text:
Ionization of amino acids
In my book it talks about amino acids. It says the general structure of an amino acid is as follows:

However, on the next page it lists specific amino acids such as glycine shown here:

My question is when comparing the first image to glycine, why does glycine or any of the other specific amino acids have an extra H atom as in "H3N+"?
Where does the extra H come from and why is it positively charged?
Also, on the general structure it shows the an amino should have a COOH but on glycine it is just C double bonded to an O and then another O- negative charge.
Does this have to do with the pH and ionization?

The two forms of amino acids that your question asks about are called zwitterions (technically IUPAC says they should be called dipolar ions but the term zwitteron is pervasive).
Since amino acids contain both acidic (the carboxylic acid) and basic (the amine) moieties the molecule is able to undergo what almost appears to be an intramolecular acid-base reaction in which the acidic proton of the carboxylic acid protonates the amine, this is what gives rise to the species on the right of the image below.

Its worth pointing out the this property of the amino acids is valid at physiological pH for the naturally occurring amino acids, generally in aqueous medium (which living organisms have. Also as Jan pointed out in the comments, other polar solvents would also allow the amino acid to show this characteristic).
At other pH values, the molecules become singly ionised, favouring either the cationic or anionic forms. The range of pH values at which the amino acid is zwitterionic will also vary depending on which amino acid you're looking at (the pKa values are different for every amino acid, even if only subtly)

Thank you so much! Very helpful. What exactly causes this "protonation?" Im assuming the water or cytosol within the cell? Or is that a much more complex theory based question? Also, The arrows indicate that this reaction is a reversible reaction? If so, do you know how/why is this reversible?

Its not a 'reaction' so much as an equilibrium between the zwitterionic and the neutral forms- they are in constant flux but the amount of zwitterion at equilibrium remains constant (the equilibrium constant). The water assists with the proton transfer between the carboxylic acid and the amine, yes.  I noticed your images were from the Campbell Biology, you'd be well advised to read it alongside something like Lehringer's Biochemistry- it explains the chemical concepts far better

Thank you again, I will be sure to checkout that material as well.

NotWoodward has answered it perfectly! The glycine which has both positive and negative (resulting neutral charge) is a zwitter ion form. This form is achieved at different pH for different amino acids depending on their R groups.
How does pH actually affect the amino acid? I always break it down when I explain it to my students as follows:
An amino acid in a very protonated solution (low pH) is going to be saturated with hydrogen ions, so all molecules that can take a proton will take it as they are abundant. Therefore, N will take them and become $\ce{NH3+}$ and the carboxyl group will take it and form COOH, but as you keep on adding a base (increasing the pH), the base would react with an available proton for neutralisation.
Therefore, N will easily give away its extra proton and as pH increases will become $\ce{NH2}$. If you continue adding a base, once all N has given away its extra proton, Carboxy group would then give away the proton, resulting in $\ce{COO-}$. So this results in different forms of the same molecule at different pH to attain equilibrium.
Certain amino acids also have different names based on their states, example: Glutamic acid (RCOOH) and Glutamate ($\ce{RCOO-}$).

The amino acid is at its 'isoelectric point' in its zwitterionic form.


text:
Convergence issue in Gaussian
I have been struggling with geometry optimization of a prostaglandin molecule (20 carbon atoms). I finally got the optimization, all right. Then I did the Frequency job on the optimized structure (I opened the Optimization job .LOG file scrolled up to the last geometry (highest number) and launched the Frequency job from there). Below is my route line from the .gjf file:
# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity

The Frequency job terminated normally and that's what I got in the Frequency job .LOG file:
          Item               Value     Threshold  Converged?
 Maximum Force            0.000005     0.000450     YES
 RMS     Force            0.000001     0.000300     YES
 Maximum Displacement     0.013516     0.001800     NO
 RMS     Displacement     0.003867     0.001200     NO
 Predicted change in Energy=-2.250043D-08

Unfortunately, there is no statement: "Stationary point found", however according to Gaussian manual: "A stationary point is found when the Maximum Force and RMS Force are two orders of magnitude smaller than the thresholds shown, regardless of the values of the displacements."
Is not it just the case? If so, why wouldn't Gaussian kindly state that the stationary point was found?

Just few suggestions for finding a stationary point. You can add some additional angular flexibility into the basis by changing it to, say, 6-31G(2df,p) or even switch to more modern Ahlrichs' def2 bases (Def2SVP to start from). Besides, you'd better use UltraFine integration grid and tight optimization criteria and do geometry optimization and frequency calculation in a single run: Opt=Tight Freq Int=UltraFine.

Generally, you need all the 4 parameters to converge. However, displacement values may be large unrelated to the question wheater your geometry is stationary or not. It is a judgement call and you should have your chemical insight into the system to see wheater these values raise a red flag or not.

Thank you so much Greg and @Wildcat. Opt=Tight and Int=Ultrafine did the trick! Stupid question: if I want to double check after an Opt-Freq job and just do the Freq job, I should of course skip the Opt=Tight, shouldn't I? How about Int=Ultrafine - should I always put it in the Freq job for this .chk or .log file? I mean, is it mandatory?

@RicardoMoreno, yes, you should use the same integration grid for geometry optimization and frequency calculation. And yes, you can skip the Opt keyword for frequency calculations when an input geometry is already optimized (at the very same level of theory, of course, and using the same settings).

But even if you include the Opt keyword in frequency calculations for an already optimized input geometry it would not harm: the program will just run 1 geometry optimization cycle and then switch to frequency calculation. Note though that a better approach for separate geometry optimization and frequency calculations is to save the checkpoint file for geometry optimization, then read the optimized geometry from it to start frequency calculations.

Thanks again @Wildcat. However, if I got you correctly, if I included the Opt keyword in a Freq job, it might change the geometry, right? And I don't t want it as my goal is to later calculate vertical ionization potential and electron affinity. So I need to stick to the exact same geometry at all cost. Thank you for such a quick response. I am new to StackExchange and with guys like you the site seems awsome!!!

It is quite unlikely that geometry optimization of an already optimized structure will make any changes to it. It certainly takes some computer time to evaluate energy & forces for the input structure, but if a structure is already optimized the program will stop geometry optimization after this single cycle and proceed immediately to frequencies calculation.

OK, @Wildcat. Thanks for sharing this one with me. I appreciate it!

What was the reported max "Cartesian Forces" from the optimization for the corresponding geometry and what did it say in the freq calculation?

Is not it just the case?

Looks like it is indeed not the case. I think the phrase OP quoted from the manual, which says that the optimization stops when

The Maximum Force and RMS Force are two orders of magnitude smaller
  than the thresholds shown, regardless of the values of the
  displacements.

can be interpreted a bit differently, but I guess that algoritmically it is implemented in a pretty simple way: the above mentioned quantities should be at least 100 times smaller that the corresponding thresholds. And this is not the case for the calculation under consideration: Maximum Force value (0.000005) is not 100 times smaller (but only 90) than the threshold (0.000450).

Thank you Wildcat. You are absolutely correct. It is 90 times less and not 100. It is all about precision. Thank you!

The actual answer to this question is that the Hessian is calculated analytically for DFT methods when you do a frequency calculation, but it is estimated when performing a geometry optimization. Therefore, in some cases, the optimization will show a converged structure but the frequency analysis shows that it is not below the convergence thresholds when the analytical Hessian is generated. The best option is to continue from the checkpoint file of the frequency calculation using freq opt=ReadFC guess=Read. This is all discussed on the Gaussian website here.
Oftentimes, the structure optimized from the analytical Hessian is nearly identical to the one from the approximated Hessian, but you simply don't know in advance how big this difference will be. In large, flexible molecules (ones that can have moderate displacements even at low forces), I have seen that the optimization can sometimes actually have a long way to go after continuing from the analytical Hessian.

As an answer to complete the previous ones, I've got this issue many times in the past for my calculations and tried several options to have full convergence for my optimizations. As mentioned by the author, an optimization job followed by a frequency job are required to find a stationary point and check it through frequencies analysis. However sometimes, the frequency job will not fully converge due to a different threshold than for the optimization, being more precise. This has been perfectly explained in the first answers.
A way to overcome this problem is to take the optimized structure obtained at this step and run another optimization job with an extra keyword:
opt=calcall

This keyword allows you to perform a freq analysis at each step of the optimization. It makes the calculation way more expensive but since your structure is close to the minimum you want to reach, convergence is in general met very quickly. It should never be used in your input from the structure you did draw in your editor but it's worth the try when you get a good guess of your optimized structure that is close to the minimum.

Frequency jobs don't converge, they just calculate the frequencies.

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text:
Balanced equation for neutralisation reaction betweeen potassium carbonate and hydrochloric acid
They seem all correct to me, is there any reason why one reaction would happen over the other(s)?
$$\begin{align}
\ce{K2CO3 + 2 HCl &-> 2 KCl + H2CO3}\\[6pt]
\ce{K2CO3 + 2 HCl &-> 2 KCl + H2O + CO2}\\[6pt]
\ce{K2CO3 +   HCl &-> KHCO3 + H2O}
\end{align}$$

Welcome to Chemistry.SE! Feel free to take the tour to learn more about this site. Your question appears to be a homework question, and in its current form doesn't comply with our homework policy, and is subject to removal. Please edit your question accordingly. Visit this link to learn about MathJax and mhchem

@getafix Was not a homework question, but I've edited it accordingly anyway, thanks.

If you read the policy, a "homework question", doesn't have to be from an actual homework assignment. You must demonstrate some effort towards solving the problem, or at least share your thoughts with us on the matter.

They are all correct and they will all stay in equilibrium with one another, whereas the products of the second equation will be the most favoured, since carbonic acid is an unstable weak acid and will likely dissociate into $\ce{CO_2}$ and $\ce{H_2O}$. At the same time potassium hydrogen carbonate will get protonated easily: $$\ce{HCl + KHCO_3 \rightarrow KCl + H_2CO_3}$$ again resulting in more carbon dioxide and water.

