text:
Stereochemical outcome after elimination followed by addition

One of the following diastereomeric chlorides undergoes racemization on heating whilst the other is returned unchanged. Explain.

From the reaction conditions I can see that there must be an elimination of $\ce{HCl}$, presumably via E2 with the base being the amine group on another molecule. Then addition back to the double bond.
However, I have a couple of questions-
Firstly, wouldn't $\ce{HCl}$ add back onto the alkene the other way around so as to form the more stable tertiary heteroatom stabilized carbocation? Which would then be attacked by the chloride ion thereby inducing racemization, and thus it should be impossible to be "returned unchanged".
Secondly, why would one isomer undergo racemization and the other remain the same?

Upon heating, there is an intramolecular $\mathrm{S_N2}$ reaction where the nitrogen lone pair attacks the chlorine-bearing carbon, which gives an aziridinium ion. The stereoelectronic requirement for a "backside attack" results in the formation of a specific diastereomer, illustrated here for the first of the two compounds:

Since there is absolutely no difference between both carbons in terms of sterics or electronics, the free $\ce{Cl-}$ ion can now attack either carbon with equal probability to give a mixture of two products:

The second compound reacts in exactly the same way:
$\hspace{20 mm}$
However, no matter which carbon the $\ce{Cl-}$ ion attacks, the same product is obtained - you can verify this yourself. The product is the same as the starting material.


text:
When is a reaction reversible?
When speaking in thermodynamic terms, when is a reaction reversible?
My attempt: A reaction is reversible when $\Delta\text{S} < 0$, because according to the second law of thermodynamics, "in every natural thermodynamic process the sum of the entropies of all participating bodies is increased" (Wikipedia.org), therefore if a reaction is caused and has negative change in entropy, a reversed reaction should be natural.
However, this is not the answer that is absolutely correct, nor a rigid solution or a rule. Can anyone help me out? (I have searched on the internet, and although thermodynamic reasons are preferred, I do not mind answers to be molar or solution chemistry, as long as a thermodynamic approach is included as well or a statement that there is no apparent thermodynamic rule that can be applied to this question.)

Also, is a spontaneous reaction reversible?

"Reversible" is not binary.  Both the forward and backward reactions always occur and the equilibrium system never has zero reactants or zero products.  Thus, irreversible reactions are called this not because they cannot be reversed - they absolutely can - but because reversal is impractical.  The equilibrium constant may be so skewed toward product that the reactant quantities at equilibrium would be undetectable.  The products may be gases that escape solid reactants, which ensures that the system never reaches equilibrium in the open anyway.  There may be a lot of side reactions on the products.
A "reversible" reaction just lacks factors like this so that we can have measurable quantities of reactant beside measurable quantities of product at equilibrium.  We add some reactant, watch the reaction go forward.  We add some product, it goes in reverse.  See?  It's reversible.  There is no exact cutoff in any measure.

This is a great answer, but the remaining question is, is there a way I could know this from looking at things like Gibbs Free Energy, Entropy, internal energy, etc. or if there isn't, are there specific limit values assigned to the equilibrium constant?

@PhyCS a question to ask would be "if we gave you a formula to mark chemical equations as reversible or irreversible based on Gibbs Free Energy, Entropy, etc., what would you do with that information?"  I can tell you that reactions deemed irreversible in a science class can turn out to be highly reversible at high pressures or temperatures in industrial manufacturing because the "useful" meaning of the word shifts based on what equipment you are allowed to use to reverse the equation.

As an example, the oxidization of aluminum into aluminum oxide is typically considered "irreversible" in most situations, because at room temperature/pressure, aluminum really likes to oxidize and does not tend to revert back to aluminum after doing so.  However, aluminum smelting factories literally make their living reversing this reaction (using 350kilo-amp electrodes to do it!)

As an addition to what Cort said, consider a color spectrum. What is the wavelength where red stops and orange begins? A similar situation occurs with reversibility--I might have it in my head that around x is where a reaction is generally irreversible, but that number can move a lot depending on the context. Numerical cutoffs are nice and easy to memorize, but they don't always accurately describe reality.

@CortAmmon is there a temperature (and pressure) where the oxide spontaneously converts into aluminium and oxygen (at some positive net rate), or does it always require an external voltage to be applied?

To add to @sqykly's excellent answer, the equilibrium constant, which represents the degree to which a reaction may be considered 'reversible', is related to the Gibbs free energy change for the reaction by: $$\Delta G_\text{rxn} = -RT\ln K$$
A very large equilibrium constant, which corresponds to a large negative $\Delta G_\text{rxn}$, essentially means that the reaction goes to completion. A very small equilibium constant, which corresponds to a very large positive $\Delta G_\text{rxn}$, means that the reaction essentially does not occur.

However, as we can see from the graph above, the equilibrium position will always have some amount of reactants and products present because the shape of the curve is the same for all reactions, only with different values. The reaction shown is exergonic (it has a negative $\Delta G_\text{rxn}$) but an endergonic reaction will have the same profile, only the position of 100% products will have a higher Gibbs energy than that of 100% reactants.
A 'reversible' reaction is one which has an equilibrium constant somewhere in the middle and so changing the concentration of reactants or products has a significant effect on the position of equilibrium. Generally, a reaction where $|\Delta G_\text{rxn}| < 30 \mathrm{kJ~mol^{-1}}$, can be considered reversible. At $\mathrm{298~K}$ this corresponds to $5 \times 10^{-6} < K < 2 \times 10^5$. Interestingly, this means that even if the equilibrium constant is less than one (i.e. $\Delta G_\text{rxn} > 0$) then you can still extract useful products from a reaction by removing them as they form.

That last bit is what I was trying to get across toward the end of the first paragraph of my answer - pretty much any reaction where you get rid of the product immediately is going to look irreversible.  Biochem is what I went to school for, and that's just full of examples of systems that are perfectly reversible - as soon as they're dead and stop eliminating this or that product.  Same is true for physically removing the product.  When you put it like that, though, the removal has a free energy too - perhaps if we calculate the whole system we will be able to define reversible more exactly?

Jeez, this answer is very detailed. Thank you all for your answers. I will return if I need further clarification. :)

You're mixing two things - from thermodynamic point of view any process would be reversible when $\Delta\text{S} = 0$ for whole isolated system; which never really happens.
What sqykly is saying is about chemical reactions, which are reversible in the sense that they can be reversed from chemical point of view concerning substrates and products, not energy and entropy.

Thank you for the clarification. What I'm getting from both your answers are that there are no exact thermodynamic quantities that can limit the reversibility of reactions? All reactions are reversible?

Yes.  All reactions are  reversible.

Another associated issue that often arises is the criterion for a reaction to be spontaneous.  The criterion that, at constant temperature, if a reaction is spontaneous, $\Delta G^0$ in going from pure reactants to pure products at 1 bar will be negative is just a rule of thumb which, over the years, has confused students to no end.  If you start out with pure reactants and mix them, spontaneous reaction will always occur to some extent.  It's just that, if $\Delta G^0$ is positive, the reaction will tend to proceed  to a lesser extent when equilibrating, and if $\Delta G^0$ is negative, it will tend to proceed to a greater extent when equilibrating.  So, in this sense, every reaction is spontaneous.

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text:
Explaining temperature change in an exothermic system.
In a neutralization reaction, is heat given out to the system, thereby raising its temperature? Suppose we put a thermometer in the acid-base mixture and there is an increase in the thermometer reading. Is this increase simply because part of the reaction's heat given out is supplied to this thermometer, or because the water solvent is actually warmer? In other words, if we measure the temperature of the system before and after the reaction, will there have been an overall decrease in the temperature, or increase (and please explain why)?

Does this answer your question? How can decreasing in temperature indicate an endothermic reaction?

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).
So in an neutralisation reaction, the system is the actual acid and base while the surroundings is the solvent (the water). Since it is an exothermic reaction, this means that energy is released from the system to the surroundings. Since the water gains more kinetic energy from the system as the reaction occurs, its temperature will increase. Also this means that the system loses energy and therefore become colder. So to answer your question, the solvent gets hotter while the system gets colder.

so then if we put a thermometer in it, we'll be measuring the temperature of the water?

Technically you will be measuring the temperature of the system and the surroundings combined, however since the solvent is in such excess, it is basically measuring the temperature of the water.

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text:
Calculating Solubility Product of AgCl from electrode potential
I have to use a measured electrode potential to find the solubility product of $\ce{AgCl}$. In our experiment, we first mixed one drop of $\pu{1.0 M}$ $\ce{AgNO3}$ with $\pu{10 mL}$ of $\pu{1.0 M}$ $\ce{NaCl}$ to precipitate $\ce{AgCl}$. For the purposes of our experiment, we are to assume that the concentration of $\ce{Cl-}$ remains $\pu{1.0 M}$, and use the Nernst equation to determine the value of $\ce{[Ag+]}$. The reaction that occurred in our cell was between $\pu{1 mL}$ of the solution of $\ce{Ag+}$ and $\ce{Cl-}$ left behind from the precipitation reaction described above, and $\pu{1 mL}$ of of $\pu{0.10 M}$ $\ce{Zn^2+}$. The reactions were all carried out at $\pu{25 ^\circ C}$, and the voltage obtained by our group was $\pu{0.85 V}$.
My Method
The half-reactions going on in this cell were:
\begin{align}
 &&\ce{2Ag+(aq) + 2e- &-> 2Ag(s)} & (E^\circ &= 0.80)\\
+&&\ce{Zn(s) &-> Zn^2+(aq) + 2e-} & (E^\circ &= 0.76)\\\hline
 &&\ce{2Ag+(aq) + Zn(s) &-> Zn^2+(aq) + 2Ag(s)} & (E^\circ_\mathrm{cell} & = 1.56)
\end{align}
From here, I used the Nernst equation to solve for $\ce{[Ag+]}$, with $n=2$ moles of electrons transferred and $Q = \ce{[Zn^2+]}/\ce{[Ag+]^2}$:
$$0.85 = 1.56 - \frac{0.0592}{2}\log\frac{0.10}{[\ce{Ag+}]^2}$$
Solving this for $\ce{[Ag+]}$ yields approximately $3.21 \times 10^{-13}$.
I then applied the law of mass action to determine the solubility product of $\ce{AgCl}$: $$ \ce{[Ag+][Cl^{-}]} = [3.21 \times 10^{-13}] \times [1.0] = 3.21 \times 10^{-13} $$
However, this number is quite different from the expected value of $1.8 \times 10^{-10}$. I am wondering if my math or approach is wrong or if my  group's data is simply wrong?

we are to assume that the concentration of $\ce{Cl−}$ remains $\pu{1.0 M}$ But why?

