it is my first cheminformatics project. I was able to discover ferrocene by searching: delocalized qsar but when it's smiles notation is converted into structure, it gives positive charge on each C. I am reaching to conclusion that probably approaching this problem via SMILES is not the optimal way.

There are general problems representing delocalized bonding and organometallic compounds in cheminformatics (generally) and SMILES in specific.
To quote the Open SMILES standard:

This simple mental model [of a connection table between atoms using valence bonds] has little resemblance to the underlying quantum-mechanical reality of electrons, protons and neutrons, yet it has proved to be a remarkably useful approximation of how atoms behave in close proximity to one another. However, the valence model is an imperfect representation of molecular structure, and the SMILES language inherits these imperfections. Chemical bonds are often tautomeric, aromatic or otherwise fractional rather than neat integer multiples. Delocalized bonds, bond-centered bonds, hydrogen bonds and various other inter-atom forces that are well characterized by a quantum-mechanics description simply don’t fit into the valence model.

Other interesting cases are agostic interactions, 3-center 2-electron bonds (e.g., diborane).
There are several suggestions for advancing beyond the typical connection-table valence bond view, including:

"Yet Another Representation of Molecular Structure" J. Chem. Inf. Comput. Sci., 1995, 35 (5), pp 787–802
"Overcoming the Limitations of a Connection Table Description:  A Universal Representation of Chemical Species" J. Chem. Inf. Comput. Sci., 1997, 37 (4), pp 705–714

That both papers were published 17+ years ago should be some indication of:

Chemists like the valence bond view of molecules, despite the limitations
How complicated bonding can be in non-organic chemistry
How far cheminformatics needs to go to properly handle organometallic and non-organic compounds

By the way... anyone interested in implementing such efforts should let me know.. Many people would be happy to help. :-)
UPDATE
I should point out that there are some practical solutions to representing such things. For example Alex Clark suggested recording zero-order bonds (and sometimes dummy atoms):
"Accurate Specification of Molecular Structures: The Case for Zero-Order Bonds and Explicit Hydrogen Counting" J. Chem. Inf. Model., 2011, 51 (12), pp 3149–3157


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How do I make PDB 1HIG's data usable for producing ball-and-stick models?
PDB: 1HIG, when opened using Accelrys DS Visualizer, Avogadro and Jmol produces a series of dots in space and cannot be displayed as 3D ball and stick models. Is there any way to overcome this problem?
This is the output from Accelrys DS Visualizer 4.0:

This is the output of Avogadro:

This is the output of Jmol:

This specific PDB file only contains the coordinates of all the $\ce{C_{\alpha}}$ atoms of the protein. It simply does not contain enough information to display an actual 3D structure of the protein.
The file does contain secondary structure assignments, so it should probably be possible to create a cartoon display of the protein, but that would require more than a simple PDB viewing program. But there is no way to create a ball-and-stick model with the data contained in that file.

There is a mmcif file in the archive, but I am unable to open it with mercury or any other molecular viewer I have installed.

To add to the answer above, the pdb file only contains Ca atoms and as such it is not possible.
However the structure 1EKU should be usable instead.


text:
Why is benzoic acid a stronger acid than 3-aminobenzoic acid?
Why is benzoic acid a stronger acid than 3-aminobenzoic acid? Obviously at meta position, only negative inductive effect will operate. But, how does that make it a weaker acid than benzoic acid?

The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.2$. The $\mathrm{p}K_\mathrm{a}$ of aniline (aniline's conjugate acid really) is $4.8$.
The two macroscopic $\mathrm{p}K_\mathrm{a}$s of 3-aminobenzoic acid (aniline-3-carboxylic acid) are $3.1$ and $4.8$.
When the microscopic $\mathrm{p}K_\mathrm{a}$s of two groups in a molecule are similar, the macroscopic $\mathrm{p}K_\mathrm{a}$s are not reflective of one particular group protonating/deprotonating. The neutral form will include both a zwitterionic and an non-zwitterionic form in the case of 3-aminobenzoic acid.
When the groups are close in space within the molecule, there will be electrostatic interaction. If the amino group is protonated, it facilitates deprotonation of the carboxylic acid group. If the carboxylic acid group is deprotonated, it facilitates protonation of the amino group.
Addition: There is a very relavent article: Carbon-13 NMR determination of acid-base tautomerization equilibriums J. Org. Chem., 1984, 49 (4), pp 691–696.  It reports that the neutral species is 59% zwitterion and 41% molecular (non-zwitterion).

3-Aminobenzoic acid exists as a zwitterion in solution (just like aminoacids, which the given compound is, just not natural) thus instead of existing as  $\ce{C6H4(COOH)(NH2)}$ it exists as  $\ce{C6H4(COO^-)(NH3+)}$,  thus instead of $\ce{-COOH}$ acting as the acid group, $\ce{-NH_3+}$ acts as the proton donor, and as we know that $\ce{NH_3+}$ is a weaker proton donor than $\ce{COOH}$, we can say that $\ce{C6H4(COOH)(NH2)}$ (3-aminobenzoicacid) is a weaker acid than $\ce{C6H5-(COOH)}$ (Benzoic acid).