The real point about equation two is that this reaction is typically taking place in aqueous solution, and the $\ce{CO2}$ escapes from liquid into gas phase.

All the equations are correct. The first two are basically the same. Carbonic acid is a weak acid. It is less stable. So it dissociates itself to give $\ce{CO2}$ and $\ce{H2O}$. And the last one is just based on the simple difference in the moles. So you can use the equation depending on the moles.

Could you please fix that caps lock first?

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text:
Shape of the P1/2 Orbital
I always thought that p-orbitals had a dumbbell shape as pictured below.

(image source)
However, I was reading an article (see Table 1, item 2) that says, "...the spherical $\mathrm{p_{1/2}}$ subshell..." (my emphasis). The implication being that the $\mathrm{p_{1/2}}$ orbital (subshell?) has a greater electron density near the nucleus than the $\mathrm{p_{3/2}}$ orbital (subshell?). Why is this and why do they call the $\mathrm{p_{1/2}}$ orbital spherical?

They  'split p shell, using sum of spin and angular moments' instead of common non-relativistic quantum chemistry concept of the two being independent variables. Spin and angular momentum may be considered indepenedent for lighter atoms, but once relativistic effects become significant (afaik, around copper), this is no longer the case. Consequently, instead of 3 pairs of spin-orbitals with only spin being opposed in the pairs we should consider 6 spin-orbitals of potentially very complex shape. In case you are interested enough in the matter (I'm not), you should delve into  relativistic QC.

The article "Contour diagrams for relativistic orbitals" indeed states that $\mathrm{np_{1/2}}$ orbitals (one per shell) are spherically symmetric, while the $\mathrm{np_{3/2}}$ orbitals (two per shell) are shaped like the familiar dumbbells. I don't particularly know why this is the case; I thought relativistic corrections introduced only small amounts of "smearing" to the non-relativistic orbitals (e.g. turning orbital nodes into pseudo-nodes, where the wavefunction is almost but not quite zero)

A more thorough explanation, along with multiple graphs, drawings and pictures of several relativistic orbitals can be found in "Pictorial Representations of the Dirac Electron Cloud for Hydrogen-Like Atoms". Unfortunately I can't grasp the matter well enough to write a decent answer.

As orthocresol mentioned, this is all about relativity, so let's talk about it. I am hardly an expert myself, but I'll try to give an answer to the best of my limited knowledge. For an interesting and accessible overview of the incorporation of relativistic effects into chemistry, I recommend the 2012 review article "Relativistic Effects in Chemistry: More Common Than You Thought" by Pekka Pyykkö (free access!), one of the great names in the field.
The theoretical basis for most of our work as chemists is "old" quantum mechanics, developed up to around 1925, and one of the hallmarks of this era is the Schrödinger equation. The whole concept of orbitals comes as the solution of the Schrödinger equation for the hydrogen atom, (i.e. a two-body system composed of an electron in a spherically symmetric electric field originating from the nucleus). The stationary states in this system are spherical harmonics, described by the quantum numbers $n$, $l$ and $m$ (or $m_l$). The first two numbers determine the shape of the orbital, but we've gotten used to calling the second number by a letter (s, p, d, f, g, ... for 0, 1, 2, 3, 4, ...). So that's what orbital names, such as $\textbf{4s}$, $\textbf{3d}$ and $\textbf{5f}$ mean.
That said, old quantum mechanics is fundamentally incomplete, in a way which very much matters to chemists. In particular, in 1922 the Stern-Gerlach experiment discovered the property of spin. While its existence could be accommodated within the prior framework by "tacking it on" and adding an extra quantum number $s$, there was no theoretical background to explain it. This changed a few years later, when Paul Dirac incorporated special relativity into quantum mechanics. Old quantum mechanics implicitly assumes the speed of light is infinite, and by adequately introducing its finitude, many new experimentally-verified phenomena appear, such as antiparticles and, indeed, spin. In other words, spin is a purely relativistic quantum mechanical property!
The equivalent of the Schrödinger equation in relativistic quantum mechanics is the Dirac equation. Much like the Schrödinger equation, the Dirac equation can be solved for a two-body system composed of an electron in a spherically-symmetric electric field created by the nucleus. The resulting solutions ("Dirac orbitals"), however, are somewhat different. In particular, the quantum numbers $l$ and $s$ are no longer "good" quantum numbers (that is, they no longer describe stationary states, due to spin-orbit coupling), but their sum ($\overrightarrow{j}=\overrightarrow{l}+\overrightarrow{s}$) is. For the electron, $s=\frac{1}{2}$, and thus for each value of $l$, there are now two possible values of $j$, given by $|l+\frac{1}{2}|$ and $|l-\frac{1}{2}|$. This means that the s, p, d, f... subshells from the Schrödinger solutions are now split according to their $j$ values. The s subshells are unique in that they don't split ($|0+\frac{1}{2}| = |0-\frac{1}{2}| = \frac{1}{2}$), but are now termed $\textbf{s}_{1/2}$ subshells. The p subshells, which are triply degenerate via the Schrödinger equation, now split into a single $\textbf{p}_{1/2}$ orbital and two degenerate $\textbf{p}_{3/2}$ orbitals. The d subshells, previously quintuply degenerate, split into two degenerate $\textbf{d}_{3/2}$ orbitals and three degenerate $\textbf{d}_{5/2}$ orbitals. The trend continues similarly for the further subshells. Here is a figure from "Relativistic quantum chemistry: The electrons and the nodes", which will hopefully make it clearer:

So what are the physical consequences of these different mathematical solutions? Firstly, the orbitals change in energy. Orbitals with a lower $j$ for a given $n$ and $l$ have lower energy (e.g., the single $\textbf{4p}_{1/2}$ orbital is lower in energy than the two degenerate $\textbf{4p}_{3/2}$ orbitals). The shape of the orbitals also change; all orbitals with $j=\frac{1}{2}$ are spherical, regardless of $l$.
According to "Contour diagrams for relativistic orbitals":

Contour plots for $^2P_{1/2}$ and $^2S_{1/2}$ are spherically symmetrical,  while those for $n = 2$ and $3$, $l = 1$, $j= 3/2$, and
  $m = \pm 3/2$ look very similar to those for p orbitals already published in this journal.

Similarly, all orbitals with $j=\frac{3}{2}$ are dumbbell-shaped, including $\textbf{d}_{3/2}$ orbitals. The article brings a helpful schematic:

As additional confirmation, "Pictorial Representations of the Dirac Electron Cloud for Hydrogen-Like Atoms" states:

As an example of these two theorems the angular charge distributions for the magnetic states $m = \pm 3/2$ for a $^2P_{3/2}$ term are not only the same but are also identical with the two magnetic states $m = \pm 3/2$ of the $^2D_{3/2}$ term. It should be pointed out that the radial charge distributions of the $^2P$ and $^2D$ terms are different. [...]
  In this connection one of the consequences of the Dirac theory is that
  a single p electron or two similar p electrons, depending on whether $j=\frac{1}{2}$ or
  $j=\frac{3}{2}$ respectively, present spherical symmetry. Not only are all S states
  spherically symmetrical, as on the Schrodinger theory, but also all one electron systems having but one valence electron and that electron in a $^2P_{1/2}$ state,
  eg., the normal states of B, Al, Ga, In, Tl.

The article also shows a table of the angular distribution functions for the wavefunction. Note how the function is equal to $1$ for $\textbf{s}_{1/2}$ and $\textbf{p}_{1/2}$ orbitals. That is, there is no angular dependence, implying spherical symmetry.

The article also brings some diagrams and pictures of simulated orbital shapes.

I wouldn't be able to tell you why this is the case any more than I would be able to explain the shapes for the solutions to the Schrödinger equation - that's just how it pans out.
Another difference is that the Dirac orbitals do not possess nodal points or planes. They are now replaced with "pseudo-nodes", where the wavefunction assumes very small values, but never reaches zero anywhere. This is nicely shown in this figure from "Relativistic quantum chemistry: The electrons and the nodes":

You wrote, "The p subshells, which are triply degenerate via the Schrödinger equation, now split into a single p1/2 orbital and two degenerate p3/2 orbitals". Could you explain this?  I find 6 j values, 4 at 1/2 and 2 at 3/2.

@ron $j$ is always a positive half-integer with at most two possible values, given by $l+\frac{1}{2}$ and $l-\frac{1}{2}$. For the s subshells, there is only one positive value ($j=0+\frac{1}{2}=\frac{1}{2}$) but for all other subshells, there are two possible values, e.g., for p ($l=1$) you can have $j=1+\frac{1}{2}=\frac{3}{2}$ or $j=1-\frac{1}{2}=\frac{1}{2}$, and for d ($l=2$) you can have $j=2+\frac{1}{2}=\frac{5}{2}$ or $j=2-\frac{1}{2}=\frac{3}{2}$.

The amount of orbitals in each split subshell is given by how many positive $m$ equal or below $j$ such that they are spaced apart by 1 (see the table in my answer). For example, for a $\textbf{s}{1/2}$ subshell, the only possible $m$ is equal to $j$, namely $\frac{1}{2}$, which means a single orbital in the subshell. The same is true for any $\textbf{p}{1/2}$ subshell. However, for $\textbf{p}{3/2}$ subshells, there are two possible values of $m$, namely  $\frac{3}{2}$ and  $\frac{1}{2}$. For $\textbf{g}{9/2}$ subshells, there are five possibilities, from $\frac{9}{2}$ to $\frac{1}{2}$.

The dumbbell shape of the $\mathrm{p}$ orbitals are actually a cartoon representation of the actual electron density. Graphed according to the Schrödinger equation, the orbitals are actually much more spherical. See here:
$\hspace{2.3cm}$
$\hspace{2.3cm}$
Being in a higher energy state, electrons in the $\mathrm{p_{3/2}}$ are naturally further away from the nucleus, and orbital on the whole will have a lower electron density.

