text:
How could scientists produce accurate chemical structures before they could capture images of molecules?
How were scientists able to produce these accurate structural diagrams long before scientists were able to capture images of the actual molecules?
Reference: Scientists capture first images of molecules before and after reaction

It should be noted that those molecules are being resolved at low resolution (we see 7 units in a molecule with 40 atoms) -- so much of the molecular structure is still being inferred the old-fashioned way.

Technically, this is a duplicate of another question, but the answers to that question only address chemical composition rather than structure...so I wouldn't say this is a true duplicate. Anyway, the first step to determining structure is to determine composition, so this other question provides part of the answer: https://chemistry.stackexchange.com/questions/771/how-were-elements-and-molecular-formulae-determined-in-the-old-days?rq=1

They ran chemical reactions on the molecule and identified the fragments.  Then like a jigsaw puzzle, using additional information derived from other chemical reactions, they pieced the molecule back together.  Some Nobel prizes were awarded for such structure determinations.


text:
Orbital Electronegativity Considerations in Resonance Structures
After consulting three organic chemistry texts (Wade, Klein, and Brown), only one actually clearly delineates a few guidelines for picking the most significant resonance structure. The others mention the same factors (maybe less one or two). So in sum, the important factors to consider are:
1) A filled valance shell is preferable to a partially filled valance shell.
2) The more covalent bonds the better.
3) Least separation of oppositely signed charges. I'm thinking this is a Coulombic effect. Some books imply formal charge through the term "charge" but what about partial charges (which are much more reflective of true charges than formal charges. Which one should I really be looking at? Least separation of opposite formal charges or opposite partial charges?
4) Negative charges should generally be assigned to the more electronegative elements; the opposite is true for electropositive elements. Again, formal charge misleads - consider the ammonium ion (positive formal charge on nitrogen; negative partial charge on nitrogen). Again, one book implies formal charge but really, isn't looking at partial charges a bit more informative?
5) Unmentioned factor in any of the three textbooks: What about hybridization? If we have a molecule with two nitrogens, one $\ce{sp}$ hybridized and one $\ce{sp^2}$ hybridized but otherwise identical, shouldn't we also take into account "orbital electronegativity" - i.e. the ability of hybrid orbitals with higher s-character to better stabilize negative charge (and conversely higher p-character better stabilizes positive charge)?
6) Unmentioned factor in any of the three textbooks: Induction? Shouldn't induction also play a role in resonance stabilization? Resonance delocalizes electron(s); can't induction further stabilize electrons? debunked.

Inductive effects can weaken resonance structures by localizing the electrons (observe the effect of various substituents on aromatic rings).  You can't have resonance on $sp^3$ hybridized molecules

Good points. I'll edit the text to reflect your catch! I should probably have written sp2 hybridization or something! And I should have caught that part about inductive effects localizing reaction; I just had a question answered earlier today about enols. I learned that the hydroxyl group of enols concentrates electron density in the C=C double bond adjacent to the alcohol group.

Both ammonia and ammonium ion have nitrogen with a formal negative charge (to balance the positive hydrogens).

Not the way I calculate formal charge; ammonia should have 0 formal charge and ammonium ion, +1 formal charge.
http://en.wikipedia.org/wiki/Formal_charge#Examples

You can have resonance involving $\ce{sp^3}$ hybridized carbon, it's called hyperconjugation.

Thank you for the comment! Another good term to research!

What is the question here?

As far as hybridization goes, the hybridization of all of the atoms must be the same in all resonance structures. In considering resonance structures, none of the nuclei should be moved in space. Since hybridization is designed to explain the arrangement of nuclei/electrons in space, hybridization cannot change between resonance structures, otherwise nuclei may be forced to occupy new positions.
In practice, this means that certain atoms may appear to be sp3 hybridized but with consideration of resonance structures actually should be treated as sp2. This often means that a lone pair is not in an sp3 orbital as expected, but an unhybridized p-orbital.

Thank you for this insight; I keep forgetting that resonance structures are not discrete!

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text:
Fourth principle of Molecular Orbitals
The fourth principle of Molecular Orbitals state that:
Molecular orbitals are best formed when composed of Atomic orbitals of like energies.

I'm not sure about how I should interpret this; which is the correct?

It means that the molecular orbitals that I need to consider are the ones formed by the same orbitals on each atom (like a pair of $2p$, two $3s$, etc).
It means that I need to evaluate the actual energy on each orbital using: $$E = \frac{-13.6 z^2}{n^2}$$ And then compare the energies of every orbital, to match each orbital on Atom 1 to the closest one on Atom 2? Meaning that I could possibly have a $2s$ pairing with a $3p$, depending on both atoms? (Note that the formula gives different results for each Atomic Orbital depending on the atom's $z$).

I known that every possible orbital combination exists, but I'm asking about the ones that are more relevant and thus need to be considered.

Both of the statements have merit. When the orbitals are of the same type, the overlap is maximized. When the orbitals are nearly the same energy, the atoms are closer together (and give better overlap) than when the energy difference is larger. You don't really need to calculate the energies of the orbitals, just look at the size of the ions. The larger the ion, the higher the energy level of the bonding electrons.

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text:
Is this textbook bonding picture of hybridization in ethene wrong?

Regarding the above picture, why is the $\ce{C-C}$ $\sigma$ bond composed of two $\ce{sp^2}$ orbitals while the $\ce{C-H}$ $\sigma$ bonds composed of $\ce{sp^3}$ hybrid orbitals? I don't see why; I thought the geometry around the carbons was trigonal planar. Why does hybridization differ across atoms in the molecule?
I'm thinking the textbook authors made a mistake; yes, there are four $\ce{C-H}$ bonds, but this does not mean there is $\ce{sp^3}$ hybridization because there are two carbon atoms, each with two $\ce{C-H}$ bonds and one $\ce{C-C}$ bond. Perhaps the authors mistakenly took there to be one carbon and four $\ce{C-H}$ bond (i.e. methane.) That would be $\ce{sp^3}$ hybridization.
Or perhaps I'm wrong altogether. Please enlighten me!
EDIT: Saw this in another text:

You are not mistaken. Your book is wrong. If one desperately wants to involve the hybridisation concept (which makes sense for carbon), then the description should be in accordance with the molecular geometry of ethene:

So the carbons can be described as $\ce{sp_{x}p_{y}=sp^2}$ hybridised, hence there are four $\ce{H_{s}-C_{sp^2}}$ $\sigma$ bonds, one $\ce{C_{sp^2}-C_{sp^2}}$ $\sigma$ bond and one $\ce{C_{p_{z}}-C_{p_{z}}}$ $\pi$ bond.
However the concept of hybridisation is no law, but yet another tool to describe bonding. In case of methane, (ethane,) ethene and ethyne it is quite simple to come to certain $\ce{sp^{x}}$ hybrid orbitals where $\ce{x}\in\mathbb{N}$ - different molecule may lead to fractional $\ce{x}$. (Not a rule, just a guideline)

In response to Dissenters comment:
The coordinate system is chosen arbitrarily to indicate the molecular plane to be $xy$, as indicated in the picture taken from the german article of wikipedia:

As the π orbitals have to be perpendicular to this plane they only can be composed of $\ce{p_{z}}$ orbitals.
The molecular plane can be chosen different, e.g. as the $xz$ plane, the the $\ce{sp^2}$ orbitals would be formed from $\ce{sp_{x}p_{z}}$ and the $\pi$ bond would consist of $\ce{p_{y}}$ orbitals.

Thank you. Do we have to use the px and py orbitals in that order in hybridization (you wrote spxpy).

Okay! And if I understand correctly the z-axis is taken to be up/down (not up/down like the y-axis in 2D) but up and down like standing up and sitting down.

And yes, I have to involve the hybridization concept since it is emphasized in all the introductory level organic textbooks. You might be appalled at the level of qualitative analysis these organic textbooks start out with (having us draw cartoons of molecular bonding; mixing and matching aspects of valence bond theory with molecular orbital theory, etc.) I guess authors are just trying not to let the math take away from the chemistry at this point.

@Dissenter The hybridisation concept makes a lot of sense for carbon in most cases. This is talking about simple organic molecules - just be a little critic about the concept (your question shows that you are) and learn where there might be the limits of that concept. I am not opposed to using hybridisation as a tool - however, the bonding description could be accurately obtained without it (It would not be such a nice picture though). I have added some more information to the answer, I hope that further clarifies the plane I have chosen.

Thank you. What other ways to describe bonding are there?

I suggest you ask another question for that case, as that would be much to long for a comment...

Okay! I will pursue this question then later! Thank you very much again for your helpful description!

Martin has provided an excellent answer to your question.  And although I've arrived late at the party I'd like to comment on the hybridization in ethylene.  All of the introductory texts I've come across use the same bonding description as your text: the carbons in ethylene are hybridized such that we have 6 carbon $\ce{sp^2}$ bonds (4 $\ce{C-H}$, 2 $\ce{C-C}$) and 1 pi bond based on overlap between the unhybridized p porbital remaining on each carbon.  Actually, if we refine this analysis to take into account the known $\ce{H-C-H}$ angle of 117 degrees, we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bonding are $\ce{sp^{2.2}}$ hybridized and the carbon orbitals involved in the $\ce{C-C}$ bond are $\ce{sp^{1.6}}$ hybridized.
I want to point out that, while staying within the MO framework where we only consider the s and p orbitals on carbon, there is an alternate way of describing the hybridization in ethylene - and these two descriptions predict the same physical reality, so they are equivalent descriptions.  The texts arbitrarily choose to hybridize only 2 of the carbon p orbitals along with the carbon s orbital.  It is just as valid to mix all 3 p orbitals with the s orbital.  If we constrain 2 of the resultant orbitals just as we did above (those that will bond to hydrogen) to produce an $\ce{H-C-H}$ angle of 117 degrees, then these two orbitals will be $\ce{sp^{2.2}}$ hybridized, just as the earlier analysis concluded.  The remaining two orbitals on carbon, however, can consequently be described as $\ce{sp^{4.3}}$ hybridized.  The picture that emerges from this alternate hybridization scheme is one of bent bonds - a 2-membered ring if you will!  Something to think about.

Thank you, are you saying there is a easily derived relationship between s character and bond angle?

