text:
Why exactly are standard potentials additive?
I don't really study chemistry so while my question may be very obvious, its not obvious to me. If we take an electrochemical reaction like
$$\ce{2Fe^2+ + Au^3+ -> 2Fe^3+ + Au+}$$
we can find its standard potential and standard Gibb's energy by summing half reactions whose are known, and can be consulted in tables.
It's intuitive for me that the standard Gibb's energy will be given by a sum of the corresponding reactions, and that I can multiply a reaction by minus one, or whatever, and use that to construct my complete reaction.
The reason I see Gibbs energy as intuitive is because you're essentially just taking
$$\sum_{\mathrm{products}} \mu_{\mathrm{products}} - \sum_{\mathrm{reactants}} \mu_{\mathrm{reactants}}$$
and this will prove the additive property of "summing up reactions".
But I don't see any reason for additivity being true when considering standard potentials. I don't even see why inverting the reaction should invert my potential if my half cell is in equilibrium and is a kind of "whole" that is compared to the standard hydrogen electrode and should therefore have the same potential no matter how I write it.
How am I screwing up here? Why can we add up potentials?

Actually standard potentials are not always additive. They turn out to be additive in the case when the final reaction is a complete redox reaction not a half cell reaction.

It helps to view this in terms of the equation
$$\Delta G^\circ = -nFE^\circ$$
Reversing the reaction reverses the sign on $\Delta G^\circ$, and therefore the sign on $E^\circ$. The sign on the potential determines the direction in which electricity flows. By convention, a galvanic cell (spontaneous reaction) has a positive $E^\circ$, while an electrolytic cell (non-spontaneous reaction) has a negative $E^\circ$. The magnitude of the potential stays the same, but a sign is added to signify the direction of current flow.
As far as adding half cells to determine the overall potential, this can be done because of the relationship in the above equation. You can see that $E^\circ$ depends on $\Delta G^\circ$ and $n$. But for instance, if $n$ is multiplied by 2, $\Delta G^\circ$ will be multiplied by 2 as well (there will be twice as much Gibbs' energy because the reaction involves twice the electron transfer). This means that you can add half-cell potentials to get the overall cell potential. Be careful to note though that if you multiply a reaction by a coefficient that you don't multiply the potentials, as you would for $\Delta G^\circ$; the potential does not depend on the stoichiometric coefficients, only on what the cathode and anode are.

nice! I get it now. Thanks. By the way whats the exact difference between $Eº$ and $E$ ? is $Eº$ the potential you get at the beginning or something, since things are in their standard states?

$\mathrm{E^\circ}$ is the standard electrode potential; i.e everything is measured under standard conditions. $\mathrm{E}$ is any electrode potential under consideration and should be specified with the conditions under which it applies.

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law.
How so?
Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new free energy will equal the sum of the individual ones.
This is a critical issue when trying to figure out the potential of a new reaction that is not the sum of an oxidation and a reduction.
For instance, take a look at these reactions (with all electrode potentials measured with reference to the standard hydrogen electrode):
$$\begin{align}
\ce{Fe^3+ + e- &-> Fe^2+} & E^\circ_1 &= \pu{+0.771 V} \\
\ce{Fe^2+ + 2e- &-> Fe_{(s)}} & E^\circ_2 &= \pu{-0.447 V} \\
\ce{Fe^3+ + 3e- &-> Fe_{(s)}} & E^\circ_3 &= \pu{? V}
\end{align}$$
It is tempting to simply add these potentials and get the third value, but we would be incurring on a major error. The proper way to proceed would be to calculate each reaction free energy and then figure out the standard potential.
$$\begin{align}
\ce{Fe^3+ + e- &-> Fe^2+} & \Delta G^\circ_1 &= (-1)FE^\circ_1 \\
\ce{Fe^2+ + 2e- &-> Fe_{(s)}} & \Delta G^\circ_2 &= (-2)FE^\circ_2 \\
\ce{Fe^3+ + 3e- &-> Fe_{(s)}} & \Delta G^\circ_3 &= (-3)FE^\circ_3 \\
\end{align}$$
Since we know that $\Delta G^\circ_3 = \Delta G^\circ_1 + \Delta G^\circ_2$, we have:
$$E^\circ_3 = \frac{E^\circ_1 +2E^\circ_2}{3} = \pu{-0.041 V}$$
Why don't we go through all this way when working with cell potentials? Because you always work your equations so that the number of electrons transferred on each of them has the same value. That way, the sum of the potentials is an indirect way of summing the free energies, which are now normalized by a common value.
This trick is evidently very useful, but sometimes it may be very misleading.

The $E_\text{cell}$ is independent of no. of of how many times you increase the reaction. If you see equation $ΔG= -nFE_\text{cell}$  you will find that increasing n also increase $ΔG$ by same times as $ΔG$ is additive and increasing n let say x times will increase $ΔG$ by x times.
Also the $n$ in eqn. means how many electrons are used( if I am right) in reaction. Therefore when you are talking about adding Standard Reduction Potential(SRP or simply electrode potential according to IUPAC) of cathode and Standard Oxidation Potential(SOP) of anode, the value of Cell potential in both case is same. And let say $E_\text{cathode}$ and  $- E_\text{anode}$ are the reduction potentials. Then if net cell potential is $E_\text{cell}$ then,
$ - nFE_\text{cell} = -nFE_\text{cathode} -nF(- E_\text{anode})
= -nFE_\text{cathode} +nF E_\text{anode}$
it gives,
$E_\text{cell} = E_\text{cathode} - E_\text{anode} $

Some suggestions; (1) text subscripts should be written as \text{cell}, otherwise, just writing cell indicates the four variables c, e, l, and l multiplied together. This applies to all the subscripts in your post. (2) Please don't use acronyms and abbreviations, it's only a few more characters for you to write in proper English. Along the same lines - check your spelling and punctuation. (3) $\Delta$ not $\delta$ is used for (finite) changes in a quantity.

I didn't know how to write $\Delta$ thanks for correcting


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Vaporized sodium hydroxide?
Some time ago, I was making a $\ce{NaOH}$ solution. The solution I had made was much too concentrated, and the heat released in the dissolution of the ions caused the water to begin steaming. Inadvertently, I inhaled some of this steam, and instantly my nose began to sting, but only for an instant. I can only assume I had inhaled some $\ce{NaOH}$, but how is this possible? How can ions be present in this steam, yet the process of distillation produces steam without ions?

When a liquid is evaporating or even boiling, minute drops of the liquid are formed by the mechanical disintegration of the surface of the liquid (bubble bursting, splashing or foaming). Some drops are carried away from the heated liquid by the steam (by convective and vapour flow).
The concentration of non-volatile dissolved matter in the entrained drops approximately corresponds to the concentration in the bulk liquid. Therefore, if the bulk liquid contains $\ce{NaOH}$, the entrained drops also contain $\ce{NaOH}$.
A quantitative parameter that describes this transport is the airborne release fraction (ARF). The ARF increases with temperature of the liquid and air velocity above the liquid. Measured values for liquids that are evaporating but not boiling range from $< 10^{-8}$ to $3\times10^{-5}$ (Mishima, J., L.C. Schwendiman and C.A. Radasch. November 1968. Plutonium Release Studies IV. Fractional Release From Heating Plutonium Nitrate Solutions in Flowing Air Stream, BNWL-931, Pacific Northwest Laboratory).
In principle, distillation and similar technical processes have the same problem. That is one reason why distillation shows only a limited decontamination factor (DF) even for non-volatile dissolved matter.

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Know any micropipettor that is acid resistant?
We pipet concentrated nitric and hydrochloric acids. our current pipettor is rainin lite but its piston is metal and thus corrode easily.  We discussed this with our inhouse technician and he suggested a pipettor with a plastic piston.  So do you guys know any?
We try to avoid using filtered tips as we need to leach the tips prior to use.
Thanks

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Approximating the concentration of ions when a salt dissociates
In my textbook written that:

Given that $K_\mathrm{sp}(\ce{Ag2CrO4})=9.0\times 10^{-12}$, consider the solubility of $\ce{Ag2CrO4}$ in a $0.100\:\mathrm{M}$ solution of $\ce{AgNO3}$.
Initial concentrations (before any $\ce{Ag2CrO4}$ dissolves) are,
  $[\ce{Ag+}]_0=0.100\:\mathrm{M}$ (from $\ce{AgNO3})$ and $[\ce{CrO4^{2-}}]_0=0\:\mathrm{M}$
The system comes to equilibrium as $\ce{Ag2CrO4}$ dissolves according to the reaction:
  $$\ce{Ag2CrO4(s)\rightleftharpoons 2Ag+(aq) + CrO4^{2-}(aq)}$$
  $$K_\mathrm{sp}=[\ce{Ag+}]^{2}[\ce{CrO4^{2-}}]=9.0\times 10^{-12}$$
We assume that $x\:\mathrm{M}$ of $\ce{Ag2CrO4}$ dissolves to reach equilibrium, which means that:
  $$x\:\mathrm{M\:\ce{Ag2CrO4}(s)}\rightarrow 2x\:\mathrm{M\:\ce{Ag+}(aq)}+x\:\mathrm{M\:\ce{CrO4^{2-}}\:(aq)}$$
Now we can specify the equilibrium concentrations in terms of $x$:
  $$[\ce{Ag+}]=[\ce{Ag+}]_{0}+\Delta[\ce{Ag+}]=(0.100+2x)\:\mathrm{M}$$
  $$[\ce{CrO4^2-}]=[\ce{CrO4^2-}]_{0}+\Delta[\ce{CrO4^2-}]=(0+x)\:\mathrm{M}$$
Giving:
  $$9.0\times 10^{-12}=[0.100+2x]^{2} \cdot [x]$$

Also, here is line I want to discuss (again, taken from the textbook):

Since the $K_{\mathrm{sp}}$ value for $\ce{Ag2CrO4}$ is small (the position of the equilibrium lies far to the left), $x$ is expected to be small compared with $0.100\:\mathrm{M}$. Therefore:
  $${9.0\times 10^{-12}=[0.100+2x]^{2}[x] \approx [0.100}]^{2}[x]$$

When can this approximation be made and when can it not be made?

