text:
On the notion of a component in thermodynamics
Wikipedia article on a component (in thermodynamics) first mentions that:

In thermodynamics, a component is a chemically-independent constituent
  of a system. The number of components represents the minimum number of
  independent species necessary to define the composition of all phases
  of the system.

A bit later it is said in the same article that:

The number of components is equal to the number of distinct chemical
  species (constituents), minus the number of chemical reactions between
  them, minus the number of any constraints (like charge neutrality or
  balance of molar quantities).

How does the second statement follows from the first one?

Well, a reaction comes with an equilibrium constant, i.e., mathematically speaking, some equation on constituents. Now we have N unknowns (constituents) and M equations (constraints and equilibrium constants). Supposedly N>M. This means we can't determine all unknowns. To do so, we must fix N-M free variables. These are our components.


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How to prove that the orientation of the atomic orbitals in the superposition $\psi= a\psi_{1} + b\psi_{2}$depends on the coefficients $a$ & $b$?
Hybridisation is a quantum superposition of orthogonal atomic orbitals; for instance
$$\newcommand{\k}[2]{\langle #1 \lvert #2 \rangle}
\psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y} + c_4\psi_{2p_{z}}.$$
It is a quantum superposition which implies the coefficients $c$ to be the probability amplitudes for finding the electron in respective atomic orbital provided that it is prepared in $\lvert\psi_{sp^3}\rangle$(provided $\sum_{i=1}^4 c_1^2 =1$) that is, $$c_1= \k{\psi_{2s}}{\psi_{sp^3}}, \\ c_2= \k{\psi_{2p_x}}{\psi_{sp^3}},\\ c_3= \k{\psi_{2p_y}}{\psi_{sp^3}},\\ c_4= \k{\psi_{2p_z}}{\psi_{sp^3}}.$$
Up to this, I could easily understand this as it was just like some other superposed state often encountered in QM like $\lvert\phi\rangle = \lvert i\rangle c_i + \lvert j\rangle c_j + \lvert k\rangle c_k$ where $c_i = \k{i}{\phi}.$
But, here in the case of hybridisation, they are much more than mere comples numbers that are the amplitudes of the electron to find it in one of the base states that is the associated atomic orbitals when being prepared at the hybridised state. It is evident from Peter Atkins' words:

The coefficients in the hybrid have been chosen to give the correct directional properties of the hybrid. The squares of the coefficients give the proportion of each atomic orbital in the hybrid.

Also, the snapshot$^1$ below shows how for different values of coefficients, the spatial orientation is changed.

That means the coefficients  direct the spatial orientation of the atomic orbitals that are the base states for the superposition or hybridisation.
Is there any mathematical deduction that shows the orientation of the atomic orbitals in the superposition depends on the coefficients $c_i$?
I'm basically not understanding how these coefficients, which are rather probability amplitudes, would involve in determining the spatial orientations of the atomic orbitals. Can anyone explain me why these probability amplitudes influence the orientation of the atomic orbitals? Also, what is the mathematical proof of this?

$^1$The snapshot, above, is taken from Principles of Physical Chemistry
 By Hans Kuhn, Horst-Dieter Försterling, David H. Waldeck.

Do not cross-post same question on different SE sites. You already asked the question on Physics.SE, so wait for an answer there (or ask the mods there to migrate it here).

I'm already deleting it. So, no offense @Wildcat .

In quantum mechanics states/orbitals are defined within the mathematical framework of the Hilbert space. A Hilbert space is a vector space that generalizes the notion of Euclidean space -  one example of the latter would be the normal two- or three-dimensional spaces you might be used to from basic geometry. So, orbitals are nothing more but abstract vectors in a Hilbert space and you can draw many analogies between them and the more familiar two- or three dimensional vectors you learn about in high school. So, let's think of quantum mechanical states as vectors for now. If you add two vectors, $\vec{a}$ and $\vec{b}$, you get another vector, $\vec{c} = \vec{a} + \vec{b}$, which if $\vec{a}$ and $\vec{b}$ are linearly independent, will have a direction different from either $\vec{a}$ or $\vec{b}$. You can also choose different coefficients for the superposition, e.g. $1$ for $\vec{a}$ and $-1$ for $\vec{b}$ yielding a vector $\vec{d} = \vec{a} - \vec{b}$ which now has a direction different from either $\vec{a}$, $\vec{b}$ or $\vec{c}$. So, the direction of a sum vector changes with the coefficients.

The same principle holds true for orbitals. If you for example think of the $\mathrm{p}_{x}$ orbital as a vector pointing in the $x$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}1 \\ 0 \end{smallmatrix}\bigr)$, and of the $\mathrm{p}_{y}$ orbital as a vector pointing in the $y$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}0 \\ 1 \end{smallmatrix}\bigr)$, then a superposition of both (with equal coefficients of $\frac{1}{\sqrt{2}}$ to ensure normalisation) will give you a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$ direction.

But as with vectors you are by no means confined to simply add them together with equal coefficients. For example, if you add $2$ parts of $\mathrm{p}_{x}$ and $1$ part of $\mathrm{p}_{y}$ you get a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}2 \\ 1 \end{smallmatrix}\bigr)$ direction, and that is clearly different from $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$.

I appreciate your effort, sir but Feynman always in his lecture cautioned that state-vectors are rather abstract stuffs & can't be comparable to three-dimensional vectors. Hmmm....Quantum superposition has no analogy with vector-addition:(

There certainly are differences but in your case you only need to use the basic properties that all vector spaces share and that thus allow you to make analogies from three-dimensional euclidean to hilbert spaces: for example every linear combination of members of a vector space must also be a member of the vector space. And those linear combinations can be done for euclidean and hilbert spaces alike. Also, if two members of a vector space are linearly independent then they have a different "direction". Those are general concepts that don't depend on the kind of vector space.

@user36790 Well, quantum mechanical states are complex functions rather than real vectors. That alone gives you some important differences. But I don't have any striking pitfalls in the back of my mind where it is actually dangerous to draw the analogy between three-dimensional vectors and quantum states.

@user36790 Btw, what you do when forming hybrid orbitals is not a quantum superposition but a change of basis or rotation of your coordinate system, in the same sense as that you can use the vectors $\bigl(\begin{smallmatrix}1 \ 0 \end{smallmatrix}\bigr)$ and $\bigl(\begin{smallmatrix}0 \ 1 \end{smallmatrix}\bigr)$ as a basis for the two-dimensional euclidean space but you can just as well use linear independent combinations of the two, e.g. ...

... $\frac{1}{\sqrt{2}}\bigl(\begin{smallmatrix}1 \ -1 \end{smallmatrix}\bigr)$ and $\frac{1}{\sqrt{2}} \bigl(\begin{smallmatrix}1 \ 1 \end{smallmatrix}\bigr)$.

Hybridisation is indded quantum superposition. Check the answers of  this post.

@user36790 Ok, I am not sure then if quantum superposition might not simply be the same as linear combination. I thought they might have subtly different meanings, but my QM course was quite some time ago. If it is nothing more than a linear combination then I don't understand Feynman's statement that Quantum superposition has no analogy with vector-addition because both are simply a manifestation of one of the basic axioms of vector spaces in general (see here).

Quantum superposition & linear combination are both same things, but in QM, they are bit-abstract like the possibility of electron being at two places at the same time but that is another topic.......

Feynman, I think, wanted to distinguish the subtle abstractness of quantum superposition from the mere linear-combination of vectors.

Ok, I am still convinced that my answer is correct and doesn't use any wrong analogies. If someone can show me where my analogies fail I would be interested to see that. It would actually be interesting to know in what sense Feynman sees a fundamental difference between vector additon and quantum superposition because the only difference I can think of is the possibility of complex coefficients.

You are overestimating his statement by saying  fundamental. Also, I've not in any way tried to offend you that your answer is incorrect; I was just ... just a bit confused, that's it. I am again saying thanks for your effort. Sorry for those arguments:(

@user36790 I am not offended. No worries. Nor is there anything to feel sorry about. Maybe you could ask the question what Feynman ment by "Quantum superposition has no analogy with vector-addition" on Physics.SE. Because I think I know some quantum mechanics and I don't really see his point. Maybe it is really just about the abstractness and the complex nature of quantum states but maybe there is actually a fundamental difference that I simply don't see. Would be an interesting question.

I am quite sure he is talking about the abstractness, though. He is an excellent explainer; I've learnt QM(introductory) from his lectures but really sometimes his English does annoy me a lot.

As a German I think I am in no position to question anyone's English (our dialect can be really terrible) :) If you want a more mathematical yet very well explained exposure to QM I can recommend this book by Shankar. The first two chaperts give you a really nice résumé of the linear algebra and classical mechanics basics you need for QM.

Yes, this book is quite famous in India after Griffiths but I preferred Feynman to them as it is more diverse & above all it was his valuable lectures at CALTECH; who would want to miss that?  His lucid, unorthodoxical way of  explanation made these lectures a classic. Though these lectures are over 45 yrs old, still they are not outdated. Every physicist has studied his lectures at least one time in their career. Thanks for the recommendation, nevertheless.

@Philipp - re: ""Quantum superposition has no analogy with vector-addition" // I would take that to be a comment on the multibody problem. So "real" atoms don't form bonds using two dimensional vectors. The "real" problem is a 3D 3-body problem.  So we use it because it provides useful insight, not because the method is "real."