Please don't post the same answer twice. Also for future reference, check the help pages (site guide) and guides to formatting chemical equations.

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text:
Thermochemistry problems?
Question 1:
A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.
3 A (s) +B (aq) ––> 2 C (aq) ∆H° = ?
A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ∆H° for the reaction as written in the equation.
Answer: -1408 kJ
Question 2:
For which of the following reactions is ∆H° = ∆H°f, the heat of formation?
i. C (s) + 2 F2(g) ––> CF4 (g) ∆H° = – 221.0 kJ
ii. H(g) + Br (g) ––> HBr (g) ∆H° = –366.2 kJ
iii. 2 C(s) + H2(g) + 3 Cl2(g) ––> 2 CHCl3 (g) ∆H° = –268.2
Answer: i only
Also, could someone explain exactly what delta H means in terms of Thermochemistry. I would like to have a better understanding of it.
Thank you very much!  :)
For the first question I attempted to use the q=m x s x (delta) T for the water and I attempted to use stochiometry for substance A but I always receive the wrong answer. Could someone help me on where to start with these types of problems?
Water calculation:
(250) x (4.184) x (26.26-24.85)
= 1474.86 kJ

H is enthalpy.
By definition H = U + PV (where U is internal energy, P is pressure and V is volume).
For question 1, your work says:

(250) x (4.184) x (26.26-24.85)
= 1474.86 kJ

Because you failed to show units in your work, you are off by a factor of 1000.
But the main reason your answer is different from the desired answer is you have calculated the amount of heat released by 0.152 g reacting, whereas the desired answer comes from calculating the heat released by 3 moles of A reacting, considering change in enthalpy of the system to be negative when heat is released to the surroundings.
For question 2:
Answer ii is wrong because the monoatomic forms of hydrogen and bromine are not the standard states of the elements.  Answer iii is wrong because the equation is written for 2 moles being formed rather than 1 mole.

So I calculated the heat released by .152 g of the substance which is -1474.86 J. But how would I get from 1474 J to 1408 kJ?

Oh, I understand know. I caculated what would be .152 g of heat released. This would be .003 moles of substance a. The question asked for three moles of substance a. So I took 1474.86 and divided it by .00314 moles to get a ratio for substance a. I got 469700 J I then multiplied it by 3 which gave me 1409101.911. Then divide the answer by 1000, I get 1409 kJ, an approximate answer. Is this the correct method to use. I want to ensure that my logic is sound and that I didn't arrive to the correct answer from luck. Essentially it takes a release of 1409 kJ of energy for 3 moles of a to interact w

@dsmith First, find how many moles of A equal 0.152 g.  Then calculate how much heat would be released if 3 moles reacted.

Almost right.  You just need to understand why there is a negative sign in the answer.  If the temperature of the water increases, the change in enthalpy of the reaction is negative.

Yes the negative sign indicates that it is an exoteric reaction. When the water temperature increased it absorbed 1474 J. That means that .152 g of a released -1474.86. This would make all my caculations negative.


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Why can't an enolate anion successfully displace a halogen from a benzene?
Why can't an enolate anion successfully displace a halogen from a benzene? I am told that the geometry is incorrect; that such SN2 substitutions take place on $\mathrm{sp^3}$ hybridized (i.e. tetrahedral) substrates.
This reasoning is circular. So what's a better reason?
I can think of one reason but it is not satisfactory; it's still a bit circular. SN2 = backside attack; how would one attack a planar bromobenzene? From within the ring? Seems like there would be a lot of possibly unfavorable interactions with the p-orbitals for something to get within the ring. Also this doesn't explain why substitution reactions with enolate anions don't take place at other $\mathrm{sp^2}$ carbon atoms - i.e. bromoethene.

The steric reason you mentioned, incombination with the $\pi$-cloud, is in my knowledge the reason that the $S_{N}^{2}$ reaction does not occur. Is this by any change also a reference to the answer I gave earlier today? link

Related: Why do SN1 and SN2 reaction not occur at sp2 centres?

explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms

As explained in this earlier answer, $\ce{S_{N}2}$ reactions do occur at $\ce{sp^2}$ carbons, they're just higher energy and consequently, less common.  So the question becomes, "why is this pathway higher in energy?"
Since the $\ce{S_{N}2}$ reaction involves "backside" approach to the carbon undergoing substitution, steric considerations play a role.  However, the attacking reagent could alter its approach by 15 degrees or so lessening steric interactions (e.g. in benzene the attacking nucleophile could approach at an angle of 15 degrees above the plane of the benzene) and not increasing the activation energy too much.
One way to explain the high energy associated with $\ce{S_{N}2}$ attack at a vinylic carbon is as follows.  Consider the ethylenic double bond as a bent bond system, that is, we have a two-membered ring (instead of a traditional sigma-pi double bond) and each carbon in the ring is $\ce{sp^{4.3}}$ hybridized.  This bent bond description is equivalent to the alternate pi bond description in all regards including electronic (btw, it makes the reason [strain] for the high heat of combustion of olefins more palpable). If possible, build a model to help visualize the situation.  As the attacking nucleophile approaches and begins to interact with one of the ring carbons, that carbon will attempt to rehybridize to an $\ce{sp^2}$ carbon (just like in the traditional case of $\ce{S_{N}2}$ attack on an $\ce{sp^3}$ carbon).  In this case we're forcing the already highly strained bond angle in our two-membered ring to try and open further - towards 120 degrees. Hence, this is a high-energy pathway that is not often observed.
Alternately, this same concept can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state.  It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons in higher energy p orbitals, not offset by the energy lowering of the electrons shifting from $\ce{sp^2}$ bonds to $\ce{sp}$ bonds) thereby increasing the energy required to reach the transition state.
Another point to consider is that since the $\ce{S_{N}2}$ reaction proceeds with inversion of configuration, such a reaction in a ring system would lead to the formation of a trans-double bond.  This would be an extremely high energy process in the case of a small ring system such as benzene.
Finally, $\ce{sp^2}$ hybridized carbon is more electronegative than $\ce{sp^3}$ hybridized carbon due to the increased s-character.  This will increase the electron density around an $\ce{sp^2}$ carbon and, as a nucleophile approaches, the increased electron repulsion will make this a higher energy pathway that nucleophilic attack at a less electron rich $\ce{sp^3}$ carbon.

When attacking a benzene ring for an SN2 reaction, I think your explanation is good enough. The $\ce{C-X}$ antibonding orbital lies in the plane and inside the ring. The nucleophile would be forced to approach through the rest of the atoms of the ring.
As for other $\mathrm{sp^2}$ hybridized organohalides, there was a good explanation recently on this site: Why do SN1 and SN2 reaction not occur at sp2 centres?

Why can't an enolate anion successfully displace a halogen from a benzene?

As you know enolate is the conjugate base of enol which is a tautomeric form of ketone. Thus being an electron rich molecule it basically acts like a nucleophile.

Aryl halides are extremely less reactive towards nucleophilic
substitution reactions due to the following reasons:

Resonance effect : In haloarenes, the electron pairs on halogen
atom are in conjugation with π-electrons of the ring and the
following resonating structures are possible.
$\ce{C—Cl}$ bond acquires a partial double bond character due to
resonance. As a result, the bond cleavage in haloarene is difficult
than haloalkane and therefore, they are less reactive towards
nucleophilic substitution reaction.
Difference in hybridisation of carbon atom in $\ce{C—X}$ bond: In
haloalkane, the carbon atom attached to halogen is $\mathrm{sp^3}$
hybridised while in case of haloarene, the carbon atom attached
to halogen is $\mathrm{sp^2}$-hybridised.
The $\mathrm{sp^2}$ hybridised carbon with a greater s-character is more
electronegative and can hold the electron pair of $\ce{C—X}$ bond
more tightly than $\mathrm{sp^3}$-hybridised carbon in haloalkane with
less s-chararcter. Thus, $\ce{C—Cl}$ bond length in haloalkane is
177pm while in haloarene is 169 pm. Since it is difficult to
break a shorter bond than a longer bond, therefore, haloarenes
are less reactive than haloalkanes towards nucleophilic
substitution reaction.

Instability of phenyl cation: In case of haloarenes, the phenyl
cation formed as a result of self-ionisation will not be stabilised
by resonance and therefore, SN1 mechanism is ruled out.
Because of the possible repulsion, it is less likely for the electron
rich nucleophile to approach electron rich arenes.


text:
Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode
I've run into trouble on this question

Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl-}$. Assume $\Delta_\mathrm rS$ is independent of $T$.

I know the reaction is
$$\ce{AgCl +e- <=> Ag(s) + Cl- (aq)}$$
$$E^\circ=0.222$$
The left cell is a hydrogen electrode making the equation
$$\ce{1/2H2 + AgCl(s) -> Ag(s) + Cl- (aq) + H+(aq)}$$
forming a nernst equation of
$$E=E^\circ-\frac{RT}{vF}\ln\left(\frac{\alpha_{\ce{Ag(s)}}\ \alpha_{\ce{Cl-}}\ \alpha{\ce{H+}}}{\alpha_{\ce{AgCl(s)}}\ \alpha_{\ce{H_2}}^{1/2}}\right)$$
which reduces to
$$E=E^\circ-\frac{RT}{vF}\ln\left(\alpha_{\ce{Cl-}} \alpha_{\ce{H^{+}}}\right)$$
I know $R$, $T$, $v$ and $F$ but I don't understand where to begin to solve for $\alpha_{\ce{Cl-}} \text{and}\ \alpha_{\ce{H}}$
How would I find the activities?

for a cell, there is two half reaction. You have the first one, you have now to find the reaction from the Hydrogen electrode side. Then combining both Nernst relation for both side maybe you'll make a step forward.