I edited this post so it contains the correct formatting. Please use this formatting in future posts, as it makes it easier for everyone to understand your question and thereby help you:)

Really, your question seems to be a math question.
You need to solve the cubic equation:
$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$
Cubic equations are exactly solvable; however, it is common to use an approximate method.
Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.
Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.

Really, I didn't realize that. Should I just delete this question?

@Ridho Don't delete it. It might be useful for other people in the future.


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Primary and secondary standard
Why are secondary standard solutions used in titrations? Why don't we always use primary standard solutions?

Can you explain briefly what "primary and secondary standard" do you mean?

These are standard terms covered in any quantitative analysis course.

KHP would be a primary standard in an acid base reaction. It can be obtained in high purity, and dried in an oven so that a given weight of KHP has a known amount of acid. You'd use the KHP and titrate with NaOH. The NaOH solution would then have a known value and be a secondary standard since its value was dependent on the KHP.

A primary standard is critical for an analytical result. If we dissolve a known mass of a material into a known volume, we must be able to know its concentration with confidence. Such a standard must be easily made very pure (> 99.95%), it should not decompose, and it must be fairly robust to say heating or vacuum. We use such standard solutions as, for example, titrants. This is surprisingly hard to come by, and so we say that secondary standards are materials that do not meet these strict criteria.
Say you want to titrate a sulfuric acid solution for quantitative work, you could use a known concentration of NaOH or KOH and volumetrically determine the concentration of the sulfuric acid solution. In fact, dissolving pellets of these particular bases in a known volume of water is not sufficient to calculate concentration with any confidence in analytical work because the solids are contaminated with carbonate from atmospheric CO$_2$. In general these carbonates contaminate alkaline solutions quite readily through this reaction: $\ce{OH- + CO2 -> HCO3-}.$ This is why we would have to titrate the concentration of $\ce{NaOH} $ solution with a primary standard before using it as a secondary standard.
Now there are acid and base primary standards available, so why would one ever need to use a secondary standard? Usually it’s a matter of specific needs and convenience. Whenever possible, use a primary standard if your results actually matter.
Let us look at a specific case where a secondary standard is useful: potassium permanganate is exceptionally useful as a titrant because it serves as its own indicator where the presence of a purple color signifies the end point, but it is unstable in light or heat due to this reaction: $\ce{4 MnO4^- +2H2O -> 4MnO2 + 3O2 + 4OH-}$. This is why solutions of it must be stored in a dark cool place; even still, we must standardize the concentration of such a solution and we can only trust its measured concentration for a short time thereafter.
A secondary standard solution of $\ce{KMnO4}$ is usually titrated by oxalic acid, a primary standard, and then used to determine the concentration of a number of other analytes.
The reaction with oxalic acid can be summarized as $\ce{5 H2C2O4 + 2 MnO4- + 6H+ -> 10CO2 + 2 Mn^{2+} + 8H2O}$ where everything except $\ce{MnO4-}$ is colorless. The end point of this titration is a marked by a purple color due to unconsumed $\ce{MnO4-}$. This allows you to make a standardized solution of potassium permanganate that can be used for a short while.
Now, how would we use this? Solutions containing $\ce{Ca^2+, Pb^2+, Ag+, Mg^2+, Co^2+, Zn^2+}$ and etc. can be precipitated as a metal oxalate. You would then digest this ppt in an acidic solution and titrate the released $\ce{H2C2O4}$, where the end point is marked by a purple color, allowing you to infer the original concentration of the metal ion in the original solution.
$\ce{KMnO4}$ was used in mineralogical surveys of uranium prior to the Manhattan project, where better techniques were developed to quantitate.

This is way too long...

As @Ridho requested, you should clarify the meaning of primary and secondary standard as those expression gets differents meaning depending on the context used. For my part I will try to answer you with this.
When you are doing a sample determination by titration, you usually use a titration compound which is easy to calculate proportion with, like NaOH or HCl for pH titration. Those two are pretty well forward, they are widely available, they are considered to be 100% dissociated when mixed in water, and 1N of NaOH/HCl is considered to neutralise 1N of anything else in the pH world. But, it comes with a catch. Those two don't give solutions that are stable in term of concentration. NaOH pellets are highly hygroscopic and HCl tend to "evaporate" as it is primarely a gas. So even if you weight 39.9971g of NaOH and you make a solution of 1000.000mL, you will not get a 1.00N solution. So you need to confirm or standardize your solution with another compound which is more stable, got an easy to work with pHI, but is useless for your first détermination. For NaOH, in our lab we use potassium hydrogen phthalate with phenolphthalein.


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Elimination from Vinylic Halides

Vinylic halides do not show nucleophilic substitution at all, because of their stability due to resonance. But elimination to form alkynes is possible. No nucleophile at all can displace the halide ion in a nucleophilic substitution, but a strong base suffices for elimination.

Why is this?

I wonder if that is really the case. Source please? I could imagine nucleophilic substitution would more difficult due to the increased bond strength, but because of resonance structures it is impossible? Idk.

But a halogen attached to a carbonyl atom, also resonance stabilised, can be removed by nucleophilic attack.

In the $\ce{sp^2}$ transition state for an $\mathrm{S_{N}}2$ reaction

the attacking nucleophile, the central carbon and the leaving group prefer a linear arrangement
the 3 remaining groups around carbon are arranged with 120° bond angles

(image source)
Now, let's look at ethylene (as a model for a vinyl halide) and let's use the bent bond description of ethylene because it makes it easy to visualize what's happening (the bent bond model has been shown to be equivalent to the pi bond model in all aspects, see below for the same explanation phrased in terms of the pi bond model).

(image source)
Replace one of the ethylene hydrogens with halide and imagine what happens as we approach the transition state.  The bond angle in the two-membered ring wants to be small, but as we approach the transition state it needs to open up wider to 120°!  This would create a lot of strain.  It is this large strain energy that causes the $\mathrm{S_{N}}2$ reaction at $\ce{sp^2}$ hybdridized vinylic carbons to be a very high energy pathway and therefore less likely to occur.
On the other hand, the elimination reaction at a vinylic carbon is quite facile.  Here as the base removes a proton we build up negative charge on an $\ce{sp^2}$ hybridized carbon.  An $\ce{sp^2}$ orbital has more s-character than an $\ce{sp^3}$ orbital; the more s-character in an orbital the lower the energy of an electron in that orbital.  Therefore, negative charge is more stable in an $\ce{sp^2}$ orbital than a $\ce{sp^3}$ orbital.  Consequently it is easier to remove a proton when the negative charge will reside on an $\ce{sp^2}$ hybridized carbon compared to an $\ce{sp^3}$ hybridized carbon (and that's why it is even easier [lower energy] to remove a proton from an $\ce{sp}$ hybridized carbon).

Explanation Based on Pi Bond Model
As expected for equivalent models, the bent bond explanation can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons are in higher energy unhybridized p orbitals and this is not offset by the energy lowering experienced by the electrons as some bonds rehybridize  from $\ce{sp^2}$ to $\ce{sp}$) thereby increasing the energy required to reach the transition state.


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Stability of half and fully filled orbitals
1) My textbook tells me that chromium's electronic configuration is $\ce{[Ar]}\space 3d^5 4s^1$ instead of $\ce{[Ar]} \space 3d^4 4s^2$, as fully filled and half filled orbitals have more stability. But how does that happen? Why is the fully or half filled orbital more stable?
2) Following this method as we go down group 6 we would expect tungsten's last two orbitals to be filled as $5d^5 6s^1$ to attain more stability. But the actual configuration turns out to be $5d^4 6s^2$. Isn't this supposed to be "unstable"?


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Why does reduction always accompany oxidation and vice versa?
We've all learnt that the reduction always accompanies oxidation and vice versa Why is this so? Why is it not possible that a species releases an electron and the electron just "floats around"? Is this related to the stability of a free electron?

@Asker123 So there are highly unfavourbale cases where this happens? Can you provide an example?

My bad, there are no highly unfavorable cases. Even then, there must be an outside cause.

If something is to be reduced then there must be a reducing agent present. Also if something is oxidized then there must be an oxidizing agent present.
You must have a reducing agent or oxidizing agent present in order for a redox reaction to take place. An element can't just dislocate it's electrons. There must be an outside force present in order to do that.

Let's say I supply a lot of heat to a sodium atom. Why can't the electron be ejected and remain as such without reducing any of the species present in the environment?

The outside force in this situation is heat. Of course this can happen, but however I doubt that the electrons will just remain present in the environment for long. Theoretically, yes it is possible.

Is there any example you can provide where electrons remain present in the environment for long (theroretical or experimental)?


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Why an asymmetric geometry with sp3d and sp3d3 hybridization?
A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce{SF_6}$), for example, acquires an octahedral shape, which can be explained by assuming that the hybrid orbitals arrange themselves to minimize repulsion amongst themselves. An octahedron makes sense, it the only way to symmetrically arrange six hybrid orbitals on an atom. The case for an $sp^3$ hybridized atom is similar which from a similar logic can be explained to have a tetrahedral geometry.
But consider the case of $sp^3d$ hybrid orbitals which arrange themselves in a trigonal bypyramidal shape ($\ce{PCl5}$). This is a gross breach of the usual adherence to symmetry. Two bonds are at the axial location and three at the equatorial location. In fact, the axial bonds are slightly longer than the equatorial ones. The case for $sp^3d^3$ hybridization is similar ($\ce{IF7}$) with five orbitals crammed into the equatorial plane and two sticking out above and below it. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?

Which alternative "simple symmetric arrangement" do you have in mind? Do you know of possible structures for $\ce{PCl5}$ or $\ce{IF7}$ which are more symmetric than their trigonal bipyramidal and pentagonal bipyramidal geometries with $D_{3h}$ and $D_{5h}$ symmetry, respectively? You may also be interested in this page which discusses a way to calculate these predicted geometries without any direct inference to Chemistry or VSEPR theory.

Hybridization involving d-orbitals is not an accurate description for these molecules, rather hypercoordinated (or hypervalent) bonding using only s- and p-orbitals provides a better explanation.  For application of this concept to $\ce{SF6}$ see here. for $\ce{PCl5}$ see here and for $\ce{IF7}$ see here.