That's a part of what I'm saying.  See the following link for the 2 relevant formulas (Coulson's theorem and sum of the s character=1) http://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test

(five sp2 bonds not six) The representation of $\ce{sp^{2.2}}$ is highly inaccurate as it only gives the relation between s- and p- character. Symmetry in ethylene dictates that the surrounding of carbon has to be planar, so it is only possible to use in plane p-orbitals, and there are only two. Hence the hybridisation is better described as $\ce{s^{0.91}p^{2}}$. And a $\ce{sp^{4.3}}$ orbital is just not possible. Also molecular geometry (and symmetry) dictates a $\pi$ bond and this is only possible by combining  the p-orbitals.

@Martin  1) I should have said 6 orbitals emanationg from the 2 carbons; 2) $\ce{s^{0.91}p^2}$ is the same as $\ce{s^{1}p^{2.2}}$, just divide 2 by 0.91, the "s" portion is always normalized to 1; 3) Do you agree that $\ce{s^{1}p^{2.2}}$ is possoible (I assume yes)?  Then why not $\ce{sp^{4.3}}$?  You seem to accept a continuity of hybridization indices between 1 and 3, truth is it runs from 0 (pure s) to infinite (pure p).  4) Google bent bonds in ethylene, Linus Pauling and Bent were proponents.

I am generally not a big fan of hybridisation, as I believe it oversimplifies most bonding situations. It is basically only another term for a certain type of linear combination. Most bonding situations are perfectly fine described without any hybridisation. 2+3) In a hybrid orbital $\ce{s^{$x$}p^{$y$}}$ I will accept, that $x\in\mathbb{R}\in(0,1)$ and $y\in\mathbb{R}\in(0,3)$, everything else is overcomplicating an oversimplified concept. 4) I love bent bonds!

The bonding model commonly accepted is neither "right" or "wrong" - it is a model, i.e. an approximation of reality using simple concepts (sort of like describing movement of trains as being linear and at constant speed, and ignoring wobbles, curves, speeding up and slowing down). There is an alternate model, although not popular, that postulates $sp^3$ hybridization only. Try to search for "banana bond theory".


text:
Rationalizing the Planarity of Formamide
In formamide, the nitrogens appear to be $\ce{sp^3}$ hybridized, implying tetrahedral geometry. However, analysis shows that the molecule is actually very nearly planar with bond angles close to 120 degrees.
EDIT: as suggested by Martin and another poster, hybridization is a rough concept. So perhaps the hybridization of nitrogen upon further analysis should best be described as somewhere inbetween $\ce{sp^3}$ and $\ce{sp^2}$. This however would still necessitate planarity, correct? Pi bonds are formed through the above and below electron-pairing in p-orbitals; effective bonding is achieved when these p-orbitals are parallel with respect to each other.
I'm thinking this has to do with the partial double bond character in the molecule (also appears to be some ionic character to the molecule - likely due to electron-withdrawing effects of the nitrogen and the oxygen).

This is the standard answer. However, would intramolecular hydrogen bonding also play a role? Couldn't there be a hydrogen bond between the peripheral hydrogen on the nitrogen and the oxygen? Couldn't this also help with achieving the 120 degree bond angles?

What strikes me as weird is your suggestion that O-H hydrogen bonding might be involved: none of the hydrogens are anywhere near the oxygen, and even if they were, I have no idea why you'd expect that to promote planarity. You might have built up some incorrect intuitions about hydrogen bonds that you might want to review and possibly try to unlearn.

Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide.
The carbon is obviously $\ce{sp^2}$ hybridised (as this concept is very well applicable here), hence organising it's ligands in one plane with roughly $120^\circ$ angles. Naturally one would assume the nitrogen to be $\ce{sp^3}$ hybridised, which is the case for most amines. However, the inversion barrier for these molecules is (depending on the substituents) very low. $$\ce{[NH3]^{pyr-top} <=> [NH3]^{TS-plan} <=> [NH3]^{pyr-bot}}$$ For the nitrogen that means going from $\ce{sp^3}$ to $\ce{sp^2}$ and back again. You can now stabilise the intermediated structure with conjugation, and that is exactley the case here. In your chart, that refers to entry 2. This causes the nitrogen to most likely be of $\ce{sp^2}$ hybridisation and the lone pair of entry one would be in a $\ce{p}$ orbital.
The conjugation happens as stated in you chart by the overlap of that orbital with the antibonding $\pi^*~\ce{C-O}$ orbital. This causes the $\ce{N-C}$ bond order to increase, while the $\ce{C-O}$ BO has to decrease.
All those resonance structures are only descriptions of extreme states, the truth lies between them. The following scheme considers the most common ones and adds a third, that might explain delocalisation (in a non-traditional Lewis way) up to a certain visual point.

In molecular orbital theory you can form 3-centre orbitals from all molecules perpendicular
to the molecular plain. If you choose this plane to be $xy$ then the contributing orbitals will be $\ce{p_{$z$}}$. The following scheme may support this claim, the dipicted orbitals were obtained by a BP86/cc-PVTZ calculation. (The HOMO is an in-plane lone pair of oxygen.)

While there is certainly no intra molecular hydrogen bond ($d(\ce{O-H}\approx 2.57$ same level), there will also most certainly be an attraction between the $\ce{C-O}$ and $\ce{N-H}$ bond helping to stabilize the planarity. But I will not go into detail about that, because it would involve breaking away from the very handy hybridisation concept.


text:
pH of a solution of acetic acid and ammonium acetate
After studying ionic equilibrium, I was just making some wild questions and I was unable to calculate the $\ce{pH}$ of this solution. Assume, 1 mole of each was added to $1L$ solution of water.
My attempt :
$\ce{CH3COOH}+\ce{NH4OH<=>CH3COONH4}+\ce{H2O}$
This reaction will happen till all $\ce{CH3COOH}$ is consumed by $\ce{NH4OH}$. More of $\ce{NH4OH} $ will be formed by Le Chatelier's principle. So, the answer must be same as that of a weak acid and weak base salt : $pH=\dfrac{1}{2}(pK_w+pK_a-pK_b)$.
But how will we calculate if there was $2$ mole of acetic acid?

You have some mistaken assumptions.
$\ce{NH4OH}$ and $\ce{CH3COONH4}$ are not molecules that exist.
The assuption "all $\ce{CH3COOH}$ is consumed" is incorrect.
Why do you say more $\ce{NH4OH}$ will be formed? How is this possibe? What can it be formed from?
Review that Henderson–Hasselbalch equation

$\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{OH-}$?

you mean + = -?

Well, that is : $\ce{NH4+}+\ce{H2O<=>NH4OH}+\ce{H+}$

Again, ammonium hydroxide isn't something that is stable. The two components undergo a large extent reaction to make ammonia. As a matter of fact, this is what Wikipedia has to say: "Ammonia solution, also known as ammonium hydroxide."
So what Dave is suggesting is that you write the net ionic equation, or in other words - the only right equation you can write for this system. Leave out the spectator ions to help you better see the chemistry.

@Awesome $\ce{NH3}+\ce{H2O<=>NH4+}+\ce{OH-}$ Again, $\ce{NH4OH}$ does not exist as a molecule.

So, how do we proceed to find out concentrations of acetic acid and acetate ion?

We start out by inventorying all the species in our system and then by considering all possible reactions. A systematic way to consider all possible reactions is to first start by reacting the strongest acid with the strongest base you have in the system (note that both the acid and base must be present in significant concentration, and at your initial molarity of 1, we can ignore things such as hydronium ion from the auto-ionization of water). Also don't confuse this with kinetics; reacting the strongest acid with the strongest base is just a bookkeeping method.

@Awesome, to a very good approximation, the concentrations of acetic acid and acetate will just be how much of each you add to the solution.  First you consider 1 mole of each, then you say what if I added 2 moles of acetic acid.

@DavePhD I want to consider both cases. Let me work on your advice.

@DavePhD Can you explain this assumption? Does ammonium acetate dissociate almost completely?

@Awesome Firstly, ammonium acetate is not a molecule and fully dissociates. Secondly, you should look up the pKa of ammonium ion. You will see that at the pH the the acetic acid and acetate will establish, the ammonium will stay ammonium to an excellent approximation.

@DavePhD Thanks. I wonder what would be the order of its Ksp. If you find a link, please share. I am unable to find it.

http://en.wikipedia.org/wiki/Ammonium pKa of ammonium is 9.25


text:
Semi-Infinite Potential Square Well: Negative Potential
I am having trouble with a problem involving a Semi-infinite potential square well:
I have written down some notes that I added to the post. I have problems understanding the physical situation. See the figure in the notes. When the potential is just shifted upwards, so that the $V(x)=0$ region is $0<x<a$, I have no trouble understanding the solution. However, I am just really confused about the negative potential.
As you will see in the notes, I have interpreted Region I as a bound state (e.g. an electron bound to a proton). Region II, I believe, is the free state (e.g. "bound" electron in energy level $n=\infty$), since $V(x) = 0$.  Is this correct?
I face some troubles with the boundary conditions, as one of my wave-functions simply disappear!