But I wouldn't call them spherical.


text:
Gibbs free energy change at equilibrium
I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable of doing but what happens to this energy at equilibrium? Why is the system unable to do work? Why is the Gibbs free energy change equal to zero at equilibrium?
Edit: Please don't simply write the derivation that we use to conclude that ∆$G$ must be zero at equilibrium. Please try to explain the physical significance behind this fact.
Please also note that I have already searched this website for an answer to my question, found only the following question but the answers only explained the Math:
Gibbs free energy-zero or minimum
Thanks ever so much in advance :) Regards.

What exactly are you looking for when you say "physical significance"?

At equilibrium, exactly what happens to the energy that had once been present, allowing the reaction to do work? What is the meaning of ∆$G$=0?

It means the chemical potential of the reactants and products is the same

Can you please elaborate that point? I don't quite understand fully.

You would have to allow the use of some mathematics then

Sure, no problem :) I just wanted to get an intuitive sense of this concept, rather than understand the derivation. Please go on.

I detect some inaccuracy in for question. I am going to try to clarify a few things since I perceive genuine curiosity in your question. The answer can not be so short.
First some clarifications:

Why is the Gibbs free energy change equal to zero at equilibrium?

The very common phrase free energy change at equilibrium is inadequate. In thermodynamics the state of a simple system is specified with state parameters.  In the matter at hand temperature ($T$) and pressure ($p$). As the system would be described by $T$ and $p$, any "change" that does not involve changes in $T$ or $p$ is outside of the scope of the theoretical treatment. That is, to say the system is at equilibrium is accurate within a theoretical framework, so change should refers to changes within the same framework. Of course they could be important in uncommon cases but they are not included in the model. So, let use the $^*$ symbol to refer to an equilibrium state. At equilibium $G = G(T^*,p^*)$. Any change that do not involve $T$ and $p$ will let the system in the equilibrium state (although in others way the system can look not the same). So, its change is zero just because is the same function with the same values for its arguments.
At equilibrium the system is in an stationary state. That is, the system (thermodynamic!) state do not change with time. So, if you inspect how $T$ and $p$ evolve in time they do not change anymore. So any spontaneous change won't change the value of $G$. To get a change of $G$ you must make a perturbation on the system by making a change in the environment or the walls that divide them up, but once it is done the system is not in equilibrium state anymore (although it can reach other equilibrium state in the future).
That is why at equilibrium there are not useful work available from the system, because it is in a stationary state in harmony with the environment. So changes won't take place. Notice that to get work from the system it is needed some kind of macroscopic movement (according to the definition of mechanical work).
You said:

I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable...

It is not exactly true, it is not a requirement that the process involve a reaction.
In the ligh of getafix commentary:
Take in mind that physical significance most times, in the core, is just an inaccurate idea about a physical theory/model/law. Although it may be useful sometimes from a practical perspective (and many times leads to catastrophic conclusions), it has not an add value. There is the nature. There are good formal descriptions of it obtained through years of hard work from experienced people. There are pictorial interpretations of those formal descriptions used when formalism turns too hard/complex (physical significance). I noticed that it is fashion to say that one get the true understanding of phenomena when one get the physical significance. I think that one truly understand the phenomena when feels natural the formal description and is also aware that it is just a description.

I'm really not getting why that post didn't answer OP's query.
Nevertheless, I'm posting the following argument as an answer.

Why is the Gibbs free energy change equal to zero at equilibrium?

Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs
Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.
Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,
$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$
where $\Delta n$ is the change in composition of the system.
Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.
$$\Delta_r G= 0\;.$$
Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$
The definition of equilibrium position sheds light on this:
A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .
This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$

First of all, that post didn't answer my query because I was looking to get an intuitive feel for this concept and not the mathematical description. What is the definition of ∆$n$ for a system? Yes, I'm aware if the first derivative test, but I'm afraid I didn't understand much of what came after that in your answer. Okay, it's spontaneous in neither direction, but what does all this have to do with energy? Please try to understand what I am asking; the term is "Reaction Gibbs energy" (please elaborate on that too) so what does it have to do with energy and why is this energy becoming zero?

@KaumudiHarikumar: The 'intuitive feel' of the query lies in that very phrase 'spontaneous in neither direction'. Do you know what spontaneous means? Try to correlate it with the Second Law in the light of Gibbs Free Energy.

I don't think I am putting out something truly novel here either; the answer by @MAFIA36790 and the discussion in the link you yourself post it seem quite satisfactory to me, yet I shall try to put a different spin on things so to speak.
Let me define the chemical potential, as the partial molar Gibbs energy of species "$j$":
$$\mu_j = \left ( \frac{\partial G}{\partial n_j} \right)_{p,T,n'}$$
and for a system of components A, B, etc. the fundamental equation of chemical thermodynamics gives us:
$$\mathrm dG= V~\mathrm dp-S ~\mathrm dT+\Sigma \mu_j~\mathrm dn_j$$
Now, I consider a simple equilibria $$R \rightleftharpoons P $$
and additionally, I introduce a parameter " $\xi$ " called the extent of the reaction (it has the dimensions of amount of substance)
for macroscopic changes, the amount of a component $j$ changes by $\nu_j\Delta\xi$ ($\nu_j$ are the "signed" stoichiometric coefficients; positive for products, and negative for reactants)
Now, if we defined $$\Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} $$
This $\Delta _r G$ still holds the meaning you might already be familiar with, namely:
$$\Delta _r G = \Delta _f G\textrm{ (products)} -  \Delta _f G\textrm{ (reactants)}$$
and I have nearly extended it to mean the derivative defined above so that i can relate it to the extent of a reaction (measured in amounts of substance consumed/produced)
and at constant temperature and pressure, from the second equation in this post:
$$ \mathrm dG = (\mu_P - \mu_R)~ \mathrm d\xi$$
so, $$ \Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} = (\mu_P - \mu_R)$$
Now, as already mentioned, if we were to plot the reaction Gibbs energy as a function of the extent of the reaction, a minimum on the plot would correspond to the first derivative defined above being equal to zero. (or, equivalently in the chemical potential of reactants and products being exactly equal). This is the condition for equilibrium.
Perhaps, stooping down to layman terminology (caution), one can say $\Delta _r G $ is like a "force" that drives a reaction forward, and the system is in equilibrium when this driving force is zero (and moves in the reverse direction when it changes sign).
An appropriate mechanical analogy could perhaps be two weights on either end of a string wrapped across a pulley. If the heavier weight is higher, it would descend raising the lower weight. The system would come to rest when they are in balance.

Thank you, but I have only just graduated high school and I'm not aware of some of the terms that you(and others in the post that I have linked in my question)have used. Nevertheless, I will try to understand what you have said as best I can. At my level, we aren't introduced to many important concepts, which poses a problem for understanding clearly. One last doubt; how can you use the analogy of a seesaw like static mechanism to help explain a dynamic equilibrium? When you say that the driving force is zero, I don't understand this because like I just said, isn't equilibrium dynamic?

It's not a perfect analogy, I admit. Which is have included (a literal) word of caution. If you want to improve the mechanical analogy, then imagine perturbing it equilibrium perhaps? Also, maybe at your level then you could just take Gibbs energy = 0 as a definition of equilibrium?

Yeah, well, this is the part that is very annoying; when I try to understand more but a barrier of some advanced science props up. I only wanted to know why "the ability of the system to do non-expansional work"(for this is how it is defined for us)reduces to zero at equilibrium and in neither of your answers have I found anything useful. Sigh. I guess that I will have to accept the fact that there is no way for me to understand it without understanding the advanced terms and must resort to not knowing for the time being :(


text:
What is active mass?
I'm aware of the fact that active mass is defined as the molarity of a substance, but my textbook states that

"Active masses are dimensionless quantities but for our purposes we generally take them with dimensions of molarity, partial pressures etc."

What point is my textbook trying to get across to me? Why do active masses have no dimensions? Why have we defined this quantity in the first place, if we use it interchangeably with the term "molarity"?
Edit: Please keep in mind that I have only just graduated high school and would be very grateful if you could explain this concept to me in simple terms.

It sounds like the book is focussing on the ratios of molecules. This is the same idea as moles or partial pressures in a gas: they measure the ratios of the compounds present.

Okay, but why? Can you please elaborate?

Well, I want to mention that activity might be intuitively understood by its usage in radioactivity: the activity of the given substance determines how fast it is disintegrating and how much radiation it is generating, analogous to how "reactive" a particular reagent is in the reaction. However, I am not sure if activity and active mass are the same thing...

The term "active mass" is a historical term.
The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage.  The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically.
Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques.

l'on peut choisir un volume arbitraire, par exemple un centimètre cube. Les
   quantités d'un corps qui se trouvent dans lcc du volume total se nomment la
  masse active du corps.

or roughly translated:

One can choose an arbitrary volume, for example a cubic centimeter. The
  amounts of a body located in 1cc of total volume are called the active mass of the body.

And as stated in The Laws of Reaction Rates and
of Equilibrium J. Chem. Ed. 1956, 33 (4), p 178 :

Guldberg and Waage in their study of chemical
  affinity formulated the law of the rate of chemical reactions
  as proportional to the "active mass" of the reagents.
  This they called the law of mass action and
  very clearly defined "active mass" as amount per unit
  volume. (emphasis added)

So, from a historical point of view, it is false that "active mass" is dimensionless.
However, in relating Gibbs energy to the equilibrium constant, it is necessary for the equilibrium constant to be dimensionless, because only the logarithm of a dimensionless quantity is logical.  For this reason, the text mentioned in the OP redefines "active mass" to be dimensionless.

Sir, in my study material, it's written that, active is actually a dimensionless quantity, and for our convenience and purposes we take them with the dimensions of molarity, partial pressure, etc.

@AdeshTamrakar "activity" is dimensionless.  "active mass" is a historical term that professional chemists do not use anymore.  The correct historical meaning of "active mass" is mass per volume, such as mass per cubic centimeter.  The term "active mass" is older than the term "molarity".

