text:
Non-bonded orbitals in water
General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $sp^3$ hybridized. Oxygen forms two sigma bonds with hydrogens, and there are two lone pair orbitals in the molecule. There are five occupied molecular orbitals: The core $1s$ of oxygen, and the four $sp^3$ orbitals, out of which two are lone pair orbitals.
Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2v}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $a_1, b_1,$ and $b_2$.

There is only one non-bonding orbital ($1b_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $\mathrm{sp}^3$ hybridized.

No. One could say that, but $\mathrm{sp}^2$ is equally possible. In fact, it may seem more likely, we’ll get to that in a second.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2\mathrm{v}}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $\mathrm{a}_1, \mathrm{b}_1$, and $\mathrm{b}_2$.

Yes, and you did a nice job of drawing them.

There is only one non-bonding orbital ($1\mathrm{b}_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

And this is where it gets interesting.
I’ll once again point to Professor Klüfers’ web scriptum for the basic and inorganic chemistry course in Munich, section about localising molecular orbitals. If you don’t understand German, all you need to do is look at the pictures and understand that wenig means a little in this context.
You see, to do a discussion such as ‘this is the bonding electron pair’ we need to localise molecular orbitals. Specifically by linear combining them. It’s almost like in the linked pictures: $\Psi_2 (1\mathrm{b}_2) + \Psi_3 (3\mathrm{a}_1)- \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_1$. Invert the signs: $\Psi_2 (1\mathrm{b}_2) - \Psi_3 (3\mathrm{a}_1)  + \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_2$. These are your two bonding orbitals to hydrogen. We have some $\Psi_1$ remaining (and, theoretically, also some $\Psi_2$ and $\Psi_3$, since we didn’t use all of them). That (remaining linear combination) is our second non-bonding orbital.

Now why is $\mathrm{sp}^3$ a bad description? You see, we arrived at one lone pair that is definitely $\unicode[Times]{x3C0}$-shaped or p-shaped. The second is somewhat p-ish but also somewhat s-ish, so maybe some sp-hybrid. But there is no mistaking that $1\mathrm{b}_1$ is in no way influenced by all the others. It is antisymmetric to the $\ce{H-O-H}$-plain of symmetry while all the others are symmetric. $\mathrm{sp}^2$ describes this a lot better.

How does one rectify $sp^2$ hybridization of the oxygen center with the H-O-H bond angle being much closer to tetrahedral?

@Blaise From an angle point of view, both are wrong and the actual situation is more of a ‘p with a little s mixed into the in-plane orbitals’. (Bond angle 104.5° < 109.5°, not larger as tetrahedral or $\mathrm{sp^2}$ would imply.) But that is a higher level of analysis. What I want to make clear is that there is an almost pure p type nonbonding orbital and only two p and one s orbital of oxygen take part in hybridisation at all. Hence ‘$\mathrm{sp^2}$’.

I understand where you're getting the sp2 from, but that's essentially assuming that the second lone pair and the O-H bonds are equivalent (so each is sp2). But they aren't. Based on calculated s and p content of localized bond orbitals, the O-H bonds are approximately sp4. The combined lone pair electron density can be described as one p and the other something like sp(2/3) or as two equivalent ~sp(8/3) hybrids. So with respect just to the O-H bonds, the general chem "sp3" is closer to reality than your description of sp2.

Hmm, this is a bit of a contentious issue: https://pubs.rsc.org/en/content/articlelanding/2015/rp/c5rp00061k As long as we limit ourselves to the gen chem sp^n models, then the question IMO isn't really which is better, it's more like which one out of a set of flawed models is least flawed.

@Andrew In the end, the closer we look, the less useful the simple $\mathrm{sp}^n$ (where $n\le 3$ and a whole number) becomes. The bond angle (104.5°) suggests a higher p content than both $\mathrm{sp^2}$ (120°) and $\mathrm{sp^3}$ (109.5°) – an estimation of $\mathrm{sp^4}$ might be good but I’m too lazy to look up and do the calculations. But the p-shape of one lone pair is obvious. General, especially introductory chem needs to strike this balance between an accurate and an easy to understand explanation. I feel that the white lie of $\mathrm{sp^2}$ is not too wrong for this area.

Having been misled by this myself, I think it's inappropriate to try to deduce localized hybrid orbitals from delocalized MOs. Although the local and delocalized orbitals necessarily describe the same total electron density, they arrive there completely differently. Each should be built up from AO's independently. When you do that for the localized description, you treat the two lone pairs as equivalent and find the same hybridization for both that gives the right net electron density. Instead of one all p and one mostly s, you get two that are both just a bit shy of sp3.

Although this is an old question with an accepted answer, I think it's worth providing a second perspective rather than continuing extended discussion in the comments.
First off, let's consider the purpose of even considering hybridization. The purpose of constructing hydrid orbitals is to create localized bonding orbitals and to see how they can be formed as linear combinations of atomic orbitals. By determining the contribution of different s and p orbitals to the hybrids, we can draw some conclusions about the bonding in a molecule.
Since delocalized orbitals such as those shown in the question are also constructed from linear combinations of atomic orbitals, one approach (as described in the other answer) is to construct our localized bonding orbitals by linear combinations of the delocalized MOs. As shown in the other answer, we can create two equivalent localized O–H bonding orbitals by the combinations
$$\Psi_2(\mathrm{1b_2}) + \Psi_3(\mathrm{3a_1}) − \text{some } \Psi_1(\mathrm{2a_1}) = \sigma_1$$
and
$$\Psi_2(\mathrm{1b_2}) - \Psi_3(\mathrm{3a_1}) + \text{some } \Psi_1(\mathrm{2a_1}) = \sigma_2$$
Since the question is about the nonbonded orbitals, however, we need to complete the process and create two equivalent localized non-bonding orbitals. These necessarily are comprised of the MOs we didn't use to make the bonding orbitals.
That means combining $\Psi_4(\mathrm{1b_1})$ [pure p] with the remaining $\Psi_1$ [mostly s] and a little bit of $\Psi_2$ [mostly p] and $\Psi_3$ [mixed s and p], since we didn’t use all of them. The result is two orbitals that are both a mix of s and p.
The problem with this approach is that it's hard to deduce the quantitative contribution of the atomic orbitals to either the localized bonding or the localized non-bonding orbitals without extensive calculations, so we don't know how well this picture matches to the general chem "water has $sp^3$ hybrid orbitals" description. I've been misled myself trying to draw conclusions from this approach (as evidence by some wrong answers I've written on this site).
Instead, it's much simpler to build the localize hybrid orbitals from atomic orbitals independently of the delocalized orbitals.
When doing so, Coulson's Theorem is very helpful for translating a bond angle into a degree of hybridization. For two equivalent bonds (as the O-H bonds here are expected to be) with a bond angle $\theta$, the hybridization of the localized bond orbital is given by $sp^\lambda$ where $$\lambda = -\frac{1}{\cos\theta}.$$
Applying this to the known bond angle of water, we find that the O–H bond orbitals are $\approx sp^4$, that is, 80% p and 20% s. So they are slightly more p in character than the simple $sp^3$ model (ie 75% p). This is consistent with our intuitive expectation that a bond angle less than 109.5 indicates more than 75% p character.
Although we don't have any bond angle between the lone pairs to work with, we can conclude by simple subtraction that if the bond orbitals are 80% p/20% s, the lone pairs must be 70% p/30% s, so $sp^{2.3}$.
The reason that these values differ from the Gen Chem simplification is that the Gen Chem approach treats the bonding and non-bonding orbitals as equivalent and provides only a single hybridization value. Is that incorrect? Yes. Does it really matter? Probably not. It's a halfway decent approximation, and most Gen Chem texts follow shortly with a discussion of $\ce{H2S}$, where the smaller bond angle indicates even further deviation from $sp^3$, to the point that the increased p contribution is far too great to be ignored.
One last important point is that the electron density overall on the molecule that is described by these localized orbitals is necessarily identical to that total electron density described by the delocalized orbitals. Much as we can describe the number 12 as 1+2+4+5 or as 3+3+3+3, the two sums are equivalent even though the component parts are completely different from each other.

"By determining the contribution of different s and p orbitals to the hybrids, we can draw some conclusions about the bonding in a molecule." - I say that no, we can't, because the resulting hybrid orbitals have neither an energy, nor a fixed shape to them. I've made that point in detail here.

I completely agree they have no energy associated with them, but disagree that they are not stringently mathematically defined, since Coulson's theorem does that, admittedly in a time-independent way, but pretty much all basic models of chemical structure and orbitals are treated as time independent, so I don't see a problem with that. If we consider time, then the canonical delocalized basis orbitals do not have defined shape either.

They may very well be treated as time-independent, but that doesn't mean that it would be correct. Also, if we're talking about canonical orbitals as they come out of e.g. a Hartree-Fock calculation, then their associated probability density for sure is time independent. (Granted, I didn't talk about the probability density before, but my point still stands.)


text:
Why is the inversion barrier larger in PH3 than it is in NH3?
The inversion barrier in $\ce{NH3}$ is approximately $5~\mathrm{kcal~mol^{-1}}$ and that of $\ce{PH3}$ is $35~\mathrm{kcal~mol^{-1}}$. This has well-known stereochemical consequences in that amines are not chiral whereas phosphines can be. How can the larger inversion barrier be explained?

related http://chemistry.stackexchange.com/questions/16190/ http://chemistry.stackexchange.com/questions/464/ http://chemistry.stackexchange.com/questions/50106

Ammonia is the classic system for $\ce{sp^3}$ hybridisation save methane. The lone pair (and each of the $\unicode[Times]{x3C3}$-bonds) has almost $25~\%$ s-character which corresponds nicely to $\ce{sp^3}$. However, the whole system can also swing around, changing its hybridisation to $\ce{sp^2}$ and back; a process during which the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.
Elements outside of the second period show a much smaller tendency to involve the s-orbital in bonding. The bonding orbitals only have an s-character of approximately $16~\%$. This also means that the phosphorous lone pair has a much higher s-character of approximately $50~\%$, while the bonding $\unicode[Times]{x3C3}$-orbitals have a larger p-character. (This also explains the much smaller $\ce{H-P-H}$ bonding angle of approximately $90^\circ$.) An orbital of high s-character has a long way to go to turn into a p-type orbital, and the three largely p-type $\unicode[Times]{x3C3}$-orbitals have an equally long way to go to give $\ce{sp^2}$-type orbitals so the interconversion and thereby the inversion of $\ce{PH3}$ is strongly hindered.