Answering your quantum mechanics question seems to become a hobby of mine :). First let me say that there is nothing wrong with shifting the potential up so that $V_{0} = 0$. This choice would indeed simplify the problem as it gets rid of negative energy values and thus avoids the following point of confusion: when $E < 0$ the wave function of a particle in region $\text{II}$ is not a plane wave $\mathrm{e}^{\mathrm{i} k x}$ ($k \in \mathbb{R}$) like that of a free particle moving in vacuum but a decaying exponential $\mathrm{e}^{-\kappa x}$ (with $\kappa \in \mathbb{R}$). So, the choice $V_{0} = 0$ is advisable. Shifting the potential back later is always possible and might be easier.
Nevertheless, of course the problem is solvable irrespective of the choice of $V_{0}$. I think your problem in this case is that you enforced an incorrect boundary condition. To avoid confusion I will start from the beginning:
I will only consider the case $E < 0$ in the following, so $E$ is always a negative number between $-V_{0}$ and $0$. For region $\text{II}$ you have to solve the Schroedinger Equation:
\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= E \psi_{\text{II}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= - \frac{2mE}{\hbar^{2}} \psi_{\text{II}}(x) \ .
\end{align}
Ok, here comes the first "problem" with the choice of $V_{0}$: In your solution you chose $\frac{2mE}{\hbar^{2}} = k^{2}$ as you would naturally do for a free particle. But since $E$ is negative $k$ will be a complex number (and not a real number as it would be for a free particle) and so this differential equation does not have an oscillatory solution $\psi_{\text{II}}(x) = A \sin(kx) + B \cos(kx)$ with $k \in \mathbb{R}$. Rather you get a decaying exponential function by doing the following: $- \frac{2mE}{\hbar^{2}}$ will be a positive number so you simply set $- \frac{2mE}{\hbar^{2}} = \kappa^{2}$ to get
\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= \kappa^{2} \psi_{\text{II}}(x) \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}
Now, this differential equation has a solution of the form
\begin{align}
  \psi_{\text{II}}(x) = A_{\text{II}} \mathrm{e}^{\kappa x} + B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}
At this point you can enforce the boundary condition that the wavefunction $\psi_{\text{II}}$ must be square integrable over the whole interval $a < x < \infty$ (Note that you don't have to consider the interval $-\infty < x < \infty$ but only the interval on which $\psi_{\text{II}}$ is defined). Since $\lim_{x \to \infty} \mathrm{e}^{\kappa x} = \infty$ this mean that the prefactor $A_{\text{II}}$ must be equal to zero, so
\begin{align}
  \psi_{\text{II}}(x) = B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}
So much for region $\text{II}$. Let's move on to region $\text{I}$: Here you have to solve the Schroedinger Equation:
\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} - V_{0} \psi_{\text{I}}(x)  &= E \psi_{\text{I}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - \frac{2m(E + V_{0})}{\hbar^{2}} \psi_{\text{I}}(x) \ .
\end{align}
Again, care must be taken: Since $-V_{0} < E < 0$ the term $(E + V_{0})$ is positive and so $- \frac{2m(E + V_{0})}{\hbar^{2}}$ is a negative number and you can set $- \frac{2m(E + V_{0})}{\hbar^{2}} = -k^{2}$ which leads to
\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - k^{2} \psi_{\text{I}}(x) \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}
This differential equation has a solution of the form:
\begin{align}
  \psi_{\text{I}}(x) = A_{\text{I}} \mathrm{e}^{\mathrm{i} k x} + B_{\text{I}} \mathrm{e}^{-\mathrm{i} k x} \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}
(You could also formulate an equivalent solution in terms of $\sin$ and $\cos$ functions using Euler's formula if that is more to your liking. But I prefer plane waves ;))
This is the point where you made your main mistake: You discarded one of the terms in your $\psi_{\text{I}}$ wavefunction because it blew up for $x \to \infty$. But $\psi_{\text{I}}(x)$ is only defined on the interval $0 \leq x \leq a$ and within this interval both exponential terms are perfectly square-integrable and you can't discard any of them.
Finally, you just have to remember that $\psi_{\text{III}}(x) = 0 \ \forall \ x$ and enforce the boundary conditions given by the system:
\begin{align}
  \psi_{\text{I}}(0) &= \psi_{\text{III}}(0) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{0} + B_{\text{I}} \mathrm{e}^{0} = 0 \quad &&\Rightarrow \quad B_{\text{I}} = -A_{\text{I}} \\
  \psi_{\text{I}}(a) &= \psi_{\text{II}}(a) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{\mathrm{i} k a} - A_{\text{I}} \mathrm{e}^{-\mathrm{i} k a} = B_{\text{II}} \mathrm{e}^{-\kappa a} && \\
  & &&\hphantom{\Rightarrow} \ \quad A_{\text{I}} \underbrace{\bigl(\mathrm{e}^{\mathrm{i} k a} - \mathrm{e}^{-\mathrm{i} k a} \bigr)}_{= \, 2 \mathrm{i} \sin(k a)} = B_{\text{II}} \mathrm{e}^{-\kappa a} \quad &&\Rightarrow \quad B_{\text{II}} = 2 \mathrm{i} \sin(k a) \mathrm{e}^{\kappa a} A_{\text{I}}
\end{align}
Edit:
Concerning your comment: You can indeed enforce another boundary condition. The probability
flux is required to be continuous across any surface, since otherwise the surface would
contain sources or sinks. This entails the requirement that both $\psi$ and $\nabla \psi$ must be continuous. The continuity condition $\psi_{\text{I}}(a) = \psi_{\text{II}}(a)$ has already been enforced but the one for the gradient is still left:
\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \ .
\end{align}
From the above discussion we know that:
\begin{align}
  \psi_{\text{I}}(x) &= A_{\text{I}} \underbrace{\left(\mathrm{e}^{\mathrm{i} k x} - \mathrm{e}^{-\mathrm{i} k x} \right)}_{= \, 2 \mathrm{i} \sin(k x)} = 2 \mathrm{i} A_\text{I} \sin (kx) = C_\text{I} \sin(kx)
\end{align}
and
\begin{align}
  \psi_{\text{II}}(x) &= \underbrace{2 \mathrm{i} A_{\text{I}} \sin(k a)}_{= \, \psi_{\text{I}}(x=a)} \, \mathrm{e}^{\kappa a} \mathrm{e}^{- \kappa x} = \psi_{\text{I}}(x \! = \! a) \, \mathrm{e}^{- \kappa (x - a)}
\end{align}
with $C_\text{I} = 2 \mathrm{i} A_{\text{I}}$, so
\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \\
  k C_\text{I} \cos(ka) &= - \kappa C_\text{I} \sin(ka) \\
  \kappa &= - \frac{k}{\tan(k a)}  \ .
\end{align}
While this equation looks innocent enough it is not easily solved for $E$ and you will need the assistance of something like Mathematica for that.

Yes, it seems that way :) I totally follow your discussion on making sure $k$ or $\kappa$ is a real value, and generating the general solution thereafter; that makes sense! I got the this for the wave-function ($with 2Ai = A_I$):
$$\Psi (x) = \begin{cases} A_I\sin (kx) & 0\leq x \leq a \ A_I e^{\kappa a} \sin (ka) e^{- \kappa x} & x>a \ 0 & x<0  \end{cases}$$
But does not continuity at
$x=a$
need me to set:
$$\psi_I' (a) = \psi_{II}' (a)$$
Based on the equations above, this just gave me
$-k = \kappa$
, which is incorrect. How would I go about determining the allowed anergy values?

@AndersMB I think you might have made a small mistake while differentiating the sin function in $\psi_{\text{I}}$. I've added a new paragraph to my answer which shows the way to the (hopefully) correct solution.

Thanks! Sometimes, when I differentiate sine and cosine functions, I forget to switch the trig functions! Just for fun, I found that $E = \frac{V_0}{\tan^{-2}(ka)-1}$. The Problem asked me to determine the allowed energies, but I am not sure how detailed I should be. If, as you say, I need software to do so, then I probably do not need to go very deep into it. Thanks again! You are a lifesaver. I must watch out for these little mistakes at the exam!

@AndersMB Yeah, it's easy to forget something like that. And there is still a problem with this equation: the argument of the tangens contains $E$ (disguised as $k$) so this is still an implicit equation for $E$ that needs to be solved computationally.

Right - I'll just stop there, then, and move on :)

I believe there are two seperate mistakes, one in region I and one in region II.
For region I, it looks like you made a sign error, and that you should have $-V_0 - E$ instead of $V_0 - E$.
While this may seen trival at first, the sign is important to the form of the solution to the differential equation (wavefunction).
For region II, the way the diagram of the question is drawn, E is negative.  Assure that this fact is properly accounted for in the form of the solution.

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text:
Is visible spectroscopy the only non-electronic method of all the spectroscopy method?
Is visible spectroscopy the only non-electronic method of all the spectroscopy method?
I know some of the spectroscopy method is possible being used because of electronics.

No, Ultraviolet spectroscopy and Infra-red spectroscopy were practiced without any use of electricity.
Infra-red and ultraviolet light were both known by 1801.
Ultraviolet absorption lines had been observed by 1843.


text:
Resonance in Cyclohexenone
I don't think that cyclohexenone should have any resonance forms. Moving the double bond around would necessitate moving hydrogens around as well. For example, I know that benzene exhibits resonance (duh) and the second picture attached shows the pi bonding electrons shifting but doesn't show any hydrogens shifting positions.

That being said, is there a easier way to determine if a resonance form of a cyclic compound is valid? Drawing out each individual atom and bond is time-consuming! Should I use Huckel's rule? Do all cyclic compounds which exhibit resonance follow Huckel's rule?

The structure you've drawn on the left does have a resonance structure, but not the one you've drawn on the right.

Drawing resonance structures out by hand gets quicker over time.  You find that you don't need to actually put in a "C" for carbon, or label the hydrogen atoms.  Huckel's rule generally applies to planar, cyclic molecules with continuous conjugation, although it has also be applied to transition states where an S or P orbital on a shifting C or H moiety completes a loop of continuous conjugation (for example, the 1,5 hydrogen shift in cyclopentadiene).


text:
What is the formula for heat of vaporization under reduced pressure?
I haven't taken a chemistry class in several years, and I have honestly been searching the web for the past hour and can't find anything. I know this is something basic we learned back in high school chemistry but maybe I am not searching for the right term...
Is there a formula? I know how to calculate distillation speed based on wattage, and the vaporization constant of the molecule, but cannot find any particular formula that can calculate how that number changes under reduced pressure.

see if this helps: http://deepblue.lib.umich.edu/bitstream/handle/2027.42/37333/690110226_ftp.pdf?sequence=1

You are looking for the Clausius-Clapeyron equation.  It comes in various forms:
The exact version is
$$\frac{\mathrm dp}{\mathrm dT}=\frac{\Delta H}{\Delta V T}$$
or, since the volume of the gas phase is so much larger than the volume of the condensed phase
$$\frac{\mathrm dp}{\mathrm dT}=\frac{p \Delta H}{R T^2}$$
or even more succinctly, assuming $\Delta H$ remains constant
$$\frac{\mathrm d \ln p}{\mathrm d \frac{1}{T}}=-\frac{\Delta H}{R}$$


text:
Why does potassium react the most vigorously with water?
Link to exam

Which metal reacts most vigorously with water?
(A) Ca (B) K (C) Mg (D) Na

The given answer is B and the reasoning is that is is the most electropositive since it the furthest left and down according to periodic trends.
However, isn't $\ce{K}$ the conjugate acid from a strong base -- $\ce{KOH}$? So isn't the conjugate base of a strong acid quite weak? So why would $\ce{K}$ react strongly?