@NicolauSakerNeto: As far as I am aware, the structure I have in mind does not have any common name. I think the best description might be an operational one. To take the example of  $sp^3d^3$, tie seven balloons at their common centre. The structure they assume is similar to the one I have in mind.

@NicolauSakerNeto: I think the answer might lie in the fact that only five platonic solids are possible. And as mentioned in the link you provided: "The vertices of the five Platonic solids give the only perfectly symmetrical distributions of points on the surface of a sphere. Therefore, if N positively charged particles are constrained to the surface of a sphere, and N is not equal to the number of vertices of a Platonic solid, then the particles must have an equilibrium configuration that is not perfectly symmetrical. "

Lowest energy states don't always have the highest symmetry. For example, in the Thomson problem, where the goal is to find the points on a sphere where the Coulomb repulsion energy is minimized, the configurations for n=3 and n=5 points have two points of one type of symmetry on opposite poles and the other points of a different type of symmetry and forming a polygon around the equator.

As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.
d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.
Conveniently, all the cases you presented in your question can be adequately explained using only s- and p-orbitals. See the links for $\ce{SF6}$, $\ce{PCl5}$ and $\ce{IF7}$ respectively. Many thanks to ron for providing them in his comments!

For a structure to be perfectly symmetric, it must satisfy this condition:

For any two bonds, the plane passing through them must divide the molecule into two symmetric halves.

Note that this condition is satisfied by trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and pentagonal bipyramidal geometry.
I'm going to show that there are no better structures satisfying this condition for sp$^3$d and sp$^3$d$^3$ hybridisation.
For sp$^3$d, we're going to first arrange any two bonds at some angle (unknown). Obviously, the two bonds form a plane ($3$D here):

Using the condition above, we have to arrange $3$ bonds around the plane, such that they are symmetric about it.
We cannot put $2$ bonds on one side and $1$ bond on the other side, as that would be unsymmetrical ($3$D here).

Therefore, for symmetry about the plane, we must keep a third bond on the plane. The instant we do that, we are left with only two options: trigonal bipyramidal and pentagonal planar. Of the two, trigonal bipyramidal is obviously a more stable structure (bond angles $90°$ and $120°$) than pentagonal planar (bond angle $72°$) ($3$D here).

When we proceed in a similar manner for sp$^3$d$^3$, we have two options: heptagonal planar (bond angle $51.43°$) and pentagonal bipyramidal (bond angles $72°$ and $90°$. Again, the pentagonal bipyramidal structure is more stable.
Thus, these are the only feasible structures that are symmetrical.

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What is the meaning of the "colon" notation in chemical formulas?
For example PEDOT:PSS and P3HT:PCBM.

All it means is that there's an intimate physical mixture of the specified components in a solid or liquid phase (no chemical bonding between the components as happens in a copolymer, for example).
In the context of solar cells, LEDs and related devices, they are often used as part of shorthand device architecture notation. Slashes ( / ) are used to separate different layers of the device, while colons ( : ) are used to enunciate all separate components in a single layer. Do note that every so often a slash is used incorrectly in place of a colon. Checking the experimental procedure should clarify whether two consecutively listed materials are in the same or separate layers.
For example, consider the notation ITO/PEDOT:PSS/P3HT:MEH-PPV:PCBM/Ca/Al for an organic polymer solar cell. There are five layers in total, deposited in sequence from left to right:

A layer of ITO (usually on top of a glass substrate which is omitted).
A layer containing a binary mixture of PEDOT and PSS, usually deposited via spin casting of an aqueous solution with both materials dissolved.
A layer containing a ternary mixture of P3HT, MEH-PPV and PCBM, often deposited by spin casting from a solution of all materials in a chlorinated solvent.
A layer of pure calcium metal, likely deposited by thermal evaporation.
A layer of pure aluminium metal, deposited on top of the calcium layer.

Sometimes the thickness, component proportion and concentration of the spinning solution for each layer is reported inside of parentheses, such as in ITO/PEDOT:PSS (40 nm)/P3HT:PCBM (1:0.8, 150 nm)/Ca (80 nm). Be sure to pay attention to small differences in notation resulting in completely different cell architecture. For example, a solar cell with a ITO/PEDOT:PSS/P3HT:$\ce{C_{60}}$/Al represents a bulk heterojunction solar cell, while ITO/PEDOT:PSS/P3HT/$\ce{C_{60}}$/Al represents a bilayer (or planar heterojunction) solar cell.


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Are there any known chemical properties of tritium water that make it unusually different from protium water?
I suppose the first question supporting the main question is, has tritium water ever been synthesized in sufficient quantity to test chemical properties?
If so, and apart from the obvious radioactive nature of the molecule, and possibly self-heating nature, what unusual properties does tritium water have over protium water?
For example, deuterium oxide is not (easily) metabolized by living things; it cannot support life in the same manner as protium oxide. Also protium water is blue and deuterium oxide is clear. What color is tritium oxide?

Yes, $\ce{T2O}$ has been prepared and is available in significant quantity.  When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil.  It must be transported in a shielded, cryogenic dewar.
A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds.  The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest.  We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a primary kinetic isotope effect. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows:
$$\frac{K_H}{K_D} \sim 6-7$$
$$\frac{K_H}{K_T} \sim 13-14$$
In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.
These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds.  The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon.  Still, as you pointed out, the effects can be disastrous in biological systems.  Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.
The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation
$$\nu_e=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$
Here $\ce{\mu}$ is the reduced mass of the system (e.g. the bond involved in the vibration that is producing the light absorption) and is given as
$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$
where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond.  We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is inversely related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration.  Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (reference, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be colorless.

Great answer! And so if I had a long enough column of $T_2O$ and, and assuming it wouldn't boil away before I viewed light coming up through it, the light would be red?

see my edit, $\ce{D2O}$ is colorless.

Can you take advantage of the tritium to helium decay to convert a tritium substituted centre to a radical ? Would be cool although not very useful.

@J.LS  Interesting, I think it would.  That might be a good question to post on its own.  The electron released carries about 6 keV of energy - that might do some local damage!


text:
Why is there a greater degree of separation in a dipole-dipole interaction?
My textbook says this:

Electrostatic forces between two ions decrease by the factor $\frac{1}{d^2}$ as their separation distance, $d$, increases. Dipole–dipole forces, however, vary as  $\frac{1}{d^4}$. Because of the higher power of $d$ in the denominator, the attractive forces diminish with increasing $d$ much more rapidly than does  $\frac{1}{d^2}$. As a result, dipole forces are effective only over very short distances.

Now mathematically I understand why this is true, but I don't understand why this is the case in reality. Can someone shed some light on this?

Like I answered my other question, this is in some ways very similar. Since electrostatic attractions are so strong, it is decreases less at a longer distances. The $+$ & $-$ are attracted together so strongly.
Dipole-dipole attractions aren't strong attractions, electrostatic attractions are stronger, since they are created by ionic bonding. Dipole-dipole occur in covalent bonding. Thus dipole-dipole attractions decrease with increasing distance faster.
This is just a piece of the whole picture.

Check the refs: 1, 2, 3. They provide more in depth detail to the entire picture.

Imagine the dipole as a partial positive ($q_+$) and negative ($q_-)$ separated by a small distance $\delta$.  The electrostatic potential due to the dipole is $\frac{q_+}{|d-\delta|}+\frac{q_-}{|d+\delta|}$ where $d$ is the distance to the midpoint between the two charges. (There is also an angle term which I left out for simplicity).
As $d$ increases $\frac{q_+}{|d-\delta|}\approx -\frac{q_-}{|d+\delta|}$, i.e. the electrostatic potential of the dipole (i.e. two very close charges of opposite charge) goes to zero faster than a single charge because at longer distances the two charges appear to be at the same position and "cancel each other out".
Feynman has given as a more thorough explanation.


text:
Why is HCl not considered to have hydrogen bonding?
A molecule that has hydrogen bonding usually follows these two premises.

1.) There is a hydrogen atom involved
2.) Hydrogen must be bonded to a highly electronegative element which are nitrogen ($\ce{N}$), fluorine ($\ce{F}$) and oxygen ($\ce{O}$).

Seeing that both oxygen and chlorine have a small difference in their electronegativity (oxygen being roughly 3.5 and chlorine being roughly 3.0), why does chlorine in a hydrogen chloride molecule ($\ce{HCl}$) have a dipole-dipole interaction, while the oxygen in a water molecule ($\ce{H2O}$) causes the water molecule to have a stronger form of dipole-dipole interaction called hydrogen bonding? I do not understand this since chlorine is just as electronegative as oxygen and nitrogen?
Edit: I should also added that nitrogen and chlorine have the same EN value (3.0)

Related. Short answer: $\ce{Cl}$ is too large.

@Asker123 Could you go a little more in depth?

Also, check out the example from this answer.

Very, very related. Check it for a more in depth answer.
Well, first-of-all you need to understand that the $\ce{H}$ bonding isn't actually bonding. It is just a covalent attraction. Also since $\ce{Cl}$ is larger than $\ce{N}$, $\ce{F}$ and $\ce{O}$ it does not make a strong $\ce{H}$ bond. The size of the $\ce{Cl}$ makes the dipole-dipole attraction weaker. However $\ce{N}$, $\ce{F}$ and $\ce{O}$ are smaller and thus have an $\ce{H}$ bond.
Although in reality, compared to other covalently bonded structures, $\ce{HCl}$ has a very strong covalent bond.

Despite its electronegativity, size of chlorine atom is large and hence, electron density of chlorine is not sufficient to form Hydrogen Bonding.
Hence, HCl does not have hydrogen Bonding.


text:
Possible Synthesis of a Polymer
Following a question I have asked previously, what I intend to inquire this time is quite similar as well, except, with a polymer.
So previously I asked whether such monomer PCDO is possible to synthesize.

That originated from this similar, but different polymer poly-PCDO. It is true that I also don't think this PCDO from my previous question is poly-PCDO after monomerization (I'm still not so sure though). But the PCDO molecule pop out when I was attempting to make this original poly-PCDO a monomer.