Sneaky little use of classic in a quantum chemical context of the explanation.

Will the sp2 hybridized intermediate be paramagnetic or diamagnetic?

@Intellex Going from diamagnetic to paramagnetic requires a realignment of spin of some some (i.e. from a singlet to a triplet state in the case of an even electron count). This is not something that happens spontaneously.

Disclaimer: This answer neglects quantum tunnelling effects, which are significant in such compounds. The gist of it is that because nitrogen is smaller and lighter than phosphorus, the rate of tunnelling and hence inversion of chirality is much faster. This issue is discussed to some extent in this question.

I've attached a Walsh diagram for the $D_\mathrm{3h} \leftrightarrow C_\mathrm{3v}$ transition below, which corresponds to pyramidalisation of a planar structure. (Image taken from Orbital Interactions in Chemistry by Albright, Burdett & Whangbo, 3rd ed., p 186.)

Now, in order for inversion of stereochemistry to occur at N or P, the molecule has to go from a pyramidal structure, through a planar transition state, and back to the other pyramidal structure. It is kind of like flipping an umbrella inside out! The energy barrier to inversion is, of course, the difference between the equilibrium geometry (pyramidal) and the transition state (planar).

Now, $\ce{NH3}$ and $\ce{PH3}$ are both 8 valence electron species. This means that the HOMO will be the $2a_1$ orbital in $C_\mathrm{3v}$ symmetry, which becomes the $\mathrm{a_2''}$ orbital in $D_\mathrm{3h}$ symmetry. As you can see, due to a loss of bonding interactions of the central $\mathrm{p}_z$ orbital with the hydrogens, the HOMO is destabilised upon planarisation.
Now, let's assume that the HOMO is the predominant orbital that determines the inversion barrier. That's quite usual in discussion of Walsh diagrams. This means that the steeper the $\mathrm{2a_1} \rightarrow \mathrm{a_2''}$ slope, the harder it is for planarisation to occur, and the larger the inversion barrier.
The question therefore becomes: how much is the $\mathrm{2a_1}$ orbital stabilised upon pyramidalisation? If it is stabilised more, then planarisation is harder and the inversion barrier is larger.
It turns out that the stabilisation of the $\mathrm{2a_1}$ orbital can be seen as a second-order Jahn-Teller distortion. This means that, in going from $D_\mathrm{3h}$ to $C_\mathrm{3v}$, there is a loss of molecular symmetry that causes two orbitals that originally do not have the same symmetry to become the same symmetry. In this case, the two orbitals are the HOMO and the LUMO:
$$\begin{array}{c|c|c}
\text{Orbital} & \text{Symmetry in }D_\mathrm{3h} & \text{Symmetry in }C_\mathrm{3v} \\
\hline
\text{HOMO} & a_2'' & a_1 \\
\text{LUMO} & a_1' & a_1 \\
\end{array}$$
It turns out that the strength of the Jahn-Teller effect is inversely proportional to the difference in energy between the two unmixed orbitals. The idea is that orbitals close in energy interact more strongly with each other - and that should be a somewhat familiar idea, because it is a central principle of MO theory. Anyway, it means that if the HOMO-LUMO gap in $D_\mathrm{3h}$ is small, there will be a large stabilisation of the HOMO upon pyramidalisation, and hence a large inversion barrier.
Finally, we have come to the last part of the explanation.

The $\mathrm{a_2''}$ orbital is a nonbonding $n\mathrm{p}$ orbital on the central atom, and the $\mathrm{2p}$ orbital of $\ce{N}$ is lower in energy than the $\mathrm{3p}$ orbital of $\ce{P}$.
The $\mathrm{2a_1'}$ orbital is higher in energy in $\ce{NH3}$ because the nitrogen $\mathrm{2s}$ orbital has strong overlap with the hydrogen $\mathrm{1s}$ orbitals, leading to a more strongly antibonding orbital.

This means that the energy gap between $\mathrm{a_2''}$ and $\mathrm{2a_1'}$ is larger in $\ce{NH3}$. As a result:

there is a smaller second-order JTE in $\ce{NH3}$
leading to less stabilisation of the HOMO upon pyramidalisation
which corresponds to a smaller energy barrier to planarisation
a smaller inversion barrier
and more rapid inversion in $\ce{NH3}$.

There is a little bit more discussion in the aforementioned textbook: Orbital Interactions in Chemistry, 3rd ed. pp 187-9.

Your question deals more with hybridization. First we have to note that $\ce{NH3}$ has a lot more hybridization than the other molecules with hydrogen of the group e.g. $\ce{PH3}$. This means for example that $\ce{PH3}$ and $\ce{AsH3}$ have bonding angles near 90°. Because the tetraedric angle needs a $sp^3$ hybridization.
The transition state of the inversion can be stabilized by a $sp^2$ hybridization which again practically happens only for the ammoniak.

Another reason is that the P lone pair is 'inert' (see also: http://chemistry.stackexchange.com/questions/32591/diagonal-relationships-in-the-periodic-table/33009#33009) as explained by Kutzelnigg and others. The 3s orbital and 3p orbitals spatially mismatch (as the 3s orbital is closer in) as opposed to the 2s and 2p orbitals. The 3s orbital doesn't quite like to mix with the 3p orbitals.

source: Stack Exchange

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

sort: votes


text:
Why we make pyrogallol alkaline to absorb oxygen?
Pyrogallol is activated by -OH groups but what is the use of making it alkaline?

Because that makes them $\ce{\dots-O-}$, which is even better for oxidation. See, this thing is just ready to give away two electrons and become the corresponding quinone without any structural rearrangement.

Thanks. Can you please explain me why it shows colour after reaction with oxygen.

Well, color in organic compounds like this is mostly a feature of conjugated $\pi$-system, and the oxidation of pyrogallol, involving breaking of cycles and condensations that form new cycles, produces quite a system (see purpurogallin).

Does this reaction yields purpurogallin or 2-pyrone?

I don't really know. Probably both, and maybe something more.

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


text:
Why is it possible to image LUMO if these orbitals are, by definition, unoccupied?
LUMO stands for lowest unoccupied molecular orbital. If it's unoccupied, how is it possible to image it?
See for example in this paper ("Pentacene imaged with STM and NC-AFM").

Why not? It's not like we poke them with our finger, anyway. All we can really see are the electron transfers. Now, the transfers from the molecule are as good as the transfers to it. If we adjust our setup so as to look at the latter, then we may see the shape of the orbital which is now empty but ready to accommodate our electron.
In one setup, the electrons tunnel from the HOMO to the tip, thus visualizing the HOMO. In another setup, they tunnel in reverse direction (hence the change in the sign of voltage), from the tip to the LUMO, thus visualizing the LUMO.

So the some of the tunnelling electrons that the STM measures pass from the HOMO to the LUMO, and only then tunnels to the STM tip?

Got it! Your phrasing made the difference...

There are quite a few people that would strongly object against MOs as objects that could be visualized. They are not in any sense QM observables (and equivalent sets of MOs could be used to describe the same states. But the pictures with canonical orbitals are quite compelling, right? Indeed the difference between HOMO and LUMO ('orbital density'?) would be the direction of the charge transfer.

OK, then what are the pictures a and b? X-ray images of a fossilized Cambrian shrimp, maybe?

A wonder often skipped by chemists too is that chemical structures - often snobbery treated, are 1:1 correspondence of what the molecule look like by means of these techniques. But they were just deduced by conducting reactions, taking melting points, and thinking wisely.


text:
Why is boron trifluoride a Lewis acid?
Why is $\ce{BF3}$ considered a Lewis acid despite backbonding with fluorine and still accepting an electron pair?

It's mesomeric stabilization not back-bonding and bonds are still polarized to fluorine.

I have a feeling that fluorine in $\ce{BF3}$ does that thing almost reluctantly, like "OK, you may use my spare electron pair for a short while, but then give it back to me a.s.a.p." Probably this has something to do with the fact that we rarely hear of fluorine in $sp^2$.

A Lewis acid is defined as an electron-pair acceptor. So for something to act as a Lewis acid, it needs to want electrons. Prime examples are $\ce{H+}$, the hardest Lewis acid around (zero polarisability, very high charge per volume ratio) and practically every metal cation out there: $\ce{Al^3+, Zn^2+, Fe^3+, Ag+}$ just to name a few.
Consider boron, a rather electropositive element — it counts as a metalloid so it is somewhere between non-metals and metals. We are binding it to fluorine, the most electronegative element, and we’re doing that three times. It should be evident that there is hardly any electron density left on boron. How happy would it be, if some other atom gladly donated their electron pair to share?
Now what are we going to do if there is no Lewis base around? Well, initially boron will still be there, depleted of all its valence electrons by fluorine (or nearly at least). This is where fluorine discovers its charity side: All three fluorines donate just a tad of electron density so that the baby boron in the middle will stop crying. This is what you referred to as ‘back bonding’ and Ivan calls ‘mesomeric stabilisation’. But the point is: That doesn’t help against the electron deficiency in any way, it’s more like boron’s final counter-measure against loosing electrons.

Chemists please forgive me for turning atoms into humans. At least I didn’t explain the whole electron-withdrawing thing with socialism …

A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.

IF you have a strong chemistry background I could go further and say that Fluorine is a more electronegative atom than Boron so the electron density in the BF3 would shift more towards the F atom. This creates the greatest electron deficiency on the B atom allowing it to accept another electron pair with ease.

Well, this is not a full answer since it doesn't talk about the $\text{p}\pi$-$\text{p}\pi$ bonding explicitly mentioned in the question. OP is asking why $\ce{BF3}$ is still a Lewis acid in spite of this bonding.

Even though there is 2pπ-2pπ backbonding in $\ce{BF3}$ molecule, it still  acts as a Lewis acid. This is because the lone pairs that are donated through this way are better and permanently donated to the boron atom than backbonding. Backbonding is just a concept for temporary internal stability of a molecule.
When any Lewis base approaches the $\ce{BF3}$ molecule, then the π-electron density of 2pπ-2pπ backbond start to shift towards the fluorine atom because there is now no need for backbonding to stabilise the molecule (as electron are now donated to boron permanently through a better way).
This situation happens many time when the molecule in which the central atom is bonded to atoms of different electronegativities through one or more multiple bonds behaves as a Lewis acid, such as $\ce{CO2},$ $\ce{CS2},$ $\ce{SO3}.$ $\ce{CO2}$ molecule acts as a Lewis acid since as the Lewis base approaches to carbon atom, the π-electron density shifts towards the oxygen atom, and carbon accepts lone pairs of a Lewis base.