@NisargBhavsar Assuming a drop of aqueous solution is about 0.05 mL, this is 0.5% of the 10 mL of sodium chloride solution volume, so 1 M drops to about 0.995 M. Reasonably safe to neglect the decrease. I guess the cell uses zinc and silver metal electrodes, not zinc and Ag/AgCl electrodes.

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text:
Why are oxyacids strong oxidizing agents?
I saw that all the oxyacids are strong oxidizing agent.
I don't know why all the oxyacids are strong oxidizing agents?
What is the reason?

That's at best an oversimplification. Yes things like nitric, perchloric, and sulphuric are thermodynamically strong oxidising agents, if sometimes kinetically slow. But then you have perrhenic acid, which is at best a weak oxidising agent, carbonic acid which I don't think I've ever though about in a redox context, and "sulphurous acid" (if it exists) which is normally though of as a reducing agent.

Your premise is incorrect. Some oxyacids are oxidising agents. Some are not. Take e.g. carbonic acid. You cannot oxidise anything with carbonic acid, really.
Also, the choice of words is bad. Sulphurous acid (or hydrogen sulphonic acid; it doesn’t really make a difference; both are structural isomers) is also an oxyacid ($\ce{H2SO3}$) however it is a pretty good reducing agent oxidising itself to sulphuric acid. Heck, technically $\ce{H4SiO4}$ is an oxyacid, too, but it won’t oxidise anything at all.
So reducing the question to where it is correct we arrive at:

Why are some oxyacids where the central atom is in its highest possible oxidation state strong oxidising agents?

And it basically answers itself. Depending on the electronegativity of the central atom it will be more or less ‘happy’ to cede all its valence electrons to oxygens and be the centre of an oxyacid. For those that don’t like that lot in life, most notably nitric acid, they will often find species whose electrons they can take back (the atoms are electronegative after all) and thus oxidise them.

Please leave my British ph, @ortho ;)

It is not about British or American. http://www.nature.com/nchem/journal/v1/n5/full/nchem.301.html http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf

@orthocresol (I'll admit, I hadn’t rolled back if you hadn’t incorrectly assumed the wrong thing about sulphonic.)

Yeah, it seems I assumed wrongly.

Oxidising agents are those which take electron from others. Let us take a example of oxyacid say chloric acid that is HClO4 .In this compound the oxidation state of chlorine is +7 which is not highly favourable for chlorine because it is a halogen and will tend to gain electron. Moreover here chlorine has +7 oxidation state so it will have more affinity for electron so it will oxidise other compound to take electron . That's why oxyacids are strong oxidising agents.


text:
Solubility of methanol and ethanol in water

Which substance is the most soluble in water?

$\ce{CH3OH}$
$\ce{CH4}$
$\ce{C2H6}$
$\ce{C2H5OH}$

I wasn't able to decide between 1 and 4, as both have hydroxides, but the answer is 1. How?

The question should have specified under what range of conditions. Ethanol and methanol are both completely miscible with water under normal lab conditions. Not so under extreme conditions.

The distinction between methanol and ethonol, otherwise miscible with water in any ratio, can be done by easiness of salting them out from the water phase. By addition of the enough salt to the alkohol solution, ethanol phase separates easier..

For primary alcohols, the trend is the longer the chain, the less soluble.
However, at room temperature, methanol, ethanol and propanol are all completely miscible with water.
At lower temperature, below the freezing point of water, there is finite solubility.
See the phase diagram that is Fig. 1 in CRYSTAL GROWTH OF WATER ETHANOL MIXTURE and that in Fig. 1 of Solid-Liquid Phase Diagram of the System Methanol-Water
Which alcohol would you choose for windshield deicing solution?

Options 1 and 4 both have polar $\ce{C-O}$ bonds which make them soluble in water. The question is, which one is more soluble? To answer that, notice that the non polar part of methanol is smaller, so it will be more soluble.

How does a alcohol get soluble in water? The $\ce{H2O}$ molecule acts as bipolar and the positive pole forms some weak hydrogen attraction (name this 'bond' if you like)  with the oxygen of alcohol. Now it's obvious, the more bulkier the group around oxygen, the less space around oxygen to form this bond. Hence $\ce{CH3OH}$ is more soluble in water than $\ce{C2H5OH}$. After all, $\ce{-C2H5}$ group is more soluble than $\ce{-CH3}$ group.

Except that, as many others have stated, both methanol and ethanol are infinitely miscible in water at room temperature.

Yes ,you are quite right. But I just wanted to say that if there is even a slight difference in their solubility( say at the 10th decimal place {solubility in mol/litre}),  then its probable reason is the one I gave.

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text:
Does solid fat float in liquid fat?
Ice floats in its liquid state, water, because it is less dense, and I know fat also floats in water BUT will a piece of fat dropped into a pot of melted fat also float? or will it sink and what is it about its molecular structure that allows this?
I've searched google for about three hours and all I can find are articles on fat floating in water, so any help is greatly appreciated!

Most likely no. Typically solid is more dense than the corresponding liquid. Water is very unusual in this respect.

Water is not as unusual as you might think. There are at least four elements that behave similarly, and an untold number of alloys, polymers, ... that do as well.

Well, I never said it is unique.

@IvanNeretin - that is true.  There is a pretty strong meme (seems to start pretty early in school) that water is unique. However, I would not go even so far as to say 'very unusual' - it is just the most commonly encountered example.

so the fat would sink?
what about when gravey or or juices from cooking meat cool down and there is a layer of fat at the top? or is that just something completely different?

@JonCuster While I generally agree, which polymer are you thinking about? I don't think I know one. ;-)

@cev When cooking you get a mixture of fat and water. The layer of fat on top is on top of what is mostly water.

I thought this question was a good idea for a home experiment to do with my 8th grade son.
Into a block of dry ice, we carved a conical mold hole with the pointed end of scissors and poured canola oil into the hole:

We covered this with a second block of dry ice for about a minute and extracted the frozen oil with forceps:

Then we dropped the frozen oil into a jar of liquid canola oil:

And it sank.

Nice experiment!!

My chemistry teacher in year ten just made oil cubes (think ice cubes) from sunflower oil and showed us the same thing. But yes, nice experiment ^^


text:
Why is phenol more polar than alcohol?
Why is phenol more polar than alcohol. I mean the benzene ring seems to be extremely unpolar. Furthermore, is the hydrogen bonding stronger in phenol than in alcohol? Are polarity and hydrogen bonding related (in hydroxy compounds)?
I am doing TLC and I was taught that hydrogen bonding can occur with the alcohol/phenol and that alcohol has stronger hydrogen bonding to the silica gel than it would have despite the lower polarity a lower Rf value than phenol.
But to be honest I am very unsure about those teachings.

When you say "phenol is more polar than alcohol", how are you quantifying polarity? Phenol has only moderate solubility in water, while ethanol and other small alcohols are infinitely miscible. Several small alcohols also have an appreciably higher dipole moment (~1.6 D) compared to phenol (1.2 D).

The OH group of phenol is indeed more polarized - and a stronger acid - than the OH group of an aliphatic alcohol - maybe that is what you are asking. There are two contributions to this:
(1) inductive: the sp2 hybridized carbon participating in the sigma bond to the OH group is more electron-withdrawing than the sp3-hybridized carbon of an alcohol bonded to OH; and
(2) mesomeric: the phenyl group allows non-bonding electrons on oxygen to partially delocalize into the ring via pi bonding.


text:
How to convert from spin orbitals to spatial orbitals in the Hartree-Fock approximation?
I need to calculate some of the more complicated self-energy terms from chapter 7 of Szabo and Ostlund's "Modern Quantum Chemistry", and I'm having trouble converting summations from spin orbitals to spatial orbitals.
Exercise 2.18 (page 85) is indicative of what I'm doing wrong, since I can't get the correct answer.
The exercise is to show that the leading correction to the Hartree-Fock ground state energy
$$
E_0^2 = \frac{1}{4} \sum_{abrs} \frac{|\langle ab || rs \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
$$
where the summation is over spin orbitals ($a$, $b$ over occupieds, $r$, $s$ over virtuals), can be converted to the following sum over spatial orbitals:
$$
E_0^2 =
\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ \langle ab | rs \rangle (2 \langle rs | ab \rangle - \langle rs | ba \rangle)}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
$$
The integrals are in "physicists notation", which means that
$$
\langle ij|kl \rangle =
\int \chi_{i}^*(\mathbf{x}_1) \chi_{j}^*(\mathbf{x}_2) r_{12}^{-1}
     \chi_{k}(\mathbf{x}_1) \chi_{l}(\mathbf{x}_2) \,\mathrm{d}\mathbf{x}_1 \,\mathrm{d}\mathbf{x}_2
$$
and
$$
\langle ij||kl \rangle = \langle ij | kl \rangle - \langle ij | lk \rangle
$$
where the $\chi_i$ are spin orbitals and the variables $\mathbf{x}$ comprise both spatial and spin coordinates.
The following is what I have, I'd appreciate any pointers to get to the right answer.
First, decompose the sum into the different possible permutations of spin for the different orbitals ($a$ and $\bar{a}$ are orbitals with opposite spin), and use the definition of $\langle ab || rs \rangle$:
$$\begin{align}
E_0^2 &=\frac{1}{4}\Bigg(
\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|rs \rangle - \langle ab|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}\bar{b}|rs \rangle - \langle \bar{a}\bar{b}|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}b|\bar{r}s \rangle - \langle \bar{a}b|s\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}b|r\bar{s} \rangle - \langle \bar{a}b|\bar{s}r \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle a\bar{b}|\bar{r}s \rangle - \langle a\bar{b}|s\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle a\bar{b}|r\bar{s} \rangle - \langle a\bar{b}|\bar{s}r \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|\bar{r}\bar{s} \rangle - \langle ab|\bar{s}\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}\bar{b}|\bar{r}\bar{s} \rangle - \langle \bar{a}\bar{b}|\bar{s}\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\Bigg)
\end{align}$$
Note that we do not need to consider terms with odd numbers of orbitals of either spin, since terms of the form $\langle \bar{i}j|kl \rangle$ and $\langle i\bar{j}|\bar{k}\bar{l}\rangle$ are zero.
Next, we use the additional rules $\langle \bar{i}j|k\bar{l}\rangle$ = $\langle i\bar{j}|\bar{k}l \rangle$ = $\langle \bar{i}\bar{j}|kl \rangle$ = $\langle ij|\bar{k}\bar{l}\rangle$ = $0$.
We see that in the above eight summations, the second and seventh cancel completely, while in the middle four, only one term is retained.
Then, expanding the squares and collecting terms, we get
$$\begin{align}
E_0^2 &=\frac{1}{4}\Bigg(\quad
4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|rs \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
-4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ \langle ab|rs \rangle  \langle ab|sr \rangle }
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\Bigg)
\end{align}$$
which is where I'm stuck (besides of course cancelling out the factor 4). Any help in moving towards the correct answer is massively appreciated!