I think the synthesis of the polymer in this structure would be easier to theorize and explain.
One can dehydrate divinyl ether with benzene (or perhaps phenol) and lithonate by condensation polymer.
$$\ce{n(C4H6O + C6H6O + HLiCO3) -> (C11H9LiO4)_{n} + nH2O + nH2}$$
Due to my limited knowledge in this area, I can only assume the phenyl group would most likely bond with lithonate being bonded to the ether already. Would someone explain to me how this will or will not work?

If you don't mind the question: What properties are you hoping to achieve for your polymer? Are you're trying to obtain a high loading and tight binding with $\ce{Li+}$?

Under acidic conditions, a part of the divinylether will probably just undergo cationic polymerization:

Could you please elaborate on what you may meant by acidic conditions? Is it the Lithium ion that acts as the cationic initiator?

@čaritisio Acidic conditions means the the presence of a Bronsted acid that releases a proton. That could even be phenol.


text:
Can gas get produced at room temperature?
Is there any reaction between solid/ liquid elements/compounds at room temperature that produces gases??

Lots, for example, solid magnesium in aqueous hydrochloric acid can releases hydrogen gas.

You seem to be implying that you don't think this should be the case. As @santiago said the complete opposite is true. Perhaps you could give us some more detail about why you are asking the question?

Yes, there are many. One you can make in your kitchen is to mix baking soda and vinegar; it makes carbon dioxide gas.

There are more than several. There are a vast number.

There is an incredible amount of such reactions. Consider the equation from thermodynamics,
ΔG = ΔH - TΔS

where
ΔG is change in Free Energy
ΔH is change in Enthalpy
T is Temperature and
ΔS is change in Entropy.
It is known that a reaction is spontaneous and favorable if ΔG is negative. We therefore see that a reaction that increases entropy (ΔS > 0) will be more likely to be favorable. Although not quite technically correct, Entropy can be thought of as a measure of disorder. Therefore, if a reaction results in products that are more chaotic than the reactants, it will be more likely to be favored.
A reaction of solids, liquids, or both that produce a gas will increase the entropy of the system. This is because gasses are much more chaotic and unordered than solids or liquids. Therefore, there will be not just many, but an unfathomably large number of reactions that will take place at room temperature that will spontaneously occur.
And if the reaction results in a release of heat (ΔH < 0), then it is guaranteed to occur at any temperature if it produces gas in this way.
Some examples include explosives, in which much gas is produced at incredible rates. Acid-metal reactions produce Hydrogen gas as well.

I doubt the OP understands the thermodynamic aspect at all, else the question would not have been asked in the first case.

@user223679 I was thinking that may have been the case so I tried to explain it as well as I could.

Yes. No doubt about your answer quality. Just that it could be a bit too much for the OP. Anyway, future visitors will benefit from it :)

@user223679 Thanks, but one question, what does OP stand for?

Good question. I use OP for Original Poster but it varies throughout SE and other communities. Some go with Original Post. Look here. Btw, where are you from?


text:
Migratory preference in a rearrangement involving carbocations
I just read about the Pinacol rearrangement and the following reaction was presented just after it. The book (Clayden) shows path 1 being taken.
My question is why is the second path not taken? I know that migratory aptitude of $\ce{Ph>H}$ but all we want is stability, so why can't the $\ce{H}$ migrate?

why is the second path not taken?

$\ce{H}$ can migrate, it is just that phenyl migrates better.  The reason that phenyl migrates better is because it is much better able to stabilize the developing carbocation center.  The phenyl group participates by what is termed neighboring group participation (see the "NGP by an aromatic ring" section).  It forms a phenonium ion.  Note the geometry of this ion, the former aromatic ring is perpendicular to the plane of the screen while the former $\ce{C-C}$ bond is contained in the plane of the screen.

(image source)
We can draw several resonance structures to describe the phenonium ion, this leads to it being especially stable.

(image source)
Such extra stabilization is not possible with a migrating hydrogen atom. Therefore the pathway involving the phenonium ion is lower energy (faster) than the pathway involving hydrogen migration.

Is there a way to compare migratory apptitudes of different alkyl/aryl groups with H and among themselves?

@user223679  Yes, many reactions (pinacol) have been used for this.  Using the present example, just make several epoxides with different substituents and use the product ratios to assign relative migratory aptitudes.  Make sure the reaction is being run under conditions to insure kinetic control.

Has someone generalized an order? Is something like this been found out yet?

@user223679  Yes.  If you search "migratory aptitude here on  Chem SE you'll find some useful information.  Otherwise just ask a question.

Coming back to this post, you said here that one can't judge between H and Ph and I feel this question is similar to the pinacol one. And another answer out there suggests H>Ph (though I can't trust that source).

@user223679 There are general trends regarding migratory aptitudes, but no rules.  The migratory aptitudes are influenced by factors such as solvent, geometric alignment of the migrating group and carbocation p orbital, early vs. late transition state (e.g. how much positive charge has been developed in the TS), etc. Dunno what else to say.

Thank you for bearing with me. I apologize for being so nitpicky. Was just confused between two things said by you in similar reactions.

@user223679  Absolutely no problem.  Confusion is often a good sign, it indicates that you're really digging in to a problem.

In both the cases that you mentioned, the lone pair of oxygen is finally going to stabilize the carbocation by donating towards the positively charged carbon atom.
Hence stability of the carbocation may not be a very good criteria of comparing the mechanism. The migratory apptitude then plays a major role in deciding the faster mechanism.
Stability is just the thermodynamic aspect while migratory apptitude is the kinetic aspect.
Both need to be considered. Since the stability brought about by the oxygen is similar in both cases, we chose the kinetic aspect as the determining criteria.
And ron's answer clearly outlines the kinetic aspect.


text:
Formation of species in electrolysis
I understand the principles of electrolysis of salts in aqueous solutions, but there are two points on which i am unsure.

At the positive electrode, how can you work out which ions will preferentially donate electrons?
If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?

If so, what decides which ions are formed?

What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

I would love to answer this question if you could make it neater and easier to read. Please bullet what you don't understand and why. Thanks.

I edited in some legibility, but the questions might be a bit broad.

just changed it for clarity

Alright, I will begin.

For the last question, I believe the latter would be true because if
  the anions produced at the negative electrode react with the cations
  that are attracted to the electrode, the cations that are attracted to
  the electrode would not be available to accept electrons and hence the
  current would become increasingly restricted until it stops all
  together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

Why the salts in a salt bridge?
Voltaic and electrolytic cells
Difficulties understanding how salt bridge works
Why is it important to use a salt bridge in a voltaic cell? can a wire be used?

Don’t be limited to those above, keep searching until you solve your dilemma.
Now for the Hard Part
To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.
$$\begin{alignat}{2}
\ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\quad&&1.23\ \mathrm{V}\\
\ce{F2_{(g)} + 2e- &-> 2F-}\quad&&2.87\ \mathrm{V}\\
\ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\quad&&{-}0.83\ \mathrm{V}\\
\ce{Ni^2+ + 2e- &-> Ni_{(s)}}\quad&&{-}0.25\ \mathrm{V}
\end{alignat}$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$).
We can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.
Now we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:
Equation 1 (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.
In general …

Eliminate which reaction cannot take place.
Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)

At the positive electrode, how can you work out which ions will preferentially donate electrons?

In an electrolytic cell, the positive electrode is called the anode, and the oxidation half of the reaction occurs here. The oxidation reactions which have less negative standard oxidation potential are more likely to occur, though most resources give the standard reduction potentials. Wikipedia is one such resource.
If you wish to know which species are more likely to donate electrons, these will be the half reactions that have the less negative standard reduction potentials.

If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?

I wrote a response to this exact question here.

What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?

In an electrolytic cell, the negative electrode is called the cathode, and the reduction half of the reaction occurs here. When an species is reduced, typically it changes phases (goes from aqueous ion to being a gas, a liquid, or a solid). Most likely if an ion is reduced to another ion here, this ion will then also be reduced.


text:
Why is tetrafluoromethane non-polar and fluoroform polar?
Consider the Lewis dot structures of the molecules fluoroform, $\ce{CHF3}$, and tetrafluoromethane, $\ce{CF4}$:

My first line of thought is that both of these molecules are symmetrical (the vectors from each individual atom naturally cancel each other out) with no lone pairs of electrons. Now I knew that just because they were symmetrical didn't automatically meant that they were non-polar. If they're symmetrical, they're usually non-polar but not always. This is one of those cases.
My next line of thought was that they both had to be polar and had dipole-dipole interactions. Then I later find out that the $\ce{CHF3}$ is actually the polar molecule with dipole-dipole interaction and $\ce{CF4}$ is the non-polar molecule with London-dispersion forces.
This bothered me a little bit because both of these molecules have a very electronegative atom (fluorine) that is more electronegative than any other element in the compound. This is why I thought they were both polar.
Why is tetrafluoromethane non-polar and fluoroform polar?

Check out this post, especially the answer concerning Avogadro.

You can build the molecules and display the dipole using the Molecule Calculator (under Polarity and solvation).

Draw the structures in 3D and then you will see why one is polar and the other not.
$\ce{CF4}$:

As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four $\ce{C-F}$ bonds cancel out, leaving no overall dipole.
$\ce{CHF3}$:

Although the molecule has some symmetries, it is not perfectly symmetrical. Since the dipoles of the $\ce{C-F}$ bonds are far larger than the basically non-existent dipole of the $\ce{C-H}$ bond, the dipoles do not cancel out and you are left with a molecule which has a notable net dipole moment of 1.649 D, with the negative end over the fluorines and the positive end over the hydrogen.
As general advice, when determining whether a molecule has a net dipole, always consider the molecule in 3D as opposed to just looking at individual bond dipoles.

I curious, what program did you use to draw those molecules?