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What role does sulfuric acid play in the titration of oxalic acid and potassium permanganate?
The titration is between oxalic acid and potassium permanganate, with $\ce{H2SO4}$ added to oxalic acid. My question is what role other than a dehydrating agent and maybe catalysing does the $\ce{H2SO4}$ play in this titration.
The question came to me while reading a book (Advanced Chemistry by Philip Matthews - Cambridge university press), I was reading about titrations and this question came to me, but it isn't mentioned in the book.

Write down the balanced equation of the reaction and you'll see right away that you need some extra acid. That's what it is for.

@IvanNeretin Why not HNO3 or HCl?

@ReeshabhRanjan HNO3 itself may oxidize something. HCl itself may get oxidized by permanganate.

how does it act as a dehyradting agent in the titratuion?

The used analytical reaction of purple permanganate to colourless $\ce{Mn^2+}$ occurs under acidic conditions:
$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$
Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$
Therefore, sulfuric acid is added to make the solution acidic. Actually, excess acid is required since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).
Sulfuric acid is used because it is stable towards oxidation; whereas, for example, hydrochloric acid would be oxidized to chlorine by permanganate.

Thanks to the comment by Ivan Neretin, I figured out that H2SO4 is actually to provide surplus H+ ions, to keep the reaction going
And also it acts as a medium which doesn't interfere with the oxidizing agent to keep the quantitative purpose of the reaction

Sulfuric acid acts as a catalyst in the reaction
$\ce{MnO4- + 8H+ + 5Fe2+ -> 5Fe3+ +  Mn2+ + 4H2O}$
It initiates a 5e transfer in the ionic equation
It used rather than $\ce{HCl}$ because $\ce{HCl}$ is volatile acid which will oxidize to $\ce{Cl}$ in presence of $\ce{KMnO4}$ (strong oxidizing agent). $\ce{H2SO4}$ is stable to oxidization. Also excess $\ce{H+}$ is used because for every 1mole of $\ce{MnO4-}$ , 8 moles of $\ce{H+}$ is consumed.

There are three reasons:

to provide acidic medium
to oxidize $\ce{Mn^{+2}}$ to $\ce{Mn^{+3}}$
to speed up the reaction

H2SO4 increase the acidic content of the solution so as to prevent MnO4 (purple) to reduced to MnO2(dark brow). Hence sulfuric acid is stable in the present of strong oxidising agent. If HCl acid is use it will be oxidised to Cl and make the end point much higher. Furtherore sulfuric acid increase the H+ as it consume.

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Does an increase in pressure necessarily lead to an increase in temperature?
Ideal gas law says that $pV = nRT$. So with compression, one way to add pressure which makes the volume smaller, can you actually increase the temperature?
It does not make sense to me that you can increase temperature, that is, add to the kinetic energy of the molecules, simply by increasing pressure, pressure meaning the force exerted on the gas from its surroundings. Can someone please explain this contradiction?

pressure does not mean the force exerted on the gas by the surroundings, it means the force exerted by the gas on the walls of the container divided by the area of the wall.

Ok fair enough feed back about wording

Karl, are you saying that, in the adiabatic compression of an ideal gas in a closed system (e.g., an insulated cylinder with a piston), there is no change in the gas internal energy or temperature?

Really.  Then please comment on the following for the case of an adiabatic reversible volume change:  $dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$.  So, $d\ln T=-\frac{R}{C_v}d\ln V$.  So, when the volume decreases, the temperature increases.

I mixed up free expansion and reversible compression/expansion.

Sorry I'm a muggle. Not sure how to work this Q and As. But in answer to the question I thought that compressed gases got colder if anything. Could be wrong of course. Thought that you slow the movements of particles when you compress things and if they are moving slower they are colder. There may come a point when they can't be compressed any further and then they'd probably explode.

I think what you are asking is "why can the temperature of the gas increase when you compress it, even if the cylinder is adiabatic so that no heat can enter the gas?"  When you move the piston to compress the gas, you are doing work on the gas at the interface with the piston.  The piston is moving toward the gas, and the molecules of gas that collide with the piston leave with a greater average velocity than when they arrived.  So their average kinetic energy is increasing.  If expansion were occurring, such that the piston were moving away from the gas, the colliding molecules would leave with lower average kinetic energy.

Who said anything about compression rate?  In the end, it is only the total amount of work that determines the change in temperature, but that's just equal to the integral of the force per unit area at the piston face times the rate of change of volume (compression rate) dt.  You know, for an adiabatic process, $\Delta U = -\int{P_{ext}dV}$, where, for an ideal gas U = U(T).

I stand by what I said.

Are you talking about an ideal gas or air?  For air at least part of the reason is that the molecules attract each other and it is not ideal.  Would an ideal gas increase in temperature?  (Either way PV=nRT, does not say wither T changes.)  A good question not answered here or anywhere else I could find.

@Tuntable I am talking about both real gases and ideal gases.  Of course, an ideal gas would also increase in temperature.  PV-nRT is not the only characteristic of an ideal gas that matters.  The first law of thermodynamics also comes into play here, and, for an ideal gas, the internal energy is a function of temperature.  Did you read my comments to the OP's post?

It is not at all clear that an ideal gas would increase in temperature, at least not significantly.  Sure, increasing the pressure increases enthalpy, but enthalpy is T+PV.  If you are sure that it increase temperature of an ideal gas, then by how much?  Do you have a formula or a reference?

@Tuntable First of all, enthalpy is not T+PV.  Secondly, the ideal gas law tells us that, for an adiabatic reversible process of any gas, $dU=-pdV$.  For the case of an ideal gas, this reduces to $$nC_vdT=-\frac{nRT}{V}dV$$So, if dV is negative, dT is positive.  This is every thermodynamics textbook.  So stop wasting my time.

The ideal gas law, PV=nRT, only tells us how to calculate the pressure given the temperature and volume.  Compressing the gas certainly increases U, which includes both temperature and pressure, not sure why you hold pressure constant.  I have never seen this explained properly, just vagueness, leaps and then differential equations.

Who said the pressure is constant?  And, for an ideal gas, U depends only on temperature, not pressure.

It’s beginning to dawn on me that you are uncomfortable because you can’t see why U would not depend on p for the case of an ideal gas.  Is that your issue?

Sorry, got my U and H mixed up.  Not sure why enthalpy is not essentially T+PV, given a non-reacting ideal gas T is essentially U + constant.  When you say dU = -pdV then you are holding p constant, changing the volume, so obviously the temperature must change.  But it says nothing about what happens to the temperature if you just compress the gas increasing its pressure, adiabaticly.  I think what I am after is a good link if you know one, rather than an explanation.

Found it at last!!  Joule-Thompson effect.
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
As I suspected, for Helium at normal temperatures, there is no increase in temperature at all.  The rise when one compresses air has nothing to do with PV=nRT, but only to do with the non-ideal nature of air.  Thanks for your help though.

If you had a way to increase pressure with no volume change, then yes, temperature would increase by the ideal gas law.  In reality, most compression take place by reducing volume or increasing N, so the temperature effect is hard to see directly because other things are changing too.
The pressure in PV=nRT is the force exerted by the gas on the walls of the container.  As the temperature increases, the particles move faster, and therefore have greater speeds, so greater momentum and therefore greater force when they collide with the walls, so the pressure increases.

I understand what youre saying and I agree. Yes the ideal gas law says it happens in theory but does it actually happen in reality without changing the volume or the number of atoms?

How can you compress a gas without changing its volume?  Compressing means decreasing its volume.

I was commenting on the idea of changing pressure without volume, not compression.

At constant volume, you need to add heat to raise the temperature so that the pressure can increase.  The temperature rise is the cause and the pressure rise is the effect, rather than the other way around.

Ok, yes thats the idea I was getting at in the previous comment. Thanks!

@chestermiller  I didn't mean one could compress s system without changing volume, or that one could do an experiment in a lab today to raise temperature by changing pressure (thats why I said "if" and "in reality") but what I was getting at is that there is a direct correlation between pressure and temperature that would go both ways if the experiment could be done with everything else fixed.  In real experiment, yes, it usually is temperature that is raised, but thats not what the question was asking.

@JiffFulcher you can increase the pressure by putting more gas into the containers. The container stays the same size and has the same volume

The original question was whether just decreasing the volume would increase the temperature for an IDEAL gas.  Obviously the pressure increases, per PV = nRT.  But what about the temperature?  For air, the temperature increases, but air is not ideal.  Please do not provide non-answers.

We all know, Solids have defenite size and obviously have defenite volume. Liquid has defenite volume but no shape. Gases have neither shape nor volume. Gas will occupy the available volume of the container. Molecules make use of the available free space for their movement.
Thus in gases, you can externally change the degree of freedom of its molecules. When you increase the container volume, you are increasing the degree of freedom of the gas molecules. And conversely also it is true.
Coming to the question, when you decrease the degree of freedom of molecules (by decreasing the container volume), due to the limitation in their mobility, the excess residual energy has to be given out (All system trend to minimize its energy state). Naturally the gas becomes hot in a big to exchange the excess energy to the surrounding.
(Most of the natural energy exchange is done by heat energy).

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Difference between calcination and pyrolysis
What is the difference between calcination and pyrolysis?
I understand they both involve heat treatment but each under a different type of atmosphere, which leads to some sort of reduction reaction? What are the differences in underlying mechanisms?

Have you read https://en.wikipedia.org/wiki/Calcination,  https://en.wikipedia.org/wiki/Pyrolysis, or other info on internet?