Associated discussion in chat.

Technical Note: This page makes heavy use of MathJax, give it time to load.

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Preamble
I need to get this out of the way first. While I think the exercises in Szabo-Ostlund are a great way to learn maths and understand how lazy chemists and physicists are when it comes to writing, I consider them all painful and of no particular pedagogical value. I don't think it is particular useful to explain everything with the dihydrogen model and leave the generalisation to the user. From my personal point of view, it should be the other way around. With that out of the way, I admire you for working with this manuscript.
It took me quite some time getting into this topic again and I spent various pencils and a lot of paper on it. As you mentioned, you have been doing the same. You also found that the same exercise will be given on S.O. page 352 again (6.8). Because you asked for hints, I will provide the key step as a single formula first, so that you have the chance of finding it without any help. At the end of the post I will explain the reasoning, too.
The Goal
Let us reconsider what our exercise was:

Exercise 2.18 (p. 85) In Chapter 6, where we consider perturbation theory, we show that the leading correction to the Hartree-Fock ground state energy is
  $$\begin{align}
E_0^{(2)} &=
   \frac{1}{4} \sum_{abrs}
     \frac{\left|\bra{ab}\ket{rs}\right|^2}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
\end{align}$$
  Show that for a closed-shell system (where $\eps{i}=\eps{\bar{\imath}}$) this becomes
  $$\begin{align}
E_0^{(2)} &=
   \sum_{a,b=1}^{N/2}\quad \sum_{r,s=N/2+1}^{K}
     \frac{\bracket{ab}{rs} \left(2 \bracket{rs}{ab} -\bracket{rs}{ba}\right)}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
\end{align}$$

And because it is the same but different:

Exercise 6.8 (p 352) Derive Eqs. (6.73) and (6.74) starting with Eq. (6.72).
  $$\begin{align}
E_0^{(2)} &=
   \frac{1}{4} \sum_{abrs}
     \frac{\left|\bra{ab}\ket{rs}\right|^2}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  \tag{6.72}\\
E_0^{(2)} &=
   \frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ab}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  -\frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ba}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  \tag{6.73}\\
E_0^{(2)} &=
   2 \sum_{abrs}^{N/2}
     \frac{\bracket{ab}{rs}\bracket{rs}{ab}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  -  \sum_{abrs}^{N/2}
     \frac{\bracket{ab}{rs}\bracket{rs}{ba}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  \tag{6.74}\\
\end{align}$$

Annotation: It is actually just simplified Dirac notation. There is nothing especially "physicist" about it. And even though they state otherwise I consider this lazy. (p. 350)
The notation $\bracket{ij}{kl}$ refers to the electron ordering 1,2,1,2.
What SO calls the chemists notation is better referred to as charge-cloud notation as the electron ordering is 1,1,2,2. It is usually denoted as $\left[ij\middle|kl\right]$.
Therefore $\bracket{ij}{kl}=\left[ik\middle|jl\right]$.
We need the following first $$\bra{ij}\ket{kl} = \bracket{ij}{kl}-\bracket{ij}{lk}.$$
We also need to consider that the indices early in the alphabet $a,b,\dots\in\mathbb{N}$ refer to occupied orbitals, hence $a,b,\dots<N$ with $N$ being the total number of electrons. Then $r,s,\dots\in\mathbb{N}$ belong to virtual orbitals.
It is further implied that summation signs without any upper limit are considered spin orbitals and to have $N$ as an upper boundary for occupied and an arbitrary number $K$ (total number of basis functions) for virtual.
A sum with multiple indices refers to as many sum signs. $$\sum_{ijkl} \text{terms} = \sum_i\sum_j\sum_k\sum_l \text{terms}$$
Regular two electron integrals
I have no idea what this actually means, so I just made the assumption, that we just consider real functions. Therefore a lot of integrals will be symmetrical.
$$\bracket{ij}{kl} = \bracket{kj}{il} = \bracket{il}{kj} = \bracket{kl}{ij} =
  \bracket{ji}{lk} = \bracket{li}{jk} = \bracket{jk}{li} = \bracket{lk}{ji}  $$
First of all we decode the shorthand and expand the square
$$\begin{align}
E_0^{(2)} &=
   \frac{1}{4} \sum_{abrs}
     \frac{\left|\bra{ab}\ket{rs}\right|^2}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
\\
E_0^{(2)} &=
   \frac{1}{4} \sum_{abrs}
     \frac{\left|\bracket{ab}{rs}-\bracket{ab}{rs}\right|^2}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
\tag{1}\\
E_0^{(2)} &=
   \frac{1}{4} \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
   \\&\qquad\qquad
   \left\{
     \frac{\bracket{ab}{rs}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
   + \frac{\bracket{ab}{sr}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
   -2\frac{\bracket{ab}{rs}\bracket{ab}{sr}}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
   \right\}
\tag{2}\\
\end{align}$$
We can further proof that
$$\begin{multline}
   \frac{1}{4} \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
     \frac{\bracket{ab}{rs}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}\\
=
   \frac{1}{4} \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
     \frac{\bracket{ab}{sr}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}\tag{3}
\end{multline}$$
because $a,b=1$ and $r,s=N+1$ and the integral permutations become identical. I will explain this at the end with an example. This is the key. Therefore we can further reduce to
$$\begin{align}
E_0^{(2)} &=
   \frac{1}{4} \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
    2\frac{\bracket{ab}{rs}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
 \\&\qquad
 - \frac{1}{4} \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
    2\frac{\bracket{ab}{rs}\bracket{ab}{sr}}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
.\tag{4}\\
\end{align}$$
We use symmetry and find (6.73)
$$\begin{align}
E_0^{(2)} &=
   \frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ab}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  -\frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ba}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  \tag{5 $\equiv$ 6.73}\\
\end{align}$$
Closed shell
This is what you tried first, me too. This is where short notation will gently kicks you into your derrière, pardon my French.
Let us consider the definitions for spin orbitals and spatial orbitals:
\begin{align}
\chi_1(\ve{x}_1) &=\psi_1(\ve{r}_1)\alpha(s_1)\\
\chi_2(\ve{x}_2) &=\psi_1(\ve{r}_2)\beta(s_2)\\
\chi_3(\ve{x}_3) &=\psi_2(\ve{r}_3)\alpha(s_3)\\
\chi_4(\ve{x}_4) &=\psi_2(\ve{r}_4)\beta(s_4)\\
\vdots           &\qquad\vdots\\
\chi_{N-1}(\ve{x}_{N-1}) &=\psi_{N/2}(\ve{r}_{N-1})\alpha(s_{N-1})\\
\chi_{N}(\ve{x}_{N}) &=\psi_{N/2}(\ve{r}_{N})\beta(s_{N})\\
\vdots           &\qquad\vdots\\
\end{align}
Since $a,b,\dots,r,s,\dots$ are arbitrary natural numbers it does not really matter which index we choose, but it can lead to great confusion if on the left side $a$ refers to a different number than on the right side.
Hence I choose, for a completely arbitrary number $酒$ of spinfunctions $\chi$ to represent them as a product of a spatial function $\psi$ with a spin component $\alpha,\beta$. The set of functions is therefore half the number of the complete set.
$$\sum_{a}^{(酒)} \chi_a =
   \sum_{c}^{(酒/2)}\psi_c\alpha + \sum_{c}^{(酒/2)}\psi_c\beta,$$
We are basically introducing just a new name of indices to avoid confusion. I will now shorten $\psi_c\alpha$ to $c$ and $\psi_c\beta$ to $\bar{c}$. And this will apply analogous to all other kinds of indices. Therefore we write symbolically:
$$\sum_{a}^{(酒)} a =
   \sum_{c}^{(酒/2)}c + \sum_{\bar{c}}^{(酒/2)}\bar{c}$$
Now we need to expand equation 5 into this formalism. We then will see that it simplifies quite a lot. Without spoiling the surprise, we can see that the number of integrals that survive spin integration is exactly twice in the first term than in the second term, just because it is symmetric.
In general we know that any integral where electron one has $alpha$ spin on the left side and $beta$ spin on the right side will be zero, hence vanish.
Therefore the integrals of the form
$$
\bracket{\bar{i}j}{kl};
\bracket{i\bar{j}}{kl};
\bracket{ij}{\bar{k}l};
\bracket{ij}{k\bar{l}};\\
\bracket{\bar{i}\bar{j}}{kl};
\bracket{ij}{\bar{k}\bar{l}};
\bracket{\bar{i}j}{k\bar{l}};
\bracket{i\bar{j}}{\bar{k}l};\\
\bracket{i\bar{j}}{\bar{k}\bar{l}};
\bracket{\bar{i}j}{\bar{k}\bar{l}};
\bracket{\bar{i}\bar{j}}{k\bar{l}};
\bracket{\bar{i}\bar{j}}{\bar{k}l};
$$
will all vanish.
I will provide the full terms at the end, so you can check your own attempt, but here is equation 5 with all non-vanishing terms. Note that we need to include the limits to the sum here, because we have expanded the spinorbitals into spatial orbitals. Because we are having a closed shell system, the number of electrons $N$ is even, while the total number of basis functions $M$ could be anything. Please also note, that I am not including the starting numbers for the indices, because it will look a bit ugly. Hence it is implied by the notation, that $a,b,c,d=1$, $r,s=N+1$ and $t,u=N/2+1$. (If we're lazy, we go the whole nine yards.)
$$\begin{align}
E_0^{(2)} &=
   \frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ab}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  -\frac{1}{2} \sum_{abrs}
     \frac{\bracket{ab}{rs}\bracket{rs}{ba}}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
  \tag{5 $\equiv$ 6.73}\\
&=\phantom{-}\Bigg(
   \frac{1}{2} \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{cd}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
  +\frac{1}{2} \sum_{\bar{c}d}^{N/2}\sum_{\bar{t}u}^M
     \frac{\bracket{\bar{c}d}{\bar{t}u}\bracket{\bar{t}u}{\bar{c}d}}
          {\eps{\bar{c}} +\eps{d} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\Bigg(}
  +\frac{1}{2} \sum_{c\bar{d}}^{N/2}\sum_{t\bar{u}}^M
     \frac{\bracket{c\bar{d}}{t\bar{u}}\bracket{t\bar{u}}{c\bar{d}}}
          {\eps{c} +\eps{\bar{d}} -\eps{t} -\eps{\bar{u}}}
  +\frac{1}{2} \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\bracket{\bar{c}\bar{d}}{\bar{t}\bar{u}}
           \bracket{\bar{t}\bar{u}}{\bar{c}\bar{d}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\Bigg)\\&\phantom{=} -\left(
   \frac{1}{2} \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{dc}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
  +\frac{1}{2} \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\bracket{\bar{c}\bar{d}}{\bar{t}\bar{u}}
           \bracket{\bar{t}\bar{u}}{\bar{d}\bar{c}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\right)\tag{6}
\end{align}$$
Now we can get rid of the spin and notice, that all integrals in the first parenthesis are equal, as wall as in the second.
$$\begin{align}
E_0^{(2)} &=
   2 \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{cd}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
     \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{dc}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
\tag{7}\end{align}$$
When we now relabel $c,d\to a,b$ and $t,u\to s,r$ we will get (6.74) and when we move $\bracket{cd}{tu}\bracket{tu}{cd}$ in front of the parenthesis we have our desired equation for exercise 2.18.
$$\begin{align}
E_0^{(2)} &=
   \sum_{ab}^{N/2}\quad \sum_{rs}^{M}
     \frac{\bracket{ab}{rs} \left(2 \bracket{rs}{ab} -\bracket{rs}{ba}\right)}
          {\eps{a} +\eps{b} -\eps{r} -\eps{s}}
\end{align}$$
Additional notes
After doing this exercise and I also spent quite a few hours on it, I still have the feeling I have not learned anything new. Surely, I do now understand the lazy notation better. I can also see why introducing the restriction of closed shell orbitals is a quite good idea from a computational point of view. But I did know that before and it seems more than logical when you consider that you slash your complete basis in halves.
Well, since I like crosswords and the likes, I actually liked thinking about this. Apart from this I don't think this exercise furthers understanding quantum chemistry.
While I did some reading along the lines, I came across a book on the shelf of my sensei, which has also a focus on implementation. I guess it doesn't hurt checking it out, if you have access to it: David B. Cook; Hondbook of Computational Chemistry; Oxford University Press: Oxford, New York, Tokyo, 1998. (New edition 2005 @ Dover publications)