@LuisAverhoff Avogadro

‘Symmetry’ is a rather generalised concept. It isn’t so much the existence of symmetry in a molecule, but much more which type of symmetry we’re encountering.
From geometry classes, you should remember that there are planes and centres of symmetry. Alongside these, rotational axes, the identity and more exotic things like improper rotation are considered elements of symmetry in group theory — in a nutshell, the part of extended chemistry that discusses symmetry and its effects. From knowing some elements of symmetry in a molecule you can deduce others, but that’s not too relevant here.
The important point is the effect that the symmetry has on your molecule. Anything that can be represented as a vector — most notably the charge separation in a dipole — needs to transform onto itself, no matter which element of symmetry you apply.
In the case of $\ce{CF4}$, the molecule not only contains a number of planes of symmetry (not two or four, but six due to the picture bon showed). These planes also generate a centre of symmetry. And the only vector that can be transformed onto itself by a centre of symmetry is the null vector, which in turn means that a dipole moment can only be zero — hence a non-polar molecule.
In $\ce{CHF3}$, we suddenly encounter the reduction of symmetry, that is already evident in your flat drawings, but only truely discussable with the 3D-model. In a nutshell, you lose three of the six planes of symmetry and also the centre of symmetry they created. The three remaining planes of symmetry share a common axis — the one which also contains the $\ce{C-H}$-bond. And a vector, that points in the same or the opposite direction as this axis will be transformed onto itself. So you have sufficient reduction in symmetry to allow a dipole momentum, creating a polar molecule.
Simplified, this means that any and all molecules that contain more than one type of atom and do not contain a centre of symmetry are polar (to some extent). However, often — especially in the case of hydrocarbons with a difference in electronegativity of only $0.4$ — the polarity is so weak that the molecules can be considered non-polar.

As you state, the CF4 is symmetrical (tetrahedral, not planar), so there is no net polar moment.

Replacing one F with a Cl, however, introduces an asymmetry because the Cl is less electronegative, about 3 V vs. 4 V for F. One can imagine the electron cloud being pulled towards the 3 F atoms and away from the Cl; so the Cl is at the positive vertex of the tetrahedron, with three F atoms opposing it across the carbon.

Those are not the Lewis dot structures. The fluorine in both of these molecules should have complete octet structure like so:

This means that there are actually more electron pairs in place than you may have been considering.

My first line of thought is that both of these molecules are symmetrical (the vectors from each individual atom naturally canceled each other out) with no lone pairs of electrons.

In $\ce{CF4}$, you are correct. The molecules is perfectly symmetrical, so every electron pair on each fluorine cancels out the electron pairs of every other fluorine. For this reason, this molecule is non-polar.
In $\ce{CHF3}$, however, the hydrogen does not have 3 other electron clouds around it like the fluorine do. These means that this side of the molecule is overall more positive than the other part of the molecule, which is dense with the fluorine atoms unbonded electrons. For this reason, this molecule is polar.

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


text:
Reading number of outer shell electrons and other properties from periodic table?
How can one read from the periodic table the number of outer shell electrons that an atom has, to predict how these atoms will make bonds with other atoms? For example to see that hydrogen ($\ce{H}$) has 1 electron free, while carbon ($\ce{C}$) has 4 and oxygen ($\ce{O}$) has 2? This allows us to infer that carbon can make four bonds with 4 different hydrogen atoms. It's not clear what in the periodic table would tell us this; it is neither in the group nor in the period of an atom, it seems.
Related to this: can someone summarize concisely all the properties (like number of free outer shell electrons) that one can read off about each element from its position in the periodic table? I was having a hard time finding a concise summary of these.

Start with a periodic table that shows the electron configurations. You might try the Dynamic Periodic Table; as you mouse over an element, its electron configuration is presented.
Then go to Electron Configurations in the Periodic Table, Electronic Structure of Atoms or Block (periodic table) to see how shells and orbitals are added. It's pretty straightforward until the transition metals, and things get much more complicated for the lanthanides and actinides, where a new shell may start before an inner shell is complete.
The general rule is: an orbital is more "stable" when filled, or half-filled, so fluorine, for example, can readily accept one more electron to finish its outer 2s2 2p5 shell and it will then have the configuration of neon, 2s2 2p6. Many periodic tables don't present the inner shells, using instead the shorthand of [Ne] 3s2 3p5 for chlorine, rather than the full 1s2 2s2 2p6 3s2 3p5.
In fact, you're not alone in trying to understand the electron structure; see the current debate on the position of Lw in the table: New data on synthetic element trigger rethink of periodic table.
BTW, note there are different ways a shell can be completed. Carbon will share its four outer electrons with four hydrogen atoms; this is called a covalent bond. Chlorine outright "steals" an electron from sodium; this is an ionic bond. See Covalent Lewis Dot Structures and Ionic Bonding and Electron Transfer for more info.


text:
Why is Omega-3 a polyunsaturated and not a monounsaturated fatty acid?
If Omega-3 is a fatty acid with a single double bond at the third carbon atom from the omega end, then why is it regarded as a polyunsaturate and not a monounsaturate? Are there different Omega-3s with more than one double bond?

The term "omega-3" just denotes that the position of the first double bond in the fatty acid will be at position "3" (counting from the omega end of the chain).  It does not mean that there is only one double bond in the fatty acid.  The most common omega-3 fatty acids contain between 3 and 6 cis double bonds (therefore they are polyunsaturated).
Here are few examples, again note how they all have a double bond at the third carbon from the omega end of the chain.
α-Linolenic acid (ALA)

Eicosapentaenoic acid (EPA)

Docosahexaenoic acid (DHA)


text:
Does a Grignard reagent react with enones to give the 1,2- or 1,4- product?

Cyclohexenone, an α,β-unsaturated ketone, is electrophilic at both the carbonyl carbon, as well as the β-carbon. So, it can either undergo 1,2-addition to give the allylic alcohol 1, or 1,4-addition to give the ketone 2.
Where does the Grignard reagent add to?

The correct answer is the 1,2-addition product (i.e. the allylic alcohol 1). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.
This exact transformation was reported by Akai and coworkers recently (Org. Lett. 2010, 12, 4900) and they obtained compound 1 in 95% yield (the supplementary information (PDF) is free).

The Grignard reaction, although very well examined, there are still arguments about the mechanism and the overall process.  The reagent is usually best described via the Schlenk equilibrium, but that is not only dependent on $\ce{X}$, but also on the solvent:
$$\ce{2RMgX <=> R2Mg + MgX2}$$
There also have been some dimeric structures reported, i.e. the symmetrical, the asymmetrical, and the alkyl bridging dimer. According to Milton Orchin[1] (and references therein) the latter is responsible for the formation of the above equilibrium.

There are many mechanism possible, two of them following a non-radical pathway are depicted below. In both cases, the 1,2-addition leads to the major product. The six-membered transition state (top route) has been proposed many times and employs the alkyl bridging dimer.[2]
The bottom route considers only the monomeric Grignard reagent. Whichever pathway is taken, the result is the same: 1,2-addition, resulting in a tertiary alcohol, with the C=C double bond retained.

Notes and References
It has been correctly pointed out by ron that Grignard agents cannot coexist with acid, as my previous draft suggested. Even water poses a big threat to the stability, as pointed out by yasir. I was a little confused about this, because adding a $\ce{Cu(I)}$ co-catalyst will result in the 1,4-product,[3] and I assumed the same would happen with an acid.

M. Orchin, J. Chem. Educ., 1989, 66 (7), 586.
See for example: E. C. Ashby. A detailed description of the mechanism of reaction of Grignard reagents with ketones. International Conference on Organometallic Chemistry, ICOMC, Organometallic Chemistry, IX, University of Dijon, France, 1979-09-03–1979-09-07. Also available for free at iupac.org.
M. S. Kharasch, and P. O. Tawney, J. Am. Chem. Soc., 1941, 63 (9), 2308–2316.

But this doesn't reason out why 1,4 wont happen. Is it that Grignard always gives 1,2?

@user223679 According to all the papers I read so far, the 1,2-addition is always favoured. The most likely second pathway is reductive elimination, but that cannot happen with $\ce{PhMgX}$. Radical pathways are favoured in presence of co-catalysts and might result in a 1,4-adduct. Also often observed is a pinacol type product.

Here is a link  to some interesting information on the subject, see p. 285 and p. 382

@ron Interestingly enough on p.382 they convert the ketone into the methyl group with the acid in the second step.


text:
How would you add saccharide residue information to PDB files?
I have multiple PDB files containing a single molecules of polysaccharides, with a single chain and a long list of all the atoms -they are not divided by residue.
The PDB files were created from ChemDraw files using OpenBabel and Corina applet form molecular-networks. The workflow was: .cdx <-> OpenBabel <-> smiles/sdf <-> Corina <-> pdb.
What i end up with is this:

How do i easily break up the list into individual residues(individual monosaccharides)? I trying to something similar to 1CAP
This is what the PDB file looks like after CORINA:
HEADER    UNK                                     15-05-08    1UNK
REMARK   1 corina 3.48 0000  08.02.2010
HETATM    1  C1  UNK     1       6.576  -2.524   4.463  1.00 20.00
HETATM    2  H2  UNK     1       6.573  -2.048   5.444  1.00 20.00
HETATM    3  C3  UNK     1       5.876  -1.614   3.449  1.00 20.00
HETATM    4  C4  UNK     1       4.411  -1.438   3.863  1.00 20.00
HETATM    5  C5  UNK     1       3.748  -2.815   3.964  1.00 20.00
HETATM    6  H6  UNK     1       3.760  -3.296   2.986  1.00 20.00
HETATM    7  O7  UNK     1       4.467  -3.621   4.900  1.00 20.00
HETATM    8  C8  UNK     1       5.829  -3.859   4.541  1.00 20.00
HETATM    9  C9  UNK     1       6.487  -4.751   5.595  1.00 20.00
HETATM   10  O10 UNK     1       5.864  -6.037   5.587  1.00 20.00
HETATM   11  O11 UNK     1       2.398  -2.662   4.404  1.00 20.00
HETATM   12  C12 UNK     1       1.655  -3.882   4.435  1.00 20.00
HETATM   13  C13 UNK     1       0.198  -3.584   4.795  1.00 20.00
HETATM   14  C14 UNK     1      -0.566  -4.899   4.964  1.00 20.00
HETATM   15  H15 UNK     1      -0.496  -5.481   4.045  1.00 20.00
HETATM   16  C16 UNK     1      -2.036  -4.592   5.265  1.00 20.00
HETATM   17  C17 UNK     1      -2.607  -3.729   4.136  1.00 20.00
HETATM   18  C18 UNK     1      -1.764  -2.458   4.000  1.00 20.00
...
...
HETATM  286  H   UNK     1      14.974  11.280   1.314  1.00 20.00
HETATM  287  H   UNK     1      15.417  11.215  -0.409  1.00 20.00
HETATM  288  H   UNK     1      15.300  13.559   0.392  1.00 20.00
HETATM  289  H   UNK     1      14.052  13.241  -0.836  1.00 20.00
HETATM  290  H   UNK     1      13.609  13.306   0.886  1.00 20.00
HETATM  291  H   UNK     1       9.253  -4.179   6.144  1.00 20.00
HETATM  292  H   UNK     1      11.359  -2.412   4.593  1.00 20.00
CONECT    1    2    8    3  115
CONECT    2    1
CONECT    3    1    4   86  120
CONECT    4    3    5   69  121
CONECT    5    4    6    7   11
CONECT    6    5
CONECT    7    5    8
CONECT    8    7    1    9  122
...
CONECT  289  114
CONECT  290  114
CONECT  291  118
CONECT  292  119
END

Does the *.pdb file only have one header (or multiple headers for each of the monosaccharides)?