If you notice my Nernst in the question includes the activity of H.

Yeah, that's why I told you that writing the Nernst equation from the hydrogen electrode reaction could help, like this, you can combine both, and guess out what is this alpha bothering you.

I'm sorry for the confusion, what i meant to imply is that the nernst i displayed is derived from including the hydrogen half reaction.

Oh ok, so you should know what is this alpha then ?

No, that is the reason for posting this question.

How can you obtain this equation combining the two Nernst relation from the two electrode if you write it by yourself without knowing what you are putting inside ?! by the way my guess is this alpha may represent the activity of both species (usually written "a" but anyway alpha could fit two) so it's just the concentration of both species multiplied by the gamma coefficient.

This discussion is being moved into a chat.

John, use \ln for $\ln$ and \left( and \right) for adjustable brackets.

Sorry, my brain's not on the ball today. Doing a bit of research, it seems that these temperature dependences are determined experimentally and cannot simply be derived. They seem to depend on heat capacity and tabulated temperature coefficients exist.

Now that I think about it, if you're allowed to assume that ∆S doesn't change with temperature, you might be able to calculate the change in ∆H using Kirchhoff's law and use ∆G to calculate the new potential.

Here's the answer from the key:
$$\begin{align}{2}\frac{1}{2}\ce{H2 &-> H+}+e^-  \tag{0V}\\
\\ \ce{AgCl(s) +e^- &-> Ag(s) +Cl-} \tag{0.22V}\\\\
&E^o =0.22V\ \text{at 298 K} \\\\
\ce{2AgCl+H2 &->Ag(s)+2HCl(aq)} \\\\
\Delta_rS &= \frac{vF(E^0_1-E^o_2)}{T_1-T_2} \\\\
\Delta_rS &=S^o_m(\ce{Ag(s)}) +S^o_m(\ce{2HCl(aq)}) - S^o_m(\ce{H2}) -S^o_m(\ce{2AgCl})\\
&=(2\cdot 42.55+2\cdot 56.5-130.68-2\cdot 96.2) \\ &= -124.98 \\ -124.98 &= \frac{1 \cdot 9.6485\ x \ 10^4 (0.22-E^o)}{298-310} \\ 0.22-\frac{1499.66}{9.6485x10^4} &=
 E^o \\\\ E^o_{(310)} &= 0.204V \end{align}$$
Thanks anyway for the attempts

There is a slight mistake to the answer key given, John. In the last step while putting in the value of v, you should put 2 instead of 1, as all calculations of entropy change were done taken into consideration of 2 mole electron transfer. So ans should rather be 0.212 approx

Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl−}$

The reactions would be:
$$\ce{AgCl +e- ->Ag +Cl-}\tag{$E_1$ V}$$
$$\ce{2H+ +e- ->H2}\tag{0 V}$$
Adding:
$$\ce{AgCl +H_2->Ag +Cl- +2H+}\tag{$E_1 V$}$$
The standard potential would be:
$$E=E^\circ -\frac{RT}{\nu F}\log\ce{\frac{[Cl- ][H+]}{P_{H2}^2}}\quad (E^\circ =E_1,\nu=1)$$
H
Now you have $E_1$, also you must have $\ce{[Cl- ],[H+ ]}$ beforehand, there's no other way, other than considering the fact that (standard) Hydrogen electrode is maintained at $\text{1M}$ and $\text{1 atm}$, and to maintain charge $\ce{[Cl- ] =[H+ ]}$.In that case,
$$E=0.222-\frac{8.314\times 310}{96500}\ln(1)=0.222$$

Just for info (from Wikipedia):
$$\begin{array}{c|c}\text{Temperature}&\text{Potential} E^\circ\\
^\circ\text{K} &\text{V versus SHE at the same temperature}\\\hline
298 &0.22233\\
333 &0.1968\\
398 &0.1330\\
423 &0.1032\\
\end{array}$$
Also when you consider $\Delta S\ne\text{ constant}$:(A.J. Bard, R. Parson, J. Jordan, "Standard Potentials in Aqueous Solution", Marcel Dekker, Inc., 1985.)
$$E^\circ=0.23695 - 4.8564\times10^{−4}t - 3.4205\times10^{−6}t^2 - 5.869 \times 10^{−9}t^3$$

@JohnSnow where did you find the question, did you had more info, have any answer to this?

The question was given to me by my physical chemistry professor. the only other statement was " assume $\Delta_r S$ is independent of T."

you asked in your edit "how can I find the activities" but I already answered to this in the caht you oppened ... so I'll say it again, for species in solution, the activity is the concentration time the so called gamma coefficient (but often approximated to be one in practical case, either way those coefficients are tabulated) and for a gaz, the activity is the partial pressure. Since you've already been told that for a NHE (Normal Hydrogen Electrode) you can assume the concentration of H+ to be 1M and the partial pressure of H2 to be 1 atm, you can answer the question now.