Active mass implies that amount of mass which is taking part in a reaction, that it is the effective concentration of a substance.
It's other names are reactive mass or chemical activity.
Units: The units of activity are nominal, rather than real, because chemical activity is formally defined as the ratio of the actual chemical activity of a substance to its chemical activity under some defined standard conditions, and ratios have no units because the units divide out. Or simply it's the ratio of concentration at required conditions to standard condition.The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of an ion is particularly influenced by its surroundings.
For example : with the hydrogen ion, we use pH = − log10 [H+ ] where [H+ ] is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).
Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).
In these circumstances we talk about ‘activity’, which is a way to correct for this. Basically it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words it uses a correction factor for the simplified laws.
Also Wikipedia says:

In a solution of potassium hydrogen iodate [KH(IO3)2] at 0.02 M the activity is 40% lower than the calculated hydrogen ion concentration, resulting in a much higher pH than expected

I'm afraid I don't understand how taking the ratio functions as a correction factor. Additionally, when using "activity" in the expression for equilibrium constant, the final value does end up having some units. How is this so? Why do we settle for using molarity in place of activity? Can you please further elaborate on all this?

@JM97 I don't think "active mass" and "chemical activity" are interchangeable terms. The idea behind them is the same though, I agree.

@KaumudiHarikumar the equilibrium constant is dimensionless

@KaumudiHarikumar "molarity" is used because it is a readily measurable physical thing, and using it is usually approximately correct.  "activity" is artificially defined mathematically to make the equilibrium constant exactly constant for all concentrations https://www.iupac.org/goldbook/A00115.pdf  http://goldbook.iupac.org/S05915.html It is as if we say, for all people (body weight)/(brain weight) = 27, and when we realize it isn't exactly true, we define something called pseudoweight to make the equation always be true.

Um, OK. I don't know about pseudoweight though.

Part of this answer appears to be a verbatim copy of http://www.3rd1000.com/chem101/chem102q.htm. Please add references if citing external sources.


text:
Why do Ionic Bonds exist?
This is a legitamate question.  If one atom wants to lose an electron and another wants to gain an electron why don't they just transfer the electrons and just part ways?

Think about electrostatic forces

To emphasize/clarify @getafix, why doesn't the electron on a hydrogen atom just 'part ways' with the proton?

It really depends if you're questioning about a solid where ionic bonds do exist, or what happens when such a salt is dissolved where ionic bonds don't exist.

To add on to @Jayanth Psy 's answer, ypu should also consider ionization energy and electron affinity. In the case of NaCl, sodium can't just lose its valence electron, it must be excited into ionisation (because of effective nuclear charge) and the electron must transfer to the chlorine. So you can't just transfer electrons because of this, without having some sort of electrostatic attraction and energy flow.

If a Na+ ion and Cl- ion exist as NaCl,(ionic) it doesn't mean that, that particular Na donated electron to Cl. Na may lose electron somewhere else and Cl may gain electron somewhwere else.
The main point of existence as Na+Cl-, ie NaCl is the electrostatic attraction between them.

"electrostatic attraction" is the gist, but in a solid crystal of $\ce{NaCl}$ there are six $\ce{Cl^{1-}}$ ions the same distance from the $\ce{Na^{1+}}$ ion. So there isn't a "particular" ionic bond. Also a particular $\ce{Na^{1+}}$ ion is attracted by all the $\ce{Cl^{1-}}$ ions in the solid crystal and repelled by all the other $\ce{Na^{1+}}$ ions. Overall there is a net attraction which is why a crystal of $\ce{NaCl}$ stays together.

Ionic bond is not a bond in real. It’s just a force attraction between two ions placed closely. It’s something similar to the van der Waals’, but it has a lot of physical significance. Many have this query and I have seen a lot troubling on ionic bond. So no transfer or such thing takes place. Only attraction to hold them tight. This is all about ionic bond.


text:
Can Fe(CO)5 adopt both square pyramidal and trigonal bipyramidal geometries?
I know the hybridization of $\ce{Fe(CO)5}$ is $\mathrm{dsp^3}$. According to my book, coordination compounds with coordination number 5 can interchange between square pyramidal and trigonal bipyramidal geometries. Is it true for this compound? Doesn't seem so according to the picture:

Was my understanding wrong somewhere?

The orbitals used in $\ce{[Fe(CO)5]}$ are 3d, 4s and 4p. If that makes it an $\mathrm{dsp^3}$ complex in your book then so be it. But then consider that orbitals such as 4p and 4d are energetically too far apart to take part in bonding together in a noteworthy manner.

Obligatory link to Why is it wrong to use the concept of "hybridization" while studying complexes?

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...).
Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, though MO theory and crystal field theory can help understand which might be more favourable.

The structure of iron pentacarbonyl is widely agreed to be trigonal bipyramidal, with three CO ligands occupying equatorial positions, and two CO ligands occupying axial positions.

Trigonal bipyramidal structure of iron pentacarbonyl

Whilst both geometries are possible (and indeed observed, IF5 is square pyramidal), I suspect that your confusion arises from the fact that iron pentacarbonyl is able to undergo a phenomenon known as Berry psuedorotation.

Mechanism of Berry psuedorotation

Berry psuedoroation is the process in which two of the equatorial ligands are switched out for the two axial ligands via a square planar intermediate. This process is sufficiently rapid that if observed by NMR, only one environment for the carbon of the CO ligand is observed (i.e. the psuedorotation is faster than the NMR timescale).

No.Usually dsp3 refers to inner orbital complex while the latter refers to outer orbital complex..Maybe you follow a different convention?

I've never known there to be a difference other than the way people choose to write it, but I'm just an organic chemist, not an inorganiker

@MathJack A hybridisation scheme involving s-, p-, and d-type orbitals is garbage. They are far too much apart to considerably interact with each other.

@NotEvans. As far as i understand, IF5 has sp3d2 hyb. Hence the comparison between IF5 and Fe(CO)5, which I think is dsp3, does not really make sense (as in your answer). The square pyramidal structure of IF5 is very easily explained using VSEPR Theory. Moreover, I don't think that IF5 can show trigonal bipyramidal geometry.

This distinction between "sp3d" and "dsp3" is a relic of an incorrect and flawed theory.

The most recent single crystal structure investigation (high precision, $R_1 = 1.9\%$, $wR_2 = 5.7\%$) of molecular iron pentacarbonyl [1] suggests trigonal bipyramidal coordination environment (also mentioning anti-Berry pseudorotation), as shown by the calculated geometry index $\tau_5$:
\begin{align}
\begin{cases}
\tau_5 &= 0 \qquad &\text{square pyramidal geometry} \\
\tau_5 &= 1 \qquad &\text{trigonal bipyramidal geometry}
\end{cases}
\end{align}
$$\tau_5 = \frac{\beta - \alpha}{60^\circ} = \frac{178.93^\circ - 121.13^\circ}{60^\circ} = 0.96$$
where $\alpha$ and $\beta$ - two greatest valence angles of the iron coordination center calculated from the CIF-file (ICSD #74197).
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#E06633}{\Large\bullet}~\ce{Fe}$.

The magnitude of the $\ce{Fe-C}$ axial vs equatorial band length distortion ($0.44\%$) is notably less than in an alogous p-block trigonal-bipyramidal species (e.g. in $\ce{PF5}$ $\ce{P-F_{ax}} = \pu{1.577 Å}$ and $\ce{P-F_{eq}} = \pu{1.534 Å}$; i.e. distortion $2.8\%$, by gas-phase electron diffraction).
There is a small angular distortion of the trigonal bipyramid in crystalline $\ce{Fe(CO)6}$ with the axial ligands closing on the unique carbonyl ($\ce{C(1)-Fe-C(2)} = 89.47(5)^\circ$, $\ce{C(2)-Fe-C(2a)} = 178.94(10)^\circ$), two equatorial angles opened from their ideal values ($\ce{C(3)-Fe-C(1)} = 121.11(5)^\circ$), and the third slightly closed ($\ce{C(1)-Fe-C(1a)} = 117.8(1)^\circ$). This distortion is of an anti-Berry pseudorotation type, i.e. in the direction away from a square-based pyramidal geometry in which $\ce{C(1)-Fe-C(2)} = \ce{C(1)-Fe-C(3)} = ca. 105^\circ$.

Bibliography

Braga, D.; Grepioni, F.; Orpen, A. G. Organometallics 1993, 12 (4), 1481–1483. DOI 10.1021/om00028a082.

Here the hybridisation is dsp3 in $\ce{Fe(CO)5}$. It is basically an inner orbital complex. So, it is a spin-free orbital. The hybridisation plays a major role in determining the shape in these cases. So, when the hybridisation is dsp3 in most of the cases you get the shape as trigonal bipyramidal, whereas in sp3d we get square pyramidal.

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


text:
Voltammetry on homemade supercapacitor
I would like to get some voltammetry curves with my homemade supercapacitor in order to be sure that I have only capacitive effects (capacitive current) in my cell and no redox reactions (faradic current).
One of my electrode is my carbon material and the other one is stainless metal.
I only know the 3 electrodes system, do I really need it in this case ? How should I process for my experiment ?

What equipment have you got?

Testing capacitors is almost always done with a two electrode setup.
There are some details here about interpreting the data, but the experimental setup is that you just connect the counter and reference electrode inputs together to one half of the capacitor and the working electrode input to the other.
A three electrode setup is used when we only care about the interfacial potential of a single electrode (the working), and thus use a measurement near it by the reference electrode to control the counter electrode accordingly.
In a capacitor, we are generally interested in controlling the voltage across the entire device and observing its behaviour as a whole, so we don't need a reference electrode. (though depending on your specific device, it may be useful to observe only one electrode for any number of reasons)


text:
Do ligands in ionisation isomerism interchange?
The text says:

This form of isomerism arises when the counter ion in a complex
  salt is itself a potential ligand and can displace ligand which can
  then become the counter ion.