$\ce{K}$, not $\ce{K^+}$.

I might be wrong here, but I believe that the question asks about the metals in their atomic (non ionic) form.
In order for a $\ce{KOH}$ bond to form, the K must be ionized (since the $\ce{KOH}$ bond is ionic). This means that when $\ce{KOH}$ dissolves in water it'll form the $\ce{K}^+$ electrolyte, which is not the same thing as a pure $\ce{K}$ atom with its 19th electron.
A $\ce{K}$ atom, then, since it's the most electropositive of that list, is the one to react "more vigorously" (which I guess it refers to more speed/energy) with water (a polar molecule).
I hope this helped you. I'm sorry if I'm wrong!

I interpret "reacts most vigorously" to mean which metal reacts the fastest.  This is a question of kinetics, not thermodynamics (e.g. the answer is not based on the enthalpy of the reaction).  Kinetic rates are determined by the height of the energy barrier that needs to be surmounted, we need to determine and compare the activation energies in this process for the various metals.  The activation energy in this process is determined by the following two steps: $$\ce{M(s) -> M(g)}$$ $$\ce{M(g) -> M^+(g) + e^-}$$ $$\ce{or}$$ $$\ce{M(g) -> M^{+2}(g) + 2 e^-}$$ depending on whether we are analyzing a group I or group II metal.  The following link provides a nice analysis and tabulation of the energies required for these two steps with the group I metals
Group I link
and this link does the same for the Group II metals
Group II link
I'll summarize their findings
Metal ...   Activation Energy (kJ/mol)
K   ...         +508
Na  ...        +603
Mg  ...       +2200
Ca  ...       +1950
CONCLUSION: Potassium has the lowest activation energy and should react the fastest.

I believe that "most vigorously" would mean the explosive heat that is given off, e.g. when throwing an ingot of potassium into water. Seems thermodynamic to me.  I prefer Martin's answer, which speaks also about the hydroxide, and points out that solvation energy is important.  There is a significant effect of bulk transport, also -- the metal-hydroxide crust has to be dissolved into the water to allow for fresh metal to react.

Another thing is that the full ionization process is never fully realized in solvent, so these values might be supportive of the observation, rather than explanatory.

Please read ron's post before continuing here (too long for a comment). Its reasoning is very good and makes a lot of sense. The activation barrier is the defining component for how vigorously components react.
His presented reaction may cover the initial activation barrier, but subsequent reactions also have to be taken into account.
The overall reaction that will occur in the case of sodium can be formulated like this:
\begin{aligned}
  \ce{2Na(s) + 2H2O(l) &-> 2Na+(aq) + H2(g) ^ + 2{}^{-}OH(aq)}&
  \Delta H &= -184~\mathrm{kJ/ mol}
\end{aligned}
So the start of the reaction has to be considered as the atomisation energy of the metal, after that the ionisation energy, both of these reactions need energy. Energy is released to the system by the hydration of the ion. For sodium this boils down to
\begin{aligned}
  \ce{Na(s)  &-> Na(g)}             & \Delta H &= +109~\mathrm{kJ/ mol}\\
  \ce{Na(g)  &-> Na+(g) + e-}       & \Delta H &= +494~\mathrm{kJ/ mol}\\
  \ce{Na+(g) &-> Na+(aq)}           & \Delta H &= -406~\mathrm{kJ/ mol}\\[2ex]\hline
  \ce{Na(s)  &->[\ce{H2O}] Na+(aq)} & \Delta H &= +197~\mathrm{kJ/ mol}.\\\hline\hline
\end{aligned}
So why would this reaction happen, when there is no energy released? The cause of this are the secondary and tertiary reactions, that then again fuel the initial reactions. So the key question here is, what happens to the electron?
\begin{aligned}
\ce{2H2O + 2e- &<=>  H2 ^ + 2{}^{-}OH}\\
\end{aligned}
While I cannot find a value for this reaction, however I strongly suspect it to be negative. It more certainly gives us a hint on what to consider else, i.e. the hydration of hydroxide ions. (source)
\begin{aligned}
  \ce{{}^{-}OH &-> {}^{-}OH(aq)} & \Delta H &= -460~\mathrm{kJ/mol}
\end{aligned}
So this releases a lot of energy to the system, helping to atomize and ionize the sodium.
A tertiary reaction also has to be considered. Because of the produced heat, hydrogen will immediately burn when coming in contact with air. (wikipedia)
\begin{aligned}
  \ce{H2(g)  + 1/2 O2(g) &->  H2O(g)} & \Delta H &= -242~\mathrm{kJ/mol}\\
  \ce{2H2(g) +     O2(g) &-> 2H2O(g)} & \Delta H &= -484~\mathrm{kJ/mol}
\end{aligned}
So all these reactions successfully overcome the activation barrier once the reaction started. This also then has the effect of multiplying the initial reaction, causing a chain reaction and hence reacting even more furiously.
Watch Sodium burn in water and at one point you see the hydrogen ignite and how vigorous this reaction is.
The same applies to all other metals, but as ron stated the kinetics depend on the activation barrier.

The enthalpy of formation of $1\ \mathrm{mol}$ $\ce{H2O(g)}$ is $\Delta H^\circ=-242\ \mathrm{kJ}$; thus, the molar enthalpy of reaction for $\ce{H2(g) + 1/2O2(g) -> H2O(g)}$ is $\Delta H^\circ_\mathrm m = -242\ \mathrm{kJ/mol}$, whereas the molar enthalpy of reaction for $\ce{2H2(g) + O2(g) -> 2H2O(g)}$ is $\Delta H^\circ_\mathrm m = -484\ \mathrm{kJ/mol}$.

@Loong Thanks for pointing that out. I sometimes have issues with all that mol stuff.... :/

The other answers explain things in terms of the activation energy barrier, but they don't explain why the activation energy barrier is lowest for K. I will attempt to do that here.
When a metal reacts with water, several things must happen (not necessarily in this order):

One or more electrons have to be removed from the metal
One or more $O-H$ bonds have to break on water molecules (one per electron)
The electron(s) must be "picked up" by the $OH$ to become $OH^-$

As @ron said, the "vigorousness" of this process depends on the kinetics of the reaction, which are in turn determined by the activation energy of the rate-limiting step.
In the links that he provided (1,2), they show that the activation energy is lowest for K, which means it would react most vigorously. (note - these are not real activation energies - the atoms are not actually atomized to the gas phase, and the electrons are not removed under vacuum to infinite distance, but they still give us a good relative estimate)
However, knowing that the kinetics of the reaction depends on the activation energy won't help you if you don't happen to have a chart of activation energies handy. On an exam, you certainly won't, and it is good to develop an intuition for these sorts of things anyhow. And so, you might ask:
Why is the activation energy lower?
Looking at the links, you can see that the activation energies are dominated by the ionization energy. In other words, when it takes more energy to remove electrons, the reaction slows down.
Why should K have the lowest ionization energy? First we need to look at the equation that governs the potential energy between charges:
$E=\frac{{\kappa}Q_1Q_2}{d}$
Here E is electrostatic potential energy, $\kappa$ is Coulomb's constant, and $Q_1$ and $Q_2$ are charges.
You can see from this that if the charges have the opposite sign, then energy will be become more negative as they move closer together (inversely proportional to distance). As either of the charges increase, the energy will become more negative as well (directly proportional to charge). In other words, big charges are harder to pull apart, and charges that are close together are harder to pull apart than charges that are far apart. As a rough analogy, imagine pulling apart two magnets - it is more difficult when they are close together.
Now compare K to the other group I element (Na). K is further down the column. Since atomic radius increases with the row number (due to the fact that orbital radius increases with principle quantum number n), the valence electron on K is further from the nucleus (on average) than it is on Na. This means that it is easier to remove a valence electron from K. As an analogy, imagine trying to move a rock up a hill. You have to do less work (put in less energy) if the rock is already halfway up the hill. Since the electrons are further from the nucleus in K, they are already "further up the hill."
What about group II elements? As you progress from left to right, the atomic radius decreases - this is because the effective nuclear charge seen by the valence electrons increases. The combination of increased charge and decreased distance leads to a steeper "hill" that must be overcome. On top of that, now we have to remove two electrons for the reaction to work, which is roughly twice as difficult (technically we don't have to remove both, but the overall reaction thermodynamics won't work out if we don't). As a result, group I elements will always react more vigorously with water than group II will.
I think that it helps to imagine reactions as sort of a "fight" between elements over electrons. Who will win the fight and how it will go is determined by the number of protons in the nucleus and the electron orbital configuration. This approach is useful because it applies to more advanced chemistry as well.
For some demonstrations of alkali metal reactions with water, check out this video:
Alkali Metals

I'm inclined to believe that the liberation of heat is a thermodynamic effect. While there may be vestiges of the thermodynamics in the kinetic barrier, I would like to point out that the discrete ionization is probably not observed, rather being conflated with transfer to the hydroxide, as well as the enormous solvation energy.

@Eric - I thought that too, until I read the references that ron posted. If you check those, you will see that potassium's net reaction enthalpy with water is less exothermic than either sodium or magnesium, both of which react much less "vigorously."


text:
What attributes cause a substance to be nonconductive or conductive
I understand what causes elements to be conductive or nonconductive. But what causes a substance say rubber to be nonconductive at a molecular level?

Conductivity requires mobile electrons. Metals tend to have loosely held valence electrons so metals are generally conductive. Metals have loosely held valence electrons for two reasons: metals tend not to be very electronegative (able to stabilize electron density) and metals tend to be somewhat big atoms by nature of many metals having a d-group of electrons. D electrons are further from the nucleus on a time average basis than s electrons or p electrons.
However, conductors do not have to be metals. Consider graphite. Graphite is $\ce{sp^2}$ hybridized. Graphite is made solely of carbon atoms - and carbon isn't a metal! The mobile pi electrons in graphite due to resonance allows graphite to be a good conductor.
I should also note that conductivity does not always imply mobile electrons; solutions can be conductive too and this is due to mobile electrolytes (ions). For example, you will find that deionized water is completely unconductive. However, regular tap water is generally conductive due to the presence of ions (e.g. chloride ions from chlorination). Gatorade is also very conductive because it is especially formulated to contain a lot of electrolytes.