Proof (deduction) of (3)
Recall what I used earlier, I already threw away the $\frac{1}{4}$:
$$\small
   \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
     \frac{\bracket{ab}{rs}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}
=  \sum_{a=1}\sum_{b=1}\sum_{r=N+1}\sum_{s=N+1}
     \frac{\bracket{ab}{sr}^2}{\eps{a} +\eps{b} -\eps{r} -\eps{s}}\tag{3'}
$$
 I am no mathematician. The only way I knew how to tackle this was to actually expand each term (completely). And yes, that is quite some work.
To make things a little easier, we can reduce the terms a little further, because the denominator on both sides is equal when $a,b,r,s\in\mathbb{N}$; $a=[1,N]$; $b=[1,N]$; $r=[N+1,M]$; $s=[N+1,M]$; and $N,M$ are the same on the left and the right side.
Therefore we "simply" need to show that
$$
   \sum_{a=1}^N\sum_{b=1}^N\sum_{r=N+1}^M\sum_{s=N+1}^M
     \bracket{ab}{rs}^2
=  \sum_{a=1}^N\sum_{b=1}^N\sum_{r=N+1}^M\sum_{s=N+1}^M
     \bracket{ab}{sr}^2\tag{A1}
$$
Now we again need to simplify our notation or we will not get anywhere. For $a=1$ we simply write $1$ and for $r=N+1$ we also simply write $1$, where the position of the number implies if it is part of the first set or the second. (I can't believe I am actually writing this.) In the integral we will from now on use a semicolon to separate the functions. Since there are so many squares and we don't need them, we also get rid of that. Let's make a practical example:
$$\text{For} a=1,b=2,r=N+1,s=N+2: \bracket{ab}{rs}^2:=\bracket{1;2}{1;2}$$
Let's tackle the left side first:
$$\small\begin{align}
  \sum_{a=1}^N\sum_{b=1}^N\sum_{r=N+1}^M\sum_{s=N+1}^M \bracket{ab}{rs}^2
  &=
  \begin{array}{l}
    \color{\green}{
    \phantom{+}
     \bracket{1;1}{1;1} +\bracket{2;1}{1;1}
    +\bracket{1;2}{1;1} +\bracket{2;2}{1;1}
    +\cdots+\bracket{N;N}{1;1}}\\
    \color{\red}{
    +\bracket{1;1}{2;1} +\bracket{2;1}{2;1}
    +\bracket{1;2}{2;1} +\bracket{2;2}{2;1}
    +\cdots+\bracket{N;N}{2;1}}\\
    \color{\navy}{
    +\bracket{1;1}{1;2} +\bracket{2;1}{1;2}
    +\bracket{1;2}{1;2} +\bracket{2;2}{1;2}
    +\cdots+\bracket{N;N}{1;2}}\\
    \color{\green}{
    +\bracket{1;1}{2;2} +\bracket{2;1}{2;2}
    +\bracket{1;2}{2;2} +\bracket{2;2}{2;2}
    +\cdots+\bracket{N;N}{2;2}}\\
    +\qquad\vdots\\
    +\bracket{1;1}{M;M} +\bracket{2;1}{M;M}
    +\bracket{1;2}{M;M} +\bracket{2;2}{M;M}
    +\cdots+\bracket{N;N}{M;M}\\
  \end{array}
\end{align}$$
This doesn't really help, because we also need the right side.
$$\small\begin{align}
  \sum_{a=1}^N\sum_{b=1}^N\sum_{r=N+1}^M\sum_{s=N+1}^M \bracket{ab}{sr}^2
  &=
  \begin{array}{l}
    \color{\green}{
    \phantom{+}
     \bracket{1;1}{1;1} +\bracket{2;1}{1;1}
    +\bracket{1;2}{1;1} +\bracket{2;2}{1;1}
    +\cdots+\bracket{N;N}{1;1}}\\
    \color{\navy}{
    +\bracket{1;1}{1;2} +\bracket{2;1}{1;2}
    +\bracket{1;2}{1;2} +\bracket{2;2}{1;2}
    +\cdots+\bracket{N;N}{1;2}}\\
    \color{\red}{
    +\bracket{1;1}{2;1} +\bracket{2;1}{2;1}
    +\bracket{1;2}{2;1} +\bracket{2;2}{2;1}
    +\cdots+\bracket{N;N}{2;1}}\\
    \color{\green}{
    +\bracket{1;1}{2;2} +\bracket{2;1}{2;2}
    +\bracket{1;2}{2;2} +\bracket{2;2}{2;2}
    +\cdots+\bracket{N;N}{2;2}}\\
    +\qquad\vdots\\
    +\bracket{1;1}{M;M} +\bracket{2;1}{M;M}
    +\bracket{1;2}{M;M} +\bracket{2;2}{M;M}
    +\cdots+\bracket{N;N}{M;M}\\
  \end{array}
\end{align}$$
Now you can probably already guess from the colour-coding where I am going with this. We can easily see now, that the sums are equivalent since $a,b$ and $r,s$ run over the same indices in both cases. The only difference we have is in which order the integrals will appear. So the green lines are identical in both cases (and zero btw.) while the red and navy ones just interchange positions.

Full expansion of (6)
Note that I actually appreciate how insane this is. I am including this, because I noticed that your expansion was incomplete. It still leads to the correct conclusion, because the terms you did not include are all vanishing.
$$%\require{cancel}
  \newcommand{\cancel}[1]{\color{\red}{#1}}
  \begin{align}
E_0^{(2)} &=
  \phantom{-}\frac{1}{2}\Bigg(
%all alpha
    \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{cd}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%three alpha one beta
  + \sum_{\bar{c}d}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{\bar{c}d}{tu}}
           \cancel{\bracket{tu}{\bar{c}d}}}
          {\eps{\bar{c}} +\eps{d} -\eps{t} -\eps{u}}
  + \sum_{c\bar{d}}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{c\bar{d}}{tu}}
           \cancel{\bracket{tu}{c\bar{d}}}}
          {\eps{c} +\eps{\bar{d}} -\eps{t} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{cd}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{cd}{\bar{t}u}}
           \cancel{\bracket{\bar{t}u}{cd}}}
          {\eps{c} +\eps{d} -\eps{\bar{t}} -\eps{u}}
  + \sum_{cd}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{cd}{t\bar{u}}}
           \cancel{\bracket{t\bar{u}}{cd}}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%two alpha two beta
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{tu}}
           \cancel{\bracket{tu}{\bar{c}\bar{d}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{t} -\eps{u}}
  + \sum_{c\bar{d}}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{c\bar{d}}{\bar{t}u}}
           \cancel{\bracket{\bar{t}u}{c\bar{d}}}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{cd}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{cd}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{cd}}}
          {\eps{c} +\eps{d} -\eps{\bar{t}} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{\bar{t}u}^M
     \frac{\bracket{\bar{c}d}{\bar{t}u}\bracket{\bar{t}u}{\bar{c}d}}
          {\eps{\bar{c}} +\eps{d} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{c\bar{d}}^{N/2}\sum_{t\bar{u}}^M
     \frac{\bracket{c\bar{d}}{t\bar{u}}\bracket{t\bar{u}}{c\bar{d}}}
          {\eps{c} +\eps{\bar{d}} -\eps{t} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}d}{t\bar{u}}}
           \cancel{\bracket{t\bar{u}}{\bar{c}d}}}
          {\eps{\bar{c}} +\eps{d} -\eps{t} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%one alpha three beta
  + \sum_{c\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{c\bar{d}}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{c\bar{d}}}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}d}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{\bar{c}d}}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{t\bar{u}}}
           \cancel{\bracket{t\bar{u}}{\bar{c}\bar{d}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{t} -\eps{\bar{u}}}
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{\bar{t}u}}
           \cancel{\bracket{\bar{t}u}{\bar{c}\bar{d}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%all beta
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\bracket{\bar{c}\bar{d}}{\bar{t}\bar{u}}
           \bracket{\bar{t}\bar{u}}{\bar{c}\bar{d}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\Bigg)\\
&\phantom{=} -\frac{1}{2}\Bigg(
%all alpha
    \sum_{cd}^{N/2}\sum_{tu}^M
     \frac{\bracket{cd}{tu}\bracket{tu}{dc}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%three alpha one beta
  + \sum_{\bar{c}d}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{\bar{c}d}{tu}}
           \cancel{\bracket{tu}{d\bar{c}}}}
          {\eps{\bar{c}} +\eps{d} -\eps{t} -\eps{u}}
  + \sum_{c\bar{d}}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{c\bar{d}}{tu}}
           \cancel{\bracket{tu}{\bar{d}c}}}
          {\eps{c} +\eps{\bar{d}} -\eps{t} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{cd}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{cd}{\bar{t}u}}
           \cancel{\bracket{\bar{t}u}{dc}}}
          {\eps{c} +\eps{d} -\eps{\bar{t}} -\eps{u}}
  + \sum_{cd}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{cd}{t\bar{u}}}
           \cancel{\bracket{t\bar{u}}{dc}}}
          {\eps{c} +\eps{d} -\eps{t} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%two alpha two beta
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{tu}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{tu}}
           \cancel{\bracket{tu}{\bar{d}\bar{c}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{t} -\eps{u}}
  + \sum_{c\bar{d}}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{c\bar{d}}{\bar{t}u}}
           \bracket{\bar{t}u}{\bar{d}c}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{cd}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{cd}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{dc}}}
          {\eps{c} +\eps{d} -\eps{\bar{t}} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{\bar{t}u}^M
     \frac{\bracket{\bar{c}d}{\bar{t}u}
           \cancel{\bracket{\bar{t}u}{d\bar{c}}}}
          {\eps{\bar{c}} +\eps{d} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{c\bar{d}}^{N/2}\sum_{t\bar{u}}^M
     \frac{\bracket{c\bar{d}}{t\bar{u}}
           \cancel{\bracket{t\bar{u}}{\bar{d}c}}}
          {\eps{c} +\eps{\bar{d}} -\eps{t} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}d}{t\bar{u}}}
           \bracket{t\bar{u}}{d\bar{c}}}
          {\eps{\bar{c}} +\eps{d} -\eps{t} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%one alpha three beta
  + \sum_{c\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{c\bar{d}}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{\bar{d}c}}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
  + \sum_{\bar{c}d}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}d}{\bar{t}\bar{u}}}
           \cancel{\bracket{\bar{t}\bar{u}}{d\bar{c}}}}
          {\eps{c} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{t\bar{u}}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{t\bar{u}}}
           \cancel{\bracket{t\bar{u}}{\bar{d}\bar{c}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{t} -\eps{\bar{u}}}
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}u}^M
     \frac{\cancel{\bracket{\bar{c}\bar{d}}{\bar{t}u}}
           \cancel{\bracket{\bar{t}u}{\bar{d}\bar{c}}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{u}}
\\&\phantom{=-\frac{1}{2}\Bigg(}
%all beta
  + \sum_{\bar{c}\bar{d}}^{N/2}\sum_{\bar{t}\bar{u}}^M
     \frac{\bracket{\bar{c}\bar{d}}{\bar{t}\bar{u}}
           \bracket{\bar{t}\bar{u}}{\bar{d}\bar{c}}}
          {\eps{\bar{c}} +\eps{\bar{d}} -\eps{\bar{t}} -\eps{\bar{u}}}
\Bigg)\tag{6}
\end{align}$$
I indicated the vanishing elements with red.
If there are any remaining questions, don't hesitate and visit the associated chat. (If this chat is no longer available, go to the main chat and ask there, please.)
And because that was quite tedious, have a coffee stain to go:

Image is courtesy of Roger Karlsson (http://www.free-photo-gallery.org/photos/coffee-stain/) obtained from flicker.
(I assume this is now quite close to my longest proof.)

Well this is the second longest lol ! :)

@Hexacoordinate-C oh, oh, five of those posts are mine.

This is the highest-quality answer I've seen on any Stack page. Came here from a web search for "spin orbitals"; coincidentally, I ran into this problem the night prior in chap 2, had no idea what to do with it, looked at an answer guide, and quickly determined it was beyond me. Your explanations are fantastic, and make explicit several things that I'd been confused about.

@TurtlesAreCute Thank you very much for the compliment. I am especially happy that it was of use to you. Good luck on your further studies!

Regarding your derivation. It should be noted that the norm-squared is not the same as the squared. $$ |f(x)|^2=|\langle x|f \rangle|^2= (\langle x|f \rangle )^* \langle  x|f \rangle =\langle f|x \rangle \langle x|f \rangle \neq \langle x|f \rangle \langle x|f \rangle $$
Therefore, I hope the following set of equations makes things clearer for anyone looking for answers.
$$E_{0}^{(2)}=\frac{1}{4}\sum_{a,b=1}^{n}\sum_{r,s=n+1}^{m}\frac{|\langle ab||rs \rangle |^2}{(\epsilon_a+\epsilon_b)-(\epsilon_r+\epsilon_s)}$$
$$ E_{0}^{(2)} =
\frac{1}{4}
\sum_{a,b=1}^{n}\sum_{r,s=n+1}^{m}
\frac{\langle ab||rs \rangle \langle rs||ab \rangle}
{(\epsilon_a+\epsilon_b)-(\epsilon_r+\epsilon_s)}$$
$$ E_{0}^{(2)} =
\frac{1}{4}
\sum_{a,b=1}^{n}\sum_{r,s=n+1}^{m}
\frac{(\langle ab|rs \rangle - \langle ab|sr \rangle)(\langle rs|ab \rangle - \langle rs|ba \rangle)}
{(\epsilon_a+\epsilon_b)-(\epsilon_r+\epsilon_s)}$$
$$ E_{0}^{(2)} =
\frac{1}{4}
\sum_{a,b=1}^{n}\sum_{r,s=n+1}^{m}
\frac
{\langle ab|rs \rangle \langle rs|ab \rangle + \langle ab|sr \rangle\langle rs|ba \rangle - \langle ab|rs \rangle\langle rs|ba \rangle - \langle ab|sr \rangle\langle rs|ab \rangle}
{(\epsilon_a+\epsilon_b)-(\epsilon_r+\epsilon_s)}$$
$$ E_{0}^{(2)} =
\frac{1}{4}
\sum_{a,b=1}^{n}\sum_{r,s=n+1}^{m}
\frac
{2\langle ab|rs \rangle \langle rs|ab \rangle - 2\langle ab|rs \rangle \langle rs|ba \rangle }
{(\epsilon_a+\epsilon_b)-(\epsilon_r+\epsilon_s)}$$
The final step is the only step that requires the use of symmetry note that summation over a and b is symmetrical and that summation over r and s is symmetrical (this is seen clearly in the equations).
Rearranging the resulting equation to equation (6.73) is trivial.
Regarding the Closed Shell variant of the equation, I won't say much, as the only real difference is that the summation changes as the spin does not need to be explicitly considered (which is why they can be considered spacial only) as all electrons are considered paired in the ground state MOs.
I hope this makes sense! Let me know if there are any mistakes!
ADDITIONAL INFORMATION
The answer given by @Martin and in the question has an incorrect starting point.
Here I shall explain why by showing that:
$$|\langle ab||rs \rangle|^2 \neq \langle ab||rs \rangle^2 $$
$$|\langle ab||rs \rangle|^2 \neq \langle ab||rs \rangle \langle ab||rs \rangle $$
The first thing that should be apparent is that the norm squared and squared have different syntaxes - this should hint that there is a difference between the two...
Infact, they are equal if you are working exclusively with real numbers, however, $\Psi$ can be real or complex, therefore when working with $\Psi$ it is important to note that any derivation must be general for both real AND complex numbers.
So I shall write down the true form of the norm-squared for a function $\Psi$ that is general for both real and complex numbers:
$$ |\Psi|^2 = \int\langle \Psi |x \rangle\langle x|\Psi\rangle dx = \int \Psi^{*}(x)\Psi(x)dx = \langle \Psi | \Psi \rangle$$
We also know that, complex conjugates in Dirac notation are related by:
$$ \langle A | = ( |A \rangle )^* $$
$$ |A \rangle = ( \langle A | )^* $$
Therefore,
$$ |\langle ab || rs \rangle |^2 = (\langle ab || rs \rangle )^*\langle ab || rs \rangle = \langle rs || ab \rangle \langle ab || rs \rangle = \langle ab || rs \rangle \langle rs || ab \rangle \neq \langle ab || rs \rangle \langle ab || rs \rangle $$
The final inequality is true for complex numbers despite the fact it is not true for real numbers. However, as stated earlier if $\Psi$ can be either complex or real then $a,b,r$ and $s$ can also be complex or real.
The same can be shown using matrices instead of real and complex numbers, when you know that:
$$\langle a | a \rangle = \bar{\mathbf{a}}^{T}\mathbf{a}$$
Where, $a$ can be a real or complex matrix.
This is the power of Dirac (braket) notation, it gives a general formalism that relates to matrices and integrals. If you are going to work within QM in chemistry I strongly advise using Dirac notation over the chemists notation.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. 1. The picture you are using has a very bad quality, I can barely make out the indexes. 2. I have nowhere in my answer taken a square root. Rather than putting an answer, a comment to my post would be preferable. 3. An untrained reader might not understand why you can write $|<ab||rs>|^2 = <ab||rs><rs||ab>$. It would be good if you could explain that. 4. You did not answer the part 2.

I really do not understand, where I make that conclusion or how I make that conclusion. I might be a bit slow here, but I really don't get it. I also do not understand how your proof would then be more rigorous, as we arrive at the same expression.

Please also note that: $$\sum_{abrs}\langle ab|rs \rangle^2 = \sum_{abrs}\langle ab|sr \rangle^2 \nLeftrightarrow \sum_{abrs}\langle ab|rs \rangle = \sum_{abrs}\langle ab|sr \rangle$$

The above causes an ambiguity in the derivation you provided when you consider it in relation to equation I presented in the comments above - also read my edited answer, I have shown why the expansion of the norm-squared you performed is incorrect.

I see. I am too slow for this to understand it. I really do not understand how you can take a square root from every term and still expect the left and the right side to be equal. Please note, that I have explicitly stated, that I am limiting the proof to real functions. Apart from this, your answer is still incomplete.

Which is why at the end I specify that you can prove the same conclusion through the use of matrices (i.e. real matrices). In that respect the starting point is still invalid. $$ |\langle ab||rs \rangle|^2 = \mathbf{A}^{T} \mathbf{A} $$ Where the elements of A are $$ A_{ijlk} = \langle ab || rs \rangle $$ and the transposed form elements are defined: $$ A_{lkij} = \langle rs || ab \rangle $$ In this case $a,b,r$ and $s$ are all real functions...

Because the above deals with the equations one element at a time... You can substitute the following into the equations of my answer $$ |\langle ab || rs \rangle|^2 = A_{ijlk}^{T}A_{ijlk} = A_{lkij}A_{ijlk} $$

I really don't get it, I don't have the time wrapping my head around it, and I am at the point where I don't care anymore. Thank you for trying though.

I am also quite confused by your answer... First of all, as stated by Martin, I can't see how you justify that $\sum_a a^2=\sum_b b^2$ implies $\sum_a a =\sum_b b$. Secondly if I remember well in Szabo and Ostlund the orbitals are assumed to be real functions. Finally, if you really insist to work with complex number, you should take the conjugate transpose, not just the transpose.

Okay I realise I am wrong with the case of the squares and square roots thang - I derped there. Sorry guys! However, it is still true to say that the proof martin put forward for (3) actually shows $$ \langle ab|rs \rangle = \langle ab|sr \rangle $$ and therefore $$ \langle ab|rs \rangle ^ 2 = \langle ab|sr \rangle ^ 2 $$ Therefore, the equation (2) still equates to zero

And also - R.M., you are very right that it should be the conjugate transpose ( I was just intending to show that the indices tranpose still for real functions) as I said in the comment "(i.e. for real matrices)" ***And in the previous comment I forgote to add the sum symbols - sorry!