For other structural isomers that I studied in coordination compounds I simply thought that those  compounds (structural isomers) separately exist. In fact I think of structural isomers in coordination chemistry this way only that these are independently existing with different properties but same molecular formula so we study them.
But the text above is confusing it seems that they convert like resonating structures or conformation isomers.
What is true? If they do interchange then isn't this a contradiction to Werner's theory that postulates that secondary valencies are non ionisable.
note:the below problems is related to one of the comments:

Please Consider this problem-"Give evidences that [Co(NH3)5Cl]So4 and [Co(NH3)5 So4]Cl are ionisation isomers".

Here we test by dissloving them in water and reacting them separately
  with silver and barium ions  to test for counter ions.One gives
  AgCl and other BaSo4 as ppt. .Now if there is equlibrium like keto-
  enol tautomerism then only one of the two isomers should be stable in water
  then why such kinds of evidences are given?

the counter will become ligand by donating its pair of electrons

@JM97 But since coordination number is fixed the existing ligand need be ionised for counter ion to become ligand

ligand exchange will not need ionisation of existing ligand first. Counter ion will attack the metal atom and simultaneously existing ligand will leave the cordination sphere. Similar to what happens in SN2 reaction. They would be equilibrium like in keto-enol tautomerism

http://chem.libretexts.org/Core/Inorganic_Chemistry/Organometallic_Chemistry/Fundamentals/Ligand_substitution

@JM97 Please Consider this problem-"Give evidences that [Co(NH3)5Cl]So4 and [Co(NH3)5 So4]So4 are ionisation isomers"

http://www.chemguide.co.uk/inorganic/complexions/ligandexch.html

Actually, in the ionisation isomerism the ligands do not interchange. They get displaced. Thus one ion becomes counter ion for the other. When one has look at the molecular formula it seems the ligands in the coordination sphere are interchanged with the ligands outside the sphere. So, this does not cross out Werner’s theory as there is displacement of ion and not just interchange.

Please fix that caps lock and use punctuation marks so your answer is intelligible. Recommending deletion since this is incomprehensible.


text:
Ammonia (NH3) and nitric acid (HNO3)
I understand that by adding Nitrogen to my water my Lucerne plants will grow better. Firtsly, is this correct? Secondly, if so, I further understand that I must add/mix Ammonia (NH3) and nitric acid (HNO3) to form nitrogen which is added to the water, is this correct? However I understand that these two (Ammonia (NH3) and nitric acid (HNO3)) is highly corrosive, therefore would this corrode my irrigation system or damage my lucerne crops, rather than increase the growth? Help will be appreciated!

Are you kidding? You simply buy NH4NO3 or other saltpeter.

Ammonia and nitric acid can be considered corrosive materials (though the real question would be "corrosive towards which material?"). The thing here is that ammonia and nitric acid are corrosive for strictly opposite reasons: ammonia is a base and nitric acid is an acid. So in a mixture of 1 equivalent(1) of ammonia and 1 equivalent(1) of nitric acid, both effects will cancel and the resulting mixture is no longer corrosive.
You do not create nitrogen (with such a mixture) for your lucerne, but you provide it with a form of nitrogen(2) which can be transformed by bacteria before being absorbed by your plant. Though, you do not want to mix ammonia and nitric acid because it is dangerous. As Mithoron said, you would better buy the product.
The chemistry and biology of soils are extensively (?) studied nowadays because it appears that monoculture exhausts soils and an agroecological approach can drastically outperform the monocultures we employ in western world.

(1) This requires a bit of calculations.
(2) Use of the term "nitrogen" can be very misleading as it can describe the "nitrogen gas" which composes 80% of our atmosphere (but is quite inert) or any source of nitrogen atoms for a plant.

Upon heating, $\ce{NH4NO3}$ can degrade to gaseous nitrogen. $\ce{NH4NO2}$ does that as soon as the ions meet in solution, I was told in my introductory inorganic chem class.

@Jan, But according to wikipedia, ammonium nitrate when heated, forms nitrous oxide and water.

@NilayGhosh Whoops, I think you’re correct. But the nitrite part of the comment should hold true.


text:
Temperature as an independent variable in an endothermic reaction
My experiment involves the reaction between sodium bicarbonate and acetic acid. I'll be changing the temperature between each trial and measuring the change in pressure by the gas produced. From here I hope to determine the rate expression, rate constant, activation energy then finally produce an Arrhenius plot.
My question is since this reaction is endothermic, how do I account for the change in temperature as the reaction proceeds? I also can't seem to find any published literature value for this reaction's activation energy or enthalpy change, and would love it if someone pointed out a database or something similar.

For this experiment to work, you need good control of your temperature profile.  In other words the temperature must be somewhat uniform.
If this is the case measure two temperatures.

$T_e$ is the isothermal temperature of the experiment.
$T_r$  is the temperature at the point of reaction.

If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 9°C, 18°C, and 29°C  then the temperature change can be assumed to have a negligible effect. (You would want to use $T_r$ in the kinetics calculations.)
If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 4°C, 6°C, and 8°C  then the effect is not negligible.  Accounting for the endothermic temperature change in the calculations is a poor option because it would add needless complication. A better approach is to keep things simple by revisiting the controls in the experimental design to see what can be done to minimize the endothermic effect.  One possibly would be using less reactants.

And does there happen to be some sort of standard for determining a negligible effect?

This is an experimental statistics question, not really a chemistry question.  I  suggest posting to cross-validated http://stats.stackexchange.com/ so an answer comes from the right type of expert.

If it were me, I would replicate each test 3 times.  Take the standard deviation at each temperature.  If 5x the standard deviation is less than the differences between $T_e$ then the rule of thumb is the effect is significant.   ... In practice it is customary to consult with a statistician on experimental design.  A chemist can tell you if the theory is sound.


text:
Why does the ionization energy decrease anytime the atom size increases?
I would like to have a better understanding of the relationship that exists between the Ionization Energy commonly known as IE and the size of an atom.
I think that the Ionization Energy can be defined as the smallest energy that is required to remove an electron from a gas phase atom.
Why does the Ionization Energy decreases when the atom size increases?

This is because as the size of atom increased the electrostatic force between nucleus and electron become weaker, thus less amount of energy is required to remove the electron from atom in a gaseous state.

The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.
There are certain factors that have an impact on the ionization energy value.
To understand the answer to your question, you must understand the effect of all three

Atomic radius (Inversely proportional to I.E. Value)
Shielding effect of electrons (Inversely Proportional to I.E. Value)
Nuclear Charge (Proportional to I.E. Value)

The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question) Even the smallest change in this distance will cause a change in ionization energy for example The first ionization energy of Magnesium and Aluminum. The ionization energy should increase as the nuclear charge increases across the period but the I.E. of Aluminum is lesser than that of magnesium because there is a change in orbital which increases atomic radius, decreasing I.E.

Now going for a bit more information.
The Shielding effect of the other electrons decreases the I.E. because it doesn't let the outer most electron feel the full force of attraction.
IF both the first factors are CONSTANT (atomic radius and shielding effect of electron) only then the last factor comes in play.
As the atomic charge/Nuclear charge increases, there is a general increase in ionization energy because of the increased force of attraction by the extra proton. for example, again across the period, the general trend is increase in ionization energy because there is no (significant) increase in atomic radius or shielding effect so the I.E. increases proportional to the Nuclear charge.
(the decrease in sulfur is because of the pair of electron in the p orbital repelling each other. Both electrons apply a force on each other. The looser of the two, is removed when energy is applied.)

Electrons are in general indistinguishable, hence the IE is the energy required to remove any electron. The atomic radius is also not the distance between the nucleus and an electron, but the probability cutoff of finding one. While the ad-hoc explanation is quite common and very graphic one should be aware, that it is a simplification and that the ionisation energy is only truly understood when applying quantum mechanics. However, your answer is still good enough for a start, hence +1.

@Martin Any chance of adding this into an answer yourself? The comment looks like the kernel of one already. This is the kind of question where readers surely benefit from seeing answers written at more than one level.

The electric field around an atom with one electron removed is that of a single positive charge.  The work done, to lift an electron away from a small radius to infinity, is greater than the work done to lift an electron from a large radius to infinity.
That work is the ionization energy.   How could it NOT go down for
larger atomic radii?


text:
Are there processes to standardize the shape of a crystal during crystallization?
I am trying to standardize the shape of europium dibenzoylmethide triethylammonium (EuD4TEA) during its crystallization process. I'm a mechanical engineering student and am not too familiar on the details of crystallization. Is there a process for creating crystals of standard size, shape, and weight? If so, would the process vary between what I am using and any other type of crystal?

All you can do is to try to create reproducible conditions. Crystals can generally take a multitude of shapes based on all sorts of factors - concentration, temperature, pressure, flow rates, other stuff in solution and so on.

First of all, please specify what size of crystals you are interested in: micrometer size? Centimeter? Smaller crystals with uniform size are much easier to achieve than large ones.

Thanks @MaxW! @vapid I'm interested more in centimeter/millimeter sized crystals.


text:
Addition of inert gas at "constant volume"?
My textbook states that

"The addition of inert gases at constant volume does not affect the equilibrium state of the reactants and products contained in that volume."

Although I am not questioning the truth behind this statement, I am very confused by it.
Let us consider a homogeneous gaseous equilibrium system contained within a closed cylinder. The cylinder has a very small hole on its side, through which an inert gas can be pumped in. This is how I visualize the addition of an inert gas to the system. Now, upon introducing a given amount of inert gas into the system, how does the volume remain constant in any way?
While deriving Van der Waals equation for real gases, I learned that the term $V$ in the ideal gas equation stands for the empty space that is available to the gas molecules for movement. Therefore, when we introduce an inert gas into the system, aren't we essentially decreasing this amount of space that is available to the reactant and product molecules for movement?
I am very confused. Please share your insights for it would be tremendously helpful for me. Thanks ever so much in advance :) Regards.