Yes but i was hoping to find out more about the chemical mechanisms at play

Essentially calcination is decomposition at high temperature with inorganic reactants such as calcium carbonate in making cement.
$$\ce{CaCO3 ->[\Delta ] CaO + CO2}$$
Pyrolysis is decomposition or oxidation at high temperatures of organic materials such as cellulose in making charcoal.
$$\ce{(C6H12O6)_{n} ->[\Delta] C7H4O + C_{x}H_{y} + H2O}$$
where $\ce{ C_{x}H_{y}}$ is volatilized byproducts.


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Trying to understand the statement: 'Resonance is not a flickering between the contributing states. '
So, the famous resonance definition: it is the weighted average of different Lewis structures....
Well, this is purely wrong.
Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'-Peter Atkins.
For-instance, to describe the wavefunction of $\ce{HCl}$, quantum superposition is applied viz.  $$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$ where $C_1$ is the probability-amplitude to find the electron-pair at $|\psi_{\ce{H^+ Cl^-}}\rangle$ provided it was prepared in $|\psi_{\ce{HCl}}\rangle$ that is, $C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$ also $C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$ and $C_1^2 + C_2^2= 1.$
As written by Peter Atkins in his book Elements of Physical Chemistry:

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H_Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $\lambda$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $\lambda$  as the relative proportion of the ionic contribution. If $\lambda^2$ is very small, the covalent description is dominant. If $\lambda^2$ is very large, the ionic description is dominant.
[...] The interpretation of the wavefunction, which is called resonance hybrid, is that if we were to inspect the molecule, then the probability that it would be found with an ionic structure is proportional to $\lambda^2$. For instance, we might find that $\lambda = 0.1,$ so the best description of the bond in the molecule in terms of a wavefunction is a resonance structure described by the wavefunction $\psi= \psi_\text{covalent} + 0.1\psi_\text{ionic}.$ This wavefunction implies that the probabilities of finding the molecule in its covalent & ionic forms are in the ratio $100:1.$

This is what is known as wavefunction-collapse in Copenhagen interpretation which means $\psi$ collapses to one of its base state after the measurement; here the wavefunction is more prone to collapse to the covalent description. Hence, in this case, the state of the molecule has mostly covalent properties.

I've pretty much conceived the upper part; but speaking at the present context of the title of the question, let me start from Peter Atkins' description of benzene in his book Elements of Physical Chemistry:

One of the most famous examples of resonance is in the VB description of benzene, where the wavefunction is written as a superposition of the wavefunctions of the two covalent Kekulé structures: $$\psi = \psi_{\text{Kek}1} + \psi_{\text{Kek}2}$$ The two contributing structures have identical energies, so they contribute equally to the superposition. The effect of resonance in this case is to distribute the double-bond character around the ring and to make the lengths and strengths of all the carbon-carbon bonds identical. The wavefunction is improved by allowing resonance because it allows for a more better description of the location of the electrons, and in particular the description can adjust into a state of lower energy. ...

The statement that baffled me comes from Atkins in the last paragraph of resonance as:

Resonance is not a flickering between the contributing states: it is a blending of their characteristics, much as a mule is a blend of a horse and a donkey. It is only a mathematical device for achieving a closer approximation to the true wavefunction of the molecule than that represented by any single contributing structure alone.

The cause of my bafflement is due to the following excerpt of Feynman's lecture The benzene molecule:

There is a mystery about this benzene molecule. We can calculate what energy should be required to form this chemical compound, because the chemists have measured the energies of various compounds which involve pieces of the ring—for instance, they know the energy of a double bond by studying ethylene, and so on. We can, therefore, calculate the total energy we should expect for the benzene molecule. The actual energy of the benzene ring, however, is much lower than we get by such a calculation; it is more tightly bound than we would expect from what is called an “unsaturated double bond system.” Usually a double bond system which is not in such a ring is easily attacked chemically because it has a relatively high energy—the double bonds can be easily broken by the addition of other hydrogens. But in benzene the ring is quite permanent and hard to break up. In other words, benzene has a much lower energy than you would calculate from the bond picture.
Now we want to resolve these mysteries—and perhaps you have already guessed how: by noticing, of course, that the “ground state” of the benzene ring is really a two-state system. We could imagine that the bonds in benzene could be in either of the two arrangements.[...] You say, “But they are really the same; they should have the same energy.” Indeed, they should. And for that reason they must be analyzed as a two-state system shown in Fig.

Each state represents a different configuration of the whole set of electrons, and there is some amplitude $A$ that the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.

In the blocked-letters Feynman told that if at one time the molecule be found in $|1\rangle$, at some time later, it may be found at $|2\rangle$ (& vice-versa) as the Hamiltonian elements $H_{12},H_{21}$ are not zero but $-A$, the probability of transition per unit time.

As we have seen, this chance of flipping makes a mixed state whose energy is lower than you would calculate by looking separately at either of the two pictures in Fig. Instead, there are two stationary states—one with an energy above and one with an energy below the expected value. So actually, the true normal state (lowest energy) of benzene is neither of the possibilities shown in Fig., but it has the amplitude $1/\sqrt2$ to be in each of the states shown. It is the only state that is involved in the chemistry of benzene at normal temperatures.

Why should one found the benzene at $|1\rangle$ at one time but at other time, at $|2\rangle?$ Because the transition-amplitudes are non-zero and 'the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.'The electron can flip from one state to the other. But doesn't this statement is at dagger's drawn with the common statement 'Resonance is not a flickering between the contributing states.'? I am completely sure both Feynman and Atkins is telling right, I am misinterpreting both the statements but don't know how. Can anyone please, please explain me how does flipping of the electrons from one dance to the other doesn't contradict the statement Resonance is not a flickering between the contributing states?? Please help.

Discussion in chat.

Related: http://chemistry.stackexchange.com/questions/51632/what-is-resonance-are-resonance-structures-real

When it comes to resonance, you should not take the notion of superposition of different states literally. Feynman mentioned it explicitly right from the get go, just read carefully what he says:

We could imagine that [...]

See, we could imagine. The superposition of states in a description of what is called resonance in chemistry is not the superposition of real states of a molecular system, it is a superposition of imaginary states. The only real state is to the left of the equality sign, everything to the right is imaginary. Thus, your analysis of this "superpositions" in accordance with the Copenhagen interpretation is plain wrong.
Atkins seems to be going a bit too far taking resonance almost literally along the road, but at the end rightfully warns us that it is "not a flickering between the contributing states".

It is only a mathematical device [...]

Concise! And here I was yammering about in http://chemistry.stackexchange.com/a/48334/28144

Why not go back to the experimental evidence that the C-C bond lengths are all the same length so that a picture of alternating single and double bonds is not correct and so this makes any combination of resonance structures as just a way of reconciling a valence bond picture with experiment. That is, we are used to single and double bonds as its a extremely convenient (and good) way of describing lots of chemistry and are reluctant to let go. In a molecular orbital approach such a resonance does not need to be invoked. Its just a change of point of view, changing a basis set if you like.


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pKa of Tris corrected for ionic strength
I have a question regarding the molar fraction calculation of Tris buffer at different temperatures.
Tris dissociates according to the following equation:
$$\ce{Tris + H2O -> Tris+ + OH-}$$
So, in order to calculate the amount of $c(\text{Tris}^+)$ and thus the amount of strong acid needed for titration, one needs to use derivation of Henderson-Hasselbalch eq.:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}' - \log_{10}\left(\frac{[\text{Tris}^{+}]}{[\text{Tris}]}\right)$$
However, $\mathrm{p}K_\mathrm{a}'$ is $\mathrm{p}K_\mathrm{a}$ corrected for ionic strength ($I_c$) at a certain temperature.
The correction of $\mathrm{p}K_\mathrm{a}$ is done via the Debye-Hückel eqation:
$$\mathrm{p}K_\mathrm{a}' = \mathrm{p}K_\mathrm{a} -0.509\times\left| z-z^{+}\right| \sqrt{I_c}$$
Now, for Ic itself one needs to calculate the molar fraction of $[\ce{Tris+}]$:
$$I_c = 0.5\times \left([\ce{Tris+}] + [\ce{OH-}]\right)$$
which requires the use Henderson-Hasselbalch equation making this problem somewhat circular. or am I missing something?

If you write also mass balance and charge balance can't get [tris+]?

First decide the Ic you want to work with. Second, you should correct the pka with the Debye Huckel equation and finally you use the pka corrected with the Henderson Hasselbach equation.


text:
Why is honey more viscous than water?
Why is honey more viscous than water? And, if we boil honey to just below its boiling point, will it become equally viscous as water at normal temperature? How does changing temperature affect viscosity?

Viscosity generally decreases with increasing temperature so honey will become less viscous at higher temperatures. Regarding the actual values, look at this article. It appears that even at quite high temperatures honey remains a couple of times more viscous than water.
I think the reason for higher viscosity in honey is the high sugar () content - hydroxyl groups in sugars can form strong, hydrogen-bonded interactions with water and each other, and the sugar molecules being quite long, they are difficult to shear.


text:
Hydrogen displaces lead?
there was a reaction in my book. It is as follows:
$\ce{PbO + H2-> Pb + H2O}$
My question is Hydrogen is below lead in the reactivity series. Then how is this reaction possible?

I looked up the enthalpy of formation of PbO and it's -219, so that reaction would be endothermic.  http://chemister.ru/Database/properties-en.php?dbid=1&id=4082

The activity series you're referring to is constructed for ambient conditions where the temperature is $25°\text{C}$ and water is a liquid.  But, of course, thermodynamic properties can depend on temperature and especially on phase.  At higher temperatures where water is ordinarily a gas, it has a lot more entropy than the liquid.  That makes oxidation of hydrogen to water more thermodynamically favorable than what appears in your ambient temperature series.
For high temperature reactions where water mixes with hydrogen in the gas phase, the entropy effect pushes hydrogen up the series so it roughly matches iron.