A lewis acid is defined as an electron pair acceptor. The boron in BF3 is electron deficient & has an empty d orbital,so it can accept a pair of electrons,making it a lewis acid.Also it contains only 6 electrons in outermost shell making it able to accept an electron pair to complete its octet.Thus it's lewis acid

This has nothing to do with d orbitals. I think you mean an empty p orbital, not a d orbital.

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


text:
Intermolecular Forces for Ionic Compound?
So in class we have learned London dispersion, dipole-dipole, ion-dipole and hydrogen bonding for intermolecular forces. Our teacher always uses covalent molecules as examples. So I was wondering which intermolecular forces ionic bonds have, if so, how are they formed?

Ionic bonds are the product of Coulombic interactions within the ionic compound.  Intermolecular forces means 'between molecules' and (unfortunately), in this case, extends to ion/dipoles.

It would be only natural to continue the series "dipole-dipole, ion-dipole" with ion-ion interaction, which is precisely the same as ionic bond.

Ions in ionic compounds are held together by electrostatic attractions, i.e. the idea that "opposite charges attract".
The strength of an electrostatic attraction is given by Coulomb's law:
$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$$
where $q_1$ and $q_2$ are the charges on the two ions and $r$ is the distance between them. In a completely ionic bond, $q_1$ and $q_2$ are multiples of the elementary charge, $e = 1.602 \times 10^{-19} \text{ C}$; however, no bond is completely ionic.

But this doesn't answer the question.....

My first sentence answers your question. I just decided to expand a bit. The fact that opposite charges attract is a consequence of the electromagnetic force, one of the four fundamental interactions of the universe.

So do ionic molecules have intermolecular forces?

I would not say so. Ionic compounds do not exist as discrete molecules anyway, so the term intermolecular loses a bit of its appropriateness. If you talk about a single "molecule" of NaCl, then it should be an intramolecular force because these forces are what hold the constituent ions together. This is why textbooks will never write about the "intermolecular forces in ionic compounds". They will write: ionic compounds are a giant lattice of ions held together by electrostatic attraction.

To me ionic bonding stand alone from both intramolecular force and intermolecular force. As it is a not interaction involving any molecules/atoms.
Other forces such as ion-permanent dipole and ion-induced dipole would be intermolecular force, because those forces involved molecules. Although the fact is that all those 3 kind of bonds involving electrostatic force/electromagnetic force(force between charge particle/poles). It’s just generally ionic bonding stronger than ion-permanent dipole interaction, and ion-permanent dipole interaction stronger than ion-induced dipole interaction.

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


text:
What is the explanation for this stereochemical product?
I have this reaction with a dr = 96:4 but I can't explain the product theoretically. I suppose it is some orbital interaction?

My hyporthesis is that there is a $\pi$-$\sigma^{*}$ interaction between the alcohol-group ($\sigma^*$) that draws electrons out of the $\pi$ orbital of the double bond? I cannot formulate it really. And why the ratio is so pronounced.

No, you’re overthinking this. It’s just sterics.
The iodine will form a iodonium ion (three-membered ring with positive charge on iodine) where the double bond was. Since iodine is huge, it will come from the less sterically hindered side which is from behind the monitor plain.
The carboxylic acid then attacks the iodonium rom above closing a five-membered lactone ring (kinetically favoured against a six-membered one). The side chain carbon retains iodine pointing backwards (you can twist along the $\ce{C-C}$ bond to push the carbon into the monitor plain and iodine backwards) while the lactone oxygen coming from above means that the methyl substituent on the ring must be pointing behind the plain, again.
Also, stereochemistry on the hydroxide should be retained since it does not take part in the reaction.

Scheme 1: A not-well drawn mechanism for the reaction.

Couldn't the hydroxide react with adjacent carbon to give the epoxide intermediate?

@wuschi It could, and that could explain the $4~%$ of the diastereomeric ratio. If the epoxide is formed, iodine will still point backwards, but the carboxylate will also attack from behin the monitor plain meaning both methyl and hydroxyl group will point forwards. Unless of course OP (or source) actually observed $96~%$ of the other diastereomer in which case the argument needs to be turned around.

You said, the iodine comes from the less sterically hindered side, i.e. from behind the monitor. But why would this side be sterically favoured? The two carbons of the double bond are both sp2, they form a plane, so no side is inherently more sterically hindered (and the hydroxy group can be turned away). Am I missing something?

@ste That alcoholic carbon next to the double bond has two further non-hydrogen residues. No matter if you draw it the way I did or have the hydroxyl group in the plain, if the hydrogen is not in the plain it will point backwards thus preferring backward. You can try to turn away the hydroxyl group, but you won’t succeed.


text:
Why is the sneaky method of smothering a grease fire more effective?
In this demonstration, a grease fire is temporarily extinguished if you put the cover on the pan, but reignites as soon as you take the cover off.  However, if you slowly slide the cover across the pan, then the fire stays extinguished.
Why does sliding the cover on make such a big difference?

By smothering the fire your starving it of oxygen. Without oxygen it can't burn.
If you cover the fire and take the cover off quick it reignites because it quickly interacts with the oxygen again because the heat of the oil is still around the same temperature as when it caught on fire.
If you slowly slide of the lid your increasing the time you've starved the fire of oxygen and also allowed the grease to cool slightly so when the oxygen comes back in contact with the grease at a slow rate the grease won't reignite.
The best way to deal with a fire is to prevent it from happening in the first place , make sure you know the boiling point and smoke point of what your using and use a thermometer .


text:
How is hyperconjugation into antibonding orbitals stabilizing?
So I learned that hyperconjugation is electron delocalization from a $\ce{C-H}$ $\unicode[Times]{x3C3}$ bond into an empty p orbital (on a carbocation for example) or an antibonding orbital. I get how it's stabilizing in the case of an empty p orbital, but how is it stabilizing in terms of delocalization into an antibonding orbital, since by definition electron density in antibonding orbitals is destabilizing?

This is nothing specific to antibonding or nonbonding orbitals. There is a very general principle behind it:

Mix an occupied and an empty orbital with each other, and the occupied orbital will be stabilised.

You’re not actually ‘transferring electron density into an antibonding orbital’. You are simply linearcombining two orbitals: one occupied one with a rather low energy and one unoccupied one with a rather high energy. With this linear combination, you receive two new orbitals, where the unoccupied one is further destabilised while the occupied one is further stabilised.
Your electron density is still in the (now more stable) occupied orbital and the antibonding/nonbonding/otherwise empty orbital (now less stable) is still empty. Only their shapes have changed a bit, because we mixed them.

What you just said contradicts the concept of hyperconjugation, because the occupied orbital is not associated with the atom that has the antibonding orbital and is being stabilized. You're basically saying you are gonna destabilize the antibonding orbital even more

@Michael No it doesn’t. It is the concept of hyperconjugation.

if that were true, then the more alkyl groups that are substituted in an alkene,  the less stable the alkene would be because what you are saying that the double bond is more unstable. The opposite is true though the alkyl groups make the enthalpy of the reaction of breaking a double bond in an alkene more positive, not negative

@Michael No, you just got the argument the wrong way around. The (stable and occupied) $\pi$-orbital of the alkene mixes with the (instable and unoccupied) $\sigma^$-orbital of the $\ce{C-H}$ bond. This interaction stabilises the $\pi$ bond and destabilises the $\sigma^$

The electron density increasing in the antibonding orbital only means that the original bond is reduced in strength. Consider propene, the hyperconjugative structure  adds electron density to the $\pi^*$ orbital of carbon. Let us say that the double bond was between C1 and C2.Now the electron density from the alpha C-H bond at C3 moves into the antibonding orbital, thus weakening the C2 -C1 pi bond, and begins to create a double bond between C3 and C2. Thus one bond is weakened and another is strengthened. The stability is only on account of the delocalization, similar to resonance stability.

In organic chemistry, hyperconjugation is the interaction of the electrons in a sigma bond (usually C–H or C–C) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding σ or π orbital, or filled π orbital, to give an extended molecular orbital that increases the stability of the statement.
No of canonical structure= no of hyperconjugable $H$ atom +1
Higher is the no of canonical structures formed Higher is the stability.
It effects are by following ways:
Bond length: Hyperconjugation is suggested as a key factor in shortening of sigma bonds (σ bonds).
The heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene
Stability of carbocations:
$\ce{(CH3)3C+ > (CH3)2CH+ > (CH3)CH2+ > CH3+}$
The C–C σ bond adjacent to the cation is free to rotate, and, as it does so, the three C–H σ bonds of the methyl group in turn undergoes the stabilization interaction. The more adjacent C-H bonds there are the larger hyperconjugation stabilization is.
In chemical bonding theory, an antibonding orbital is a type of molecular orbital that, if occupied by electrons, weakens the bond between two atoms and helps to raise the energy of the molecule relative to the separated atoms. Such an orbital has one or more nodes in the bonding region between the nuclei.
Antibonding molecular orbitals (MOs) are normally higher in energy than bonding molecular orbitals. Bonding and antibonding orbitals form when atoms combine into molecules.
as the spacing between the two atoms becomes smaller, the electron wave functions begin to overlap. Each energy level of the isolated atoms slits into two molecular orbitals belonging to the pair, one lower in energy than the original atomic level and one higher. The orbital which is lower than the orbitals of the separate atoms is the bonding orbital, which is more stable and promotes the bending.
A molecular orbital becomes antibonding when there is less electron density between the two nuclei than there would be if there were no bonding interaction at all. When a molecular orbital changes sign (from positive to negative) at a nodal plane between two atoms, it is said to be antibonding with respect to those atoms.
For example, the He2 molecule, in which both the 1sσ and 1sσ* orbitals are filled. Since the antibonding orbital is more antibonding than the bonding orbital is bonding, the molecule has a higher energy than two separated helium atoms, and it is therefore unstable.
negative hyperconjugation is the donation of electron density from a filled π- or p-orbital to a neighboring σ*-orbital.This phenomenon, a type of resonance, can stabilize the molecule or transition state.  It also causes an elongation of the σ-bond by adding electron density to its antibonding orbital.And hence helping in  delocalization of orbitals.
Delocalizationelectrons are electrons in a molecule, ion or solid metal that are not associated with a single atom or a covalent bond . its also the Distribution of electron density beyond a fixed place such as a single atom, lone pair, or covalent bond via resonance or inductive effects.
For example ,
1.The oxygen lone pairs of ethoxide ion are not delocalized.
2.The oxygen lone pairs of trifluoroethoxide ion are delocalized via the trifluoromethyl group's electron-withdrawing inductive effect.
Negative hyperconjugationis most commonly observed when the σ*-orbital is located on certain C–F or C–O bonds,and does not occur to an appreciable extent with normal C–H bonds.
In negative hyperconjugation, the electron density flows in the opposite direction (from π- or p-orbital to empty σ*-orbital) than it does in the more common hyperconjugation (from σ-orbital to empty p-orbital).
decolalization of electron is though conjujation of $\ce{CH σ-bond}$ with π - orbital or atomic orbital( π - orbital which does not participate in resonance) in hyperconjugation
Increase in delocalization, increases the stability, by decreasing the total (potential) energy.
I hope this will help.