Short answer: your workflow doesn't include residue perception, so you would need to do this manually.
I don't use CORINA, but I suspect the answer is the same as if you generated coordinates using Open Babel.
You're using ChemDraw, which as far as I know, doesn't include any sort of residue information in the file, then creating SMILES and/or SD files, which definitely don't include residue information.
Then you create 3D coordinates with CORINA.
But nowhere along that chain is there a step to create / perceive / retain residue information. So I'm not surprised that you get one large unknown residue.
Open Babel does have code that does amino acid and nucleic acid residue perception when translating to PDB or Mol2 or other format that allows storing residue information. It does not have support for perceiving saccharides. Indeed, I'm not even sure what algorithms exist for automatically perceiving sugars.
My suggestion would be to either:

Number / label the saccharides by hand (yuck)
Write some sort of script in Python to label the residues - since presumably the atoms come in a particular order
Use a software builder that creates the residue information with the 3D coordinates.

For the latter case, a quick search suggested a few possibilities, although I haven't tried any of them:

Glycam
Yasara

Thanks. I feared your short answer was the only way. I plan to use LPS structures in lipidbank.jp in CHARMM simulations and that is why i need the PDB split in individual residues.

@babu Please consider accepting the answer. I suggest taking a look at the Glycam builder, which is designed for MD simulations.

Yes, Glycam looks very promising and it just might be the tool i'am looking for. Also looked to CHIMERA. It has an option of creating and modifying similar to what i want but not quite. It can create residues from selected atoms but in doing so, it modifies the atom types.

Make sure you have different residue numbers for each residue. Here's an example:
HETATM    1  C1  NAG A 600      19.340   9.045  41.827  1.00 23.83           C
HETATM    2  C2  NAG A 600      20.265  10.267  41.992  1.00 25.36           C
HETATM    3  N2  NAG A 600      20.722  10.713  40.685  1.00 21.83           N
HETATM    4  C3  NAG A 600      21.468   9.913  42.881  1.00 23.92           C
HETATM    5  O3  NAG A 600      22.240  11.098  43.157  1.00 31.96           O
HETATM    6  C4  NAG A 600      20.974   9.305  44.194  1.00 29.02           C
HETATM    7  O4  NAG A 600      22.087   8.931  45.028  1.00 32.10           O
HETATM    8  C5  NAG A 600      20.112   8.085  43.874  1.00 25.92           C
HETATM    9  O5  NAG A 600      18.973   8.502  43.103  1.00 26.24           O
HETATM   10  C6  NAG A 600      19.585   7.353  45.095  1.00 29.15           C
HETATM   11  O6  NAG A 600      18.639   8.141  45.801  1.00 30.55           O
HETATM   12  C7  NAG A 600      20.897  12.007  40.404  1.00 19.02           C
HETATM   13  O7  NAG A 600      21.415  12.386  39.350  1.00 26.03           O
HETATM   14  C8  NAG A 600      20.437  13.046  41.410  1.00 12.16           C
HETATM   15  C1  FUC A 601      23.465  11.198  42.483  1.00 40.56           C
HETATM   16  C2  FUC A 601      24.142  12.529  42.836  1.00 45.83           C
HETATM   17  O2  FUC A 601      23.243  13.605  42.609  1.00 43.55           O
HETATM   18  C3  FUC A 601      24.578  12.521  44.304  1.00 52.54           C
HETATM   19  O3  FUC A 601      25.277  13.719  44.606  1.00 54.93           O
HETATM   20  C4  FUC A 601      25.476  11.312  44.578  1.00 53.83           C
HETATM   21  O4  FUC A 601      26.684  11.438  43.843  1.00 59.66           O
HETATM   22  C5  FUC A 601      24.755  10.026  44.164  1.00 52.07           C
HETATM   23  O5  FUC A 601      24.344  10.110  42.778  1.00 46.32           O
HETATM   24  C6  FUC A 601      25.630   8.795  44.302  1.00 50.48           C
HETATM   25  C1  NAG A 602      22.004   9.334  46.354  1.00 36.94           C
HETATM   26  C2  NAG A 602      23.011   8.547  47.198  1.00 40.27           C
HETATM   27  N2  NAG A 602      22.674   7.137  47.173  1.00 35.92           N
HETATM   28  C3  NAG A 602      23.018   9.060  48.643  1.00 41.77           C
HETATM   29  O3  NAG A 602      24.056   8.420  49.372  1.00 41.95           O
HETATM   30  C4  NAG A 602      23.233  10.575  48.670  1.00 44.23           C
HETATM   31  O4  NAG A 602      23.108  11.054  50.004  1.00 45.34           O
HETATM   32  C5  NAG A 602      22.198  11.260  47.766  1.00 44.11           C
HETATM   33  O5  NAG A 602      22.295  10.741  46.420  1.00 38.50           O
HETATM   34  C6  NAG A 602      22.376  12.768  47.690  1.00 43.86           C
HETATM   35  O6  NAG A 602      23.507  13.121  46.907  1.00 44.99           O
HETATM   36  C7  NAG A 602      23.592   6.247  46.811  1.00 37.09           C
HETATM   37  O7  NAG A 602      24.596   6.552  46.167  1.00 36.50           O
HETATM   38  C8  NAG A 602      23.367   4.802  47.226  1.00 36.20           C

The column contains the values 600/601/602 refers residue number.


text:
What is the purpose of the solution in a galvanic cell?
In other words: What is the purpose of the solution that the electrodes are immersed in, in a galvanic cell, and is it possible to generate a current without any solution?

related: http://chemistry.stackexchange.com/q/5477/4945

also related: http://chemistry.stackexchange.com/questions/2644/why-dont-the-electrons-move-through-the-electrolyte-instead-of-the-circuit-in?rq=1

The solution completes the circuit allowing the cathode and anode to be separated and forcing the electron exchange thru the wire.

The solution provides ions to the electrodes and a solution where the oxidized anode material can dissolve when it looses it's electron.
When the anode electrode oxidizes it looses it's electron  which moves through the wire to the cathode electrode and the resulting ionized anode material dissolves in the solution.
The electron moving through the wire then combines with the cations in the cathode solution so the cations are then reduced to a metal.
So the solution gives something to be reduced at the cathode and something where the oxidized anode material can dissolve into or form a salt with.


text:
Disproportionation with aqueous NaOH

Which of the following undergoes disproportionation with $\ce{NaOH}$?

$\ce{SiO2}$ (silicon dioxide)
$\ce{CO2}$ (carbon dioxide)
$\ce{SO2}$ (sulfur dioxide)
$\ce{SO3}$ (sulfur trioxide)
$\ce{NO2}$ (nitrogen dioxide)

From the above we can conclude that $\ce{SiO2}$, $\ce{CO2}$ and $\ce{SO3}$ do not undergo disproportionation as they are all in the maximum oxidation states. $\ce{NO2}$ is in the +4 oxidation states so it disproportionates in to +3 and +5 upon addition of aqueous $\ce{NaOH}$. But what about $\ce{SO2}$? It can also disproportionates because it is in +4 oxidation no. But the answer states only $\ce{NO2}$. Can anyone please explain the reason?

Disproportionation is a redox reaction. Since
$$\Delta G = -zF\Delta E$$
you may compare the redox potentials for both half-reactions in order to determine whether disproportionation occurs spontaneously ($\Delta G < 0$).
Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$)
$$\ce{2 NO2 <=> N2O4}$$
and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):
$$\begin{alignat}{2}
\ce{N2O4 + 2 e- \;&<=> 2 NO2- }\quad &&E_1^\circ = +0.867\ \mathrm{V}\\
\ce{2 NO3- + 2 H2O + 2 e- \;&<=> N2O4 + 4 OH-}\quad &&E_2^\circ = -0.86\ \mathrm{V}
\end{alignat}$$
The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.
In equlibrium:
$\begin{align}
E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\
E_1^\circ - E_2^\circ &= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\
\left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\
\left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\
{10^{58.5}} &= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\
{10^{29}} &= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}}
\end{align} $
Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation:
$$\ce{N2O4 + 2 OH- -> NO2- + NO3- + H2O}$$


text:
Radical formation via tritium to helium decay
Following on from this question about tritium water, I was wondering whether the decay of tritium to helium-3 has ever been used in organic synthesis. Such a transition would presumably result in no strong $\ce{C-He}$ interaction and thus radical or cation formation at $\ce{C}$.