The enthalpy of formation of H2O is significantly larger than the enthalpy of formation of PbO, so this reaction is exothermic.  Reactivity series are not perfect predictors.  Results depend on what an element is being reacted with.

you contradicted your initial comment that the reaction is endo-thermic.

Look at it from a redox reaction perspective. I'm not sure what the parameters the reaction is taking place under but the Pb is going from Pb2+ to just Pb. This does make sense because lead oxide is commonly used as an oxidizing agent and hydrogen gas is considered to be a powerful reducing agent.
Hope this helps
Just verified this with my advisor. It's an odd reaction but it does work.
All the best

Lead oxide which is commonly used as an oxidizing agent is another lead oxide, $\ce{PbO2}$.

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


text:
How is a molecular orbital a 'quantum superposition' of the atomic orbitals?
Let me quote from Feynman's lectures the concept of superposition:

(1)The probability of an event in an ideal experiment is given by the square of the absolute value of a complex number $ϕ$ which is called the probability amplitude:
  $$\begin{equation}
\begin{aligned}
P&=\text{probability},\\
\phi&=\text{probability amplitude},\\
P&={|\phi|}^2.
\end{aligned}
\end{equation}$$
  (2)When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately. There is interference:
  $$ϕ=ϕ_1+ϕ_2.$$

The second point is the core of quantum superposition.
Also, in Dirac's language from The principles of Quantum Mechanics,

[...]It requires us  to assume that between these states there exists peculiar relationships such that whenever the system is definitely in one state we can consider it as being partly in each of two or more other states.[...]

These all imply that when a system is in a state of superposition, it is 'partly' in both of the states that contribute to the superposition( Here 'partly' means possibility which implies the system has the possibility to exist in both states that contribute in the superposition).
For instance, let's take $\ce{H_2^+}$ to illustrate the concept of superposition. Firstly, the words of Feynman:

A positively ionized hydrogen molecule consists of two protons with one electron worming its way around them. If the two protons are very far apart, what states would we expect for this system? The answer is pretty clear: The electron will stay close to one proton and form a hydrogen atom in its lowest state, and the other proton will remain alone as a positive ion. So, if the two protons are far apart, we can visualize one physical state in which the electron is “attached” to one of the protons. There is, clearly, another state symmetric to that one in which the electron is near the other proton, and the first proton is the one that is an ion. We will take these two as our base states, and we’ll call them $|1⟩$ and $|2⟩.$ They are sketched in Fig.

]1
Now, the general state of $\ce{H_2^+}$, say, $|\psi_{\ce{H_2^+}}\rangle$ is in quantum superposition of the two base-states $|1\rangle$ & $|2\rangle.$
Now using the point no.(2) of Feynman, we can prove the superposition of the states as :
What should be the amplitude for finding the electron in, say, $x'$ from the center of the molecule $\ce{H_2^+}?$  (We would use each-coordinate as our new base-state, that is, $|x\rangle$ is a base-state; this means we would now treat continuum of base states).
We would've to find $\langle x'|\psi_{\ce{H_2^+}}\rangle.$ This can happen in two ways & thus applying the point no.(2), we get $$\begin{equation}\langle x'|\psi_{\ce{H_2^+}}\rangle\\=\int \langle x'|x\rangle \langle x|\psi_{\ce{H_2^+}}\rangle dx\\= N\left(\int \langle x'|x\rangle \langle x|1\rangle dx +\int \langle x'|x\rangle \langle x|2\rangle dx\right)\end{equation} ;$$ where $ N= \text{normalisation constant}.$ From this, we can easily shew, $$|\psi_{\ce{H_2^+}}\rangle = N(|1\rangle + |2\rangle).$$ Thus, quantum superposition is used when there is more than one way an event can happen; in this case, our event was finding the electron at $x'$ ; we superpose the alternative amplitudes to get the total amplitude of the event of finding the electron at that coordinate.

Let's consider $\ce{H_2}$; we describe first the VB description; again starting with Feynman's words:

As our next two-state system we will look at the neutral hydrogen molecule $\ce{H_2}.$ It is, naturally, more complicated to understand because it has two electrons. Again, we start by thinking of what happens when the two protons are well separated. Only now we have two electrons to add. To keep track of them, we’ll call one of them “electron a” and the other “electron b.” We can again imagine two possible states. One possibility is that “electron a” is around the first proton and “electron b” is around the second, as shown in the top half of Fig. We have simply two hydrogen atoms. We will call this state $|1⟩.$ There is also another possibility: that “electron b” is around the first proton and that “electron a” is around the second. We call this state $|2⟩.$

So, we can write the general state $|\psi_{\ce{H-H}}\rangle$ as the superposition of $|1\rangle$ & $|2\rangle.$ We can again prove the superposition from point no.(2) but this time we've to use $|x_1,x_2\rangle$ instead of $|x\rangle$ that is, $$\begin{equation}\langle x'_1,x'_2|\psi_{\ce{H-H}}\rangle\\=\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|\psi_{\ce{H-H}}\rangle dx\\= N'\left(\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|1\rangle dx +\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|2\rangle dx\right)\end{equation} ;$$ where $N'=\text{normalisation
constant}.$ This evidently proves the superposition $$|\psi_{\ce{H-H}}\rangle= \frac{1}{\sqrt 2}(|1\rangle + |2\rangle)$$  where $N'= \frac{1}{\sqrt 2}.$
Come to the MO description of $\ce{H_2}$(only the bonding orbital is concerned here to shorten the query). This is where my problem begins...
This is quoted from J.D.Lee's Concise Inorganic Chemistry:

Consider two atoms $A$ and $B$ which have atomic orbitals described by the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}.$ If the electron clouds of these two atoms overlap when the atoms approach, then the wavefunction for the molecule(molecular orbital $\psi_{(AB)}$) can be obtained by a linear combination of the atomic orbitals $$\psi_{(AB)}= N(c_1 \psi_{(A)} + c_2\psi_{(B)})$$
[...]Suppose the atoms $A$ and $B$ are hydrogen atoms ; then the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}$ describe the $1s$ atomic orbitals on the two atoms. [...]

Now, wavefunctions are nothing but probability amplitudes; since _molecular orbital is a quantum superposition of the atomic orbitals; it must satisfy the point no.(2) like all other superposition described above.
But I'm not understanding what base states are used in this case; in VB theory, $|1\rangle$ represented the state of having one electron around one $\ce{H}$ atom & the other electron around the other $\ce H.$ So, $\langle x_1,x_2|1\rangle$ meant the amplitude to find electron $a$ at $x_1$ & electron $b$ at $x_2.$
But what should I use in this case? How to represent $\langle x_1,x_2|\psi_{\ce{H-H}}\rangle$ using point no.(2) in MO theory?
I'm not getting how to apply the point no.(2) here; $\psi_{(A)}$ represents the amplitude of only one electron; same with $\psi_{(B)}.$
For applying point no.(2), there must be two ways for the event to find the electrons at $x'_1,x'_2$_ so that point no.(2) yields the result $\psi_{(\ce{H-H})_{(g)}}= N( \psi_{(H)_{1s}} + \psi_{(H)_{1s}}).$ But I'm not getting how $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes of the same final event of finding the electrons at $x'_1,x'_2$; if they were not representing the amplitudes of getting to the same final event, how could they be superposed by point no.(2)? In VB theory, the base states are defined so that the alternative amplitudes define the same final event & so they were superposed; however how should I interpret the base-states in MO theory, so that $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes to the same final event? How should I apply point no.(2) in case of molecular orbital to prove the superposition?

Same story as with the previous question: quantum superposition is always expressed mathematically as a linear combination, but the converse is not necessarily true. Not each and every linear combination expression has something to do with real physical quantum superposition. In many cases it is just a mathematical trick.
In the question about resonance we've already faced a situation when we could only formally think about a linear combination as a superposition of some imaginary states. Same story here: molecular orbitals are just linear combination of atomic orbitals. As for the case of resonance, the states you think are superimposed (atomic orbitals this time) are not even states of electrons in a molecule. The only "real" (though approximate) states of electrons in a molecule are molecular orbitals, not the atomic one, thus, strictly speaking, there is no superposition out there.
Besides, you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule. To construct (approximate) many-electron wave function you need to make (antisymmetric) product of $n$ molecular orbitals, where $n$ is the number of electrons. And since each and every molecular orbital describes just one electron, all your analysis of molecular orbitals for two-electron case of a hydrogen molecule is plain wrong.

P.S. And to be honest, I don't understand what you did for one-electron $\ce{H2+}$ case as well. To me it looks like you have too many misconceptions in your head currently. I suspect, you're jumping the gun by trying to learn the applications of quantum mechanics to chemistry having no really solid background in theory itself.