Why does adding electron density to the antibonding orbitals increase stability though? That is what the question is about. Most of this answer is only vaguely relevant and it could also do with using better grammar.


text:
Determining solubility of silver sulfate in its saturated solution
Question:

Calculate the molar sulfate concentration in a solution that is $0.200$ M in $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$.

My attempt:
Looking through old notes and googling lead me to this webpage. Example #2 looked similar to my problem, and my notes from general chemistry involved something called an ICE table to deal with stoichiometry and concentrations of ions. This is what I did:

Write the dissociation equation  $\ce{Ag2SO4 + <=> 2Ag+ + SO4^2-}$
Write the solubility constant formula $$K_\mathrm{sp} = \frac{\text{products}}{\text{reactants}}$$

I was given the value of $K_\mathrm{sp}$ as $1.6 \times 10^{-5}$ and I assume these values are listed in an appendix in my text or online.
Also $\ce{Ag2SO4}$ is supposed to be a solid, I think? Because the question states the solution is "saturated" with $\ce{Ag2SO4}$ and does "saturation" not imply that the $\ce{Ag2SO4}$ is going in and out of solution as a solid and as an ion? And therefore since it is a solid we do not consider it for the equation (not sure why, but this is similar to what I have in my notes)?
$$1.6 \times 10^{-5} = \text{products}$$
$$1.6 \times 10^{-5} = \ce{[Ag+]^2 * [SO4^2- ]}$$
Here is an image if that is acceptable of the ICE table I used (note: $\text{I+C=E}$).
$$\begin{array}{|c|c|c|}\hline
&\ce{Ag+}&\ce{SO4^2-}\\\hline
\text{I}&\pu{0.200M}&x\\\hline
\text{C}&\frac x2&0\\\hline
\text{E}&\pu{0.200M}+\frac x2&x\\\hline
\end{array}$$
Now I substitute the values I obtained for concentrations from the ICE table into my formula:
$$1.6 \times 10^{-5} = [0.200~\text{M} + \frac{1}{2}x]^2 \times [x]$$
Now since $K_\mathrm{sp}$ is less than $1 \times 10^{-5}$ we can ignore the $\frac{1}{2}x$? I guess because it is very small and therefore negligible?
$$1.6 \times 10^{-5} = [0.200~\text{M}]^2 \times [x]$$
$$ \frac{1.6 \times 10^{-5}}{[0.200~\text{M}]^2}  = x = 0.0004$$
$$ x = 4.0 \times 10^{-4}~\text{M}$$
Therefore the concentration of $\ce{SO4^2-}$ is $ 4.0 \times 10^{-4}~\text{M}$?

Solubility constants
You correctly state that they are defined as:
$$K_\mathrm{sp} = \frac{[\text{products}]}{[\text{rectants}]}$$
And in your case ($\ce{Ag2SO4}$) this would be:
$$K_\mathrm{sp} = \frac{[\ce{Ag+}]^2 \cdot [\ce{SO4^2-}]}{[\ce{Ag2SO4}]}$$
And concerning the reactand, the ‘concentration’ of a solid is always defined as $1$, so it is removed from the equation (technically this is an activity not a concentration but that doesn’t matter in this case).
The problem itself
What do you have? Let’s take another look at it:

Calculate the molar sulfate concentration in a solution that is $0.200\,\mathrm{M}$ in:
a) $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$

So we have dissolved $\ce{Ag+}$ with a concentration of $[\ce{Ag+}] = 0.200\,\mathrm{M}$ in aquaeous solution and also a solid $\ce{Ag2SO4}$ precipitate. You also state you are given the value of $K_\mathrm{sp} = 1.6 \times 10^{-5}$. And you are supposed to calculate $[\ce{SO4^2-}]$.
You have all the information you need and all you need to do is solve a single equation.

Did I solve it correctly though?

Yes you did. But you took a long detour ;)


text:
Convert Concentration (particles/ml) to (mg/ml)
I am struggling with converting concentration from (particles/ml) to (mg/ml). I have the concentration for latex particle (particles/ml) and I want to convert it to (mg/ml).
I worked  out the number of moles since I know the the pressure and volume. But I am not sure what to do after that.

First lets see how to get from particles $N$ per volume to mass $m$ per volume.
Dividing the particle number by the Avogadro constant gives moles.
$$
\frac{N}{V}\cdot \frac{1}{N_\mathrm A}=\frac{n}{V}
$$
Multiplying moles with molar mass gives mass.
$$
\frac{n}{V}\cdot M=\frac{m}{V}
$$
Concluding:
$$
\frac{N}{V}\cdot \frac{M}{N_\mathrm A}=\frac{m}{V}
$$
Now your problem arises because you do not know the molar mass. Because you probably have some distribution function of particle size and not just one size. I just assume now that you made an emulsion polymerization in order to get the latex particles or bought them commercially. Then this distribution is very sharp and you can easily take the averaged molar mass as the molar mass of the most probable particle size.

Don't think pressure matters here but if you know the molar concentration you can work out how many g/L which is the same numerically as mg/ml


text:
What do the molecular orbitals of a single water molecule look like?
Some time ago, (a few years back) I heard that the orbitals of a single water molecule were solved analytically (or was it just numerically). What do they look like?
I am interested in the shape (and equations) of those orbitals. There should be an exact solution around and that is what I am looking for (no LCAO or similar please).

@Jan yes, I meant molecular orbitals, but analytical solution. However, it seems that there is none (by 2015). I'll fix the title.

Check them here

If I am interpreting this question correctly, you mean experimentally? You may be thinking of this Nature Materials paper
I'm not sure that solved would be the correct word, but the frontier orbitals of water were certainly imaged with STM.

Wau, cool. I didn't know about the experimental part, but that's really cool. However, I meant the exact analytical solution for the orbitals.  In the nature paper you linked they are comparing the experimental results to theoretical ones. These theoretical ones I am after. There should be analytical calculation starting from Schrödingers equation somewhere. In this paper they use DFT because of the NaCl surface and its an approximate solution (although it probably is pretty close to exact one).

All theoretical calculations are approximate. The exact analytical solutions are available for H-like ions and $\ce{H2^+}$.

Oh, I see what you meant Juha. Ivan is right. There is a nice list of QM systems with analytical solutions here: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions ... Is there a reason why DFT is insufficient for you?

Exact calculations are difficult I admit. I just wondered how far in the exact calculations they were. I even thought helium was solved. Ok, the answer to my question is that there is no exact solution, yet. The experimental result and DFT is then perfectly fine.

@Juha The problem is that the $\hat{V}_\mathrm{el-el}$ operator does not permit an analytical solution; if I recall my quantum chemistry class correctly, it has been proven to be unsolveable.

@Jan: I'd really like to see the prove for that being unsolvable. If you have a link it please add it here.  Or as answer, because this is what I am after.

@Juha Unfortunately, all I can do is forward you to professors who teach quantum chemistry. Mine in this case was Prof. Bräuchle from the LMU Munich but there should be one at every major uni. It is eight years ago for me, after all.

First things first, orbitals are not real: neither atomic, nor molecular. Except for, probably, one-electron atoms (also known as hydrogen-like atoms), where one might still argue that to get the orbitals you need to treat the nucleus as a classical point particle that just creates a spherically symmetric electric potential which can be already considered as an approximation. For many-electron systems, one have to necessarily make the so-called mean-field approximation (plus the Born-Oppenheimer approximation prior to that for the case of molecules).
Second, in principle, no further approximation is done in LCAO-MO procedure; it is a purely algebraic trick of expanding a function over a complete set. Such expansion is in principle exact, but the problem is that a complete set is infinite, so in real-world calculations we have to truncate the linear combination at some point which indeed is yet another approximation.
Thirdly, analytical expressions for orbitals are known for one-electron systems only (such as $\ce{H}$ atom, or $\ce{H2+}$ molecular ion) since the electronic Schrödinger equation was solved analytically only for such systems.
And finally, strictly speaking, it is electron density that was imaged with STM, not orbitals. The fact that it closely resembles the density corresponding to orbitals we get in our calculations justifies the quantum chemistry as a whole: at the end of the day, with all the above mentioned approximations we still get something that is close the reality!

What do you mean by non-relativistic approximation only? Dirac equation for electron in Coulomb potential is solvable exactly, so you can have an orbital for this case too (with all its caveats, but still true).

@Ruslan, honestly speaking, now I have no idea what I meant by saying so. You're right, of course: Dirac equation can be solved for a one-electron system resulting in a set of 4-component wave functions, which are also called orbitals. May be I just meant that these orbitals are different from their usual non-relativistic counterparts, i.e. 2-component wave functions we get when solve the Schrödinger equation. But anaway, currently, this part of answer is wrong, as you noticed.

OK, I removed that wrong statement plus provided a link to Wikipedia article on hydrogen-like atom that mentions, among other things, 4-component solutions of Dirac equation for that system.


text:
Organic Chemistry Inductive and Basic effects
Do all basic groups show +I effect?
Or do they simply exhibit an electron donating tendency which has nothing to do with Inductive effects?
I can't decide if the electronegativity and basicity are completely decoupled phenomena. Does one have an effect on the other?
Or do I always have to consider them separately.
(note: I'm talking about lewis bases not just ones with -OH groups)

The basicity is determined by the amount of -OH ions present in solution, pka values are associated with electronegativit y and net charge.

Can you give an example of these "basic" groups that you are talking about? Functional groups or substituents aren't usually classified as Lewis basic.

@orthocresol really? I heard amines being classified as Lewis bases often enough … or thioethers …

@OP: $\ce{NH2}$ is a basic group but shows a $-\mathrm{I}$ effect. These two have nothing to do with each other. Electronegativity and basicity are also two only distantly related concepts.