The inert gas that you add does not act like a piston, denying part of the volume to the other molecules in the mixture.  For a constant volume system comprised of an ideal gas mixture, the partial pressures of the reactants and products do not change when you add the inert gas.  It only increases the total pressure, and that change is only because of its own partial pressure.  However, if you are not operating in the ideal gas regime, the addition of the inert gas will affect the equilibrium.

OK, so what you're saying is that according to the assumptions of KTG for ideal gases, the volume of the reaction mixture will remain the volume of the container since the molecules of the inert gas have negligible volumes themselves, yeah?

Additionally, if the reaction mixture is at constant pressure instead of constant volume, these inert gas molecules will exert pressure on the piston, moving it up and hence, increasing the volume, yeah? While studying Le Chatelier's principle for changes in volume, I didn't relate the volume change to pressure change. Instead, I imagined that on decreasing the volume, the equilibrium shifted in that direction which produces lesser number of molecules so that the volume would increase and not so pressure would decrease. I am wrong to think of it like this, right?

I agree to your first comment.  I don't understand your second comment.

You should not :)
"does not affect the equilibrium state" means that the thermodynamical constant is still the same. If you add an inert gas $\ce{G}$ in your system it will not react with other chemical coumponds.
For example at the beginning you have $\ce{A} \rightleftharpoons \ce{B}$ and after you add you inert gas you have $\ce{A + G} \rightleftharpoons \ce{B + G}$. So it not afftects the value of $\ce{K_{eq}}$ because $\ce{K_{eq}}$ depends only of $\ce{T}$ the temperature.
If you use a solid container the volume will not change but the pressure will.
Also we have $$\ce{\mu_i(T,P_i)=\mu_i^°(T)}+\ce{RT\ln\frac{P_i}{P^°}}$$
Then $$\left(\frac{\partial\Delta_r\ce{G}}{\partial n_{inert}}\right)_{\ce{T,V}}=0$$
If you prefer calculus.


text:
How long did the most difficult compound to synthesize last?
I heard that the most difficult compound to produce was a molecule that included Argon because it is a noble gas. I was wondering how long it was stable for.

*coughs*Maitotoxin*coughs*

@Jan.. just a few more months ;)

There is no such thing as the most difficult compound to produce. Instead, there are some hundreds of thousands of "stable" compounds of which we know pretty much everything - melting point, boiling point, heat capacity over a wide range of temperatures, refractive index, speed of sound, magnetic susceptibility, electrical conductivity, crystal structure, and many more. Around that field is a much wider "grey zone" of the not-so-stable compounds (millions of them), which miss some of these properties because they can't be measured.
As we reach out to the far outskirts of that zone, we meet more and more compounds with less and less known properties in each one, because they are increasingly frail and short-lived, to the point that the very words "produce" or "synthesize" (or even "exist", for that matter) do not quite apply to them anymore. Some of them last but a fraction of second; some live longer, but require very peculiar conditions. There is absolutely no meaningful way to arrange them in order of increasing "difficulty".
Also, welcome to Chem.SE.


text:
What determines whether iodination or reduction will occur?
Alkyl Iodides are formed by reaction of alcohols with iodine and red phosphorus.  I already am well aware that this forms $\ce{PI3}$ in situ, which acts similarly to $\ce{PBr3}$ and allows alkyl iodides to be prepared from alcohols.
However, when the same reaction is performed on ephedrine, the hydroxyl group is reduced rather than iodinated.  What determimes whether iodination or reduction occurs?
I have a few theories, one being temperature may be kept lower to form the alkyl iodide.  Or is it to do with the benzylic position of the hydroxyl group?  Or perhaps the red phosphorus is in excess to favor reduction?  Any incite will be helpful.

You might want to look at http://chemistry.stackexchange.com/questions/8877/what-are-the-mechanisms-for-the-reduction-of-organic-compounds-employing-red-pho

@vapid Thanks!  I don't know why that didn't show up when I googled it.

see this:- http://chemistry.stackexchange.com/questions/38513/why-pi3-is-not-used-in-hvz-reaction


text:
How does Fick's Law apply to complex diffusion and diffusion within living cells? How can Fick's Law be modified for unsteady-states?
I am trying to understand how diffusion in an unsteady system such as a living cell occurs and if Fick's law can be modified to predict the change in the concentration in the cell over time.
Fick's Law is used to determine the diffusion flux to the concentration in steady state systems and is represented by the equation
$$ J = - D \frac{\partial\varphi}{\partial x} $$
where $J$ is the diffusion flux, $D$ is the diffusion coefficient, $x$ is the position/dimension of length, and $\varphi$ is the concentration.
How can Fick's Law and Fick's second law be used to determine how the concentration of a substance within a living cell will change over time? Is this possible? Or is there another equation used in a case like this?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with this markup converting your graphic. If you want to know more, please also have a look  here. We prefer to not use MathJax in the title field, see here for details.

I think that the question is too broad, the answers is very context dependent.


text:
Why does the barium ion in barium titanate move in its tetragonal crystal structure?
Barium titanate is known in the materials science world to be a great dielectric. This is known to be because the $\ce{Ba^2+}$ ion sits off center in the middle of the tetragonal crystal structure.
How does this $\ce{Ba^2+}$ ion, or even the crystal structure itself, move in order to accompany a larger amount of charge? How much freedom of movement does this ion have in a solid crystalline structure? Additionally, does the $\ce{Ba^2+}$ ion naturally sit off-center in this configuration, or does it only behave that way when introduced to a charge?

"The remarkable properties of this material arise from the cooperative behavior of the Ba2+ centres.[citation needed]" Thanks, Wikipedia! Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

This is known to be because the $\ce{Ba^2+}$ ion sits off center in the middle of the tetragonal crystal structure

First of all this is not correct. It is the $\ce{Ti^4+}$ ion that sits in a tetragonal off center configuration (black atom in perovskite structure below). Also this hold for any material with perovskite structure with the $\ce{ABC3}$ general formula where $\ce B$ is your center site, $\ce A$ is the corner sites and $\ce C$ is your face site counter ions $(\ce{BaTiO3, LaAlO3, NaMgI3, ...})$.

Additionally, does the $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion naturally sit off-center in this configuration, or does it only behave that way when introduced to a charge?

It does sit off center all of the time in any of 6 directions on a $(001)$ plane but the many centers are randomly oriented which produces no net dipole moment. When charge is introduced, the $\ce{Ti^4+}$ are attracted toward the negative plate and will shift toward that plate creating a net dipole moment.

How much freedom of movement does this ion have in a solid crystalline structure?

Very little, the $\ce{Ti^4+}$ is located between the plane of 4 oxygens (red atoms) and an oxygen atom perpendicular to the 4-atom plane. It can move between the 6 possibilities of this shift $(\pm x, \pm y, \pm z)$ but not between cells.

How does this $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion, or even the crystal structure itself, move in order to accompany a larger amount of charge?

When the electric field is applied, the titanium ion receives enough force to squeeze between oxygen atoms to a new lower energy position (e.g. $-x$ to $+y$).


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The effect of pressure on boiling point?
My textbook states the following:

Qualitatively, I understand why the boiling point of a substance increases when the pressure is increased. However, I learned that if the pressure is increased when the reaction system is at equilibrium, the volume of the system decreases and hence, the reaction will proceed in that direction in which the volume of the system increases. In the above scenario, however, it has been given that water in the gaseous form occupies more volume than its liquid form and so, shouldn't the given reaction proceed in the forward direction, hence reducing the boiling point?
Please note that I am convinced and I believe that it makes sense for the boiling point to increase when the pressure increases but can somebody please explain the flaw in my reasoning with regard to the given equilibrium?
Much thanks in advance :) Regards.

what's the reaction in boiling of water?

I'm afraid I don't understand. What do you mean by "reaction of boiling of water"?

There is no reaction in boiling water, why do you apply le chatliers principle?

There is no chemical reaction but there is a physical reaction, no? An equilibrium does exist between the gas phase and the liquid phase of water(or any liquid, for that matter).

at equilibrium ratio of concentration of products to concentration of reactants remain constant. (K_c) . How would you apply this to a physical change?

In this case, we use Kp=Pressure(of water vapour). At equilibrium, this will be equal to the vapor pressure of the water at that temperature.

Yeah then what, As you decrease volume partial pressure of water increases and to maintain same K_c the partial pressure of water has to decrease this is by its dissolution into liquid water.

The equilibrium will shift in a direction in which the pressure decreases, yes? So it shifts in the direction of the liquid phase for which pressure is...not defined?

Your flaw in the logic lies in that you think that the volume of the system includes the liquid. Le Chatelier's principle on pressure equilibrium only applies to when the compound involved is in gaseous phase, in an ideal sense. Therefore if you increase the pressure (decreasing the volume), then the reaction will progress towards the direction where gas volume DECREASES, which is towards liquid phase. Overall, the entire apparatus increases the vapor pressure on the gas (gas pressure on the external pressure), which increases the boiling point.

OK, but on decreasing volume, why does the reaction progress in that direction in which the volume further decreases?

Le Chatelier's principle states that a reaction that undergoes a change in pressure, temperature, or volume, the equilibrium shifts towards the direction of the reaction that counteracts and subsists with that change. In this case, the change in volume causes an equilibrium shift that subsists by decreasing the volume

The shift in equilibrium should be such that it counteracts the stress(change) applied, yes? So why, on decreasing the volume, will the equilibrium shift in a direction in which the volume further decreases?!

Decreasing the volume increases the pressure. The volume decreases because the pressure increase causes the gas to contract. Look at this link for more info: https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle#Effect_of_change_in_pressure

So if I decrease the pressure, the volume increases and the reaction will shift in that direction in which the volume further increases? Seems very absurd sir/ma'am. I think it should shift in that direction in which the volume decreases, causing the pressure to increase.

I really don't know what to say. Its Le Chateliers principle.