I would also like to note that proofs have to be true for both complex and real basis functions, we are always working in complex functions - as defined by the two-electron integral form:$$ \langle ab|rs \rangle = \int \chi_a^(\textbf{r}_1)\chi_b^(\textbf{r}2)(r{12}^{-1}) \chi_r (\textbf{r}_1) \chi_s (\textbf{r}_2)d\textbf{r} $$ It's just so happens for real numbers $$ a = (x+0i)^=x-0i=a^ $$ In reality the real numbers we are using are complex, it's just practically we decide to equate the complex bit to zero, but the equations used have to be derived in their complex forms.

I am only showing that $$\sum_{abrs}\langle ab|rs \rangle^2 = \sum_{abrs}\langle ab|sr \rangle^2,$$ but because I am lazy I don't write down the square - I should probably make that clearer.

Well is is also true that $$ \sum_{abrs} \langle ab|rs \rangle = \sum_{abrs} \langle ab|sr \rangle $$ it is clear from the symmetrical sum over r and s. Also, despite all this - you have not expanded the norm squared correctly. It is not the same operation as a square and yes we have to work in complex numbers. Out of interest what is not understood about my answer?


text:
Equilibrium constant. Can it be reached?
Consider the reaction below:
$$\ce{A + B <=> C}$$
Suppose that the equilibrium constant for this reaction is $K = 10$.
I then prepare a reaction vessel with volume of $\pu{1 dm^{-3}}$ which contains $1$ molecule of $\ce{A}$, $500$ molecules of $\ce{B}$ and $1$ molecule of $\ce{C}$. Thus the initial concentration of each species (in $\pu{mol\cdot dm^{-3}}$) will be $1/N_\mathrm{A}$, $500/N_\mathrm{A}$, and $1/N_\mathrm{A}$ respectively, where $N_\mathrm{A}$ is the Avogadro constant.
As a result the initial reaction quotient is $N_\mathrm{A}/500$.
Now, am I right in saying that this system cannot reach equilibrium, since the only possible values of the reaction quotient are $+\infty$, $0$, or $N_\mathrm{A}/500$?
Does this arise because there simply is not enough of each reactant/product in order to reach equilibrium? I feel like this system must reach equilibrium but I'm struggling to see how the value of $K$ can be reached in this closed system.

In a bimolecular reaction step (such as the one shown above), the rate constant for the forward reaction has units $\pu{M-1\cdot s-1}$. I have learnt that if the concentration of $\ce{B}$ is increased significantly, then the Gibbs free energy of the $\ce{A + B}$ mixture will increase because the activation energy has a concentration dependence. Is this true, and is equilibrium only reached once this Gibbs free energy decreases again?
This is confusing as I am used to the Gibbs free energies of the reactants and products remaining constant while the relative amounts of each change throughout the course of the reaction.

Dave's mathematical answer is perfectly valid, but I feel like another, perhaps more chemical point of view is based on the the law of large numbers, i.e., the "thermodynamic" limit.
"Equilibrium" is a statistical concept that only applies to large numbers of molecules.  Another way to say it is that equilibrium is a thermodynamic concept and thus only applies in the thermodynamic limit.  What is "large" depends on the particular system of interest.  In particular it depends on the equilibrium constant.  For your reaction of interest, $K_{eq} = 10$.  Thus the amount of $A$ and/or $B$ at equilibrium will be smaller than the amount of $C$.  For the equilibrium concept to apply to a system of $A$, $B$, and $C$ molecules, the calculated equilibrium amount of $A$ or $B$ molecules must be very large, say tens of thousands or so for precise work, or perhaps as low as hundreds for the equilibrium concept to be "approximately" correct.
If the equilibrium constant were enormously large, say $K_{eq} = 10^{100}$, then by the same logic, the system would need to contain on the order of $10^{102}$ total molecules for the equilibrium to be meaningful.  (And by the way that is larger than the number of particles in the universe!)
Speaking loosely, if $K_{eq}$ is $10^{100}$ and there aren't close to that many molecules, the reaction is irreversible.
A probabilistic view of equilibrium for small numbers of particles, such as that mentioned in DavePhD's answer, is occasionally useful but not always valid.  For example, consider if a second reaction consumed $A$ irreversibly, converting it to $D$ at some rate.  If you take DavePhD's probabilistic view of the $\ce{A + B <=> C}$ equilibrium, you would assume that the probability distribution of persistence times for $A$ is exactly the same as in the thermodynamic limit.  However, that is not necessarily true with small systems.  Statistical fluctuations in the $\ce{A + B <=> C}$ system will mean that some of the time, $A$ persists far longer than in the thermodynamic case because due to a fluctuation, more than the "expected" amount of the molecules were in the $C$ form.  Some other times, $A$ will resist degradation to $D$ for far shorter than the thermodynamic equilibrium calculation would suggest.  To treat these cases, more complex equations such as the chemical master equation are required.

If you pick an arbitrary real number, there is zero probability that it is a rational number. Therefore, there is zero probability that a finite sample of molecules A,B,C can exactly represent a arbitrary real number equilibrium constant.
However, the more molecules you have in the system, the closer a given real number equilibrium constant could be represented by the finite number of molecules.
In your example, the system would spend some of the time in the initial state, and some of the time in A=0, B=499, C=2, and some of the time in A=2, B=501, C = 0.


text:
Why there is a non-linear trend on the melting points for period 2 elements, especially Na to Si?
Using the knowledge of periodicity from high school cannot explain this, what it tells me is just metallic bonding.
The model of delocalised electron from a metallic lattice can explain the increase of melting point from Na to Si, but it fails to explain why the relationship is non-linear?

Why should it be linear in the first place?

What I mean is the m.p. of Al and Mg is almost same, and Si is almost doubled the amount, why? The high school textbook says this "period 2 element has linear increase on m.p. from Na to Si because of the amount of delocalised electrons", but I don't think this works as Al has more delocalised electron according to this but same m.p.

Not surprisingly, high school textbooks tend to simplify things, and offer simple explanations even for things that they should not try to explain. Looking for simple explanations for non-existent trends may only lead to madness... (hopefully just being mad at the textbook, mind you).

The number of delocalized electrons hardly matters at all. See, in molten metal these electrons are still delocalized and feel pretty much the same.

If your textbook says that the melting point increases linearly, that is incorrect, as you point out. Generally high school textbooks can only make broad, qualitative statements about characteristics like melting points since the physics involved can depend to a larg degree on very complicated (quantum mechanical) interactions between the atoms (and their electrons) in a solid.
See for instance http://www.ptable.com/, and try playing with the slider on the top right to see melting points for different elements. There is a general trend of heavier metals in the same period having higher melting points, but there are some exceptions, such as manganese (Mn) around 1570 K.
As for why specifically Si's melting point is so much higher than Al's, the only truly correct answer is something along the lines of "that's just because of the differences in the way Si and Al atoms interact with each other". Because that's not a very satisfying explanation, I can try to paint a more intuitive picture: (UPDATE: it has been pointed out in the comments that even this simplified picture is wrong, see the linked wikipedia article for more details.) Aluminium has three valence electrons, which means that it would need to form five covalent bonds with its neighbours in order to reach the favourable octet of electrons. Unfortunately there is no regular crystal structure where every atom has exactly five nearest neighbours, so what ends up happening is that one electron from each atom delocalises, and the resulting postively charged ions are weakly (i.e. noncovalently) bound to each other, resulting in a low melting point (in other words, the atoms can be shaken apart easily).
Silicon on the other hand, has four valence electrons, only one more than Al, but this means that it only needs to bind to four neighbours to get an octet, and it turns out that there is a regular arrangement of atoms such that each atom has exactly four nearest neighbours (called the diamond structure). So, in solid Silicon, each atom is covalently (and thus strongly) bound to four neighbours, (and no electrons delocalised) which means that the atoms are hard to shake apart, which leads to a much higher melting point.
Now, this might be a more pleasing explanation (at least I hope so), but note that it is only meant for illustrative purposes: the same logic, dealing with 3 vs. 4 valence atoms can also be applied to for example Thallium and Lead, but these two metals have virtually the same melting point. Why is that? Well, there are other effects at play for these heavier atoms. In other words, it really does come down to how groups of the same atom interact with each other, and much of that can only be understood through the framework of quantum mechanics, and a rigorous treatment of solid state physics is probably too much to expect from a high school textbook :-)

Your picture for Al is wrong.

Your response is unilluminating.

Al wouldn't get octet with 5 neighbors. Boron has such situation but with elecrtodeficience. https://en.wikipedia.org/wiki/Allotropes_of_boron

Fair enough, I've edited the answer to add a disclaimer.


text:
Why is the square of the wavefunction a probability density?
Can anyone explain it in simple words? I tried to read about the Born rule on Wikipedia but it seems it is just describing the rule. Is there any proof that is suitable for person with no solid math background?

Same (nearly same) question (s ) has (have) been asked now & then at Phys.SE: Born rule and unitary evolution ; Born's Rule, What is the Reason?; Born Interpretation of Wave Function. So, don't think there is a need of same question here.

First, I would not call the Born rule a law as it is done in the Wikipedia article linked by OP in the question. It is, of course, the question of the definitions, but I would call the Born rule rather a postulate. A law in physics is a principle (usually universal) deduced from observations, while a postulate is an assumption, a statement that we can not prove or derive but which is assumed to be right and tested to be consistent with experimental observations.
The Born rule is a postulate, thus, there is no any proof of it; it is assumed to be right. The only thing we can do is to justify a postulate: we can make experimentally verifiable predictions on the basis of it and test them. And it happens so that so far the experimental outcomes are always in full agreement with the predictions done using the Born rule. Thus, we know it provides a correct description of physical systems.

The Born postulate, as you call it, is also widely called the "Born interpretation". Is one preferred over the other?

@Goat'sMilk, I wouldn't call it an interpretation, since the word "interpretation" has a well-defined meaning in quantum theory: a set of statements attempting to explain the philosophical (primarily, ontological & epistemological) meaning of quantum theory beyond the rigorous physical theory. Born rule, on the other hand, is a part of the quantum theory itself, it is one of the fundamental principles of it, one of the key components of its mathematical formalism. Thus, I think one would better avoid calling it an interpretation.