@babounet if they are both equal to one then ln1= 0 canceling out the rt/vf term.

and ? isn't it an acceptable result to have E = E° ?

@Babounet it could be acceptable unless $\Delta S$ was constant (see post for more info)

yeah but assuming $\Delta{S}$ indepedant of $T$ doesn't allow to assume if it's a constant of not no ?

@Babounet I'm not getting what you are saying, conntinue in chat

I'm saying that "it's acceptable unless $\Delta{S}$ was constant", but nothing states that it's not constant.

see third comment on this post by john snow


text:
Determine which orbitals will form hybrids with one another
I've been teaching myself chemistry, so any help is greatly appreciated.  I've been reading an online tutorial that claims the two orbitals that merge in Aluminum trihydride are  1 orbital of 2s and 2 orbitals of 2p.  This confuses me, because I would expect the orbitals of the electron configuration would form a hybrid with one another, that is, Al electron configuration is [Ne]3s^2 3p^1.  Therefore, why do the lower orbitals form a hybrid? Also, what is the rule in general to determine which orbitals will form hybrids?

Can you provide a link to the tutorial?

Here is a link. http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals

Let's take your questions in reverse order.

what is the rule in general to determine which orbitals will form
  hybrids?

The "general" rule is that you mix one atomic orbital for each bond you want to form.  In methane, carbon has 4 valence electrons and is forming 4 bonds so it mixes 4 atomic orbitals (1 2s and 3 2p) generating 4 $\ce{sp^3}$ hybrid orbitals, and it places 1 electron in each orbital.
In the case of aluminum trihydride, aluminum has 3 valence electrons and will form 3 bonds, one to each of the 3 hydrogen atoms.  Therefore, aluminum must mix 3 atomic orbitals (1 3s and 2 3p) generating 3 $\ce{sp^2}$ orbitals and places 1 of its electrons in each of these orbitals.

why do the lower orbitals form a hybrid?

A 2s orbital is relatively close in energy to a 2p orbital and the same applies for a 3s and 3p orbital.  The closer in energy 2 orbitals are, the more effective is their mixing.  Therefore, 3s and 3p orbitals can mix (form new hybrid orbitals) effectively.  The reason orbitals mix in the first place is to form new orbitals that have greater directionality and can therefore form stronger bonds.  After all, we could form 3 bonds to aluminum trihydride by just bonding to the 3s orbital and separately to 2 3p orbitals.  Two of these bonds would be oriented 90 degrees from one another and the third bond would be fluxional, moving around the surface of a (3s) sphere. But hybridizing the aluminum so that we can form 3 $\ce{sp^2}$ orbitals and use them for bonding with hydrogen produces a more stable molecule and one with the observed geometry (3 bonds in a plane with a 120 degree angle between them).  If hybridization doesn't produce stronger bonds and a molecule that is more stable, then the molecule wouldn't hybridize. $\ce{PH3}$ is an example of such a molecule, the central phosphorous atom remains unhybridized.

Thanks for the detailed answer.  It still does not answer why the outer orbits 3s and 3p are not becoming hybrids though.

My bad, I should have written 3s and 3p (not 2s and 2p) when discussing aluminum.  I've edited my answer.

The tutorial does mention 2s and 2p for aluminum hydride though, which is why I was confused.  Is that a mistake in the tutorial?

Yes, that is a mistake in the tutorial.


text:
Why isn't water an ionic compound?
If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic?
This is what I'm thinking:
$$\ce{(H+)2O^2-}$$

In other words.. Is water ionic or molecular/covalent?

It seems water can be even "superionic" under high pressure and temperature: http://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound

First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.

What do you mean that one size of a polar bond is stronger?

Side, not size. One atom is more electronegative and has the electron more often.

This video from Khan Academy explains this gradual move from ionic to covalent bonding due to differences in electronegativity: https://www.youtube.com/watch?v=126N4hox9YA

No compound is purely ionic. Water is about 33% ionic.
Linus Pauling discusses the ionic character of water in his famous "The Nature of the Chemical Bond" [1] at pages 100-101.
Pauling says as a first approximation water should be considered as having contributions from four resonance structures:
\begin{array}{lc}
&37\% \qquad & \ce{H-O-H} \\
&24\% \qquad & \ce{(H-O)^-~H+} \\
&24\% \qquad & \ce{H+~(O-H)-} \\
&15\% \qquad & \ce{H+~O^{2-}~H+}
\end{array}
However, he goes on to explain that this is based upon considering each $\ce{O-H}$ bond independently, and the fact that the two $\ce{O-H}$ bonds are not independent reduces the contribution of $\ce{H+~O^{2-}~H+}$ somewhat and increases the other three contributions.
In any case, the purely convalent form $\ce{H-O-H}$ should be considered as contributing only somewhere between 37% and 44% according to Pauling.
References

Pauling, L. The Nature of the Chemical Bond and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry, 3rd ed.; Cornell University Press, 1960. ISBN 978-0-8014-0333-0.


text:
Why they have different flame colour?
At school, I learnt about the metal and their flame colour. Here's my observations.
$\ce{K^+}$- Lavender
$\ce{Ba^2+}$- Yellow-Green
$\ce{Ca^2+}$- Yellow-Red
$\ce{Li^+}$- Flame Red
$\ce{Na^+}$- Bright Orange
$\ce{Cu^2+}$- Blue-Green
So, my question is why do the each metal has its own flame colour?