@Jan well, it's quite a trivial argument but I never hear the functional group itself being called a Lewis base, it's usually the molecule being called a Lewis base. I imagine OP was referring to -NH2, but I couldn't quite figure it out.

Btw, organic compounds with -OH groups aren't Lewis basic!! They do not generate $\ce{OH-}$ in solution. In fact, phenols and alcohols are weakly acidic.

@orthocresol They can be protonated, so they can act as basic groups, even ring itself can be.

@Mithoron oh dear, I'm being stupid (again)... I should have just said they don't generate $\ce{OH-}$ in solution. Arrhenius acidity/basicity. They can most certainly act as Lewis bases.

@orthocresol I just wanted to know if I have to take electronegativity into account while comparing while comparing different lewis bases. And if I have to, how much of an effect electronegativity has on basicity. Sorry if I confused you.

Don't worry, it's my bad. My brain hasn't really been working recently... Anyway, electronegativity determines the +I/-I effects. As Jan said, basicity often has very little to do with it. Inductive effects are about how strongly the group donates/withdraws electrons from adjacent atoms (usually carbons) in the same molecule, whereas (Lewis) basicity is a measure of how strongly the group donates electrons to an external Lewis acid, such as $\ce{H+}$. Superficially they may seem similar, but the electrons that play a role in these two phenomena are different.

@orthocresol Yeah that makes sense. Thank you! :)
(and cool name by the way)

In Inductive effect due to the EN difference between the adjacent atom or group, polarisation of sigma bond occurs, which causes a partial + and -  charge on them. It is a permanent effect, which is effective upto 4-6$C$ only.
+I: polarisation occurs away from the atom or group, i.e., the adjacent group attached will act as electron pusher or donor or will have EN less than the carbon.
Example
$\ce{CH3-(C=O)OH }$,   $\ce{-CH3}$ will acquire a partial + charge and $\ce{-(C=O)OH }$ will acquire partial - charge.
-I: polarisation occurs towards from the atom or group, i.e., the adjacent group attached will act as electron puller or acceptor or will have EN more than the carbon.
Example
 $\ce{CH3-CH2-NO2}$,  $\ce{-CH2-CH3}$ will have partial + charge and  $\ce{-NO2}$ will have partial -  charge.
As far as Lewis base is concerned its generally an electron pair donor,  which makes it electron pucher most of the time , so, such lewis base would show +I effect.


text:
Work done by an expanding gas, how to solve this particular example?
I just need help with one of these types of questions and I can handle the rest.  It is from a practice quiz on Coursera, Physical Chemistry.

How much expansion work is done on the system when exactly 1 mol of solid ammonium chloride, $\ce{NH4Cl}$, decomposes completely
  to yield gaseous ammonia, $\ce{NH3}$, and hydrogen chloride, $\ce{HCl}$, at
  a temperature of 1280 K. Treat the expansion as
  irreversible and the gases formed as perfect.

I'm not sure how to compute the two volumes required to use the formula:
$$W = n R T \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V}$$
Do I assume $1\text{ atm} = 101.325 \text{ kPa}$.  I did that and was able to use the ideal gas law: $V = \frac{nRT}{P}$ to get the final volume of the 1 mole of gas, but I'm unsure how to choose the initial volume (0 won't work here as it leads to $-\infty$).

Is this the correct approach?
If 1., then how do I also get the initial volume.
If not 1., then how should I approach it?

I think that the question is missing what the pressure is but the formula you are using is correct.

The initial volume is assumed to be 0 as the volume of the solid is negligible to the volume of gas produced.

That was my initial understanding, too. Then I realized that plugging zero initial volume in the formula produces an infinite result.

You have system where 1 mole of solid (0 moles of gas) decomposes into two moles of gas at a constant temperature (given) and likely a constant pressure (not given).
Your formula for work is not correct for this problem. You cannot have $n$ outside of the integral since $n$ is not constant.
Let's write the total differential of the ideal gas equation:
$$\begin{align}
\mathrm{d}(PV)&=\mathrm{d}(nRT) \\
P\mathrm{d}V + V\mathrm{d}P &= nR\mathrm{d}T + RT\mathrm{d}n
\end{align}$$
Since $T$ is constant and $P$ is constant, we know $\mathrm{d}T=0$ and $\mathrm{d}P=0$, so this simplifies to:
$$P\mathrm{d}V = RT\mathrm{d}n$$
Can you do something with this relationship that avoids integrating with respect to (unknown) volumes?

This is kind of a weird question.  Keeping the temperature constant and uniform like that in an  irreversible expansion would be problematic.  But, if you could somehow do it, it would seem that you would also have to assume that the external pressure is kept constant at, say, 1 atm throughout the process to solve the problem.

On second thought, maybe this isn't as weird a problem statement as I thought in my previous comment.  In this particular problem, it doesn't matter whether the temperature is constant or uniform throughout the irreversible expansion.  The only thing that matters is the temperature in the final equilibrium state (1280 K) and the fact that the external pressure is held constant.  This is sufficient to determine the amount of work done by the gas on its surroundings.

I am currently taking this course and now understand this question after seeing my initial mistakes. I have a background in chemistry at the graduate level but this is a thermodynamics question that requires some basic knowledge of undergrad stoichiometry.
One mole of ammonium chloride at $1280\ \mathrm{K}$ decomposes completely to yield gaseous ammonia ($\ce{NH3}$) and hydrogen chloride ($\ce{HCl}$). This expansion is irreversible and the gases formed are perfect, which I think is to say they are not in some equilibrium or reversible state with the original solid.
The first law of thermodynamics is what is needed here and the equation needed is in the form of $\Delta U= q_\mathrm{in} + W_\mathrm{on}$.  Work on $W_\mathrm{on} = -P \times \Delta V$. This is not the reversible isothermal expansion work problem so you do not need the $W_\mathrm{on}=-nRT \ln (V_\mathrm{f}/V_\mathrm{i})$ equation as I saw in some explanations this was eluded to.  Since volume from the ideal gas law equation is $nRT/p$ the equation becomes $W_\mathrm{on} = -p (nRT/p)$.
You also need to know that one mole of the ammonium chloride yields one mole of ammonium and one mole of hydrogen chloride for a total of 2 moles of gas so:
$$W_\mathrm{on}= -p\times 2\times 8.314 \times 1280/p=21283.84\ \mathrm{J} = 21.3\ \mathrm{kJ}$$

I edited your post with MathJax for the chemical and physical formulae. I may have introduced a few errors while doing so because I wasn’t always sure what you meant. Please check if I did it correctly and, if necessary, improve. We have a few introductions to MathJax on meta and the basics can be found in the [help].


text:
Stoichiometry consumed/generated
NaHCO3(aq) + H3C6H5O7(aq) = H2O(l) + CO2(g) + NaC6H5O7(aq)
9 NaHCO3(aq) + 8 H3C6H5O7(aq) = 14 H2O(l) + 3 CO2(g) + 9 NaC6H5O7(aq)
List at least 2 types of reactions that the equation in question #2 can be categorized as?
If 0.950 grams of citric acid  (C6H8O7) reacts
How many moles of sodium bicarbonate will be consumed
How many moles of carbon dioxide will be generated
How many grams of carbon dioxide will be generated
One type of reaction is combustion  What would be the second?
Convert grams to moles
.950 g  citric acid x (1 mole/ 192.123 grams C6H8O7) = .005 moles of  C6H8O7
how many moles of sodium bicarbonate will be consumed
 .005 moles of  C6H8O7 x (9 moles NaHCO3/1 mole of C6H8O7) = .045 moles of NaHCO3
how many moles of carbon dioxide will be generated
 .005 moles of  C6H8O7 x (14 moles H2O/1 mole of C6H8O7) = .07 moles of H2O
how many grams of carbon dioxide will be generated
.07 moles of H2O x (18.015 grams of H2O/1mole H2O) = 1.261 grams of H2O

It's refreshing to see a homework question in which someone has included the work they've done so far, but do you have a question?  There are a number in the problem you've shown, but not one that you appear to be asking us.  In any case, I would start by rechecking the balanced chemical equation you've got.  You should need a 3:1 ratio of sodium bicarbonate and citric acid.

This is a question I have to study for a test   I do not know the type of reaction I have to list two.  I think the first one is combustion but I do not know the second type of reaction.  Also I want to know if my answer is correct of .45 moles of sodium bicarbonate consumed, .o7 moles of H2o produced and 1.261 grams of H20 produced.  If the answers are correct can I leave it or do I have to convert it to scientific notation.  Thank you for your help.

I thought about what you said 3:1 ratio and I changed it to this would this be corrected?
If 0.950 grams of citric acid  (C6H8O7) reacts
.950 g  citric acid x (1 mole/ 192.123 grams C6H8O7) = .00495 moles of  C6H8O7
how many moles of sodium bicarbonate will be consumed
.00495 moles of  C6H8O7 x (3 moles NaHCO3/1 mole of C6H8O7) = .01485 moles of NaHCO3
how many moles of carbon dioxide will be generated
.00495 moles of  C6H8O7 x (3 moles CO2/1 mole of C6H8O7) = .01485 moles of CO2
how many grams of carbon dioxide will be generated   .653 grams of CO2

would above be correct and what two types of reaction?  I think one is combustion

The correct balanced equation would be $\ce{3 NaHCO3 + H3C6H5O7 -> Na3C6H5O7 + 3 H2O + 3 CO2}$.  Your stoichiometry is correct with a corrected balanced equation.  The only reaction type I see here is an acid-base reaction.  I guess your instructor might be looking for a decomposition reaction as well; I'm not sure.


text:
Vapor pressure and boiling point for solutions in open containers
Question 1: If we have an open container which has water in it and is heated. If there is no lid over the container then how can this water ever boil as there will be no vapor pressure created over the water surface.
Question 2: What is the logic in the fact that boiling starts when vapor pressure becomes equal to atmospheric pressure.

related http://chemistry.stackexchange.com/questions/23566/will-the-water-at-the-bottom-of-the-tube-boil-at-the-same-temperature-as-the-wat

as there will be no vapor pressure created over the water surface.  This part is not really true.  There will be vapor pressure created over the water's surface.  It might not ever reach atomspheric pressure, but it will rise as the temperature of the pot of water is raised.  Exactly how will depend on a complex balance of convective and conductive fluid transport mechanisms (e.g. is there a fan blowing across the surface of the water or not).