There's actually quite an interesting review article about the effects of decaying tritium atoms in organic compounds. The formation of neutral radicals is very unfavourable, and almost always neutral helium is ejected leaving a carbocation behind instead. Compare the two processes:
$$\ \ce{R-He^+ -> R^. + \ \ He^{+}}$$
$$\ce{R-He^+ -> R^+ + \ He}$$
The latter is far more plausible, as the helium cation has an extremely exothermic electron affinity ($\rm{2370\ kJ\ mol^{-1}}$), the largest of any atomic and likely molecular monocation.
Thus, the article is titled "Tritium for generation of carbocations". The usage of tritium for carbocation formation seems to have been first investigated in 1963. This was perhaps especially important back then because carbocation chemistry had not been yet thoroughly explored until George Olah and investigation of superacidic media.
Here's an interesting excerpt of the article, from the section "Outline of the decay technique":

To minimize the extent of unwanted radiolytic
  processes, the activity of the tritiated precursors used
  in the decay experiments must be kept as low as possible,
  typically below 10 mCi per mol. As a consequence, the
  number of the decay ions, hence the amount of their
  neutral decay products formed in any reasonable period
  of time, is comparatively quite small. Even in a storage
  period as long as 1 year, the decay of 10 mCi of a
  monotritiated precursor in 1 mol of an inactive gas
  (“bulk gas”) yields only about $\rm{10^{16}}$ nucleogenic ions.
  Under the favorable assumption that all ions give a
  single end product, its concentration would be of the
  order of only a few parts per billion (ppb), posing
  formidable problems even to very advanced, highly
  sophisticated analytical methodologies. Even more
  seriously, the decay products would be swamped by
  those arising from the radiolysis of the bulk gas,
  promoted by the $\beta^-$ particles of tritium, since each decay
  event produces, together with a single nucleogenic ion,
  over 200 radiolytic ions.

While the reactions triggered by nucleogenic carbocations will be different from those of radiolytic ions (secondary ions formed by impacts with the beta decay electron), given that the overwhelming majority of reactions will be caused by the latter, precise application of such tritium-based carbocation production for synthetic purposes seems all but futile. It would be far simpler to just target an energetic electron beam at the sample, simulating the beta decay electron and obtaining essentially the same products.

Edit: Here's my previous answer, which is not quite as related but I thought interesting enough to share.

Such a transition would presumably result in no strong $\ce{C-He}$ interaction.

While in practice you are correct, I thought I'd point out something interesting. Some computational investigations have found that certain compounds such as $\ce{He^+ -C\equiv C-H}$ have a considerable bond between helium and carbon, with a bond dissociation energy of up to $\rm{92\ kJ\ mol^{-1}}$ at the B3LYP/6-31G** level of theory. Even the neutral molecule $\ce{HHeCCH}$ is expected to be bound, though with a small kinetic barrier to decomposition into $\ce{He + HCCH}$ of approximately $\rm{63\ kJ\ mol^{-1}}$. Such calculations have motivated the usage of tritiated molecules as a possible (and perhaps a bit desperate) synthetic pathway to helium compounds, which curiously are expected to be more stable than neon compounds.
However, as has been mentioned, the slow decay of tritium (only 0.015% of a sample decays during the course of a day) precludes synthetic usage of its transmutation except perhaps for extremely specific catalytic purposes. Increasing the tritium content in a sample could quickly make costs exorbitant, and all sorts of crazy secondary reactions pathways (having their activation energies met from the abundant energy of nuclear decay) would soon turn organic reactants into a goopy radioactive mess.

Here are two reasons why this approach would be impractical for organic synthesis:
(1) The half life of tritium is 12.3 years, so the decay is too slow to be practical as part of a synthetic method
(2) While on a scale of nuclear reactions, tritium decay is relatively low in energy emitted, it still dwarfs a typical chemical reaction (18.6 keV or 429,000 kCal/mole) and may cause degradation of the molecule (or the bulk material) in unpredictable ways.
But it could be useful for other types of chemical studies. I will defer to others in the forum to elaborate.

Regarding the first point I guess you would be using catalytic tritium both due to cost and the desire not to photolyze your material/molecules to bits. I can't think of any applications where this sort of extreme radical initiation would be useful though.


text:
Can 4-pyridones be synthesised analogously to 2-pyridones?
I understand that 2-pyridone derivatives can be synthesized by oxidizing an N-methyl pyridinium salt with $\ce{[Fe(CN)6]^3-}$ under basic conditions. Can a similar method be used to produce 4-pyridones ?

A reliable and old reference on the synthesis of 1-methyl-2-pyridone (Org. Synth. 1935, 15, 41, DOI) does not mention the formation of 1-methyl-4-pyridone as a side product and I doubt that it and can be formed that way.
The synthesis of 1-methyl-2-pyridone starts by reacting equimolar amounts of dry pyridine and dimethyl sulfate to form the N-methyl pyridinium salt. Then, the mixture is diluted with water and solutions of $\ce{NaOH}$ and $\ce{K3[Fe(CN)6]}$ are simultaneously added.
This seems to do the trick: the $\beta$-amino alcohol formed by adding $\ce{OH-}$ ortho to the nitrogen atom is immediately oxidized, which prevents the opening of the pyridinium salt.
For the synthesis of 1-alkyl-4-pyridones, a somewhat classical approach would be to react the corresponding 4-pyranone with a primary amine under more or less drastic conditions, as shown in Can. J. Chem., 1976, 54, 3377.
Altenatively, Grignard reaction of a 2-substituted 4-benzyloxypyridine-N-oxide, followed by debenzylation/methylation with methyl triflate and subsequent treatment with $\ce{NaOH}$ will yield the desired a 1,2,5-trisubstituted 4-pyridone (DOI).

Many thanks ! Do you know of any way to make 4-pyridone from pyridine (via n-oxides or otherwise) ?

@J.LS Do you mean the unsustituted parent 4-pyridone? I would buy its tautomer, 4-hydroxypyridine, and dissolve it in a somewhat polar solvent. In solution, the pyridone should dominate.


text:
What do the quantum numbers actually signify?
I know how to calculate them and such stuff, but I wanted to know what they actually signify. I have a vague idea that they have something to do with an electron's position in an atom but what do all of them mean? Any help would be greatly appreciated!

Quantum numbers give information about the location of an electron or set of electrons.  A full set of quantum numbers describes a unique electron for a particular atom.
Think about it as the mailing address to your house.  It allows one to pinpoint your exact location out of a set of $n$ locations you could possibly be in.  We can narrow the scope of this analogy even further.  Consider your daily routine.  You may begin your day at your home address but if you have an office job, you can be found at a different address during the work week.  Therefore we could say that you can be found in either of these locations depending on the time of day.  The same goes for electrons.  Electrons reside in atomic orbitals (which are very well defined 'locations').  When an atom is in the ground state, these electrons will reside in the lowest energy orbitals possible (e.g. 1$s^2$ 2$s^2$ and 2$p^2$ for carbon).  We can write out the physical 'address' of these electrons in a ground-state configuration using quantum numbers as well as the location(s) of these electrons when in some non-ground (i.e. excited) state.
You could describe your home location any number of ways (GPS coordinates, qualitatively describing your surroundings, etc.) but we've adapted to a particular formalism in how we describe it (at least in the case of mailing addresses).  The quantum numbers have been laid out in the same way.  We could communicate with each other that an electron is "located in the lowest energy, spherical atomic orbital" but it is much easier to say a spin-up electron in the  1$s$ orbital instead.  The four quantum numbers allows us to communicate this information numerically without any need for a wordy description.
Of course carbon is not always going to be in the ground state.  Given a  wavelength of light for example, one can excite carbon in any number of ways.  Where will the electron(s) go?  Regardless of what wavelength of light we use, we know that we can describe the final location(s) using the four quantum numbers.  You can do this by writing out all the possible permutations of the four quantum numbers.  Of course, with a little more effort, you could predict the exact location where the electron goes but in my example above, you know for a fact you could describe it using the quantum number formalism.
The quantum numbers also come with a set of restrictions which inherently gives you useful information about where electrons will NOT be. For instance, you could never have the following possible quantum numbers for an atom:
$n$=1; $l$=0; $m_l$=0; $m_s$=1/2
$n$=1; $l$=0; $m_l$=0; $m_s$=-1/2
$n$=1; $l$=0; $m_l$=0; $m_s$=1/2
This set of quantum numbers indicates that three electrons reside in the 1$s$ orbital which is impossible!
As Jan stated in his post, these quantum numbers are derived from the solutions to the Schrodinger equation for the hydrogen atom (or a 1-e$^-$ system).  There are any number of solutions to this equation that relate to the possible energy levels of they hydrogen atom.  Remember, energy is QUANTIZED (as postulated by Max Planck).  That means that an energy level may exist (arbitrarily) at 0 and 1 but NEVER in between.  There is a discrete 'jump' in energy levels and not some gradient between them.  From these solutions a formalism was constructed to communicate the solutions in a very easy, numerical way just as mailing addresses are purposefully formatted in such a way that is easy that anyone can understand with minimal effort.
In summary, the quantum numbers not only tell you where electrons will be (ground state) and can be (excited state), but also will tell you where electrons cannot be in an atom (due to the restrictions for each quantum number).

Principle quantum number ($n$) - indicates the orbital size.  Electrons in atoms reside in atomic orbitals.  These are referred to as $s,p,d,f...$ type orbitals.  A $1s$ orbital is smaller than a $2s$ orbital.  A $2p$ orbital is smaller than a $3p$ orbital.  This is because orbitals with a larger $n$ value are getting larger due to the fact that they are further away from the nucleus.  The principle quantum number is an integer value where $n$ = 1,2,3... .
Angular quantum number ($l$) - indicates the shape of the orbital. Each type of orbital ($s,p,d,f..$) has a characteristic shape associated with it.  $s$-type orbitals are spherical while $p$-type orbitals have 'dumbbell' orientations.  The orbitals described by $l$=0,1,2,3... are $s,p,d,f...$ orbitals, respectively.  The angular quantum number ranges from 0 to $n$-1.  Therefore, if $n$ = 3, then the possible values of $l$ are 0, 1, 2.
Magnetic quantum number ($m_l$) - indicates the orientation of a particular orbital in space.  Consider the $p$ orbitals.  This is a set of orbitals consisting of three $p$-orbitals that have a unique orientation in space.  In Cartesian space, each orbital would like along an axis (x, y, or z) and would be centered around the origin at 0,0.  While each orbital is indeed a $p$-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space.  Therefore, for a set of $p$-orbitals, there would be three $m_l$, each uniquely describing one of these orbitals.  The magnetic quantum number can have values of $-l$ to $l$.  Therefore, in our example above (where $l$ = 0,1,2) then $m_l$ would be -2, -1, 0, 1, 2.
Spin quantum number ($m_s$) - indicates the 'spin' of the electron residing in some atomic orbital.  Thus far we have introduced three quantum numbers that localize a position to an orbital of a particular size, shape and orientation.  We now introduce the fourth quantum number that describes the type of electron that can be in that orbital.  Recall that two electrons can reside inside one atomic orbital.  We can define each one uniquely by indicating the electron's spin.  According to the Pauli-exclusion principle, no two electrons can have the exact same four quantum numbers.  This means that two electrons in one atomic orbital cannot have the same 'spin'. We generally denote 'spin-up' as $m_s$ =1/2 and spin-down as $m_s$=-1/2.