See e.g. http://www.amazingrust.com/Experiments/background_knowledge/Spectroscopy.html

https://chemistry.stackexchange.com/questions/20516/why-they-have-different-flame-colour

First, remember that electrons cannot exist between principle energy levels (distances). In a flame, electrons from lower principle levels (n=1,2,3... etc) will be knocked up to higher principle levels from the absorbed energy. Electrons do not like to stay in this high energy state and will travel back down to their ground state, and in the process release that stored energy in the form of radiation (often visible light). The frequency of the radiation (color of the light) depends upon how big the fall back to ground state was. Different metals have different attainable energy levels, and so will display different colors when burned.
Some elements only have a few possible emission colors, while other elements are a mixture of dozens of colors ! This principle helps us tell what the composition of stars are, as well as allows us to test for metal contents in food products or other applications.


text:
What is the proper Lewis structure of carbon suboxide?
What is the proper Lewis structure for carbon suboxide, $\ce{C3O2}$?

According to "Electronic structure of carbon suboxide":[1]

Indeed, the results of these electron population analyses
  are consistent with the following resonance forms contributing
  to the ground state of $\ce{C3O2}$:
  \begin{align}
\mathrm{A}&: \ce{O=C=C=C=O} \\
\mathrm{B}&: \ce{O=C=C^\ominus-C#O^\oplus} \\
\mathrm{C}&: \ce{^{\oplus}O#C-C^\ominus=C=O}\\
\mathrm{D}&: \ce{O=C=C^\ominus-C^\oplus=O}\\
\mathrm{E}&: \ce{O=C^\oplus-C^\ominus=C=O}\\
\mathrm{F}&: \ce{O=C=C=C^\oplus-O^\ominus}\\
\mathrm{G}&: \ce{^{\ominus}O-C^\oplus=C=C=O}\\
\mathrm{H}&: \ce{^{\oplus}O#C-C^{\ominus\ominus}-C#O^\oplus}
\end{align}

[1] John F. Olsen and Louis Burnelle, J. Phys. Chem. 1969, 73 (7), 2298–2304. DOI: 10.1021/j100727a033

I noticed the structure $\ce{^{\ominus}O-C#C-C#O^\oplus}$ is missing, is there a specific reason on why it wasn't introduced?

@AndreaL. The authors' list is based upon the molecular orbital calculations they performed.  They find that pi and total bonds are less for C-C and more for C-O, and that the central C and the Os have negative change while the outer two Cs have positive change.  This list reflects these findings.  The structure you proposed would not fit well with those findings.

Well, looks like you haven't been looking for that long time on the internet since the answer is in the Wikipedia article about Carbon Suboxide

The bending about the central carbon occurs too easily for this to be an accurate representation of the structure.

True, but a Lewis's structure isn't supposed to described such a bending since the Lewis's structure isn't "accurate" by definition.

@Babounet what's "accurate" by definition?

As close as possible to the reality ?


text:
What is the systematic name of a "double benzene"?
Consider the molecule representation below. What is the systematic name of that molecule?

Could it possibly be 3,4-hex-2,6-anebenzene?

The commonly used name is naphthalene, the systematic name is bicyclo[4.4.0]deca-1,3,5,7,9-pentaene

According to current IUPAC recommendations (2013), the preferred IUPAC name (PIN) is naphthalene.

Chemical Formula:

$\ce{C10H8}$

Common Name:

Napthalene (also accepted by IUPAC as per 1993 recommendations)

IUPAC Name: (refer numbering in figure)

Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene

For a general guideline:

Start with one of the bridgehead carbons and number it 1.
Proceed round the longest chain of carbons to the second bridgehead.
Number the second bridgehead carbon and continue on round the next longest chain of carbons back towards the first bridgehead carbon.
Pass over the first bridgehead carbon (it already has the number 1) and along the shortest chain of carbons to the second bridgehead carbon again.
The generic name for compounds such as this is

bicyclo[x.y.z]alkane

where x, y, and z are the numbers of intervening carbons on the three paths between the two bridgehead carbons cited in decreasing numerical order, and alk refers to the total number of carbons in the ring systems.

Did you really mean to use equilibrium arrows instead of the resonance arrow?  Please read the other answers and comments to see why the usage of "pentene" is incorrect.

Numbering of atoms in picture is wrong and everyone can edit Wikipedia (well everyone with access to internet...)