Boiling is not about the upper surface at all. What happens there is but of little interest to us. Boiling means that bubbles are forming at the bottom and probably within the bulk of the liquid. Guess what is inside these bubbles? It is pure water vapor. That's where the equilibrium between vapor and water actually comes into play.
As for the second question, the bubbles must withstand the external pressure, otherwise they would not be able to form in the first place (or if they will, they will be instantly squeezed into nothing). That's why they actually appear only when vapor pressure becomes equal to atmospheric pressure.

Vapour pressure itself is defined as the pressure that the vapour of a substance apply on the liquid phase of the substance. And you are saying that vapour pressure help those bubbles to come to the surface. How ? Doesn't it sound like a contradiction.

I think OP’s confusion is how to relate Q. 1 and Q. 2. Equilibrium vapour pressure is defined in a closed container whereas for boiling, the vapour pressure of bubble of vapour formed (the pressures exerted on surrounding liquid) must be more than the ambient pressure. This is pretty confusing!

Question 1: If we have an open container which has water in it and is heated. If there is no lid over the container then how can this water ever boil as there will be no vapor pressure created over the water surface.

It's the other way around: in an open container, boiling will not cease provided you (1) continue to heat the solution so as to keep it at the boiling temperature and (2) insufficient liquid boils to allow equilibration of the vapor with the liquid. In the end there is no liquid left, precisely because the vapor never achieves equilibrium with the liquid, with the pressure of the vapor remaining throughout below the saturation vapor pressure.
The second question is nicely addressed in another answer.


text:
Is a quinone ring aromatic?
I'm trying to work out which are the aromatic ring systems in the drug aklavinone:

I am trying to follow Hückel's rules that the compound must be cyclic, be planar, possess a p-orbital on every atom in the ring, and have a π-system with $4n+2$ electrons.
I see that the benzene rings are obviously aromatic. But does the quinone ring (in red) also count as part of the 'aromatic ring system'?

related http://chemistry.stackexchange.com/questions/22100/is-the-aromaticity-broken-in-some-resonance-structures-of-para-nitro-aminobenzen

Strictly speaking, once we have anything more than a single cycle without conjugated side chains, the Hückel's rule does not really apply, and the very definition of aromaticity becomes somewhat muddy. But that's not an answer they expect from you, anyway. So just point to the first and third rings and call them aromatic, and that would be it.
Why not the second? Well, because the quinone fragment is not aromatic. Its bond length are different from each other. These $\ce{C=O}$ bonds are always double, and the $\ce{C-C}$ bonds next to them are always single; you can't draw a resonance structure where that would be otherwise.

But I can ;) Sorry but they are still conjugated.

And more importantly that quinone ring only has $4\ \pi$ electrons.

This is too much of a simplification imho. Either the entire system is aromatic through all of its conjugation, or the conjugation is broken somewhere (cyclooctatetraene). It’s not like the bond lengths of the two phenolic rings be equal whether single or double.

@Mithoron, I'd like to take a look at these. The quinone system is conjugated, I won't deny the obvious, but aromatic it is not.

@Jan, of course my answer contains a good deal of simplification, but I don't agree with your concept either. Surely this whole system is conjugated to some extent, but still we may point out some rings that are "more aromatic" and others that are less so.

@IvanNeretin It's not aromatic but still your last sentence is wrong 'cause it suggests no conjugation.

@Mithoron  So you are judging conjugation based on resonance structures? What kind of structures should those be, say, for butadiene? Something with the double bond in the middle and loose radicals dangling on both ends? Well, something like that you can have here, too.

Whether the ring is "aromatic" or not is of no importance.  You can't ever fit a structure with pendant pi bonds into the "4n+2" rule.  But you can draw molecular orbital structures.  In para-quinone you find an alternant system with four bonding pi orbitals and four antibonding ones.  So the important number is eight pi electrons being the optimal number for the system, not any number of electrons in the ring or whether the ring fits some rule for aromaticity.  And we have that both here and in the parent quinone.

1,4-benzoquinone is NOT aromatic even if it could satisfy Huckel's (4n+2) pi electron rule. The carbonyl pi electrons do not resonate within the ring = no proper conjugation, just cross-conjugation.

That is wrong. And I do not have any more words for that.


text:
How to find sulfites in dried fruits?
I am a student in my last year of high school and currently making a research on how to find sulfites in dried fruits. I've collected data and different methods online and combined few of them into one.
In my method I use hydrogen peroxide and barium chloride to create a white precipitate (result should be barium sulfite).
But the problem is that I never get any precipitate.
During my research it mentioned that I was to leave the dried fruit in distilled water over night (I used dried apricots).
I had 26g, 30g, 40g of dried apricot in 100mL of distilled water.
Once done, I filter out all of the liquids into a different beaker. Then filter it using filter paper. Once done I separate them into different beakers to test. And this is where I find the problem, I've done 6 trials so far, and all 6 have failed. Because I don't know what amount of H2O2 or what amount of the concentration I need to put into the apricot liquid for it to properly react, or for how long I should leave it in for it to react (I've waited 3 and 5 minutes). I'm then supposed to drop a single drop of barium chloride into the solution to produce the white precipitate. But so far I haven't gotten any sign of any precipitate. I've tried different amounts of H2O2 (1mL, 5mL, 10mL, 15mL, 20mL & 25mL with a 35% concentration in the first trials but rest of the trials were with a 3% concentration). Each with a drop of barium chloride.
Seeing that they did not react to a drop, I tried to use 5 drops each on the first few trials. Yet I still got no reaction.
Could I get some help or advice please?
Thank you

Try a spiking experiment: add a small amount of sodium sulfite to a typical apricot and distilled water sample. Process as per your protocol. See if you get a precipitate with the barium addition. If so, try it again, adding an even smaller amount of sodium sulfite. Eventually, you get to the point where no precipitate can be seen to occur. That tells you the approximate  “limit of detection”. It also tells you the sulfited dried apricots do not have much sulfite content. That may be fine: the sulfite is supposed to get oxidized in order to keep the apricot from getting oxidized/stale.

The colorimetric method using rosaniline and formaldehyde is more sensitive and it is well documented.
A somewhat safer version of the test (avoiding use of highly dangerous $\ce{HgCl2}$) is documented at MDPI - Open Access Publishing:

"A concentrated pararosaniline stock solution (3.088 × 10 −3 mol/L) was made by dissolving pararosaniline hydrochloride in a 10% aqueous ethanol solution. The concentrated stock solution was used to make... [a] working solution contain[ing] pararosaniline hydrochloride, hydrochloric acid (HCl) and formaldehyde in the molar ratio of 1:1255:125 respectively. For example, a working solution with PRA concentration of 4.68 × 10 −4 mol/L in 25 mL contained 3,790 μL of concentrated PRA stock (3.088 × 10 −3 mol/L), 1,439 μL of 32% HCl, 189 μL of 36% formaldehyde solution and 25 mL with water."

Pararosaniline is available from Sigma-Aldrich, though it seems a bit pricey at ~$US82/25 g; perhaps your school would obtain some for use by yourself and by other students, as you would need perhaps a gram for your experiments.


text:
Guidelines for Solubility in Water of Organic Compounds in Relation to Carbon Chain Length
Is there a way to generalize the solubility of any organic compound based on their functional groups and the length of their carbon chains? I know that beyond 3 carbons, alcohols are insoluble, and beyond 4 carbons, carboxylic acids are insoluble. But what if the alcohol had 2 hydroxyl groups? What if the carboxylic acid had 2 carboxyl groups? How many carbons would they each need for the compound to become insoluble in water again? Are there any quantitative guidelines for determining this?
And what about the solubility of esters and haloalkanes? How many carbons does an ester and haloalkane need to have for them to be insoluble? Again, what if there are multiple ester or halide groups?
Lastly, how do you determine solubility when you’ve got a mixture of functional groups? Take aspirin for example, whose structure contains a benzene ring of 6 carbons, a carboxyl group attached to one corner, an ester group attached to another corner, and one methyl group attached to the ester. How would you systematically determine if this compound was soluble in water or not?

(Image credit: http://www.rxlist.com/images/rxlist/aggrenox2.gif)

It's much too broad. And there's not much point - in practice experimental data are used.  More polar groups can offset non-polar, but they can interact with each other and steric effects can be important.

There are some situations where you need a way to estimate a chemical's water solubility--for example when EPA reviews a new chemical premanufacture notice. The submitter may not provide data on solubility and the substance itself is not available. So you use software tools to develop an estimate. Some of the tools use a group contribution approach as you suggest in the question.

To determine the solubility of some compound, a chemist would systematically throw it into water and see what happens. You may draw some correlations like "plus six for two hydroxyls, minus five for five carbons, should be soluble"; they are somewhat useful, but mostly qualitatively.
There is a whole hierarchy of computational methods for prediction of solubility, but these are highly complicated.
BTW, haloalkanes are almost as insoluble in water as alkanes themselves.


text:
Hamiltonian for a two electron system
What is the hamiltonian for an atom with two electrons and a nuclear charge $Z$?
The Hamiltonian always takes the general form: $\hat H=\hat T + \hat V $
The kinetic energy of each of the electrons needs to be taken into account so: $\hat T=-\frac{\hbar^2}{2\mu}(\nabla_1^2+\nabla_2^2)$
The potential operator is where I am finding trouble. There are two terms for the attraction towards the nucleus - one for each electron: $-\frac{Ze^2}{4\pi\epsilon_0r_1}-\frac{Ze^2}{4\pi\epsilon_0r_2}$. However, the potential energy involved in the electron electron repulsion needs to be taken into account. My question is: are there two terms for this? To clarify do I need to calculate the increase in potential of electron 1 caused by electron 2 AND the increase in potential energy of electron 2 caused by the presence of electron 1? Or would adding: $\frac{e^2}{4\pi\epsilon_0r_{12}}$ take into account all of the potential energy involved in the electron electron interaction?

Potential is a characteristic of the system as a whole, and not of individual electrons. There are three charges, hence three pairwise interactions. Two of them, namely nucleus-electron1 and nucleus-electron2, you took into account once. Why in the world would you want to add third such value twice?