Another factor that defines a metal is the availability of unoccupied orbitals that are very close in energy to the filled orbitals.  Because the filled and empty orbitals are so close in energy, the loosely bound electrons are able to "hop" from atom to atom with a very low activation energy.  This low activation energy for the hopping process means that electrons can travel through such materials very rapidly.

Yes! I think I heard of it called "ionization" energy somewhere before.

Metals have a high density of states and many delocalised orbitals. The energy difference between occupied and unoccupied orbitals (band gap) may be measured and can be used to determine conductors, semi-conductors and isolators. In a conductor it is not necessary that there are mobile electrons. There have to be mobile charge transporters, e.g. ions. (Graphite is a 2D conductor.)

True, mobile ions or electrolytes if we're talking about solution does the trick. Pure, deionized water has 0 conductivity but tap water is conductive usually due to the presence of chloride ion and the like.


text:
Electronegativity Considerations in Assigning Oxidation States
I have never seen anything other than a set of rules like these when textbooks present how to assign oxidation numbers. Such as these:

However, if we keep in mind that oxidation numbers are simply imaginary numbers which suppose all bonding to be ionic - i.e. not electron sharing - and if we keep in mind simply relative electronegativity, we can easily work out the oxidation state of any element in any compound.
For example, take water. Bonding order: $\ce{H-O-H}$. Oxygen has two lone pairs. Oxygen is more electronegative than hydrogen. So we suppose that oxygen takes both electrons in both bonding pairs. Oxygen has 8 electrons. Its valence is 6. 6-8 is -2; oxygen has two more electrons assigned to it than what it has in its valence, so naturally its oxidation state is negative 2. No rules needed.
Now hydrogen peroxide. $\ce{H-O-O-H}$. Here we have an oxygen-oxygen bond so in this case, neither element wins the electronegativity battle. Electrons are split between the oxygen and the oxygen. However, since the oxygens are much more electronegative than the hydrogens, we assign both electrons in both the $\ce{O-H}$ bonding pairs to oxygen. Oxygen has 7 electrons assigned to it; oxygen has a valence of 6; oxidation state: -1. No need to memorize exceptions as stated in the table above.
We could go on. However, I am curious:
1) Has anyone been taught to assign oxidation states this way?
2) If not, what do you think of this method?

I have, the table came afterwards as an aid to quickly assign alkali and alkaline earth metals, oxygen and halides.

I don't think I was taught that way but it's the way I use. I think oxidation numbers are a lot easier than how they are taught in high school.

I agree. Memorization is crap. EN isn't that hard and I remember I was taught to memorize the EN trend in high school. However, we were still taught how to assign oxidation numbers through memorizing a bunch of rules. Unfortunate that teachers can't put two and two together.

Please note that the definitions have finally changed, as has my answer.

In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below.
The electronegativity battle scheme is most helpful for all kinds of compounds since it is one of the most generic ways to derive oxidation states. The table represents just a cheat sheet that might be very helpful in the beginning. If you are spending most of your time with chemistry, this table will be present as some sort of muscle memory - usually referred to as chemical intuition.
The IUPAC gold book now defines oxidation state as follows:

oxidation state
gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

The underlying principle is that the ionic sign in an $\ce{AB}$ molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear $\ce{AA}$ bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:

Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:

Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:

NOTE:

Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.
Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.
The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).
When used in chemical nomenclature as a symbol, the OS value is in roman numerals.
At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating $\text{OS} = +1$ for $\ce{H}$ and $−2$ for $\ce{O}$ yields correct OS in oxides, hydroxides, and acids like $\ce{H2SO4}$; with coverage extended to $\ce{H2O2}$ via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.

Source:

Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081 DOI: 10.1515/pac-2013-0505.
Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839 DOI: 10.1515/pac-2015-1204.
Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from old.iupac.org

I recommend having a look at the sources cited above, they contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)
Prior to 2016 the IUPAC defined oxidation states in their gold book (via the Internet Archive) differently. I am including this definition, since it will be still present in prominent text books and it is what lead to the postulates given in the table of the original question.

oxidation state (deprecated, pre-2016 definition)
A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

the oxidation state of a free element (uncombined element) is zero;
for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
hydrogen has an oxidation state of $1$ and oxygen has an oxidation state of $-2$ when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of $-1$ in hydrides of active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of $-1$ in peroxides, e.g. $\ce{H2O2}$;
the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$, $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and $\ce{H2SO4}$ are, respectively: $-2$, $0$, $+4$, $+6$ and $+6$. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.

What we see is, that the table in the original question actually reflects some of these rules. However, this set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions. With only these rules it is impossible do determine the oxidation states for $\ce{BF3}$ and many, if not most, compounds.
The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in Expanded Definition of the Oxidation State, which resembles the currently used definition quite well.

The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.

So he basically came to the same conclusion as Dissenter in the question. Another statement from this article is

This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3).
(3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. Republished by Courier Dover Publications, 2012.

My apologies for bumping an old question; however I'm not sure that the accepted answer provides an adequate answer to the stated question:

1) Has anyone been taught to assign oxidation states this way? 2) If not, what do you think of this method?

For part 1, I can say  personally that I use this method in teaching a 3rd year Inorganic Chemistry course.  It is covered in Rayner-Canham's Descriptive Inorganic Chemistry textbook (Chapter 8 of the 5th edition).  After teaching this method for a number of years, I have anecdotal evidence that students get a better understanding of what the oxidation state is, and when it is suitable to use this type of electron bookkeeping instead of formal charge (for example).  In addition to relying on a periodic trend as opposed to a list of rules, this method helps to explain the differences in the oxidation numbers of inequivalent atoms in ions/molecules (such as sulfur in $\ce{S2O3^2-}$), which can't be done with the list-of-rules approach.
For part 2 of the question, the biggest drawback of using the electronegativity approach to determining oxidation numbers is that it requires a knowledge of lewis dot structures.  With the current (American) General Chemistry track, students learn about oxidation numbers prior to drawing structures, and therefore must be taught using the list-of-rules approach.  I suspect a shuffling of the content to allow students to use the electronegativity approach would have some pedagogical benefits; instead of sitting at the bottom of Bloom's Taxonomy with remembering a list of rules, we can move up to applying the electronegativity trends to analyze the degree of electron deficiency of an atom has in an ion or molecule.

I do not understand your last paragraph. Are you saying that students learn how to assign oxidation numbers before learning about the structure of the molecule which they are assigning the oxidation numbers to? How can that approach even remotely work?

@Martin-マーチン students typically learn how to determine the oxidation number of atoms in polyatomic ions based on the "rules".  They know the chemical formula of the ion, but they have not encountered Lewis dot structures.  So they can use the rules to determine that the N in nitrate has an oxidation state of +5 before they learn how to determine whether it is trigonal pyramidal or trigonal planar.

But then I don't see how considering the electronegativity instead of the rules would change anything. The procedure mostly stays the same.

@Martin-マーチン That's my point, gen chem students learn all that is needed to apply electronegativity to ox numder assignments, just in the wrong order.  Therefore, it is only useful (with the current standard US curriculum) later on in the chemistry program.

I don't want to go into a deep conversation about the topic in the comments, but I really do not understand. What is the wrong order? Do students learn the set of rules to assign ox numbers prior to electronegativity? I just want to understand why there is a problem teaching the far superior approach instead of a fixed and incomplete set of rules. I am not even starting to question that they are taught a concept of bookkeeping of electrons before learning about molecular structures. If you want to talk about this, consider creating a chat room where we can talk.

@Martin-マーチン I will think about it a bit and get back to you.

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text:
Confusion regarding specific conductivity
Is there any difference between the terms specific conductivity and conductance. If yes, please explain.

probably belongs to physics segment

It could be answered here, since conductivity is a measurement often applied to solution chemistry and materials science.

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

Let's take a wire inside cable for example. That wire's conductance is just the inverse of the resistance this cable is making when electricity passes through it. This depends on things like the length, what is made of, the maximum/minimum width of certain areas of the wire, etc.
On the other hand, the conductivity of that wire is a direct property of the material it's made of. If the wire is made of copper for example, it'll have (at the same temperature) the same conductivity as any copper wire in the world (because they are all made of copper), but not the same conductance (because they are not physically equal)
I hope this helps, I'm sorry if I was wrong!


text:
TiO2 nanotube application in cement
I am perparing for a presentation on the application of TiO2 nanotubes in cement which is a recent topic in cement chemistry. Do you happen to have any papers (or know of any accessible on Internet) on the topic from the recent years that you can share with me? Or even if you have any information/experience to share, I would be grateful.

This is very broad. What is this topic do you say? Appliances? Synthesis? What?


text:
Elemental Silver Colour
What is the colour of elemental Ag? I have doubt whether it is black or silver in colour.

In Tollens' test, we get a silvery precipitate of elemental silver - hence silver in colour.
I have read somewhere and also in the following links as a reference that it was not mere speculation

http://books.google.co.in/books?id=AFY2gKwh4AsC&pg=PA553&lpg=PA553&dq=elemental+silver+precipitate&source=bl&ots=nzfsQcBAm6&sig=jTCQk7-k0HzAByI25ESI6OSclJs&hl=en&sa=X&ei=S1d7U6aXIY-OuASHp4HIDQ&redir_esc=y#v=onepage&q=elemental%20silver%20precipitate&f=false
(towards the end)
and
http://books.google.co.in/books?id=cZL_-hyw98sC&pg=PA675&lpg=PA675&dq=elemental+silver+precipitate&source=bl&ots=DDhdhikKhU&sig=zIq-qQM7VI4i5enohZVi3oIagHw&hl=en&sa=X&ei=S1d7U6aXIY-OuASHp4HIDQ&redir_esc=y#v=onepage&q=elemental%20silver%20precipitate&f=false
Please confirm this. Thanks in advance.

@LievenB Thanks was kind of suspecting this 'colour' definition issue....

Silver:

Silver sulfide precipitate is black.  You are pobably confusing a precipitate in a test for silver, with actual silver.  The precipitate in a test will be a compound of silver, such as silver sulphide which is black.
Nanophase metal particles are optically different from bulk in many cases, and can be black, but this doesn't seem to be what the question is referring to.