So, let's look on what you're saying on $\ce{H2+}$ case. Following Feynman, we write the general state of the system $\lvert \psi \rangle$ as a superposition of the two basis states $\lvert \psi_1 \rangle$ and $\lvert \psi_2 \rangle$,
$$
  \lvert \psi \rangle = c_1 \lvert \psi_1 \rangle + c_2 \lvert \psi_2 \rangle \, .
$$
Fine. And then you decided to "reverify the superposition of the states as". What do you want to "reverify" out there? Superposition is asserted, you don't need to "reverify" it whatever it means.
Yes, you can switch from the state vector formalism to the wave function one by taking the inner product of both sides with position eigenvectors $\lvert x \rangle$,
$$
  \langle x \vert \psi \rangle = c_1 \langle x \vert \psi_1 \rangle + c_2 \langle x \vert \psi_2 \rangle \, , \\
\psi(x) = c_1 \psi_1(x) + c_2 \psi_2(x) \, ,
$$
where the wave function is defined as usual, i.e. $\psi(x) = \langle x \vert \psi \rangle$. That's fine, though trivial: you indeed get the (weighted) sum of the corresponding wave functions which are the probability amplitudes. But that is because it just an alternative form of a superposition expression: it tells the same thing as before, just in different words. We didn't "reverify" anything, merely restated the superposition one more time.
Anyway, that would be the thing I at least understand. But look what you did! A complete mess! First, you forget the coefficients in the linear combination. Secondly, you used the resolution of identity for absolutely no reason. Thirdly, normalisation constant just appeared out of the blue. Finally, what you think you get at the end is in fact the starting point of a journey: it is an assertion Feynman did.
So, for me all you algebra appears as random manipulations of symbols you've seen somewhere but don't really understand.

+1; I am not bothered at you critisizing me; you've every right  but what I'm bothered is why you didn't understand the $\ce{H_2^+}$ case; I've just used the point no.(2) & the continuum of base states the same way Feynman explains in chapter 16. It deduces the correct result. So, what is the wrong there?

@user36790, no, I do not criticise you. In fact, I like your questions. You ask a questions on things which many do not bother about but just take for granted. The problem is that your questions are not well suited for Q&A sites, such as Chem.SE: they are too long, not sharp focused, and usually consists of few questions rather then one.

Now, let me explain what in particular I do not understand about $\ce{H2+}$ case right in my answer.

You are right, sir that  you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule.: in Atkins' Physical Chemistry book, he wrote that MO orbitals are one-electron wavefunction . But I didn't pay heed as every book  says that each molecular orbital is occupied by two electrons; so if it were a one-electron wavefunction, how could it be occupied by two electrons?

@user36790, it is a long story. In short there are two kinds of orbitals out there: spin orbitals and spatial orbitals. For now, just think of each spatial orbital as consisting of two spin orbitals. So, when it is said that a particular spatial orbital is doubly-occupied, what is actually meant is that both its consisting spin orbitals are singly-occupied.

As I said, you don't quite understand what you're talking about. When you use the resolution of identity by introducing $\sum_{i=1}^{2} |i⟩⟨i|$ in the middle of the bracket $⟨x|S⟩$ you clearly assert that ${ |i⟩ }{i=1}^{2}$ is a basis. And since, it is a basis, I could write $|S⟩ = \sum{i=1}^{2} c_i |i⟩$ without working out any algebra out there. You already asserted superposition by introducing the resolution of identity, you don't need to prove anything out there.

I suggest the following: stop reading few book concurrently. Read just one of them, then, once you finished, start the next one. First, you should build a solid foundation of quantum theory in your head. I didn't read Feynman myself, but many recommend it, so I think you could continue to read it. Or, even better, since you have so many misconceptions, start from the very beginning.

Then you are right, sir; but see this; there the author introduced superposition in the same way as I've written. Nevertheless, my main point is  I can write these stuffs for $\ce{H_2}$ in VB description as the base states are defined so. However, I couldn't write those in case of MO description as I considered $\langle x_1,x_2|$ to be the base-state of position. But molecular orbitals, as you said, are one-electron wavefunction.

Now you see the main issue with your questions: they are not focused. As it said on the Tour page, "This site is all about getting answers. It's not a discussion forum. There's no chit-chat." You better break you questions apart. For instance, you could easily make a better question based exclusively on your analysis of $\ce{H2+}$ case. We already discuss it in the comments, but they are not suited for it. You should ask your question peace by peace: one question - one particular concept.

The problem is that you think your analysis of $\ce{H2+}$ is perfectly valid, so that you move forward to discuss few other cases, but already at that point I doubt you really understand what's going on.

We talked last in the chat that every quantum superposition is LC but every LC is not quantum superposition; I just want to know how to determine whether a particular LC is quantum superposition or not. How to determine this? It is conclusive that MO is not quantum superposition . If I'm given a LC, how to say that it is not quantum superposition?


text:
Trying to understand the necessity of orbital approximation or the independent electron model
I'm reading the approximations used for solving the Schroedinger's equation for molecules; the first is the the Born-Oppenheimer approximation:

Freeze the molecule with a fixed internuclear separation $R;$ then carry
  out calculations to obtain the total energy, $U,$ and wavefunctions for that $R$ value.$^1$

Then another approximation is used which is known as the orbital approximation to get rid of the inter-electronic repulsion as said by the same source.

Consider each electron to move in some sort of "average potential" which incorporates the
  interactions with the two nuclei and an "averaged interaction" with the other electron.
The electronic Hamiltonian can then be separated into two parts:
  $$\hat{H}_e = \hat{H}_1 + \hat{H}_2 \\ \text{where}:\\ \hat{H}_1 \;\text{is dependent only upon the properties of electron (1) and upon R .}\\ \hat{H}_2 \;\text{is dependent only upon the properties of electron (2) and upon R .}$$

I couldn't understand for what purpose this approximation is meant for. Also, I couldn't able to conceive what average potential actually means.
I then googled the term finally to come before a pdf file where it is written that

$\color{red}{\bullet}$ Electrons repel each other according to Coulomb's law,
  with the repulsion energy $r^{-1} _{ij} .$
$\color{red}{\bullet}$ Hartree-Fock replaces this instantaneous electron-electron
  repulsion with the repulsion of each electron with an
  average electron charge cloud.

Prior to that it is written:

$\color{red}{\bullet}$ It assumes that each electron interacts with an average
  charge distribution due to the other electrons. This is
  wrong!

I've spent a hard time understanding this approximation.
Could anyone please explain why this approximation is taken & how it actually solved that problem? The source cites the reason of inter-electronic-repulsion as the problem; how does this term in the Hamiltonian cause problem & how is it solved by the approximation? Also, what is meant by average potential?
An intuitive explanation would be appreciated.

$^1$ http://www.chem.qmul.ac.uk/software/download/mo/3.pdf

I'm tempted to repeat my advice here to stop switching from one source to another in a hope to find a simple explanation. Pick one (good) book and read it (from cover to cover). It also looks like OP might be trying to jump the gun: it looks just to early for him/her to dive into the quantum chemistry. This question is just another indication of OP lacking a good foundation of quantum theory in his/her head.

The question though is a good one, so I made an attempt to answer is as shortly as possible. :)

You may want to go and check the equations, try to write them down and see what you should put where, instead of fiddling with every single paragraph in your book. Try to understand the actual tools, not the pictures or simple explanations textbooks try to give you. Textbooks often oversimplify, or work ony as a metaphor, math clears things up.

Setting the stage
Once the Born-Oppenheimer approximation is introduced, we have (first) to solve the so-called electronic Schrödinger equation that could be written as follows,
\begin{equation}
    \newcommand{\el}{_{\mathrm{e}}}
    \newcommand{\elel}{_{\mathrm{ee}}}
    \newcommand{\elnuc}{_{\mathrm{en}}}
    \newcommand{\nucnuc}{_{\mathrm{nn}}}
    \newcommand{\op}[1]{\hat{#1}}
    \op{H}\el \psi\el(\vec{r}\el) = E\el \psi\el(\vec{r}\el) \, . \tag{1}
\end{equation}
Here $\op{H}\el = \op{T}\el + \op{V}\elel + \op{V}\elnuc$ is the electronic Hamiltonian, $E\el = U - V\nucnuc$ is the electronic energy, and $\psi\el(\vec{r}\el)$ is the electronic wave function. The electronic Schrödinger equation, which is a partial differential equation, is fairly complicated. One of the most frequently used methods of simplification of such equations is technique known as the separation of variables.
Separation of variables technique
The separation of variables is aimed to broke a partial differential equation of many variables into a set of simpler partial differential equations of fewer variables. Note though, that the separation of variables is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.
The most common way of separation of variables for a partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is by writing its solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
\begin{equation}
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) = \prod_{i=1}^{n} f_i(x_i) \, , \tag{2}
\end{equation}
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed, one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into (2) obtain the solution of the starting equation.
Separation of particles coordinates
It is fairly well known that an attempt to separate the coordinates of individual particles in the time-independent Schrödinger equation
\begin{equation}
    \op{H} \psi(\vec{r}) = E \psi(\vec{r}) \, ,
\end{equation}
by making a substitution $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$
will be successful only if the potential energy function can be expressed as the sum of the separate potentials for each individual particle
\begin{equation}
    \op{V}(\vec{r}) = \sum_{i=1}^{n} \op{v}(\vec{r}_{i}) \, ,
\end{equation}
which implies that the particles do not interact with each other. For such a system of non-interacting particles the time-independent Schrödinger equation, partial differential equation of a function of $n$ variables, breaks up into a system of $n$ independent partial differential equations of functions of $1$ variable only,
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\op{H} = \sum\limits_{i=1}^{n} \op{h}_{i}$ and $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$. All this is relatively easy to establish, and besides, this derivation is done in many textbooks in great details, so I leave at as an educational exercise to OP.
Separation of electrons coordinates
Electrons, on the contrary, interact with each other, so that a simplification of the electronic Schrödinger equation by means of separation of coordinates of individual electrons won't work. Mathematically, it is the interaction potential $\op{V}\elel$ that will prevent the separation of the electronic coordinates,
\begin{equation}
    \psi\el(\vec{r}\el) \neq \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
But what if we can approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? Obviously, such an approximate potential should have the form of the sum of one-electron potentials each of which depends only on the coordinates of $i$-th electron as we have already seen. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence the name of the approximation, the mean field approximation. The corresponding approximate potential can be formally written as follows
\begin{equation}
    \op{V}_{\mathrm{MF}}
    =
    \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
The electronic Hamiltonian $\op{H}\el$ is then approximated with the sum of effective one-electron Hamiltonian operators $\op{h}(i)$
\begin{equation}
    \op{H}_{\mathrm{MF}}
    =
    \sum_{i=1}^{n} \op{h}(\vec{r}_{i}) \, ,
\end{equation}
where
\begin{equation}
    \op{h}(\vec{r}_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{α=1}^{ν} \frac{Z_{α}}{ r_{α i} }
    + \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
And in contrast to eigenfunctions $\psi\el(\vec{r}\el)$ of the true electronic Hamiltonian $\op{H}\el$, eigenfunctions $\psi_{\mathrm{MF}}(\vec{r}\el)$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ defined by the corresponding eigenvalue equation
\begin{equation}
    \op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el)
    =
    E_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) \, ,
\end{equation}
can be written as products of one-electron wave functions
\begin{equation}
    \psi_{\mathrm{MF}}(\vec{r}\el) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equation
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
\end{equation}
The main problem with the mean field approach described above is that so far we have just assumed that $\op{V}_{\mathrm{MF}}$ exist in one form or another, but we have no idea of how it actually might look like.
There is a well-known workaround, but its a different story.