The electron-electron repulsion potential is an interaction potential, therefore it has one term per interaction. If you have two electrons, you will have only one term, representing the interaction between them. The number of potential terms will be, in your case:

Attraction between electron 1 and the nucleus (one interaction)
Attraction between electron 2 and the nucleus (one interaction)
Repulsion between electron 1 and electron 2 (one interaction)

In total, you will have three interactions and therefore only three potential terms (two for electron-nucleus and one for electron-electon interactions).
In general, an interaction potential that only depends on the distance of two particles can be written as
$$
U = \sum_{i<j} u(|\vec{r}_i - \vec{r}_j|),
$$
where $u$ represents the interaction between only two particles. You see that for two electrons you will have only one term, since $i<j$. For three electrons you will have three terms (interaction between 1 and 2, between 1 and 3 and between 2 and 3) and so on.
Maybe your confusion comes from the fact that, sometimes, this interaction potential is written as
$$
U = \frac{1}{2}\sum_{i\neq j} u(|\vec{r}_i - \vec{r}_j|)
$$
In this case, for two electrons, you will have two terms. However there is the factor $1/2$ which cancels out the double counting, since
$$u(|\vec{r}_i - \vec{r}_j|)=u(|\vec{r}_j - \vec{r}_i|).$$


text:
Is 2-oxobicyclo[2.2.1]heptan-1-ide resonance stabilized?
Is the following compound resonance stabilized?

I think its not resonance stabilized because the p-orbital of carbanion and pure p-orbital of carbonyl carbon are not in same plane. So they can't overlap. (Pure means that the p-orbital is not a part of the $\ce{sp^2}$ orbital. One p-orbital of carbonyl carbon is pure).
Am I correct?

Makes sense to me.

Let’s start by saying that the anion is not centred in an $\ce{sp^2}$ orbital but in an $\ce{sp^3}$ one. But yes, you are correct.

I think, in contrast to cyclohexanone, a resonance stabilization by

is less likely to happen as you already start with a bi-cyclic compound where the cycles involved are small to moderate. The then additional strain introduced by a (partial) double bond at the bridge-head carbon atom would destabilize the molecule too much.  I extrapolate here from Bredt's rule.

The reason for this is, as Buttonwood suggested, the strain induced by the bicyclic norbonyl system. I already hinted in the comments that the anion is in a $\ce{sp^3}$ type orbital. What makes it impossible is that that orbital is almost orthogonal to the carbonyl’s $\unicode[Times]{x3C0}$-bond. I attempted to draw two pictures below, one being the complete, the second a Newman projection along the anion-carbonyl carbon bond. The exact angles might be different ones, but these visualisations can serve as a starting point.

No,it's not resonance stablised, the reason is , the bridge head carbon with negative charge is sp3 hybridised and has tetrahedral geometry but for overlapping with parallel orbitals of carbonyl  needs parallel p orbital  and it's not possible due to its tetrahedral geometry.


text:
Is a pyrylium ring aromatic?
I am trying to figure out which rings in callistephin are aromatic:

I know that the benzene rings definitely are aromatic, but I'm not sure about the ring with the positively charged oxygen (highlighted in red).
How does the Hückel $4n+2$ rule apply to this ring?

If you take both the rings together, you get a hetero-naphthalene which is aromatic. Other than that you should have asked your former question in a more general way that would have allowed answering both in the same way.

@Jan can I ask how many pi-electrons are in this hetero-napthalene?

10, just like in normal naphthalene. (Substitute the $\ce{O+}$ with $\ce{CH}$ and you get naphthalene; that was what I was playing at.)

Pyrylium is aromatic. Following Hückel's rule, pyrylium is uninterrupted (no $\mathrm{sp^3}$ hybridization), all atoms have p-orbitals and there is 6 π electrons. the 6 π electrons indicate that Hückel's rule $(4n+2)$ is agreeable for pyrylium, $n = 1$.

Hückel's rules only apply to monocycles.


text:
How could chlorine dioxide be stabilized in aqueous solution?
There is a water-based mouthwash product which is advertised as having the ingredient "stabilized chlorine dioxide." The only other listed ingredients are trisodium phosphate and citric acid. I assume this is a buffering solution or something.
In the company's documentation, it says it contains

a stabilized molecular form of chlorine dioxide.

But in another location, the company claims that

the solution is naturally activated by the amino acids in saliva,
  creating a small and very safe amount of chlorine dioxide.

So that sounds to me like there's actually something else in the solution, and the that the $\ce{ClO_2}$ is being produced ("activated") during use.
I've done some reading, and I have learned that $\ce{ClO_2}$ can be produced in several ways, but all of these methods seem to involve unpleasant reagents like chlorine gas, HCl, or oxalic acid.
I'd like to know what actual compound might be referred to as "stabilized chlorine dioxide" and what reactions might be involved.
I'm not asking about the effectiveness of the product, or what value $\ce{ClO_2}$ might have as a disinfectant, I'm just curious as to whether the claims made about the ingredients are plausible, and if so, what reactions might account for this?

Your assumption is correct. The term “stabilized chlorine dioxide” is misleading since the solution does not contain relevant amounts of chlorine dioxide $\ce{ClO2}$.
The solution actually contains sodium chlorite $\ce{NaClO2}$, which is the salt of the unstable chlorous acid $\ce{HClO2}$. At low pH, chlorite solutions contain chlorous acid
$$\ce{ClO2- + H+ <=> HClO2}$$
which decomposes to form chlorine dioxide approximately according to
$$\ce{5HClO2 -> 4ClO2 + HCl + 2H2O}$$
or
$$\ce{4HClO2 -> 2ClO2 + HClO3 + HCl + H2O}$$
At higher pH, decomposition of sodium chlorite slows down. Therefore, sodium chlorite is mainly stabilized by using a buffered solution. In your product, the buffer consists of trisodium phosphate and citric acid. Thus, a better description would be “stabilized sodium chlorite” instead of “stabilized chlorine dioxide”.

Thanks for the explanation.  Can you shed any light on how amino acids would serve to activate this solution? Is that likely? I'm wondering if someone saw "amino acids" and assumed it's an acid so it lowers pH or something.


text:
What is a sigma complex?
According to Wikipedia, "sigma complex" refers to an arenium ion.
The suffix "-ium" denotes a positively charged ion.
So, it is correct to call the resonance structures in the below picture depictions of a sigma complex? It's distinctly anionic rather than cationic.

Yes, it's anionic sigma complex, more precisely Meisenheimer complex. They are much more stable then arenium ions (Wheland intermediates) and can be isolated much more easily.
There was actually synthesized a sigma complex which is a zwitterionic combination of both Meisenheimer and Wheland complexes, which shows how well they can be stabilised:

(picture created by Wikipedia User:V8rik)
The term "sigma complex" has also other meanings. Organometallic compounds with sigma bonds connecting with aromatic rings are called like that, too. There are also sigma complexes of cyclobutadiene with aluminium chloride.
Generally, molecules are called sigma complexes to stress the difference between them and pi complexes which can also be created from cyclic annulenes. First ones are bonded using sigma bonds, the other ones use pre-existing pi orbitals.


text:
Why do we take the compressibility factor equal to one when graph is plotted?
I saw the graph of compressibility factor against pressure I was confused to see that all the gases are started by 1 unit of conpressibility factor when pressure is zero. So why do we take the compressibility factor equal to one when the graph is being plotted?

I think that at $p = 0$ the compressibility factor ($Z = pV_m/RT$) itself is not particularly well-defined since $p = 0$, $V_m = \infty$ and $T = 0$. Zero pressure is not a state that we can reach.
However, if you use any equation of state and take the limit as $p \to 0$ (or $V_m \to \infty$, which is equivalent), you will find that $Z \to 1$. Try it with the virial equation of state:
$$pV_m = RT\left( 1 + \frac{B}{V_m} + \frac{C}{V_m^2} + \cdots \right)$$
This happens with all gases because the individual gases are identified by their virial coefficients $B, C, \ldots$ and the terms involving these constants vanish as $V_m \to \infty$.
A simple qualitative explanation is that in the limit of $p = 0$ all gases behave ideally; for an ideal gas, $Z$ is of course equal to $1$ because of the ideal gas law $pV_m = RT$.


text:
Is the heat of reaction always negative?
For example, I have a question below that I know the correct answer to, but I don't know why it's positive rather than negative.

When $\ce{NaOH}$ is neutralized by $\ce{HCl}$:
$$\ce{NaOH(aq) + HCl(aq) + H2O(l) -> NaCl(aq)}\qquad{\Delta H = \pu{-56.2kJ/mol}}$$
How much heat is produced when $\pu{195 mL}$ of $\pu{0.667 M }\ce{NaOH}$ are mixed with $\pu{125 mL 1.750 M }\ce{HCl}$?
a. $\pu{−14.6 kJ}$
b. $\pu{14.6 kJ  }$
c. None of these.
d. $\pu{7.31 kJ}$

I know that the answer is (d), but I'm not sure why it is correct. I thought the correct answer was $−7.31$, but there was no answer choice for $−7.31$.

I suspect that your instructor was trying to include the semantic difference between enthalpy change and amount of heat produced.  It's kind of like if you asked what the temperature was and I told you -10°.  You could then ask how many degrees below zero it is, and I would answer 10°.  Both answers mean the same thing, but the negative sign is included verbally rather than arithmetically.

An enthalpy change of -7.31 kJ means that this is the amount of heat you need to remove from the system so that the final temperature of the system  is equal to the initial temperature.  So that must be the amount of heat produced.

@Jason is entirely right. An alternative way of putting it is: You can have a negative enthalpy change (ΔH), and you can have a negative bank balance. But you can't produce a negative amount of heat, just like you can't owe someone a negative amount of money. The negative sign of ΔH is already implied by the word "produced", because if ΔH were to be positive, the appropriate word would be "absorbed".

I want to restate my previous comment because, as written, it's a little ambiguous.  When we say that the enthalpy change of a reaction is -7.31 kJ, it means  that we have to add -7.31 kJ to the system (i.e., remove +7.31 kJ from the system) in order for the final temperature to be equal to the original temperature.  So, if we have to remove 7.31 kJ, this must be the amount of heat produced by the reaction.


text:
Spectroscopy and Excitation of Electrons
As metals are heated, they glow white and then blue. How might you explain this change in color as the temperature of the metal is increased?
My answer consist of the idea that with increased thermal energy more electrons are excited and as they decay they emit wavelengths of the color blue. I am not sure how to connect the idea of white light to this answer because white light consist of all the wavelengths of the visible spectrum.

Do you think that all the electrons in the metal are excited to the same energy?