But it's not! Please read my comment again; doesn't it seem like the situation that I'm describing is the exact opposite of the outcome given by the principle?!

Can I get some help here? The volume of the system DECREASES, which means the systems gas' pressure MUST INCREASE. To counteract this, aka reduce the increased pressure, the volume of the GAS mudt decrease

What are you even saying?! "To counteract the increased pressure, the volume of the gas must decrease?!" However are you counteracting the change if you're decreasing the volume even further?! $P$ is inversely proportional to $V$!

Please calm down. The changes in VOLUME is not the actual thing that affects the equilibrium. The change in VOLUME, changes the pressure to be HIGHER, which is what actually shifts the equilibrium. Le Chatelier's principle is about how external forces affect the equilibrium point.

OK, sorry (:P). Since the decreased volume increases the pressure of the gas, according to Le Chatelier's principle, the equilibrium tends to shift in that direction in which its pressure decreases, yes? For a decrease in pressure, the volume increases.  Hence, on decreasing the volume of a gas mixture at equilibrium, it tends to shift in that direction in which its volume increases.

It's because when we increase pressure, we are actually increasing external pressure so according to Le Chatelier's principle, to counteract the effect it should increase the internal pressure. This is the reason that the reaction shifts where internal volume decreases, so that internal pressure increases.

Could you please edit your post so that you are usually complete words? It is difficult to understand the content of your answer with the incomplete words.


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Oxidation state of fluorine in HOF
What should be the oxidation state of $\ce{F}$ in $\ce{HOF}$.
As fluorine is the most electronegative element in the periodic table, it should be $-1$. But when I googled it, I found that many sources say it is $+1$ based on the fact that it is an oxidising agent.
Is there any theoretical way in which we can predict its oxidation no. or at least explain this anomaly.

Any compound involving both oxygen and fluorine is likely to be an exception to the normal rules of thumb — for everything, not just oxidation states.

Also, keep in mind that "oxidation state" of anything in a covalent molecule is merely an approximation to what is really going on.  I'd expect the electron distribution in $\ce{HOF}$ to be basically that the O and the F have ganged up on the H, stolen its electron, and are sharing it between them with some, but not a lot, of polarization toward the F — on the order of $\ce{H+}$, $\ce{O^{-0.4}}$, $\ce{F^{-0.6}}$.

Related: What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

While I agree with the outcome of Jan's assignment of oxidation numbers, it is very necessary to point out that this is not the only possible assignment, (probably still) not even the most common one.
I also agree that sources on the Internet can be wrong, but at the time Jan was writing his answer, the most authoritative source, the IUPAC themselves, was wrong in their definition. In 2016 the official rules changed, which I have discussed in an answer to Electronegativity Considerations in Assigning Oxidation States.
Prior to 2016 the official rules of the IUPAC indeed assign $\mathbf{+1}$ to fluorine. This is counter-intuitive, as you stated, but that is the way the strict application of the rules give it to us. For me this is one of the most disturbing flaws of these rules. What you called an anomaly is indeed due to an incomplete set of rules.
I have laid out the different possible assignments in my answer to What should be the oxidation state of oxygen in HOF (hypofluorous acid)?.
Luckily this flaw has (recently) been addressed in the new more comprehensive definition. In brief the it is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.
With this you would indeed assign $-1$ to fluorine, $+1$ to hydrogen, and $0$ to oxygen.
Nevertheless, one should be aware, that there are two systems out there, which produce contradictory results. I expect the old rules to be present in popular text books for quite a while, so this answer may serve as a reference for the issue.
A general flaw of all systems assigning oxidation numbers is that there is not really a correct way of doing it. This is especially true if you try to  consider fractional oxidation numbers, which are still not really covered in the new definition. Oxidation states/numbers are nothing you can measure, hence there is also no unique way of correctly predicting them. They were introduced as a tool to better understand and keep track of the changes in the charge density. They usually do not come close to measured or calculated partial charges and they often oversimplify bonding situations of molecules.
Does that mean we should abandon the concept? No. It is a very helpful tool, especially when considering redox reactions. It gives you a formal framework to track a reaction and to make simple predictions. With the newer, much more comprehensive framework, the assignment became less ambiguous and more in line with chemical intuition.

@OscarLanzi Is this kind of tone your comment tries to convey really necessary? I am sorry that my posts don't automatically update as soon as a rule changes, I was quoting those still in place in 2014 (see https://web.archive.org/web/20141102154648/http://goldbook.iupac.org/O04365.html), and considering those rules, as I have reported, the assignment is +1 for fluorine. The newer definition doesn't make this ridiculous assignment anymore, so I guess I should update all those posts, but I actually won't have time for that.

@OscarLanzi I don't think this answer is anywhere close to obsolete, as with many outdated concepts, they will linger in textbooks for probably more than a decade to come. Additionally, I have already stated that I should update all posts that are referencing the old rules, but that will take time. Time that I am currently neither able nor willing to invest. So, it seems like you still have all the other usual tools at your disposal to make your objection heard...

@oscar hopefully you are now satisfied with this version of the answer.

There is no anomaly in here. Fluorine’s oxidation state in $\ce{HOF}$ is $\mathrm{-I}$ as the theory says. Hydrogen’s is $\mathrm{+I}$. This leaves oxygen with an oxidation state of $\mathrm{\pm 0}$.
That final fact is where the oxidative power comes from. Oxygen, the second-most electronegative element in the periodic table, has the same oxidation state as in molecular form. It is also bonded to an even more electronegative fluorine. Even in hydrogen peroxide oxygen would me more reduced. Thus, any reaction of $\ce{HOF}$ involves oxygen gaining electrons, not fluorine.
Remember that many sources on the internet can be wrong.

thanks for the explanation. Can you show any reaction of HOF where it acts as an oxidising agent so that the oxidation states might be justified?

Well... the official instructions of the all-mighty IUPAC do indeed dictate an oxidation state of +1 for fluorine. Only the extended version (see here) will come to the more logical conclusion that you are reaching, which is technically not the common convention.

@Martin-マーチン There are instances when I don’t give a damn about what IUPAC says — e.g, they also propose double bonds in sulphones.

I think I made it clear what I think about these rules, but saying that sources on the internet may be wrong is absolutely misleading here. Especially because the common convention is the one proposed by IUPAC and found in almost all textbooks. And in that set of rules, fluorine's oxidation number is +1.

You might be happy/interested to see that the IUPAC has updated their definition... but then again, you might be completely ambivalent, because you hate them.

@Martin-マーチン I don’t hate them, I just run every recommendation or formalism through my critical mind to decide whether I agree or not ;) If they updated and now fluorine is $-1$ I highly approve.

Yep. Fluorine is -1, oxygen 0 and hydrogen +1. The sum of all oxidation states for a neutral molecule has to equal 0.


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Gaussian: Warning - assumption of classical behavior for rotation may cause significant error
I am playing around with a prostaglandin molecule (23 heavy atoms) in Gaussian09. My goal is to calculate the vertical ionization potential and electron affinity for the molecule. I already managed to converge the geometry of a neutral molecule (with the invaluable help of @Wildcat and Greg at this forum) using Opt=Tight and Integral(Grid=Ultrafine) as well as Opt(Maxcycle=1000) and SCF(Maxcycle=1000). In the log file I got the long-awaited line "Stationary point found" :) However there are also two warning messages there that make me worry a bit:
1. Warning -- assumption of classical behavior for rotation may cause significant error
2. Warning -- explicit consideration of  37 degrees of freedom as vibrations may cause significant error
Now, the zero-point energy that I am interested in has been reported as:

SCF Done:  E(RB3LYP) =  -1004.62842050     A.U. after   14 cycles

However, right below the warning messages I found something like:

Zero-point correction=                           0.442805
  (Hartree/Particle)
Sum of electronic and zero-point Energies=          -1004.185615

Now, my specific questions are:

Should I worry too much about these warnings if my resultant Ionization Potentials (IP) and Electron Affinities (EA) look reasonable?
Should I do my IP and EA calculations taking the Zero-point correction (see above) into account and use the value of -1004.185615 as my "true" zero-point energy?
What can I do to eliminate the warning messages mentioned above? I read something about a keyword "freq=HinderedRotor" that could be used in such a case.

I am eager to know how to interpret this whole mess and how to proceed. Thank you for any help you may offer!

Which basis set are you using? What do you mean by "true"? Maybe you can retain only the electronic contribution

freq b3lyp/6-31g(d) geom=connectivity scf(maxcycle=1000) integral(grid=ultrafine)

Well, I mean "true" as closer to real zero-point energy that I want to assess for IP and EA calculation. Sorry if I cannot make myself clear enough. I am new to the field of computational chemistry.

Given the inaccuracy of the method, I would go for neglecting the vibrational rotational parts.

It is ok. I find hard to justify the usage of these contribution to energy if there are few confidente of their accuracy. Most of they will cancel out.

Dear @user1420303,

Dear @user1420303. What specifically do you mean by "inaccuracy of the method"? Is my basis a kind of lousy choice? :) I have gone through a few papers in the field of computational chemistry and it looks like people use it quite a bit, don't they? Thanks a lot for guiding me here!

B3LYP/6-31g(d)  is ok for many applications. Sometimes it returns in large errors but its usage is a reasonable because it provides good balance cost-benefit. Giving the size of your molecule I think that no one will complain.
The basis set is large of small depending from the place you come. I heard people that commonly uses semiempirical methods and "feels"   B3LYP/6-31g(d)  like a very accurate method.  People trying to reach chemical accuracy (using normally 5$\zeta$ basis sets and high correlated wave function based method will perceive it as very inaccurate or unreliable.

B3LYP is by far the more used method/functional out there. Pople's basis sets are also very used (years ago almost the only one choice).  So,  you are playing safe. But do not expect to obtain very accurate results for energy differences nor hessians. So, if you have additional inaccuracies because of the assumptions for the statistical physical treatment are not well satisfied, then maybe it is better neglet those contributions. Esto hubiese sido más fácil en español.