@Wildcat There are different interpretations of QM, and they can have different nuances about what the wave-function mean and how is it tied to physical reality (as the wave function itself is not physical). In that sense, the "interpretation" is a much better term than "postulation".


text:
How derive g and u symmetry labels for orbitals?
When asked whether a molecule has an inversion center, we "invert" the coordinates of all atoms; i.e. move each atom from its position through the center of symmetry and to a new position equidistant to the center of symmetry compared to its initial position, and see if the final configuration is indistinguishable from the initial configuration. Simple enough.
I am confused when trying to decide whether molecular orbitals have g or u symmetry. Should I move the atoms through the center of symmetry, should I move "each lobe" through the center of symmetry, or some other way? What is the most general way of describing the symmetry operation for orbitals?
In Figure 1, a bonding $\pi$ orbital is depicted. This is supposed to be antisymmetric with respect to the inversion center in the middle of the bond. It seems therefore I need to invert the lobes, somehow, or move in an "X" fashion across the bond. But this seems different than for the molecular case, where we simply invert atomic coordinates. Figure 2 should be symmetric under inversion, and inverting the lobe positions lead to this result.
And what when the two overlapping orbitals are not of the same type, e.g. a $\pi$ overlap between d and p orbitals? Will the MO ever have g symmetry?
I just fail to get the unifying link between the molecular case and the orbital case.
Edit: Something just hit me: Should I always think of it as "inverting all coordinates within the entity", regardless of whether the entity is an atom or an orbital?

Figure 1. Bonding $\pi$ orbital

Figure 2. Antibonding $\pi$ orbital

Related : How to determine gerade & ungerade symmetry of a MO orbital?

Short answer:
You should move everything through the centre of symmetry. That’s why it’s a centre of symmetry.
On the other hand, note that classifying orbitals just by g and u does not make much sense. You should always consider the entire molecule’s point group and thus always consider which (set of) irreducible representation(s) an orbital or a set of orbitals has/have in said point group.
For example considering a set of six atoms arranged in an octahedron around a central atom; each of the surrounding atoms has three p-type orbitals and we are only interested in those that are perpendicular to the outside atom – central atom axis. These twelve p-orbitals transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$. Omitting the $\mathrm{t}$ and the subscripted numeral is a non-complete description of the orbitals. Only those in exactly the same irreducible representation will mix in a non-zero fashion.

g and u labels are only used when the molecule itself possesses the inversion symmetry element. For a diatomic molecule, this means that both atoms have to be the same, i.e. it has to be a homonuclear diatomic molecule $(D_{\infty\mathrm{h}})$.

How do I carry out the inversion?

$\hspace{25 mm}$

Let's consider the $\ce{O2}$ molecule.
Let's consider the $\pi$ bonding MO. (Or one of them, to be precise, since they occur in degenerate pairs.)
Find the point of inversion of the molecule. Clearly it has to be the midpoint of the O=O bond, i.e. exactly in the middle.
Draw a straight line from one lobe to the point of inversion.
Extend the straight line through the point of inversion. You will see that you have gone from a shaded lobe to an unshaded lobe, indicating that the phase has changed. Therefore, the $\pi$ bonding MO is ungerade, or u.

If you start from the bottom-left lobe, you would draw a line through the point of inversion and end up in the top-right lobe, and you would draw the same conclusion.

And what when the two overlapping orbitals are not of the same type, e.g. a $\pi$ overlap between d and p orbitals? Will the MO ever have g symmetry?

The parity of such MOs cannot be classified as g or u. However, if we are talking about a homonuclear diatomic, then this wavefunction is not acceptable. You cannot have a wavefunction that has only d orbital character from one atom, but only p orbital character from the other. By symmetry, the contribution from both atom's d orbitals have to be equal.
This arises because the Hamiltonian is invariant to the permutation of identical particles. Essentially, the idea is that both atoms are exactly identical, so there cannot be any way for them to be distinguished from each other. If you had such a wavefunction, and if it was populated, then there would be different amounts of electron density on the two atoms, which would mean that the atoms could then be distinguished. This is quantum-mechanically forbidden.
Such MOs may arise in heteronuclear diatomics. However, heteronuclear diatomics $(C_{\infty\mathrm{v}})$ do not possess the inversion symmetry element, and accordingly their MOs cannot be classified as g or u. Since inversion does not preserve the molecule, there is no requirement that it should preserve the wavefunction.

What if one atom does not have any d orbitals?

There is no such thing. Every atom has d orbitals.
Whether they have the same energy, or whether they are occupied, is another thing altogether. If the energies are different, or if one atom has occupied d orbitals and the other does not, then it is a heteronuclear diatomic and as I said earlier, the g and u labels are not appropriate in such a case.


text:
Suggestions for simulating transport through membranes with Molecular Dynamics (MD) and Monte Carlo (MC)?
I'm intending to perform a simulation of the transport of simple ions or water through perm-selective membranes, using Molecular Dynamics (MD) and/or Monte Carlo (MC). I've read around that, in general, with Monte Carlo the polymerization of the membrane can be modeled starting from linear chains and monomers, and get the pseudo-membrane well cross-linked. Once the polymer structure is optimized, MD is performed with explicit solvent, the membrane putted in between two different solutions, so transport properties through membranes can be explored.
I never used MD nor MC. This is where I stand in the knowledge:
Monte Carlo (stochastic approach):
Useful to calculate thermodynamic macroscopic properties. The simulation follows the Markov Process, where the steps are independent one from each other (Jan Labanowski).

The stochastic approach, called Monte Carlo, is based on exploring the energy surface by randomly probing the geometry of the molecular system. The iterations are independent of one another (i.e., the system does not contain any "memory'').
  Is accepted more by physicists than by chemists, probably because MC is not a deterministic method and does not offer time evolution of the system in a form suitable for viewing. (But) for Markov chains, there are efficient methods for deriving time related quantities such as relaxation times (…) Many chemical problems in statistical mechanics are approached more efficiently with MC, and some (e.g., simulations of polymers chains on a lattice) can only be done efficiently with MC.

Molecular dynamics (deterministic approach):
To find trajectories and get an insightful movie of the molecular movement, with parameters of interest. Normally the system composes of around 5000 atoms with explicit solvent (water molecules generally). Periodic conditions are employed. The Verlet algorithm and the SHAKE routine are commonly used; here the bond lengths are fixed and the whole molecules are allow to move between certain potential range given (in position and energy), this repeated successively through a minimum path over the potential surface. It's a leap-frog method.
Considering this, I basically don't know where to begin. I would like to know the basic howto's and softwares for MD (GROMACS, CHARMM, LAMMPS, AMBER, etc.) and ways to perform Monte Carlo for polymers. Also, besides the fundamentals, I would like to know if there is like a Handbook for the features of each MD software or modes of application (such us potentials/ functionals used, force fields), or the most important practical keys to be taken into account before simulating.
Thanks in advance,

I would recommend "Understanding Molecular Simulation [http://www.amazon.com/Understanding-Molecular-Simulation-Second-Edition/dp/0122673514 ]" by Frenkel and Smit  for the theory part

Another enlightening book is "Computer Simulation of Liquids", by Allen and Tildesley.

Hi @Orr22, your link seems broken (in the comment above), could you fix that?

Great thanks for the alert, finally I found this as good introduction http://amzn.to/2jK5RuS. But, and, especially the documentation of each molecular dynamics software, they are very self-explanatory.

https://en.wikipedia.org/wiki/MARTINI

I always find AMBER to be the most streamlined for conducting MD, however it (as well as CHARMM) are proprietary.  Note that both are Langevin dynamics based, whereas LAMMPS is brownian dynamics.
That being said, an MD simulation is only as good as the parameters for the force field; the engine is of no matter (unless we're discussing speed).  Basically, I would use a QM program (Gaussian is my goto, but nwchem or psi4 are excellent open source options) to optimize the geometry of the membrane.  Because this is rather heavy, you'd most likely be best to go with a semi-empircal method as ab-initio could take weeks (depending on your computational power).  Once optimized, use the QM software to derive force constants as well as ESP charges (partial charges) for each atom.  Take this data, and parameterize a force field with it (I'm partial to using the Amber FF's...google GAFF or FF99SB... Theres a paper out there from Cornell, Merz, et. al. that should give you insight).
I hope this helps give you an idea of what you're looking at for a respectable MD simulation.

It really guide me through :)


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The amount of the DMF in the Sonogashira coupling - why it is of importance?
I was looking for some protocols for Sonogashira coupling and have found that in many papers the amount of the DMF used is exactly 2 ml per mole of the substrate.
Is there any reason for that? Usually, for other reactions, we can see, that articles report more-less random amount of solvents, like 10, 50, 100 ml - some "simple" numbers. But in this case I see numbers like 68 ml (why not 70?). Why must it be exactly 2 ml/mole?
some refs:
http://pubs.acs.org/doi/suppl/10.1021/ol049474j/suppl_file/ol049474jsi20040410_035107.pdf
http://www.sciencedirect.com/science/article/pii/S0040402000008772
http://pubs.acs.org/doi/suppl/10.1021/jo8020264/suppl_file/jo8020264_si_001.pdf

I for my part did a Sonogashira reaction with very good yields and without any DMF at all.

Did you use any solvent, or just the liquid reactants? Do you have any procedure that I could read? Thanks!

Triethylamine was the solvent. Source: Y. Fan, Y.-M. Zhu, F.-R. Dai, L.-Y. Zhang, Z.-N. Chen Dalton Trans. 2007, 3885. DOI: 10.1039/b707797a.


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How to clean copper(II) sulfate crystals from sulfuric acid?
I am going to prepare copper(II) sulfate by electrolysis with sulfuric acid for crystal growing.
How would I separate the copper(II) sulfate from the sulfuric acid left over from the electrolysis process in solution?
I think, one way would be to make copper(II) oxide and mix it with the solution so that I get more copper(II) sulfate out of the solution by using left over acid. Then separating copper oxide would be easy.

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Well, this is for my own safety not a homework.  I plan to grow copper sulfate crystals. But I have doubts that I could properly clean them by household items.

Crystals do not contain acid. Small leftovers of acid won't be a problem for crystal growth.

They are wet at first. Not sure how well sulfuric acid evaporates.

It doesn't, you need to rinse them, BTW what you want to do  electrolysis for?

@Mithoron I will make copper(II) sulfate by electrolysis.

I got that the first time... You're oxidating metallic Cu in the presence of diluted H2SO4?

Yes. Both electrodes will be made out of copper wire.

I doubt it's a good idea

Using those electrodes or making this salt in general?


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what is the importance of Benzophenone in distillation
when we realize a distillation of solvent we use benzophenone ,
what is the importance of this latter , Is is considered as an initiator of this operation, that's true ?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Your question is rather unclear and as such may well be closed. Please [edit] it to include more detail as to what you actually mean. (I have a suspicion which would make a good question but I don’t know if it’s true.) Learn more about editing and how the site works in the [help].

The importance of benzophenone in organic reactions

Sodium is insoluble in most solvents. When it reacts with benzophenone, it produces dianion that is moderately soluble and is a strong reducer, cleaning the solvent of anything reducible.