Note that I ignored spin completely in the discussion above which is, of course, fundamentally wrong, but spin won't make a big difference with respect to the key concept discussed, the mean-field approximation.

First of all, again thanks for answering my question; I've understood that in order to separate the variables, we need to avoid the inter-electronic distance term. But I've not conceived how by not instantaneously acting with each other, we can separate the variable.  Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons: why should there be no interaction term  in the average field?

Coulomb interactions between electrons prevents the separation of their coordinates, but if we simply get rid of the corresponding potential, we'll obviously get qualitatively wrong result. So, we proceed more wisely: we not just get rid of Coulomb interactions potential, we also introduce a one-electron additive potential which has to at least approximately recover the energy due to Coulomb interactions between the electrons.

Now, what this one-electron additive potential could mean physically if its aim is to account for energy due to Coulomb interactions? The only sensible meaning that we can give to it is that it is a potential of Coulomb interactions between each and every electron and the average field created by all other electrons in the system. So, the form of the mean-field potential and the role it is supposed to play dictates its physical interpretation.

Sorry, sir, didn't contact as I was in holiday. Okay, I want to ask you that the trial wavefunction $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}{i})$ is the Hartree product  of $n$ orbitals, but we know that electrons are identical fermions which must be antisymmetrised by using Slater determinant $\psi= |\det(\psi_1\updownarrow_1,\psi_2\updownarrow_2,\ldots,\psi_n\updownarrow_n)\rangle$? So, would the calculation above be different if  the later wavefunction is used instead of $\psi(\vec{r}) = \prod\limits{i=1}^{n} \psi_{i}(\vec{r}_{i})?$

@user36790, antisymmetric product of orbitals is just a linear combination of ordinary products, so the mean-field separation is still possible.

That means antisymmetrisation deosn't effect the separation of variable technique & energy which is be given by $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$? That means I could work with Hartree-product wavefunction & after working, I would then antisymmetrise using Slater dterminant, is it so?

@user36790, no, that's wrong. I've mentioned that an approximate mean-field potential is in general unknown, but what is known is that using the variation method you get a different expression for the mean-field potential when antisymmetry is taken into account comparing to the case when its not taken into account.

So, the electron in question doesn't interact with each & every electron but rather it interacts with the smeared out  field created by the other electrons. But could you tell why is it called 'average' field? What is the 'average' there?

Also, is this the shielding effect that decreases the effective nuclear charge that is $(Z- \sigma)e?$

Oh! I think what average meant here is  the smeared spherical charge density around the nucleus. Peter Atkins called this average as spherical mean.

@user36790, you're confusing few unrelated concepts out there.

@user36790, the mean (or average) electrostatic field each and every electron "feels" due to all other electrons has absolutely nothing to do with nuclei.

Yes, I do know that but I've not told anything about nuclei! Just saying that the average field is actually spherically averaged charge density; that's it. You can check this pic; this is what I'm referring to.


text:
What is the physical quantity/magnitude of pH?
I have a database to classify variables/parameters in water resources as follows:
Variable: Precipitation (P); Quantity: Length; Units of measurement: mm (millimeters).
Following the example, what would be the properly quantity of variable pH?
Variable: Potential Hydrogen (pH); Quantity: ?; Unit of measurement: pHU (pH unit, unit less, dimensionless).
Maybe concentration or activity?
[I know what pH is, It is not necessery explain, I'm looking for the most appropriate word to describe its quantity]

My question is about quantity not unit.

If you don't want to know what pH is, and you don't want to know what it's units are, then it's not clear to me what it is that you do want to know.

@SimonW know the most appropriate physical quantity, if possible, to pH.

@Artur_Indio surely, using that definition, pH is a physical quantity, of various substances. So the answer to your question is the same as the answer to "what is pH"? I think you need to (a) think carefully about how to ask the question, as at present it's very unclear; (b) most probably, ask it over at chemistry.stackexchange.

I disagree with the answer that says "pH measures acidity or basicity" (no matter what is in Wikipedia).
pH is the negative of the log of the hydrogen ion (H+) activity. Activity is a thermodynamic concept that is related to but not the same as concentration. Activity is similar to chemical potential. The same concentration of hydrogen ions will have different activities, depending on what else is in solution. In weak solutions, the activity coefficient that relates the concentration to the activity depends mainly on the charges of the other ions in solution and not on the particular ion species. But at higher amounts of other species, the particular chemicals and chemical interactions become increasingly important.
On the other hand, acidity is the ability to provide hydrogen ions to a reaction and considers not only the hydrogen activity but also the amount of acid-base reaction that can occur. You may have heard of pH 'buffers'. These are solutions that are designed to have a balance of chemical species so that the pH is very stable, even if small amounts of acid or base are added or if the ionic strength of the solution changes. If you add a bit of base to the buffer solution, the solution reequilibrates to provide acid to react with the base. You have to add a fair bit of base before the buffer capacity is used up and the pH changes very much. Then, with further addition, the pH will change a very quickly until it approaches the pH of the base that is being added. So a pH buffer has high acidity and alkalinity, keeping the pH stable.
In groundwater geochemistry, acidity is measured by reacting the solution with strong base until the acid is essentially used up. This is done by titration either to a fixed endpoint pH or to the inflection point in the curve of base added vs pH curve.
Bottom line, pH doesn't measure acidity. In terms of the original question, I would say:
Variable: pH, Quantity: negative log of the hydrogen ion activity, Units: dimensionless pH units.

I agree with everything except for the last two words (which should be removed imho).

I left in 'pH units' because it is often used. One problem with dimensionless numbers in equations (IMO) is that it can be unclear where they came from. For example it is sometimes hard to tell if someone is using a weight ratio or a volume ratio. In the case of pH, you could say that it is based on the ratio of the activity of H+ to the activity of H+ in the standard state (in this case, at infinite dilution, and defined as 1).  But yes, pH units is not strictly correct but is descriptive.

pH measures the activity (for practical purposes, this is the same as concentration) of H+ cations in an aqueous solution (i.e water).
At standard room temperature and pressure, and at conditions not too far from it, a value of 7 means that the water is neutral. Lower values means acidic (lemon juice, sulfuric acid, rain water). Higher values mean basic (ocean water, lime water).
It is unit less, therefore the term "quantity", as I understand from your question, does not mean much.

So, "concentration" seems a good word to be the physical quantity/magnitude of pH.

pH measures acidity or basicity: https://en.wikipedia.org/wiki/Acidity_function.
There are several acidity measuring scales and pH is one of them. Another example is the Hammett acidity function.

So, Quantity: acidity or basicity, Variable: pH, Unit Measument: pH units?

Quantity: acidity or basicity, Variable: pH, Unit Measument: pH units (or better, unitless)

authors: 04 ก้องภพ เจริญพันธ์


text:
How does the solvent determine whether mono- or tribromination of phenol occurs?
In phenol, the $-\ce{OH}$ group activates the benzene ring towards electrophilic substitution. But why is it that the electrophilic bromination of phenol gives trisubstitution (forming 2,4,6-tribromophenol) when aqueous bromine is used, but monosubstitution (either 2- or 4-bromophenol) when a $\ce{CCl4}$ solution of bromine is used? How does the solvent control the extent of bromination?
Related but does not explain completely: How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

In aqueous medium, phenol is deprotonated to a certain extent, forming the phenoxide ion in which the ortho and para positions are even more activated than in phenol itself. Hence, trisubstitution occurs here.
$$\ce{PhOH + H2O <<=> PhO- + H3O+}$$
In an organic solvent, phenol is not deprotonated and monosubstitution tends to occur.
ron also pointed out in a comment that:

The concentration of $\ce{Br^+}$ is higher in water than in $\ce{CCl4}$ making the reaction faster in water.
The dielectric constant of water is much greater than that of $\ce{CCl4}$, so the former solvent will better stabilize the charged intermediate, again accelerating the reaction.