This is quite helpful, but do you think a little more on the significance of these numbers might be even more helpful? Such as, the energy levels for the principle quantum number or the bonding implications of the angular quantum number? (I don't know enough about the implications of the last two to generalize that much). Also, I feel somewhat like the OP. I can calculate these numbers and I understand that they give us a way to annotate 3D info for an electron, but what does that enable us to do as a result? Why are quantum numbers important for chemistry?

@Cohen_the_Librarian I've extensively edited my post to try and address your questions/suggestions.

Consider l = 1 (i.e. p orbitals), do the px, py and pz orbitals correspond to ml = -1, 0 and 1 respectively? Is there any correspondence that can be done for the d and f orbitals as well? I understand that it is a matter of perspective but is there a particular convention to assign each value of ml to a particular orbital, be it px or dx-y or what not.

The Schrödinger equation for most system has many solutions $\hat{H}\Psi_i=E_i\Psi_i$, where $i=1,2,3,..$. In the case of the hydrogen atom the solutions has a specific notation, which are where the quantum numbers come from.
In the case of the H atom the principal quantum number $n$ refers to solutions with different energy.
For $n>1$ there are several solutions with the same energy, which come in different shapes ($s$, $p$, etc with different angular quantum numbers $l$) that can point in different directions ($p_x$, $p_y$, etc with different magnetic quantum numbers $m$)
These quantum numbers are also applied to multi-electron atoms within the AO approximation.
So the quantum numbers are a way to count (label) the solutions to the Schrödinger equation.

authors: ЮРИЙ ГАВРИЛЮК


text:
Variation in atomic radii of elements in different blocks?
If we look at the values for the atomic radii (look at the table here), we can see that they rapidly decrease across the period initially. Looking at the second period,

The graph is pretty steep early on. But further down the period, as we enter the p-block elements, the graph levels out. This trend is repeated across the third and higher periods as well. Why is that?
The variation of atomic radii across the d-block and f-block elements is even more gradual. Why is that again?

Valence electrons experience an electrostatic force from the nucleus. The nucleus has its character positive charge, but due to shielding by core electrons the total positive charge is not completely felt by the electrons, so the actual net positive charge felt by the valence electron has its own name and is called the effective nuclear charge $Z_{eff}$. This effective nuclear charge, largely effects the size of the atom. The general trend of the periodic table is that as you go down a group, and go from right to left, the $Z_{eff}$ decreases, and you see an increase in the size of the atom. The easy way to think of it, is that as you go from left to right, the number of valence electrons, and  protons increases but the number core electrons stays the same; so, you see an increase in $Z_{eff}$.
$Z_{eff}$ is calculated using Slater's Rules. Where
$$Z_{eff}=Z-s$$
Where $Z$ is the atomic number and $s$ is the shielding constant.
The rules for calculating are as follows (from wikipedia):
Firstly,(1)(4) the electrons are arranged into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l, except that s- and p- orbitals are kept together.
$\mathrm{(1s) (2s,2p) (3s,3p) (3d) (4s,4p) (4d) (4f) (5s, 5p) (5d)}$ etc.
Each group is given a different shielding constant which depends upon the number and types of electrons in those groups preceding it.
The shielding constant for each group is formed as the sum of the following contributions:

An amount of 0.35 from each other electron within the same group except for the (1s) group, where the other electron contributes only 0.30.
If the group is of the (s p) type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

If the group is of the (d) or (f), type, an amount of 1.00 for each electron "closer" to the atom than the group. This includes
i) electrons with a smaller principal quantum number n and
ii) electrons with an equal principal quantum number and a smaller azimuthal quantum number l.

I know about Slater's rule. I just don't see how it applies to my question.

What's confusing? I explain why atomic radii size is affected by its $z_{eff}$ and the rules explain how d and f block electrons contribute differently than s and p block..


text:
Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?
If we make a resonance structure for the anilinium ion, with positive charge at either the ortho or the para position, we get a pentavalent nitrogen, which is not possible. So, how is the $\ce{-NH3+}$ group meta-directing in electrophilic aromatic substitution reactions, e.g. in nitration of aniline under acidic conditions?

how is anilinium ion meta directing for electrophiles?

Actually, anilinium is not meta directing (I know it is often taught that way), but rather it inductively deactivates the entire aromatic ring.
To explain the electrophilic substitution pattern observed with any aromatic system we must consider both resonance and inductive effects.
Resonance Effects: You are correct that you can't delocalize the positive charge on the anilinium nitrogen into the benzene ring through resonance.  As you point out, resonance structures with 5 bonds to nitrogen aren't allowed.

Therefore resonance effects are not involved in explaining the reactivity of the anilinium ion towards electrophilic aromatic substitution.
Inductive Effects: Inductively we expect the anilinium nitrogen to strongly withdraw electron density from the ortho position since it is closest to the positively charged nitrogen, this makes the ortho position the most deactivated.  Inductive effects die off quickly with distance, so we would expect the meta position to also be deactivated, but less so than the ortho position.  Finally, we would still expect some deactivation at the distant para position, but slightly less than we saw at the meta position. Overall, in terms of reactivity we would expect para > meta > ortho, with the difference between para and meta being much smaller than the difference between meta and ortho due to the effect distance has on inductive electron withdrawal.
If we examine the nitration of aniline in strong acid ($\ce{HNO3}$, $\ce{H2SO4}$), under these conditions the strong acid will completely protonate the aniline and reaction will proceed through the anilinium ion.  We see that, as expected, there is very little ortho product and slightly more para product than meta product.  This is pretty much as we expected and is consistent with the electron withdrawing inductive effect of the positive anilinium nitrogen.

Thanks for your answer! Can you please also comment on the directing nature of $\ce{-NR3+}$ (not $\ce{-NH3+}$)? Would here the inductive effects of the three alkyl groups make this group meta directing somehow? Is there experimental evidence related to this? Thanks!

@GaurangTandon Since what you ask is a different question from the OP, you may want to post it as a new question.  I haven't searched the $\ce{Ph-NR3^{+}}$ case, but my intuition tells me that it would behave similarly to $\ce{Ph-NH3^{+}}$

No @ron, it doesn't ... (Source: March's Advanced Organic Chemistry) only $\ce{NH3+}$ is an exception rest all ($\ce{NR3+ , NHR2+}$) etc are meta directing...I am unable to rationalise this.

@Abcd Did you happen to post a question about this exception? (And which page in March?)


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Thymol Blue - molar absorptivity
I did a lab about spectrophotometry and right now I need the molar absorptivity(ε) of thymol blue in acidic solution and also in basic solution. I have tried looking online but I haven't found the data I need.
What is the molar absorptivity of thymol blue in acidic solution and in basic solution?
 information
The specific wavelength is 540.27 nm for the acid and 601.88 nm  for the base.

Are you certain that the purpose of the lab was not to find the molar absorptivity by plotting absorbance as a function of concentration of thymol blue in both acid and base? The slope of that line will be the molar absorptivity times the path length: $$\begin{align} Abs&=\epsilon \ell c\ y&=mx+b \ m&=\epsilon \ell\end{align}$$

that is exactly what I did and I have my own results but now  I need to calculate the percent error .

First of all, molar absorptivity is dependent on wavelength, so you either need to specify a wavelength or you need the entire spectrum.
When searching for absorptivity, it's often effective to search for the older name 'extinction coefficient'. Using that search term in Google, I found an absorption spectrum for thymol blue here (pdf).

The wavelength of choice has a maximum absorbance. The molar absorptivity is not usually calculated for other wavelengths.

It's often calculated at the isobestic wavelength.

If you use the isbestic wavelength, you only find out how much thymol blue is in the solution; you don't find out the pH of the solution, which is the purpose of using the pH indicator.

I added the wavelengths in which the experiments have been done. And also,I was not able to find the information I need in that pdf document.

Just read them off the graph at Fig. 2. For the acid at 540 nm (are you sure that's the right wavelength? It's a weird wavelength to measure the acid form at) it looks like it's about 0.05 L.(mol.cm)^-1, and for the base at 600 nm it looks like it's about 0.85 L(mol.cm)^-1. You may want to measure that graph more precisely than my eyeball did.


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Atomic emission spectroscopy
There is a sentence in my book I don't understand:

Every atom or ion, which has one valence electron, will have double lines in the atomic emission spectrum. For example $\ce{Na}$ will produce lines at $589.0\:\mathrm{nm}$ and $589.6\:\mathrm{nm}$.

Why does this happen?

When an electron in any element absorbs energy it will be promoted to an excited state.  How much energy we supply determines how many \ which excited states will be created.  As the electron returns from these various excited states to the ground state light can be emitted, this is the basis of atomic emission spectroscopy.

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In the case of sodium (or any other atom with just one valence electron) many excited states are possible, so many emission lines are observed.  However, a very likely transition will be the one from the s orbital to the next highest energy orbital, the p orbital.
The magnetic energy of an electron spin is split into energy states with different total angular momentum due to the electron spin's interaction with the internal magnetic field set up by the orbital motion of the electron, this is termed spin-orbit coupling. In the case of a p orbital, it is split into two states (the total angular momentum j is given as j = l +/- s, where l is the angular momentum quantum number [l=1 for a p orbital] and s is the electron spin (1/2), therefore j has two values or states for an electron in a p orbital, 1/2 and 3/2). It is the light emission that occurs when the excited electron returns from these two p orbital energy states to the s orbital ground state that produces the doublet seen for all atoms with a ground state involving one electron in an s orbital.

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