As @ron already said in his comment, the systematic name of the naphtalene is Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene since you can start the carbon numerotation from 2 chemically equivalent carbon.
You can find all the references here and here

I thought you always use the lowest possible numbering system.  It is not either or, or am I wrong? Is it really -pentene?  You don't see -tetrene used, it is always tetraene. Google "pentaene" and see how many molecules with 5 double bonds are returned.  I think -pentene is an error.

So you think it's pentaene ?

I do.  Look at this analogous example:  http://webbook.nist.gov/cgi/cbook.cgi?ID=C2443461&Mask=4

Well, looks like you are right ;) I'll edit my answer then ! Thank you ron !


text:
Magnetic moments of tetrahedral Cobalt (II) and (III) complexes
From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$
This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling "together"). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases.

The question isn't really clear exactly what complexes you are referring to.  However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin  while the 3.87 value is for high spin (3/2).  I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar?
For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.
If those are not the reasons, see Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments Quarterly Review Chemical Society volume 22, pp 457-498.

The cyanide complex is indeed square planar, which explains the magnetic data - my textbook didn't make that clear. As for Co(III) alkyl, assuming an e(3) t2(3) configuration in a tetrahedral ligand field you would have a magnetic moment of 4.89 - but the observed values are much lower. I assume square planar geometry would be out of the question with a ligand such as norbornyl (tetra Co(III) complex magnetic moment 3.15).

norbornyl complexes were the first ever discovered low-spin tetrahedral complex of first-row transition metals.  http://pubs.rsc.org/en/content/articlehtml/1986/c3/c39860001491

but it is Co(IV) norbornyl complex that is low spin tetrahedral

@J.LS magnetic moment of 3.15 would mean cobalt (III) norbornyl complex is low spin (s=1) tetrahedral, so it looks like both Co(III) and Co(IV) norbornyl complex are both extremely rare low spin tetrahedral complexes

@DavePhD we may also go the other way: metal(V) complexes "should be" low spin ... unless they not only are tetrahedral but also have an exceptionally pi-donating ligand. Like this case, with cobalt at that.


text:
What's wrong with using tin in medicinal chemistry?
I just read this In the Pipeline post and I was slightly confused by a statement on the use of tin. Lowe reports on this paper, which describes a synthetic route to spiro heterocycles using tin compounds, and caveats the finding with the comment that

No one's crazy about using tin, but the transformation is too useful to pass up.

Why is the use of tin undesirable in synthetic chemistry?

Organotin compounds are rather toxic.  They are also persistent in the environment and have a long biological half-life.
The problem is that trialkyltin byproducts from your reaction are difficult to separate from the product.  In the lab, this is painful and (usually) involves multiple columns, but when you want to get a drug past the FDA into the clinic you have to meet purity standards for heavy metal contamination.  To a medicinal chemist, a transformation that won't make it into pilot plant and production is useless.

While I fully agree with the answer you provided, Tin really is hard and important to get rid of in some cases, I would like to add that the Stille Coupling is widely used in industry. This uses organo-Tin reagents in a Palladium coupling. General link on Stille Coupling

There are a few points in the answer and in comment that I feel could use some clarity. I've never met a process chemist willing to scale a Stille reaction up for production, but med chemists use it all the time on relatively small scales. Organotin waste is a huge problem at scale, but if you're just aiming for milligrams of something to test, a Stille coupling may be perfectly reasonable. You can try to switch to a less distasteful route if the molecule survives the initial testing.


text:
Best x-ray dopant?
An x-ray computed tomography scanner has been used in an attempt to look at the pore space of a limestone rock sample (a carbonate, $\ce{CaCO3}$) saturated with a $\mathrm{12\ wt\%}$ sodium iodide ($\ce{NaI}$) solution as the x-ray dopant.
The idea is that the sodium iodide would absorb the x-rays more so than the grain matrix of the rock and thus the fluid in the pore space of the rock can be seen, by contrast in x-ray absorption. Therefore, the primary entity that is measured through CT scanning is x-ray absorption, defined as the attenuation coefficient.  This is determined both by mass density and elemental composition of the material.
Unfortunately, this doped fluid wasn't able to be seen. Next tests will use a chalk (composed ~90% or greater of $\ce{CaCO3}$). Typically for tests of this nature sodium iodide is preferred because its interaction with sorts of clay is similar to that of sodium chloride.
The questions are:

What is the highest attainable concentration of a sodium iodide solution?
It is known that a solution of barium chloride has been used an x-ray dopant. Would this solution be compatible with chalk, i.e. non-reactive with chalk?  What would its highest attainable concentration be?
What other possible solutions can be used as x-ray dopants?

NIST has values of mass attenuation coefficients by element and provides a very useful tool for looking up the specific attenuation coefficient for mixtures.
My quest is to understand what x-ray dopant solutions are possible and what their highest concentrations achievable are. Then to rule out ones that are reactive with the rock.

How much pore space is there?

for this particular core it was estimated to be ~13%

Try to increase the concentration first. NaI has quite high solubility. Hopefully, it would be enough.