Cool @user1420303! It is very encouraging to hear from an expert who you obviously are that I am playing it safe this way. I'll do as you say then and neglect the warning about zero-point energy correction as long as my results seem reasonable enough comparing a bunch of prostaglandin derivatives. Muchas gracias! :)

De nada! Sorry, I tried to say that I you could use the value SCF Done: E(RB3LYP) = -1004.62842050 instead of  -1004.185615. Maybe the best you can do is compare the electronic energy difference with the zero point energy difference. They should be quite similar. You can improve the accuracy of the calculation, but I doubt that it pay the price. I do not know about the application of your calculation but, I guess that you calculated for an isolated molecule at 0K to represent the behavior in the human body (condensed phase at ~310K). If so it is not reasonable to invert too much in accuracy.

Dear @user1420303. Now you got me. Forgive my ignorance, but I just used the default settings of GaussView not even thinking much about the temperature set for the calculation. I kid of thought that it was room temperature 298 K... And you say the default is 0K, right? Well, where can I check the temperature in the LOG file? And second, where can I set it in the GJF file? Well, yes I am interested in the biological system, and specifically the human body so ~310K in the lipid membrane. So what do you think, what would be the optimal settings for this kind of calculation?

Yes, it is at 0K as you are solving searching for the minimum energy state. At higher temperatures excited states become available. The zero point "correction" is the lower energy due to vibrations that the molecule can have (assuming harmonic oscillations for nuclei), so also implies 0K.
You can not check, it must be at 0K. Except in the thermodynamic analysis that  is done at 298.15K by default. It is indicated in this part of the output file.
I never calculated energies to be used in biological analysis. But I am pretty sure that due to the difficulties above, you can only ...

expect to get a qualitative behavior. To that end, I suspect that your method reasonable (that is, it should not be necessary the minor corrections by temperature, differences in basis set and even changes in vibrational energy)


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Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene?
Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene?  Shouldn't the former have a higher boiling point because it possesses a dipole moment? I think the van der Waals forces should be equal in both cases.

Related: Why is the melting point of p-Dichlorobenzene higher than those of o-Dichlorobenzene and m-Dichlorobenzene?

I wonder if pi-stacking is in play here...

The difference between the boiling points is small: ortho 180 °C, meta 172 °C, para 174 °C, so they are basically the same. It is just not possible to argue bigger or smaller with hand-waving arguments about dipoles or polarity etc. with something so complicated. However, when the thermal energy is increased, whatever intermolecular interactions exist become less important relative to the average thermal energy, and which now has a dominating and disruptive effect. Thus I would expect all isomers to have a similar boiling point.

This is a guess, but maybe the spread of the chlorine atoms, which has high electronegativity, for the para-dichlorobenzene causes it to have a higher local polarity, causing it to pull the electron cloud more towards each chlorine atom, which may cancel out in total for the atom but when regarding it locally, it has a higher polarity, whereas if you place the two chlorine atoms closer together, the dipole spread is actually smaller, therefore causes it to have actually a smaller effect.
You may argue that m-dichlorobenzene has a larger polarity, but I think the effect of the larger dipole moment is smaller than it seems. The atoms is pretty big.

So basically just learn it . When in rome dance like the romans . There is no possible way that YOU can with knowledge of molecular weight , structural configuration , the bond nature , if the electron is electron donating or electron withdrawing , the dipole-dipole or van-der-waal force can identify the boiling points as it is a complex combination of all of these factor no precidence has been identified , no pattern to make it theoretically identifiable has been identify so just sit there and learn it all ?

Disclaimer: I am not sure about the following reasons but they might have a certain amount of merit in them.

p-Dichlorobenzene essentially has no dipole moment whereas the resultant dipole moment vector comes out to be 1.72D in m-dichlorobenzene. Hence certain dipole-dipole interactions are generated in m-dichlorobenzene compared to the London dispersion forces generated in its p- counterpart. It is known that dipole-dipole interactions are stronger than the London dispersion forces thereby making the boiling point of m- dichlorobenzene greater than that of p- dichlorobenzene.

If you are not sure, then it is not an answer to the question.


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Ions Na+ and Cl-
Fundamental question on ions: If, for instance, $\ce{Na}$ and $\ce{Cl}$ react to produce $\ce{NaCl}$ salt, then $\ce{Na^+}$ and $\ce{Cl^-}$ are obviously formed. What I don't understand is, for example, if this compound ($\ce{NaCl}$) is broken down by water, do the $\ce{Na}$ and $\ce{Cl}$ atoms retain their character of ions (remain $\ce{Na+}$ and $\ce{Cl-}$) or do they convert back to neutral $\ce{Na}$ and $\ce{Cl}$ atoms with no charge?

They float around as Na+ and Cl-. Why would they convert back? Neutral sodium reacts with water, and that quite violently.

http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set6/Image43.gif

@IvanNeretin first time taking chem bro. How is this a bad rating of a question? Lack of knowledge and i get a negative rating? Lmao

@Ignacio We have the mhchem package installed, i.e. chemistry markup. That'll make future edits a bit easier for you. If you want to know more, please have a look here and here.

Great, thank you! I didn't know that, actually I tend to use mhchem in my own latex stuff, so I'll feel at home =)

Well $\ce{Na+}$ and $\ce{Cl^{-}}$  ions don't really float around as simple ions in an aqueous solution. Since water is polar (has a + and - orientation) the ions attract oriented water molecules into a cluster around the ions. The first "layer" of water molecules will be highly oriented, the second layer less so, and this continues until the number of layers is so great that the water molecules are randomly oriented with respect to the ion.

They remain as ions, it takes energy to reduce $Na^{+}$ and oxidate $Cl^{-}$.
$$\ce{Na+(aq) + e–  ->   Na(s)} \space \Delta E -2.714$$
For oxidating $Cl^{-}$:
$$ \ce{2 Cl–(aq) -> Cl2(g) + 2 e–}  \space \Delta E -1.358$$
However solvation structure of Na+ is quite difficult to predict so what you have eventually are hydration structures. J. Chem. Theory Comput., 2012, 8 (10), pp 3526–3535


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Is there an edible, colour-less, taste-less, odour-less, water-soluble, cheaply-producable gel, solid or other substance?
Suppose, for example, that some oatmeal is fixed in gel, or some other substance. A portion of the substance-with-oatmeal is then removed and added to cold milk or some similar dairy product, where it quickly dissolves, yielding a standard bowl of milk and oatmeal. Is there such a substance?
Ideally, the substance should be unobtrusive to the eating experience. This also means that it should not dilute the milk significantly. Otherwise, some taste, colour or odour may be permissible, provided it does not clash with the taste of the meal.
The substance should also not soak the oatmeal, which is why it may not be a liquid.

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

Water, below 0 Celsius.

Are you talking about maltodextrin? It's a solid powder assembled from glucose units with no taste smell or color. It's commonly used as a carrier; it's used to dilute the actual sweet -flavored molecule in Splenda.
If you're dealing with Hot oatmeal, you could add methyl cellulose to gel it up. Upon mixing with cold milk it will melt, acting like reverse jello; solid when hot and liquid when cold.
You asked a rather strange question, I'd recommend instant oatmeal to prevent a possibility of bacterial/fungus growth when storing oatmeal for extended periods of time.

Gelitan is unflavoured – it is usually colorless, tasteless, and is a solid if you melt its powdered form, and then let it cool.


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Does a certain "Badecker reaction" actually exist?
A book I read, mentions a certain Badecker reaction under coordination chemistry. It goes like this:
$$\ce{Na2[Fe (CN)5 NO] + Na2SO3 -> Na4[Fe (CN)5 (NOSO3)]}$$
I googled up the name, but I didn't get anything even remotely related to chemistry.
That particular book does have a pretty long history of typos and a couple of factual errors here and there (but I still use it because its got quite a few useful tips …)
So the question is; is there such a thing called the Badecker reaction (I even tried googling it up with an umlaut, Bädecker reaction, because, well … Badecker looks a little German, but then I was directed to a Wikipedia page on cadmium oxide. So what is the reaction’s name?
Also even if it doesn't go by that name, does the reaction actually even exist?

Searched for you in my mother tongue german, but had no success as well. Maybe try finding this specific reaction searching for the complex sodium nitroprusside.

The reaction in question is actually called the Boedeker reaction. How the book came from Boedeker to Badeker I do not know, but maybe there was one instance of Boedeker being written Bödeker which was later misparsed by a human or a robot and mistook for an a. (Replacing oe for ö is a common mistake with German names since the latter is typically used in orthography but names don’t adhere to orthography.)
The reaction is well-known and often used as a qualitative analysis for sulphite ions in solution, much like the same nitroprusside ion can also be used to detect sulphide ions. In chemical formulae, it is:
$$\ce{[Fe(CN)5(NO)]^2- + SO3^2- <=> [Fe(CN)5(NOSO3)]^4-}$$
The reaction in itself is also rather interesting since for over 100 years the structure of the addition complex was unknown. Bottomley et al. were finally able to obtain crystal structures of the probably structurally similar ruthenium complex $\ce{[RuCl(bpy)2(NOSO3)]}$.[1] These show that the addition product is — very much like in the sulphide case — formed by nucleophilic attack of the sulphite sulphur’s lone pair to nitrosyl nitrogen creating a nitrosylsulphite anion $\ce{O3SNO-}$. Thus, the complex formed should be named pentacyanidonitrosylsulphito-κN-ferrate(II).

Reference:
[1]: F. Bottomley, W. V. F. Brooks, D. E. Paéz, P. S. White, M. Mukida, J Chem. Soc., Dalton Trans. 1983, 2465. DOI: 10.1039/DT9830002465.

Slightly OT: "but names don’t adhere to orthography.": this is a misconception of orthography, the orthography of names like of anything else is only a matter of definition. The definitions might show some regular behaviour, but thats essentially random.