I assume it is a homework question...
You may want to into the concept of blackbody radiation, as an explanation. It is discussed in most PChem / Quantum Chem books in the first chapters. Feel free to update your question according to.
Generally, we do not say that photons are excited. Photons do not have ground and excited states in the same chemical sense as electrons. Excited states (of e.g. electron) may emit photons during de-excitation, but those are the excited state of the electron (before the photon even existed) not the photon.
Also, we generally do not say photon decay: technically you can argue that absorption is something like that, but it is definitely not a common way to phrase it, and obviously absorption is absorption, so it will not emit any radiation.
Note: After this answer, the question was changed to talk about electronic excitation instead of "photonic excitations", therefore a large part of my answer is irrelevant.


text:
Bromination of alkenes
In regard to the mechanism of halogen addition, consider the situation when we add a halogen (say, bromine) to an unsymmetrical alkene about the double bond (say, propene) in the presence of $\ce{CCl4}$.
I knew the mechanism would be something of this sort:

A cyclic bromonium ion would be formed with the two carbons of the double bond. The second bromide ion (acting as a nucleophile) attacks on the more substitued carbon from the side opposite to the bromonium ion to cause ring opening and formation of a vic-dibromide.

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.
I have two questions in this regard:

Which of the two is the correct mechanism?
If the former is indeed the correct mechanism, then would the nucleophile  attack the least substituted carbon? (seeing as how the partial positive charge on the more substituted carbon is nuetralised to a greater extent due to $+I$ effect of attached methyl/ethyl groups).

If only ethene forms a cyclic bromonium ion, then there would be no stereospecificity in the bromination of the 2-butenes. Secondly, the active brominating agent is Br3+ Br-.

The first mechanism is more likely unless specific conditions of the experiment dictate otherwise. In the presence of $\ce{CCl4}$, the addition of $\ce{Br2}$ takes place via the formation of a cyclic halonium ion as intermediate as follows:

That's a bromonium cation and fishing hook arrows are for radical mechanisms.


text:
Caffeine dissolution
I am looking to add caffeine powder in pure form to water that has a pH level of 7 or higher. What would the pH level be if I do this? Is Caffeine itself acidic?

Possible duplicate of Best consumption-safe liquid for caffeine solubility?

The possible duplicate does not ask the same question at all.

Question is the diffrent but the answer is broad enough to answer also this.

Caffeine is not acidic. In fact, it is basic. It will raise the pH of water to some degree. It is a weak base, so the free base will exist in equilibrium with its protonated counterpart.
Due to its basicity, it will be more soluble in low pH aqueous solutions than in high pH solutions. This is because in low pH solutions, caffeine would exist completely as a protonated salt, making it more hydrophilic. Despite this, caffeine is very polar and is still quite soluble in water at higher pH levels.
The six-membered pyrimidinedione can exist in an aromatic, zwitterionic form where both amide nitrogens have formed double bonds to respective adjacent carbonyls. This six-membered ring had been observed as being planar. Therefore, it has been concluded that caffeine does indeed exist primarily as its zwitterionic resonance form, with two positive charges and two negative charges. The negative charges exist on the two carbonyl oxygens. These may be slightly basic, too, but significantly less so than a structurally similar hydroxyphenol (e.g. resorcinol) due the electronegative nature of the attached aromatic ring. Therefore, the basic imidazole nitrogen likely has largest effect on the pH.
The fact that there is possibly more that one weakly basic atom on the caffeine molecule complicates calculations as far as determining a theoretical pH. Also, the calculation would require the quantity of caffeine you will add. If you need an exact number, you will have to measure it. You can be sure, however, that the pH of the water will rise as caffeine is added.

I was previously corrected about caffiene being a zwitterion.  Apparently it is not and due to there being no removeable proton it does not form a trivial anion.

There are ways for things to be zwitterionic without a proton migration -- in this case, it is through resonance.

I appreciate you explaining that @user23444 I will research further.  I originally stated it had zwitterionic properties in my answer below but if you read the comments I was corrected that it was not a zwitterion.  I had used my notes from caffeine isolation from tea leaves as the resource but I couldn't find anything to support my zwitterion ion comment directly so i altered the answer and stood corrected.  My notes were correct in the first place. Should I change my answer back to what it was ?

I personally think that would be reasonable. I should put a picture in showing the aromaticity and the dominant resonance form. I will when I get back to my desktop.

Amide bond isn't zwitterionic! There are only partial charges. Also mesomeric forms don't exist, but are a way to describe molecule when one lewis structure is bad.

A typical amide bond is not zwitterionic, but in the case of caffeine, the increased stability of the aromatic six-membered ring pyrimidine results in increased electron density on the oxygens leading to a substantial negative charge on the two oxygens

Caffeine (1,3,7 trimethylxanthine) is not acidic by itself in its freebase form it is an alkaloid which behaves as a base, although it can be ionic or non-ionic. Resonance or mesomerism allow resonance structures to form due to delocalized electrons within the compound, because of this caffeine can be a zwitterion.
In water ph 7》caffeine will be in its freebase form so the solution will remain alkaline .
Use a ph indicator to determine the pH before and after the addition of the caffeine and analyze the change in pH.

Caffeine is not zwitterionic.

@Mithoron , in the answer above it is stated that caffeine can form an an anion without proton migration through resonance? What are your thoughts on this?

@Jan , what are your thoughts on the above statement also?

Caffeine cannot form a macroscopic anion, since there is no proton that can be abstracted. What the other answer meant (and was badly phrased) is that a resonance structure where two nitrogens each bear a formal positive charge and two exocyclic oxygens a negative charge each (for an overall neutral molecule) may be the better description of caffeine. However, that would be a formal zwitterion, not an anion. (CC @Mith ).

@Jan , so is my answer ok the way it is I have become confused?!

Your answer still says ‘caffeine […] has anionic and ionic forms’. Apart from the weird duality (why mention anionic forms separately from ionic froms?) it is wrong to say caffeine can ever have an anionic form. Maybe you meant non-ionic, but that’s not what is written there.

So is it classed as an ionic compound that acts as a base in solution?

@Joel Why would an anionic particle be attracted to an anion? Same charges repulse one another; anions are attracted by cations not by other anions. Also, the freebase is generally classed a neutral molecule that acts as a base.

Let us continue this discussion in chat.


text:
Green Tea and Caffeine
Is the caffeine from green tea that is extracted acidic?
This goes for powder and liquid form.
If I add Alkaline water (pH: 7 or higher) will it change?

The caffeine is in a free base form unless prepared otherwise . Caffeine is an alkaloid and a base , slightly alkaline. If you add base like naoh to it the solution will become more alkaline.

Do you know where I can purchase the items in order to do a home test?

Do you mean to test the pH or extract the caffeine?

where can I purchase both of those items?

They sell bromocresol purple or bromothymol blue as fish tank or soil ph testers. To extract the caffeine is more technical. A non polar extraction using Dichloromethane is how I achieved this in a laboratory I don't recommend to do this anywhere else. Pharmaceutical grade caffeine can be bought online or if you just want to know the result download an msds or call a company that sells the product and ask them the technical data.

Thank you, that was very helpful. I don't know how to extract it myself, so I will buy it in pharmaceutical grade.

No worries glad I could help you out :)


text:
How high in (mmH2O) is the column of water in the tube?

A long glass tube, sealed at one end, has an inner diameter of 23.4 mm. The tube is filled with water and inverted into a pail of water. If the atmospheric pressure is 735 mmHg, how high (in mmH2O) is the column of water in the tube (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL)?

This problem really confuses me, I know I have to use the ratio of heights of the liquid columns is inversely related to the ratio of the densities of the liquids:
hH20/hHg = dHg/dH2O. I just can't seem to put it together. I would appreciate any help on this problem.

Consider the air pressure in mm of mercury given. Is water more or less dense? Would an amount of water of the same mass as the mercury in the column take up more or less room?
So what would you need to do to change the 735 mm of mercury to an equivalent mass of water, given the densities cited in the problem?
The diameter of the column is red herring; don't go off on the kipper.


text:
How is this sugar classified as an aldose?
I have made the following image in paint, so forgive me if it isn't the most appealing sugar. I have a hard time visualizing how the following sugar is an aldose.

My speculation for the Fischer projection of the alleged carbohydrate is:

My speculation is as a result of the lack of primary alcohol at the end of traditional carbohydrates like D-glucose for example means that there is not carbon group alpha or beta to carbon 4 in the furanose ring. The fact that there is still a CHO group on Carbon 1 would make this sugar an aldose? Is this logic correct, and is that the Fischer projection for the sugar shown above in the Haworth projection?

The middle carbon in your Fisher projection should have $\ce{HO}$, not $\ce{OH}$.

Yes its an aldose it is classified as an aldose because of the aldehyde group (cho). The compound is actually an aldotetrose called L-Threose. Try using acd Chem sketch program for the projection it is a free download for the freeware version...


text:
Is a polycarbonate beer mug appropriate for serving alcohol?
Are polycarbonate & acrylic beer mug, whiskey glasses safe for serving alcohol? I mean does it leach or dissolve? What are the harmful effects in long run if any?

It should be fine if used correctly! Most drinking alcohol wouldn't be strong enough to denature the polycarbonate which is very resistant anyway hence why they make whiskey/beer cups out of it. Some moonshine might warp it! Straight spirits may over time start to dissolve the polycarbonate. The harmful effects would be intoxication I'd say and maybe alcoholism !...

There might be some question about using polycarbonate for any food contact: bis-phenol A (BPA), used as a plasticizer for a number years, has been linked to "possible health effects of BPA on the brain, behavior and prostate gland of fetuses, infants and children." Alternatives to BPA, such as bis-phenol S (BPS), may be just as toxic.
So, in short, adults probably have low risk from polycarbonate vessels and utensils, but you might prefer to use an alternative. Acrylic plastic, however, tends to craze from solvents, stress and even normal washing. Polyethylene, polypropylene and other plastics might be a better choice.
BTW, much of the BPA with which we come in contact is from the thermal printing paper used for cash register receipts. If you handle these continually in business, you might consider gloves or a barrier cream.


text:
Convert benzene to isopropylbenzene

Which of the following reagents will convert benzene to isopropylbenzene? Choose all that are appropriate.

I know for certain that (b) can be used, but I also think the following mechanism with (d) is also possible:

Is my line of thinking correct with this one?

You are right, both abstracting $\ce{Cl-}$ from the alkyl chloride with Lewis acid as well as dehydration of isopropanol with $\ce{H2SO4}$ are suitable ways to generate the 2-propyl carbocation, your electrophile in this reaction. Option (c) will not work because $\ce{AlCl3}$ reacts with alcohols to form aluminium alkoxides.
Source: Master Organic Chemisry

Out of curiosity: wouldn't d) produce a significant amount of $\ce{PhSO3H}$ ?