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


text:
what is the difference between a redox reaction and an ionic bond?
I am new into chemistry...
I'd like to know what the difference is between an ionic bond: which involves a bond with an electron transfert, and a redox reaction; which seems to be exactly the same to me.
Thanks

Redox reaction and$,\ldots$ ionic bond?!

I know that one is a reaction and one is a bond, if that is the point you are trying to make. But at the end they both seem to do the same: transfert an electron. So could we say that redox reactions creates ionic bonds?

A redox reaction is where one compound/molecule  is reduced and the other oxidized , one gains electrons and the other loses electrons . An ionic bond is the force of attraction between two oppositely charged ions in which electrons are shared between the ions and the net charge is balanced.

In short, redox reactions may or may not create ionic bonds. There are examples of redox reactions which actually destroy ionic bonds, and others that have no relation to anything ionic whatsoever.

I know that this is how an ionic bond is first presented to beginners, but you should not think of ionic bonds as bonds involving electron transfers. Ionic bonds are merely "electrostatic attractions between oppositely charged ions". Now in the case of $\ce{NaCl}$, these oppositely charged ions are $\ce{Na+}$ and $\ce{Cl-}$, and these are obtained from neutral $\ce{Na}$ and $\ce{Cl}$ by transfer of one electron from $\ce{Na}$ to $\ce{Cl}$. However, the electron transfer is not part of the formation of the ionic bond.

The ionic bond only has to deal with the positively and negatively charged species. Where those positively and negatively charged species come from is not relevant to the formation of the ionic bond. For example, I could mix $\ce{AgNO3}$ and $\ce{NaCl}$, which causes $\ce{AgCl}$ to precipitate out - you could say there is an ionic bond between $\ce{Ag+}$ and $\ce{Cl-}$. However, this ionic bond was not formed by electron transfer from $\ce{Ag}$ to $\ce{Cl}$.

@orthocresol Perfect explanation, that clears it up for me. /me wonders why we can't award bounties on comments

An ionic bond is formed by electrostatic attraction of electrons between one atom and the other whereas in redox reactions the electrons are transferred from one atom to the other. Hope this helps.

But... the electrons of both atoms are negatively charged, so surely they should electrostatically repel each other?

"$\ldots,$electrostatic attraction of electrons$,\ldots$" -- Huh? And BTW, this seems like an attempt at providing the definitions, not similarities or differences. Not much of an answer.


text:
Oxidation state of metals in complexes with bridging ligands
How do I find the oxidation number of $\ce{Co}$ in a coordination complex like the
tetraaminecobalt(2) $\mu$-amido $\mu$-oxo tetraamine cobalt(2) ion?
I searched all over the net but can't find it anywhere. Please help!

I meant oxidation number..sorry

Maybe this can help you. http://chemistry.stackexchange.com/questions/22307/how-do-i-find-the-oxidation-number-of-a-coordination-complex?rq=1

But what should I consider as the oxidation states of oxo and amido ? @Anders MB ?

It’s no different from complexes with non-briding ligands.


text:
Orthogonal Wavefunctions
My current understanding of orthogonal wavefunctions is: two wavefunctions that  are perpendicular to each other and must satisfy the following equation: $$\int\psi_1\, \psi_2\, \mathrm{d}\tau =0$$
From this, it implies that orthogonality is a relationship between 2 wavefunctions and a single wavefunction itself can not be labelled as 'orthogonal'. They must be orthogonal with respect to some other wavefunction. However I have seen some textbooks refer to single wavefunctions as being orthogonal.
Am I missing some crucial thing in my definition of orthogonal wavefunctions? Could anyone provide me with a more complete definition of orthogonal wavefunctions from a chemistry point of view?

Could you provide a little bit more context i.e. examples of said claims in said textbooks?

@orthocresol Well, it said something along the lines, 'all wavefunctions are orthogonal', and in this question, http://chemistry.stackexchange.com/questions/7029/why-are-wave-functions-orthogonal, they talk of single wavefunctions as being orthogonal

It is not really talking of a single wavefunction as being orthogonal, but rather a set of wavefunctions. By having a set of orthogonal wavefunctions $\psi_1 , \psi_2 \ldots$ it means that any pair of wavefunctions that you pick will be orthogonal to each other. This will always happen if the wavefunctions are eigenfunctions of a hermitian operator (it is one of the standard "elementary" proofs in QM). I think the question that you linked does use the word a little loosely, although I still understood it to refer to a set of wavefunctions.

@orthocresol Right , that makes a lot more sense. So is my understanding of what orthogonal wavefuntions correct? Is there more of a chemistry way at explaining orthogonal wavefunction, rather than using mathematical expressions?

Somewhat. Have you done MO theory? Say in HF, the H 1s orbital doesn't overlap with the F 2px or 2py orbitals. That's because those two wavefunctions are orthogonal to each other.

Note that the term "perpendicular" has no meaning for (wave) functions. Two are orthogonal if, and only if, said integral is zero.

@orthocresol Yes, I am familiar with MO theory. So any set of  wavefunctions that don't overlap each other are orthogonal to each other?

Yeah, pretty much. In MOT, there is an expression called the "overlap integral" and denoted $S$: $S = \int \psi_1^* \psi_2 \mathrm{d}\tau$. Of course, that's just the same expression as above (apart from the complex conjugate which you missed out). The difference in energy between the MOs and the constituent AOs can be said to be proportional to $S^2$. You will notice that, if two wavefunctions are orthogonal (such as the H 1s and F 2px), $S = 0$ and therefore the constituent MOs are no different in energy from the AOs - exactly what you would expect since their symmetries do not match.

The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.
First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if
$$
  \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, ,
$$
where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1
So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$
$$
  \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, .
$$
In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.

I wanted to add one other important thing in my answer but you were faster; feel free to include if you think its important ;). It often occurs that textbooks write orthogonal set but mean orthonormal set and normalisation is indeed a property of a wavefunction on it's own.

@Wildcat Does  $\int \overline{\psi}_1 \psi_2 , \mathrm{d} \tau = 0 \Rightarrow \int \overline{\psi}_2 \psi_1 , \mathrm{d} \tau = 0 $?

@adosar yes you can conjugate the expression.

@adosar, yes. In fact, $\int \overline{\psi}_i \psi_j , \mathrm{d} \tau = \int \overline{\psi}_j \psi_i , \mathrm{d} \tau$.

As described in Wildcat's answer, a single wave function cannot be orthogonal, but a set of wave functions can all be mutually orthogonal.
To address the second part of the OP's question, the physical meaning of orthogonality is that a pair of mutually orthogonal wave functions are mutually exclusive; observing one precludes the possibility of observing the other unless the system is changed.

The same wave functions can not be said to be orthogonal. the comparison must be between two different function.

This is not true. You can measure the overlap of any wavefunction with itself. If it's normalized, it should equal 1.


text:
Why are absolute enthalpies of reactants not available?
The standard enthalpy of reaction is given as $$\Delta_r H = \sum\nu\;  H_m(\text{products})- \sum\nu\;H_m(\text{reactants}).$$
Peter Atkins in his book Elements of Physical Chemistry then writes regarding this equation:

The problem with this eqn, is that we have no way of knowing the absolute enthalpies of the substances . To avoid this problem, we can imagine the reaction as taking place by an indirect route, in which the reactants are first broken down into the elements. [...]
With the introduction of standard enthalpies of formation, we can write $$\Delta_r H = \sum\nu\;  \Delta_f H(\text{products})- \sum\nu\;\Delta_f H(\text{reactants}).$$  The first term on the right is the enthalpy of formation of all the products from their elements; the second term on the right is the enthalpy of formation of all the reactants from their elements..the fact that the enthalpy is a state function means that a reaction enthalpy calculated this way is identical to the value that would be calculated from the former eqn, if absolute enthalpies were available.

I'm not understanding why we can't calculate the absolute enthalpies. After all, enthalpy of a substance at a certain pressure (which is constant as we are talking about standard state & that is 1 bar) is given by the defining equation:
$$H= U +pV$$.
Yes, can't we calculate the internal energy taking a certain state as the zero of the energy? Why isn't it possible to calculate the absolute enthalpy?

If you "take a certain state as the zero of the energy", then it is no longer an absolute internal energy, it is a relative internal energy. But I see what you mean. After all entropies are also defined to be 0 at absolute zero and we call those "absolute". I imagine it has more to do with the internal energy of a molecule not being easy to quantify. It is probably possible, but not worth the effort, which makes the values "not available".

@orthocresol: I do count it as an answer but let the question take some time more to see what others think.

Nah, it doesn't deserve to be an answer.

@orthocresol: It's your opinion but your argument has logic, after all.

Related: Why it is not possible to find absolute value of internal energy?


text:
How does heat energy interact between two bodies of different heat capacities in order to attain thermal equilibrium?
Suppose there are two objects $A$ & $B$ of same mass; the former has specific heat capacity $C_A$ and the later has $C_B$ such that $C_A\gt C_B.$ Let $A$ be at temperature $T$ while $B$ be at $T+\Delta T.$ So, in order to attain thermal equilibrium, they have to attain the same temperature $T+ \frac{\Delta T}{2}.$ That means there would be a change of $\frac{\Delta T}{2}$ in both bodies. For that, $A$ must gain heat energy $Q_A= m\;C_A\cdot \frac{\Delta T}{2}$ while $B$ has to loss heat energy $Q_B=m\;C_B\cdot \frac{\Delta T}{2}.$ Now, $Q_A-Q_B\gt 0$ since $C_A\gt C_B.$
Then in order to attain the equilibrium temperature, $A$ has to get $Q_A$ energy which it receives partly from $B$ viz. $Q_B$ but where does the additional heat energy $Q_A-Q_B$ come from? It can't come from $B$ as that would deviate it from achieving the thermal temperature; $B$ can lose only $Q_B$ in order to attain $T+\frac{\Delta T}{2}.$ But where does $A$ get the extra energy $Q_A- Q_B$ from?
I'm confused as I thought so-far that when two bodies achieve thermal equilibrium, they do it by exchanging heat energy between themselves - the cooler body receives heat energy fro the hotter body till the equilibrium is reached.
But here, it seems that $A$ in order to achieve thermal equilibrium, needs energy $Q_A$ which cannot be provided by $B$ since the energy $Q_B$ it releases  is less than $Q_A.$ Where would then the extra energy $Q_A-Q_B$ come from?
Let's say, somehow $A$ gained that energy from the surroundings; however what would happen if they were isolated from the surroundings?
I don't know where I'm mistaking. Is it true that the hotter body itself provide the required energy to the cooler body to attain thermal equilibrium?
Please help.