Ethereal solvents are often distilled from metals such as sodium In the presence of benzophenone as an indicator.
When the still is good, the benzophenone is blue/purple, when it goes yellow/brown the still is bad.
Classically we thought that the colour was related to moisture (ie. A purple colour indicated dryness) however its now generally agreed that the colour is to do with the absence of oxygen rather than moisture.
The distillation can be done without this indicator, indeed short path distillation almost never uses indicators. The benzoophenone is just useful in recycling stills as it allows one to see when th still needs attention.


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Why can't carbon and nitrogen be detected in ICP-MS and ICP-OES techniques?
Why can't carbon and nitrogen be detected in ICP-MS and ICP-OES techniques?
My book says this technique is efficient for detecting multiple elements, with exception of halogens and carbon. I found somewhere that ICP is an open system, therefore it doesn't allow carbon detection. But why is that exactly?

As you say, air is allowed to enter the plasma. Air consists primarily of $\ce{N2}$ and $\ce{O2}$, and would clearly interfere with any oxygen or nitrogen in the sample. The problem is not that ICP-MS or -OES cannot detect them, the problem is air.

As an aside, the lack of carbon and nitrogen detection by ICP-OES does not propose practical problems because there are other methods to detect them.  Combustion analysis is one such alternative.

Like the previous answers said, nitrogen and carbon content in the air is a contaminating factor. However according to this resource I found from 2005, it is theoretically possible to analyze nitrogen via ICP-OES by making it into a closed system.
https://www.spectroscopyonline.com/view/determination-nitrogen-and-other-essential-elements-fertilizers-icp-oes
Using combustion analysis is fine, but it's time/labor/cost intensive, where ICP-OES is relatively cheap and easy to operate. Another trade-off is that combustion analysis is fairly reliable, while ICP-OES may have to be calibrated 2-3 times over 10 hours.


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Relativistic effects in element 137 (Feinmanium) and above
Here I saw that if an element above atomic no 137 has to exist, it must have electron speed greater than speed of light. My question is , has this calculation  been done keeping in mind Einstein's relativity? {I am just asking, I have not done this calculation.}

Electrons, under quantum mechanics, do not have 'speeds' in the Bohr sense. But, if you asking if relativistic quantum mechanics is possible, it most certainly is.

The article you cite is bad, and those calculation couldn't be relativistic

Possible duplicate of The last element's atomic number

@Mithoron they could be relativistic: "Dirac showed that there are no stable electron orbits for more than 137 electrons" http://physics.info/atomic-models/

Already for elements in the fifth (or even earlier?) period, calculations of the inner electronic structure make no sense at all if relativity is not taken into account.

@DavePhD Geez, he's talking about Bohr. With Dirac you'd have oscillating ground state not "exceeding c". and with more precise analysis you have problem at about 173, but of another kind https://en.wikipedia.org/wiki/Extended_periodic_table#Feynmanium_and_elements_above_the_atomic_number_137

@Mithoron but the same number 137 (from the fine structure constant) is the limit for Dirac or Sommerfeld relativistic theory.  You get imaginary numbers in the energy equation above 137.  Other changes in the model beyond considering relativity are needed to avoid the 137 limit.

Yes, the value 137 occurs even considering Einstein's theory of special relativity.
137 comes from considering Sommerfeld or Dirac relativistic theories, when the nucleus is modeled as a point.
See equation 1 of A new method for solving the Z > 137 problem and for determination of energy levels of hydrogen-like atoms


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Bromine Homolysis Using Handheld UV Lamp
I'm trying to investigate the homolysis of bromine for an IB Chemistry (high school) Internal Assessment. To do so, I need a UV lamp capable of splitting diatomic bromine into bromine radicals. Would a handheld UV flashlight off of Amazon (395 nm) or a 4W UV tube (265 nm) work to do this?
I'm trying to react bromine radicals with methane to produce other gases. I need to produce enough of these gases to create a noticeable change in mass of my bromine sample (0.05 g or more).

Bromine has a rather high vapour pressure, 50 mg of it will be gone very quickly with a radical reaction or without. This does not sound like a well researched project. Usually not he pupils fault, but his inept teacher's.

If I use 1 mL samples in a dark room, will I see a difference with a UV light? How can I make my experiment work?

To show that bromine reacts, you should use very litte of it (a drop at most) in an excess of the other educt, and then show how the colour fades as you put on the UV lamp. Find a suitable experiment in a book, and use that. No pupil or even undergrad student is expected to invent a new experiment by himself. That's something for your PhD programme, not for school. As you teacher should know and have told you. Just "trying it out", without extended experimental experience, is bound to fail, possibly in a dangerous way, and you'll have no idea why it failed.


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What does this curve over the C-B bond mean?

I have come across this notation in a question I am trying to answer and am unsure of what exactly it means, can anyone clarify what this represents?

That there are two such groups...

Why not mention name and author of this book?

That ‘curve’ is a bracket, and the whole thing is a shorthand notation for ‘two of the same group’.
Thus, the molecule could be termed bis-(3-methylbut-2-yl)borane.

The trivial or common name is disiamylborane, with the di- confirming the above, that there are two alkyl groups attached to the BH.


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What is the connection between vapor pressure and intermolecular force explanations of boiling?
I've had boiling explained to me in two ways, and I'm having trouble understanding how they connect.
The first way is that a liquid is held together by the intermolecular forces, and boiling involves breaking those bonds/interactions. This explains why, for instance, the boiling point of water is higher than that of methane, but not why boiling point is pressure dependent.
The second way is in terms of vapor pressure. Boiling occurs when bubbles form, and in order for a bubble to form and not be crushed, its internal pressure must be equal to the atmospheric pressure. From this view, the boiling point is the temperature at which the vapor pressure equals the atmospheric pressure. This explains why boiling point drops as pressure drops.
My issue is understanding how these coexist. The way I think of it, the strength of the intermolecular interaction should be the same no matter the pressure, so the temperature to break those interactions should be constant. Also, you can boil a liquid just by lowering the pressure, but I don't see how this affects the intermolecular interactions in any way.

Consider boiling as the rapid escape of molecules.
In slow evaporation and in rapid boiling, only those molecules with sufficient kinetic energy are able to leave the liquid. As each "fast" molecule leaves, it subtracts its high KE from the average KE pool of the liquid, i.e. cooling it.
For evaporation to take place, there has to be a chance for molecules to escape and not to be replaced by those in the gas above the liquid. In fact, if the pressure is high (below the critical temperature), then the gas can condense to the liquid, faster than the liquid can evaporate to the gas.
For boiling to take place, there is the further restriction that localized temperature rise takes place faster than in the main body of the liquid, so that specific locations overheat, e.g. where heat flux is highest. Where pressure is a factor, then boiling may be prevented until the pressure is relieved, as in "black smokers". The water exiting the vent may be at 200 C or more, but does not boil unless the critical temperature is exceeded.


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Kinetic energy of an ideal gas
I've heard that the kinetic energy of any different gases are the same

All gases at a given temperature have the same average kinetic energy.

But if we have 2 moles of hydrogen and 4 moles of helium in a container at a specific temperature will the kinetic energies become the same?
I think it cannot be as the n is different..
As the kinetic energy is
3nrt/2
This might be a small question but this has been a conflict between my friends..

What has more energy?  Two stampeding bulls or four stampeding bulls?  If you mixed two brown stampeding bulls and four mottled stampeding bulls, which bulls would have more total energy?  The four?  Or the two?

Lol then I'm correct..

Yeah, but only if you're interested in total energy.  If you're interested in per particle energy, then the gases have the same energy.  The per-bull energy of the two stampeding bulls is about the same as the per-bull energy of the four.

So what about the average kinetic energies of these gases ?

The average would be the same because the four mole gas would have more energy in general but if you divide it by four moles, that would give you the average kinetic energy of the gas per particle, and it would be equal to the two mole gas' total kinetic energy divided by two moles.

The answer is in front of you. $E_v$ is the Total translational  kinetic energy not the Average kinetic energy. If you want average translational kinetic energy per mol this is given by: $E_v/n$ or $3/2RT$.


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How do I know the valence number by looking at the groups?
Groups say the valence number. For example Hydrogen is on group 1, has 1 electron on it's last and only level. So we can know the valence number on that are on the groups 1-2 and 13-18. But how am I supposed to know the valence number on transition metals(3-12)?
PD: You could tell me to use the oxidation number but that is completely wrong since they are not the same.


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analytical Lab Conundrum
We have a need for the following types of analysis.  I am not sure what analytical devices would be most suited to the needs.  I'm considering FT-IR and Ion Chromatography.  What do you think?
For the analysis of oil we require the following:

An analytical device to measure TBN total base number.
An analytical device that can measure the additive metals.

For the coolant:

I want to make sure  that in addition to determining chlorides, sulfates, glycolates, nitrates, phosphates, etc.  we also have a requirement to measure Reserve alkalinity under ASTM D 1121.

I look forward to any help you can give.


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Where does the thermal energy come from in a chemical change when two substances react with one another?
We're learning about chemical reactions in class and we recently did an experiment where we combined copper and nitric acid. Both were at relatively room temperature and when they were combined the temperature rose very quickly. Isn't this contradicting the first law of thermodynamics? If the thermal energy is neither created nor destroyed but only transferred than where did all that extra thermal energy come from? please help!

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works. I think, your question may answer itself in a few class days, if you don’t mind me saying.

Thermal energy is constantly created and destroyed (or rather, converted to and from other kinds of energy) in immense amounts. Think of electric heater of any sort, for example.

The energy was in the chemicals, in the form of bonds. Energy is liberated when there is bond formation. As the reaction between $\ce{Cu}$ and $\ce{HNO3}$ is feasible at room temperature so it happened.
And another thing, room temperature is a nice temperature for that reaction because first we need to break bonds of $\ce{HNO3}$ to make new bond, which is easily done at this temperature.
Now coming to your question, the answer is 'no'. The first law is still safe. You know it wrong. The truth is (from wikipedia) : "the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed".
Give special attention to the word 'total energy', you said the 'thermal energy' in your question. And the total energy (atleast in this case) is conserved in the form of summation of thermal energy and bond energy.


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What is the meaning of ‘oxyphilic’?
I read in the paper that:

The reduction of diphenyl diselenide using lithium aluminum hydride generates a selenolate ion having Lewis acid character due to the oxygenophilic nature of the aluminium.

Can any one explain me the meaning of the bold expression, please

It likes to form bonds with oxygen.

Thank's Sir , but what is the type of these bonds , and which oxygen do they mean in this medium ???

Nadine if you want someone to get notified about your comment then you should add '@'  before their user id...for example @orthocresol

Ok, Sir, thank you for remarkable advice @Manshu

There is indeed no oxygen in this medium. To reveal its Lewis acid character, this compound must be transferred someplace else. Also, the word "oxygenophilic" sounds kind of old-fashioned to me.

I think it just means its attracted to oxygen, or oxygen-loving?