I knew that silver sulphide is black and just another article I found- http://questions.sci-toys.com/node/130 (3rd paragraph) (mainly metallic Ag also some Ag2O produced) staining by AgNO3, please verify. Anyway thanks...

I don't see that statement in the reference, instead it is saying silver chloride is responsible for black stains on skin.

just after the mention of silver chloride -- "Silver nitrate will react with the skin and the salt on the skin to form silver chloride, which will them break down in strong light to form tiny particles of silver. These silver particles stay stuck in the skin, and look black".

ok, I see now.  The only reason I know of for an elemental metal to be black is if the particles are on the nanometer scale.

What about the nanometer scale, could you please elaborate, I don't even have a hint of any relation of its colour to any 'nanometer scale', anyway I have a feeling towards the next answer by LievenB, suspected that only but asked here just to confirm, thanks...

see slides 14-16 of this presentation: http://www.nisenet.org/sites/default/files/catalog/uploads/2474/stainedglasscart_classpresentation_jan10.ppt.pdf

and http://www.sigmaaldrich.com/materials-science/nanomaterials/silver-nanoparticles.html

The first nisenet link shows the colour for different silver nanoparticles sizes but doesn't have any 'black' colour for any size, while the second link just mentions- "have a color that depends upon the size and the shape of the particle", (wikipedia says silver nanoparticles have a size from 1nm to 100nm and the article gives colour for most in this range) also from wikipedia- "Exposure to silver nanoparticles has been associated with "inflammatory, oxidative, genotoxic, and cytotoxic consequences" ", and as far I have searched (after dropping AgNO3) that such stains are harmless.

see the discussion (and photo) of silver nanoparticles that are colored in solution being black upon drying, aggregation discussed here: http://www.researchgate.net/post/Why_does_the_deep_brown_color_of_the_silver_nanoparticles_plasmon_is_turned_black_upon_drying

Also while searching for "Mulliken Barker Test", I came up with this - http://books.google.co.in/books?id=1OgRECl_nwMC&pg=PA169&lpg=PA169&dq=mulliken+barker+test&source=bl&ots=34UH8mlPUs&sig=JJUETWAXBcvuKIZRuJ3zUOaguzQ&hl=en&sa=X&ei=YtR8U7O-CYy3uASIkYLQBg&sqi=2&ved=0CDEQ6AEwAQ#v=onepage&q&f=false

There the same Tollen Reagent is said to give a grey black ppt, so i think that both your suggestion of nanoparticles and the fact of colour definition are the possible reasons, thanks for the help...

(this is just to illustrate with an image, even though the answer was already formulated by DavePhD )
It depends on what you define as colour - a silver bar will be metallic-silver, however silver powder may seem gray  (or even black in water) (see image: http://www.nbchangjing.com/upload/2009031714521679.jpg )
I recently performed this silver-compound reaction : $\ce{Zn + 2AgNO3 -> Zn(NO3)2 + 2Ag}$ After a minute you could clearly see silver 'growing' from the zinc wire that I added to the silver nitrate solution. The silver was powdery black, however when I 'flattened' the silver powder, I had a silver-metallic surface.
I even found an image of a similar reaction involving copper wire :

You clearly see that there is a grayish-black silver powder forming on the wire, yet this doesn't have the same colour as a silver coin, for example.


text:
Bond Angles of Hypofluorous Acid - Gas and Solid Phases
This is the gas-phase structure of hypofluorous acid:

The solid-phase structure however has a bond angle of 101 degrees. Why is there a difference in the solid and gas phase bond angles?
Is this due to packing in the solid-phase? In the solid-phase the molecules should be much closer together than in the gas-phase, so I expect more intermolecular forces and interactions. However, what exactly would be responsible for the angle expansion in the solid phase? From one book I read that bond lengths often differ among the phases.
So what exactly is behind the bond angle expansion in the solid phase?

It is due to hydrogen bonding in the solid but not gas phase, in analogy to what occurs for the water bond angle.
From a VSPER point of view, the oxygen atom of one molecule would form a hydrogen bond with the hydrogen of another molecule, decreasing the repulsion of the lone pairs to be closer to that of the covalently bonded pairs, so the geometry would be closer to tetraherdral in the solid phase than in the gas phase.

Thank you; I had suspected hydrogen bonding or some kind of intermolecular force. So how does hydrogen bonding decrease lone pair repulsion? I'm guessing that the partial positively charged hydrogen on one molecule, by forming a "bond" with a lone pair on another molecule, will "dilute" the strength of the lone pair to repel; I think of physics and point charges and vector addition of all the forces on a point charge. Also couldn't the partially positively charged hydrogen bond with a lone pair on fluorine too? I know this isn't strictly hydrogen bonding but this is another factor, right?

The crystal structure is solved in this reference: http://onlinelibrary.wiley.com/doi/10.1002/anie.198803921/abstract The hydrogen bond does go between two oxygen atoms.  Fluorine can certainly participate in hydrogen bonds, but doesn't in this case.  There is discussion of both possibilities in the reference.

Thank you very much! One of my former chem profs had a near obsession with bond angles and now we both do (hence my question). I'll enjoy the paper (edit: once I figure out how to access it).

You can download the document from this link : http://booksc.org/book/479656 (however i'm not sure if it is the complete document)


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Inorganic chemistry reaction mechanisms
Are there well developed reaction mechanisms in inorganic chemistry like those in organic chemistry? If yes, please suggest an easy to follow reference.

Have you looked at organometallic reactions and metal-catalytic cycles?

Just search google for "inorganic reaction mechanisms". This looked quite interesting: http://web.archive.org/web/20160303222226/http://www.chem.ox.ac.uk/icl/dermot/mechanism1/

@LievenB  I just wanted now if simple reactions reaction mechanism can be predicted or not

@rch   no      SIR I Don't know

Nope. Inorganic chemistry is much more diverse in this part, so no really good systematization occurred.

Are you asking about mechanisms in relation to organometallic complexes?

yes I know organometallics has pretty much organised way of depicting reaction like in organic chemistry but particulary I want to know about element of p - block ( obviously not carbon) and s- block

You might find this book interesting: Arrow Pushing in Inorganic Chemistry by Ghosh and Berg.

For JEE aspirants check JD LEE for advanced by sudarsan guha, it discusses inorganic reactions from a mechanistic point of view

If this is in regard to mechanisms with organometallic complexes or coordination compounds, there are a set of specific mechanisms that are well defined. These are:

Ligand dissociation / association
Reductive elimination / oxidative addition
Intramolecular insertion (e.g. CO migratory insertion) / elimination (e.g. $\beta$–hydride elimination)
Metallacycle processes (e.g. $\sigma$-bond metathesis)
Nucleophilic / electrophilic addition and abstraction are also possible, similar to organic chemistry.

These mechanisms should be in inorganic textbooks but can also be viewed on the first page here. The later slides in this pdf cover more of the details of each specific mechanism.
In many reaction mechanisms with organometallic complexes, it is a process of a number of these smaller mechanisms occurring, leading to the overall chemical transformation that is stated in a reaction equation. These overall mechanisms can be complicated and need to be evaluated for each case. Knowing the basic underlying mechanisms will certainly help clarify processes that seem quite complex at first glance.
There are subtleties in all of these mechanisms to consider beyond the basic types above. Significantly more detail can be found in these textbooks that I have previously utilized which you may also find informative:
Collman, J.; Hegedus, L.; Norton, J.; Finke, R. Principles and Applications of Organotransition Metal Chemistry. 1987. University Science Books. Mill Valley, California.
Spessard, G.; Miessler, G. Organometallic Chemistry. 1997. Prentice-Hall. Upper Saddle River, New Jersey.


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Calculating the mass in g of a molecule
If I have the amount of the molecule in moles, how do I calculate the mass of an individual molecule?
For example, if the molecule was water?

The "amount of molecules in moles", such as 7 moles, provides no information concerning the mass of an individual molecule.
This is like asking, "if I have the amount of eggs in dozens, how do I calculate the mass of an egg?"

Can you really not? :O
So if I knew I had one mole of water molecules, then 2.02g of the 18.01g would be hydrogen atoms. If I divide this by $6.02\times10^{23}$ wouldn't I get the mass of a H atom?

Firstly, the question doesn't say anything about atoms, just "the mass in g of the a molecule" and "the mass of an individual molecule".  Secondly, your comment is not using the amount of molecules in moles to calculate anything.  Instead, it is adding additional information, that the substance is water, and assumptions about the chemical structure of water.

It sounds like you're trying to calculate the mass of a water molecule.  Water has a molecular weight of $18\:\mathrm{g/mol}$.  In a mole of water there will be Avogadro's number of molecules ($6.02\cdot 10^{23}\:\mathrm{molecules/mol}$).  Therefore, $$\text{mass of a water molecule} =  \frac{18}{6.02 \cdot 10^{23}} = 2.99 \cdot 10^{-23}\:\mathrm{g/molecule}$$


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How can I check if this mystery compound is strontium bromide?
Trying to identify a mystery compound, here are some of my notes, I'm trying to get a second opinion on what it might be!
Observations
Appears as a white powder, some chunks are formed, sticking to side of tube slightly
Solubility
3 drops of distilled water, all of substance dissolved
Flame Test
Red Flame
Sodium Hydroxide Test
A white ppt was formed when mystery liquid was combined with sodium hydroxide
Concentrated sulfuric acid test
A yellow colored gas formed when 4 drops of sulfuric acid were added with substance in test tube
My thoughts based on these results narrow the cation down to lithium and strontium. After performing a precipitate test again with my two knowns (lithium, strontium)strontium was the only one to form a ppt. So I'm almost 100% sure that my cation is strontium.
I have the anion narrowed down to bromide, nitrate, and nitrite. It has to be one of those three! I'm leaning towards bromide as the anion, just interested on what you guys think from my notes.
Here are the options again, with my choice in bold:
Cation

Lithium
Strontium

Anion

Bromide ?
Nitrate
Nitrite
Iodine

To summarize:
I have a book with 16 possible cations, and 11 possible anions. Based on my results and the information given in the book, I've narrowed it down to 2 cations, lithium, and strontium, and 3 possibly 4 anions. I'm almost certain the cation is strontium, based on the fact that it formed the same white ppt. that the mystery substance did when mixed with sodium hydroxide. Lithium did not. As for the anion, I'm not sure, I'm leaning towards bromide, but was looking for a second opinion. I've been doing some research on how bromide reacts with sulfuric acid, and the results online have been almost identical to what happened with the unknown substance, however in my option book it says bromide and nitrate have very similar reactions.
If you have any more thoughts on how I could identify to anion, please let me know!