According to your description and choice of symbols, $C_\mathrm A$ and $C_\mathrm B$ are heat capacities, not specific heat capacities.

$Q_A$ has to equal $Q_B$. Your mistake comes from the fact that at thermal equilibrium, the temperature of the two objects is not at $T+\Delta T/2$ if $C_A \neq C_B$.

@swenger:  I was guessing this reason but didn't have any logic to explain the point. Could you elaborate it  on why the thermal temperature shouldn't be $T+\frac{\Delta T}{2}?$

@Loong: Sorry for that.

Say the initial temperature of $A$ is $T_1$ and the initial temperature of $B$ is $T_2$. To determine the temperature ($T_f$) of $A$ and $B$ at equilibrium, we start with the equation $q_A=-q_B$, where $q_A$ and $q_B$ are the heat gained by $A$ and $B$ respectively as they reach equilibrium. We have $q_A=mc_A\Delta T_A=mc_A(T_f-T_1)$ and $q_B=mc_B\Delta T_B=mc_B(T_f-T_2)$. Thus the equation becomes $$mc_A(T_f-T_1)=-mc_B(T_f-T_2)$$ From here, we can solve for $T_f$. We can also note that if $c_A=c_B$, $T_f=\frac{T_1+T_2}{2}$ , which in your example is $T+\frac{\Delta T}{2}$. Otherwise, $T_f$ will not be the midpoint between $T_1$ and $T_2$.

You've pre-assumed that $q_A=-q_B.$ Is this just?

Since the system is isolated, $q_A=-q_B$ because of conservation of energy


text:
How much worse is methane than CO2 as greenhouse gas?
I can't find a coherent answer to this question. The numbers are all over the place.

25 times greater
84 times more potent

The reason I'm looking is because I can't find a reason in physics why this would be so*:

There is nothing “special” about methane that makes it much powerful than $\ce{CO2}$ (link to source)

What makes $\ce{CH4}$ absorb more infra-red than $\ce{CO2}$? What experiments have shown this?
Methane absorbs infra-red at different bandwidths than $\ce{CO2}$ (see here). Does the earth radiate more heat in those bandwidths?

The short answer is that all such "facts" are dependent on the planetary climate model which is being used. Such models are very complex. The sort of situation where you push something in here and it bulges out over there. //  In general most of the heat the earth transmits (via electromagnetic radiation, i.e. light) to space is done via infrared radiation. So having strong IR absorbers in the atmosphere stops infrared light from earth from getting into space.

See also [EarthScience.SE] Stack Exchange.

The reason for the discrepency is that the excess methane is removed from the atmosphere much quicker - about 12 years for methane compared to centuries for CO2. Source 1 says methane is "25x over 100 years", source 2 says "84x over 20 years". These are both true, it's just a difference of the timescale they're considering.

Let's take a look at this graph just so we have a reference.

This shows the IR absorption bands of various gases found in the atmosphere. For this answer, let's focus in on the absorption bands of $\ce{H2O}, {CO_2}$, and $\ce{CH_4}$.
First, the answer to why methane is considered a worse greenhouse gas than carbon dioxide is largely because methane absorbs in regions where carbon dioxide does not absorb at all. That is, methane absorbs different wavelengths of light than carbon dioxide.
Another thing to note is that methane's absorption peaks overlap with some of the absorption peaks of water vapor. This is an interesting thing to note because the strength of methane's absorptions in those overlapping regions are much stronger than the absorptions of water.
So, we have two things which contribute to why methane is sometimes called a worse greenhouse gas than carbon dioxide. First, it absorbs in areas where $\ce{CO2}$ does not absorb. Second, these absorptions are very strong.
Something you might be interested in, however, is the fact that there are some people who make the argument that methane essentially doesn't matter as a greenhouse gas. This would be due to the overlap in absorption between water vapor and methane. For instance, methane has an average concentration of 1.7 ppm (parts per million) in earth's atmosphere. Compare this with the fact that the earth's atmosphere can be composed of 1% water vapor at sea level and .4% across the whole atmosphere on average.
Thus, the amount of water vapor in the atmosphere is much more than that of methane at any given time. So, it can be argued that we do not need care much about methane as a greenhouse has because water vapor will do all the IR absorption that methane would have done.
I don't know enough about atmospheric chemistry to really say if I agree or disagree with those claims. One thing I do know, however, which is interesting is the fact that the idea of global warming would seem to present more of a problem with water vapor than methane or carbon dioxide because as temperature increases, the ability of the atmosphere to hold water vapor increases which would accelerate the warming from water vapor, which would increase the amount of water vapor the atmosphere can hold...
Also, I believe that the metric which says methane is more dangerous than carbon dioxide is a comparison which does not account for other things which might absorb that light. In other words, the figure you quote do not account for the fact that water vapor might make methane irrelevant as a greenhouse gas.
Interesting Read: http://www.nasa.gov/topics/earth/features/vapor_warming.html


text:
What is the reason behind optically activity of compounds
I am currently being taught optical activity of compounds and I had this question:

Optical rotation (optical activity) is the turning of the plane of linearly polarized light about the direction of motion as the light travels through certain materials.

What happens inside these compounds (for example sucrose) that causes them to behave on this manner. By that I mean to say that what takes place in the compound due to which they turn the plane of light when it travels through these materials?

Look at your two hands. They are mirror images of each other. That is also why you can't put your right shoe on your left foot. Then see: https://en.wikipedia.org/wiki/Chirality_%28chemistry%29

It's a good question, but asking about physical reason how optical rotation really works could be better for Physics SE - it seems many chemists are satisfied with practical side of this phenomenon.


text:
Is it really appropriate to measure entropy change using enthalpy?
One statement by Peter Atkins in his book Elements of Physical Chemistry confused me quite a bit:

If a chemical reaction or phase transition takes place at constant pressure , we can identify $Q$ with the change in enthalpy of the system, and obtain
For a process at constant pressure: $$\Delta S = \frac{\Delta H}{T}.$$

Is it really correct to measure the entropy change using change in enthalpy?
I'm telling so because enthalpy change only measures how much heat energy enters or gets expelled from the system. It doesn't measure how much energy gets expelled out of the system when the system does $PV$ work.
As Peter Atkins points out:

If $10~\text{kJ}$of energy is supplied as the heat to the system that is free to change its volume at constant pressure, then the enthalpy of the system increases by $10~\text{kJ}\; ,$ regardless of how much energy enters or leaves by doing work and we write $\Delta H = +10~\text{kJ}\; .$

I know it is the heat energy that is concerned in the definition of entropy change. But, suppose the system is heated which means it is in more disorder; however if it does work, then the energy the system got as heat would decrease which would therefore decrease the chaos & disorder of the system. If we use only the enthalpy change, wouldn't we exclude that work which decreases the heat energy gained by the system?
I'm rather confused on the use of enthalpy for defining the entropy change. Could anyone please help me explain this how enthalpy change actually measures the entropy change?

Please articulate your understanding of how the change in entropy of a closed system is determined, starting from thermodynamic equilibrium state A and ending at thermodynamic equilibrium state B.

@Chester Miller: It should be $$\int_{A}^{B}  dS = \int_{A}^{B} \frac{Q_\text{rev}}{T} dT $$ where $Q$ is the net heat energy received by the system reversibly.

Yes.  So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B.  If it is a phase change, you start out with liquid  at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B).  This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor.  During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor.  Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$.  From the first law, the change in internal energy is equal to the heat added minus the work done:  $(U_v-U_l)=Q-P(V_v-V_l)$.  Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T).  Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$.  This completes what I wanted to say about the phase change situation.  I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.

Firstly, thanks for the answer, sir.  I conceived the maths. However,  I've a query: Basically what is entropy? Entropy is the chaos or the disorderliness in the system caused due to the heat energy supplied. First the liquid got thermal energy $Q$ which increased the chaos in the liquid (soon to be vapour). But in the course of doing so, it does work which decreases the heat energy available in the system to $Q- P\Delta V.$ And it is heat which determines the chaos & now there is only $Q- P\Delta V$ left in the system which determines the chaos & not $Q.$ ...

... So, why should the definition of entropy takes $Q;?$ As the heat energy available for the chaos is $Q- P\Delta V$ & not $Q; ?$

Entropy is a measure of the number of quantum mechanical states available to the system.  This is what we mean when we talk about chaos or disorderliness.  The thermal energy of the system is vastly increased when we convert a liquid to a vapor because we have overcome the attractive forces holding the molecules together, and many more quantum mechanical states are now possible.  We have also greatly increased the volume of the gas which also allows for vastly more quantum mechanical states.