This is just me shooting bull because I don't know what has been done, but in the distant past, mineralogists used concentrated solutions of heavy metal salts for density determination by flotation.  Clerici's solution is a solution of equal parts thallium malonate and thallium formiate in water.  That may just be the ticket.
Also, those with access to appropiate protective gear (i.e. fume hood and proper gloves) may consider methyl iodide.

Methylene iodide is even better.


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Does metal oxidize (faster) in dirt?
I had an interesting event occur recently: I stuck a screwdriver in a potted plant, for it is the only way that I am able to unlock a certain door. It was left in there for about a month before I needed it again, and, when I took it out, it had what I thought to be rust. I am doubting my original assumption.
I know for a fact that it is not dirt itself, as I vigorously attempted to clean it with suds (though I'm not entirely sure whether or not that was the best approach...).
Question
Does iron a) oxidize in dirt and b) if a), does it oxidize faster in dirt?

Was the dirt wet or moist?

@Michael I'm not entirely sure, though I'd assume moist, for I did water the plant a bit.

Moisture greatly increases the rate of corrosion. See theo's answer.

What is "dirt" ?

Iron can corrode faster in soil than in air, particularly if partly immersed in the soil. Moisture and electrolyte are retained in the soil, making it a good anode. Also, microbes in the soil can accelerate corrosion.

Unless it is under cathodic protection like a pipeline.


text:
Where does the extra electron come from in the Lewis Dot Electron structure of say for example nitrate ion?
As most of you know, this is the diagram for nitrate ion:
To draw this, i have to make all bonds with electrons from nitrogen, nitrogen only has five valence electron so ONE of the bonds consists of one electron from nitrogen and the other from the MYSTERIOUS electron, that comes out of no where.
My book tells me that's if simply how you draw Lewis diagram for N, O, F, Cl. When the oxidation is positive, take away electron and when it is negative, add.
In any covalent compound, bonding atoms share their electron to complete octet.( yeh, there are exceptions)
In this case, N and O share their electron. But when I count the number of valence electrons, a negative charge means a extra valence electron. NOW, where does this extra valence electron come from?

Unfortunately your Lewis structure for the nitrate ion, $\ce{NO3-}$, is wrong. You did not include any charges and as it is an ion there is at least one. Let's first have a look at how an ion may be formed and then hopefully you see, that there is no mystery at all about it.
Take for example ordinary table salt, which has the chemical formula $\ce{NaCl}$. If you dissolve it in water the lattice will be broken and the ions will by hydrated and are now able to move more freely.
$$\ce{NaCl_{(s)} -> Na+{}_{(aq)} + Cl- {}_{(aq)}}$$
The same happens when you dissolve potassium nitrate, which releases the nitrate ion:
$$\ce{KNO3_{(s)} -> K+{}_{(aq)} + NO3- {}_{(aq)}}$$
Here you can see, that the nitrate ion has an extra electron which it has taken from the potassium. See ionic bonding for more information.
For the Lewis structure of the nitrate ion you first have to count all the valence electrons of the involved atoms:
\begin{array}{lclr}
1 &\times &\ce{N} & 5\\
3 &\times &\ce{O} & 6\\
1 &\times &\text{charge} & 1\\\hline
  &       &              & 24\\
\end{array}
In the Lewis structure you have to distribute 24 electrons in a fashion, that you do not violate the rule of two, which is better known as the octet rule. Therefore your structure may include a nitrogen oxygen double bond, a positive formal charge on nitrogen and a negative formal charge at each of the singly bound oxygens.
Another representation is that all oxygens are singly bound and therefore the nitrogen has a $+2$ formal charge and each of the oxygens has a negative formal charge. The problem with this structure is that nitrogen does not have a full octet. Nevertheless, it is a strong contributor to the actual electronic structure. The bonding situation is much more complicated than can be expressed through a Lewis structure. The concept of resonance structures has to at least be included. Better in this case would be molecular orbital theory and or valence bond theory. The molecule itself is a y-aromatic system, the charge is delocalised through the extensive $\pi$ system. The blue structure is an attempt of showing this feature.

@Doeser I have never stated that bonding within the nitrate ion is anything but covalent. Your statement is correct, in covalent bonds electrons are shared. I am a little confused by your remark about the bonding situation, could you please tell me a little more about what you are referring to?

@Doeser the extra electron is from the counter ion. In my example that would be the potassium ion. The bonding here is ionic.


text:
Which is the correct numbering of locants for ethyldimethylcyclohexane?
Can anyone kindly help me to name this compound?

Is the above compound 3-Ethyl-1,1-dimethylcyclohexane or 1-Ethyl-3,3-dimethylcyclohexane?

Related: Nomenclature of ether and locant position enumeration

3-Ethyl-1,1-dimethylcyclohexane
Following the IUPAC rules, the correct name will have the lowest numbers for the substituents. Therefore, 1,1, and 3 is lower than 3,3 and 1.