@23k find silver nitrate, it will have very distinct reactions.

Relying on the fact that the compound is almost certainly a strontium compound, I'll look at the reactions with sulfuric acid of the specific anions:

$\ce{SrBr2}$

$$\ce{SrBr2 + H2SO4 ->  2HBr + SrSO4}$$
Both compounds $\ce{HBr}$ and $\ce{SrSO4}$ are white or colourless. BUT! concentrated sulfuric acid oxidizes $\ce{HBr}$ to bromine gas:
$$\ce{2 HBr + H2SO4 → Br2 + SO2 + 2H2O}$$
And bromine gas has a yellow/red colour. So your compound is strontium bromide.

$\ce{Sr(NO2)2}$

This is a yellow compound. As you mentioned, the compound was white, so it isn't strontium nitrite.

$\ce{Sr(NO3)2}$

$$\ce{Sr(NO3)2 + H2SO4 ->   2HNO3 + SrSO4}$$
Both compounds $\ce{HNO3}$ and $\ce{SrSO4}$ are white or colorless. After looking some more stuff up, it turns out that sulfuric acid and nitric acid don't react (at room temperature) to produce any yellow/brown $\ce{N2O4}$ gas.


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Enantiomer of Amphetamine

I saw this graphic in Organic Chemistry by Klein (2nd edition too!) and this has to be wrong, correct?
The mirror images should have the double bond on the far side of the ring in both images, right? I am sure, unless my visualization skills have totally failed me.

Here's the result when I flip the entire image using an online mirror effect creator:

In short, formulae themselves are not to be mirrored, they are just representation of discussed objects – molecules.

The graphic from the book is correct, those two molecules are mirror images.  Same for your drawing, those are the same two mirror images as in the book.  Remember that the electrons in the aromatic ring are delocalized.  The double bonds in the aromatic ring are not fixed.  Where you choose to draw the double bonds in aromatic rings is your choice and does not play any role in determining if molecules are enantiomers.  Bottom line, the following three structures all represent the same thing.

Ah, looks like I still lost my wits. Thank you!


text:
Protoporphirinogen IX formation in Heme synthesis

This picture shows one of the step of heme synthesis. In this step, two propanoate groups of Coproporhyrinogen III are decarboxylated to from protoporphyrinogen IX. The enzyme used is oxidase. Oxidase enzymes catalyze the removal of hydrogen in the presence of oxygen to from water or hydrogen peroxide.
So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?

Please don't cross post your question.  You had posted this one originally on Cognitive Sciences where it's way off-topic.

So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?

No.
Firstly, note that there are 4 more hydrogen atoms in the reactant than in the product (one from each carboxylic acid group and one where each double bond forms).
However, at physiological pH, the two protons on the carboxyl groups will not really be there.  The equation in the question also ignores the 4 electrons transferred.
As for the other two protons, a 5’-deoxyadenosyl radical abstracts a hydrogen atom (proton + electron) from each propionate side chain.
See Coproporphyrinogen III Oxidase Enzyme and references cited therein for more information.


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Assessing fire hazard of compounds
Following up on a somewhat too broad question about amateur chemist safety, this will be the first to address some of the issues discussed there.
I often use the NOAA MSDS database (material safety data sheet) to investigate the chemicals I'm working with (they also have a nice reactivity predictor). As an example, let us examine the compound cyclohexylamine (I chose this as leading example for these followup question, because of the availability of extensive research in my country)

sometimes used as a plasticizer or catalyst. As an amateur I would carefully go through the MSDS and notice for example that it is flammable, caustic compound, scoring a 3 on flammability and health in the NFPA 704 hazard diamond.
As an amateur hobby chemist without all comfort of a basic lab (aside from safety goggles, gloves, lab coat, fire blanket), how would I judge if I can safely handle this compound with regard to its flammability, aside from conforming to the Protective Clothing section?
For example, I could imagine that flammable liquids with high vapor pressure (11 mm Hg for cyclohexylamine according to NOAA, I guess this is measured at a standard temperature as vapor pressure varies with temperature) or low auto ignition points (560 ° F) or low boiling points (274.1 °F) are unsafe for amateurs. Are there empirical (or theoretical) recommendation considering vapor pressure and ignition points, i.e., starting at what value will these factors become too dangerous for the compound to be handled in a relatively amateurish lab.
(I understand that the answers will be partly subjective, as for all answers regarding safety, but please try to give an objective as possible answer, preferably referring to external sources)

@jonsca No problem. I totally agree with you that it is much more useful for me as well as for other potentially interested home chemists in this fashion. Some other related questions will follow.

If you do not have a fume cupboard or an extractor hood, or whatever you might call it, do not make experiments inside, do them in an open space outside. It's like working with paints - always have an open window.

I guess I don't completely agree with the premise of your question. Gasoline is highly flammable, even potentially explosive, yet "amateurs" deal with it constantly. The key is that it is stored and used properly, from the pump to your gas tank. It becomes less safe if you pump it into a gas can, but gas cans also have certain requirements for storage safety.
So I think the issue may be what sorts of flammables would an amateur have a high chance of storing or using inappropriately. For example, any pyrophoric compound would be a good candidate to avoid, as most require a way to exclude air and moisture, which often requires specialist equipment.
Solvents with low flash points and high vapor pressure should also be treated with respect, but rubbing alcohol (isopropanol) or acetone might qualify for that designation. Storage in airtight fireproof containers in cool and dry places is often sufficient. Better yet, use a flammables cabinet. This simple step is something an amateur might skip for cost, but it may be the most sensible precaution. I have flammables cabinets, and I use them for ordinary hardware solvents along with experimental chemicals.
Finally, work safely! No flames, as they say in organic chemistry. Use hotplates or heating mantles rather than burners. If you must use flames for some reason, put away all flammables and keep them a safe distance. Pay careful attention to any spills and attend to them immediately before resuming work. And perhaps key for amateurs, small quantities! No need for an amateur to have gallons of flammables on hand. Even professional labs are trying to reduce quantities in the name of green chemistry.
EDIT: My go-to source for using and disposing of hazardous chemicals is Hazardous Laboratory Chemicals Disposal Guide by Margaret-Ann Armour. In addition to disposal techniques, each chemical has an extensive incompatibility section about what combinations can make a chemical even more hazardous, and all these entries have literature references. I use this book literally every time I do an experiment. It deals with much more than flammables, but flammables are definitely present.

http://dx.doi.org/10.1201/9781420032383 quite pricy, but maybe available through a library.

Here is an illustrating link: http://www2.galcit.caltech.edu/EDL/public/flammability/USBM-680.pdf I hope it has a good analysis of what you want.

Seems interesting, but an 80-page pdf link isn't really an answer. I suggest you remove it (or transform it into a comment) or improve it (for example by providing a good excerpt of the information in the document provided).

Two good comments others made:

Avoid pyrophorics
Avoid open flames
Work in a very well ventilated space (outdoors if necessary)

I'd add the following:

Use the smallest quantities necessary
Use the lowest concentrations necessary
Keep a fire extinguisher nearby if working with anything flammable

Also, don't get too caught up in the NFPA labeling. The information there is primarily meant to communicate hazards to first responders. Instead, read through the Safety Data Sheet and look for information on the types of hazards associated with the chemical, as well as the recommended personal protective equipment.
A potentially useful resource is the ILO Chemical Toolkit. This is meant primarily for small and medium enterprises, but will help you to get a sense of the controls that would be considered necessary in industry. At the hobby level, you can certainly scale down accordingly - just be aware that working with hazardous substances is inherently risky and it's up to you to be aware of and to manage/mitigate that risk.


text:
π Bonding in Carbon Dioxide
I am fairly sure the first diagram I drew for carbon dioxide is wrong in terms of showing π bonding. This is because we use a π orbital twice, which isn't possible. The second diagram corrects this by realizing there are two unused p orbitals on the carbon.

EDIT: This is what I got for allene. Again we have an central $\ce{sp}$ carbon and this implies that two p-orbitals are free to form π bonds.

What’s the actual question?

In terms of a simple bonding concept the answer of ron is correct, however due to symmetry constraints it is not possible to distinguish the two $\pi$ systems. The reason for this is, that these $\pi$ orbitals are degenerated and perpendicular to each other.

Carbon dioxide
The molecule is very symmetric and one of only a few with the point group $D_{\infty h}$. That also imposes constraints on how the spatial distribution looks like.
Along the molecular axis ($\ce{O-C-O}$) is a rotational axis of infinite fold, $C_\infty$, as the molecule is linear. Perpendicular to that axis there is a horizontal mirror plane $\sigma_h$ passing through the centre of mass, i.e. the carbon. Additionally there is an infinite number of two-fold symmetry axis $C_2$ lying in that plane, which is the criterium for $D_{xh}$ point group. In line with the main axis there is also an infinite number of vertical mirror planes $\sigma_v$ (which is implicit due to the horizontal mirror plane).
For the sake of convenience we can define an upper and a lower half of the molecule (taking the cartesian $z$ axis as the main symmetry axis). These two halfs are indistinguishable, hence their orbitals must reflect that.
Having a look at the $\pi$-orbitals, clearly shows that (enlarge):

So your first drawing was not too wrong after all. The only thing you forgot is, that there is an identical orbital perpendicular to that.
$\ce{CO2}$ is also a nice example to see the boundaries of the hybridisation concept. In this example, due to symmetry, all atoms have to be $\ce{sp}$ hybridised. And that matches all the $\sigma$ orbitals (enlarge).

Allene
The allene on the other hand is not linear and has a much lower symmetry. There is the main symmetry $C_2$ axis (for convenience) along the $\ce{C-C-C}$ axis, accompanied be two $\sigma_v$ mirror planes. There are also two diagonal symmetry planes $\sigma_d$ right between them. Perpendicular to the main symmetry axis we have two more $C_2$ axis (also perpendicular to each other). The point group is therefore $D_{2d}$ and has a lot less symmetry constraints.
However, the orbitals are actually (surprisingly) quite similar though. We can define $\sigma$ and $\pi$ orbitals accordingly. The biggest distinction is still in the $\pi$-orbitals (enlarge).

The Lack of symmetry and therefore less constraints make is possible, that the orbitals combine in a different manner. The top and the lower half of the molecule does not have to be the same anymore, hence the orbitals are more localised to one bond in particular. A notably thing is, that hydrogens take part in $\pi$ bonding (Hyperconjugation).
The sigma orbitals are quite the same as in $\ce{CO2}$ (enlarge).

However, the traditional bonding picture of localised orbitals is quite valid here. The central carbon is still being $\ce{sp}$ hybridised and the moieties are $\ce{sp^2}$ hybridised.

Ketene is a combination of the two above, with even lower symmetry ($C_{2v}$). Therefore the conventional bonding picture should hold.
For Oxygen it is usually always good to check, if hybridisation is a good concept to describe the bond. When bonded to more than one partner, the answer is usually yes. If less, then there is a good chance, that hyperconjugative effects outweigh the information gained from the hybridisation concept.

Additional fun note
One could argue that the carbon suboxide ($\ce{C3O2}$) is not too different, but here again you will be surprised, as it has a very shallow bending potential. (crystal structure, vibrations)

In your second comment to ron's answer below, did you mean a 4 electron 3 center bond instead of 2 electron?  BTW very good and clear explanations here -- I wasn't aware of the subtlety you called attention to, but it makes sense

@SilvioLevy Thank you. Yes, i think i meant that. Although there is not really a point in differentiating these. The four electron bond has to be doubly degenerated mo, while the two (and one) electron case does not have this restriction.

I wish to synthesize a new answer by combining Martin and ron's answers.  I think the real issue is to ensure that the cylindrical symmetry of the electron density of $\ce{CO_2}$ is preserved: that is, $\ce{CO_2}$ should be like a "rod".
In the image below, I have drawn two "resonance structures" (valence bond configurations) that must be combined in order to generate the cylindrical symmetry of the electron density, while maximizing the conceptual relationship with resonance structures.  I have chosen to use a) unhybridized C and b) two ${sp}^2$ fragments for the O's.
A small missive: Of course, multiconfigurational MO theory and multiconfigurational VB theory are equivalent.  However, certain problems can be describe most economically in either MO theory or VB theory.  This is a case where the electronic structure is probably easiest in MO theory, since it requires only one configuration to have cylindrical symmetry (+1 for Martin's answer).
But your question is about "resonance structures" for $\ce{CO_2}$ (+1 for ron's answer), but the constraints on the density mandate the use of the admixture of two resonance structures.  If we use only one resonance structure, it is easy to see from the lone pairs that the density would not be cylindrically symmetric!
(Please note that I am not sure about the sign of the combination, it may be - since we are assuming a singlet state.  And maybe I am overthinking it. Perhaps it should be a double-headed arrow for resonance structures.)

In terms of VB-Theory these will be the most contributing configurations and the plus would be correct. In VB-Theory there are configurations, which can be described as resonance structures. The total wave function is a linear combination of all configurations (coefficients may also become negative).

This is a pretty good picture although I'd note that there are antibonding orbitals as well which (I think) some correspond to the left out p orbitals and which would end up being in the place a bond would but because there's already been a bond there's now an antibond.

Your second drawing is correct.   The bonding scheme in carbon dioxide is the same as that in allene and ketene in that the two pi systems are orthogonal.  Draw or build a model of allene and notice what effect the orthogonality has on the two planes the hydrogens lie in.

Thank you! I have edited the text of my post to include a pi-bonding diagram for allene. Same concept it seems like.

Yep, same concept.  In allene, the terminal hydrogens are in different planes, 90 degrees to one another, just as you've drawn.

It is only correct when one considers the hybridisation to be a restriction, rather than a tool of describing the bonding situation. While for the allene this works very well, it does not for carbon dioxide.

@Martin  Specifically, how does it fail for carbon dioxide?

@ron See my answer, but in short: Due to symmetry restrictions the pi orbitals are delocalised over the full molecule - they cannot be assigned to one or the other bond (2 electron 3 centre bonds). Your bonding picture would also not match the electron density.

@Martin 1) Do you agree that carbon dioxide and allene can be viewed as analogous cases in terms of structure and hybridization?  2) Are each of the two pi orbitals in allene delocalized, as you say above for carbon dioxide,  over the full molecule?

Hybridisation is a concept that may be twerked into fitting any bonding description. In terms of canonical molecular orbitals it is quite obvious, that $\sigma$ orbitals are very similar, while $\pi$ orbitals still are similar. In terms of structure, I would go as far as to state that the central carbon is very similar. However the values for the $\pi$ orbitals are significantly different. Carbon dioxide is linear and therefore very different (from most other molecules). 2) I do not quite understand the question, sorry.

@Martin and ron: I'm really not following this conversation very well.  If we were to assume a multiconfigurational valence bond description, then I believe the answer is a linear combination of the bottom ("right") O-C-O with the complementing (quarter turn along the molecular axis) structure.  The admixture would properly conserve the Dinfh electron density.


text:
Proper Resonance Structures
Would this be proper resonance structures? I know there's a lone pair on the top carbon. Having a carboanion isn't favorable as opposed to a negative formal charge on the more electronegative carbon so I flipped a bonding pair to oxygen's valence. I then proceeded to move the pi electrons inside the ring as well. This seems to satisfy all octet requirements.

For the second resonance diagram I saw the negative charge on the oxygen and decided to take one of its lone pairs and make it a bonding pair so that oxygen has 0 formal charge. This in turn necessitates pushing around a few pi electron pairs inside the ring. Same thing for the third one; I tried giving nitrogen a positive formal charge.

Yes, they are all valid resonance structures.  In the first series, there is a third resonance structure you could draw with the negative charge alpha to the carbonyl carbon. In the second series, the phenoxide anion, you can draw two more structures.  One with the negative charge in the other ortho position and one with the negative charge in the para position.  This significant resonance stabilization of the phenoxide anion explains why phenol is such a strong acid.  In your third series there is also a third structure with the positive charge situated alpha to the nitrogen.

Yep! I noticed that the second is the conjugate base of phenol. The delocalization of negative charge in the conjugate base of phenol provides the conjugate base stability and this is what allows phenol to be an unusually strong acid as far as hydrocarbons go.

Phenol can cause severe burns.  It is also rapidly absorbed through the skin leading to a host of other problems, including death.

Sounds like hydrofluoric acid, which isn't a strong acid per se, but dissociates to a significant extent so that skin contact can cause significant absorption of fluoride ion (which has a pretty high electron affinity) and can cause heart attacks.


text:
Would a carbamate be attacked twice by an alkyl lithium cuprate reagent?
For example, say you react ethyl carbamate with methyl-$\ce{LiCu}$; would two methyl groups displace both the amide and the ester, creating an isopropyl alcohol?

I don't think so, carbonyls don't typically react with Gilman's reagent (dimethyllithium cuprate) unless the carbonyl happens to be $\ce{{\alpha}-{\beta} }$ unsaturated. So I would expect no reaction under normal reaction conditions.

Wouldn't it at least attack once to remove the ethanol group? Because, I know that a Gilman's reagent attacks carboxyl derivatives, such as esters, acetyl chlorides, and amides, but it only does so twice. That's why I was asking if GR would attack twice on a disubstituted acyl derivative.

I don't think so.  I know it reacts with acid chlorides and a,b-unsaturated carbonyls, but maybe under special conditions it will react with other carbonyl compounds.


text:
How would you name this organic compound?

Because there are at most seven consecutive carbon atom, and four alkyl groups attached to the chain of carbon, I would name it 2,4-tetramethyl heptane.

Yes, it's tetramethyl heptane, where are the 4 methyls? 2,4,4,6-tetramethylheptane.

You are nearly right. You forgot to include that there is also a methyl group on carbon atom 6 and the name should also reflect that there are actually 2 methyl groups on carbon 4, so it's 2,4,4,6-tetramethylheptane.

It is 2,4,4,6-tetramethylheptane. The parent chain is circled in red. And there are 4 methyl substituents.


text:
Determine Cu+2 and SO4-2 concentration using Potentiometry?
The title says it all. I've been researching as well but all I could come close was solid state ion selective electrodes. Could someone please help me out?
There should be two methods(or one) to measure the concentrations.

What's wrong with electrodes?

You certainly need electrodes to measure a potential, like a potentiometer for a potentiometric titration. (While Copper is better measured via photometrics and sulphate by gravimetrics)

What is the range of the components? This is very important because this will determine if titration or photometry is used (or a combination of both).
Also interfering components should be known. For eaxample. Barium is known to react with a lot of other components (solubility table). Also EDTA reacts with oter components (Also depending on the pH (EDTA complexation constants)
Titration:
Copper and sulfate can be determined in one analysis using and copper ISE. This is done by adding an NH3 buffer (pH of solution = 10) and adding an exxess of Barium nitrate. Barium will react with SO4 and the free barium is titrated. Titrate the solution with EDTA. This first inflection point will be copper, the second one the free barium.
Keep in mind that EDTA reacts with a lot of other components. Also if the concentration difference (molar) between copper and barium is large, high titrant usages are to be expected
Photometry:
Copper can be determined using an variaty of chemical substances (e.g. cuprizone, Bathocuproine). The choise depends on range, interfering components ect..
Sulfate is possible. However, titration is preferred.

@Tharindu Any progress on the application?


text:
CFSE/LFSE Explains Hydration Enthalpy
Time and time again textbooks and websites state how wonderful it is that the deviations of Hydration Enthalpies from a linear relationship can be accounted for by the CFSE parameter for the particular complex.
My question is what is the mechanistic explanation in terms of water and the metal ion interacting ? Why does CFSE lead to much more negative hydration enthalpies ?
EDIT : A simplie overview of what I'm talking about : http://sst-web.tees.ac.uk/external/u0000892/Web%20thing/CoordChem/CFSE5.htm

What is CFSE? I am not familiar with all english acronyms and never heard that before.

Crystal Field Stabilisation Energy also known as Ligand Field Stabilisation Energy (LFSE).