text:
If A reacts with B (aq), does that mean it is soluble in it?
For example, ethanol is soluble in water, but doesn't react with it. On the other hand, H2SO4 is said to be soluble because it dissociates (reacts) in water.
Would ethanol be soluble in HCl (aq)? It won't react with the HCl, but it can form hydrogen bonds with the water, right?
How about ethylamine or ethyl methanoate? Would they be soluble in HCl (aq)? They react with the HCl, so I guess they will be. Is this correct?

Summary, TL;DR:
No, A reacting with B does not mean that A is soluble in B. Solvation and chemical reaction are two different processes, although the distinction can in some cases seem less than perfectly clear. All of the example compounds you discussed are soluble or miscible with both water and hydrochloric acid. The IUPAC definition of a chemical reaction is "A process that results in the interconversion of chemical species", which is fundamentally different from solvation, and does not apply to any of the examples in the question.

As you suggest, solvation and chemical reactions are two fundamentally different concepts. For example, if compound A reacts with water to form a homogeneous liquid solution, that is a different processes than if compound A were to simply be dissolved by water.
Solvation is defined by the IUPAC Gold Book as follows:

Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

Whereas the IUPAC Gold Book defines a chemical reaction as:

A process that results in the interconversion of chemical species.

One aspect of solvation vs. reaction that may seem confusing is the solubility of ionic species. Even though $\ce{NaCl}$ becomes $\ce{Na+}$ and $\ce{Cl-}$ in an aqueous solution, this does not constitute a chemical reaction as defined above, and we say that $\ce{NaCl}$ is soluble in water. The same is true of your example of $\ce{H2SO4}$; even though it dissociates in water, it is not converted to a new chemical compound. One way to think of this is if you remove the solvent, the solute should typically resume to it's original form. From our previous example, if we evaporate the water from our aqueous solution of $\ce{Na+}$ and $\ce{Cl-}$, we just get the solid $\ce{NaCl}$ back.
Your examples of solubility in hydrochloric acid are a bit complex because that is a two-component system of water and $\ce{HCl}$. All of the compounds you discuss are water soluble and it doesn't matter if the $\ce{HCl}$ is there or not. One slight exception in your examples is ethylamine. Ethylamine itself is miscible with water, but many higher molecular weight amines are not water soluble, but are soluble in hydrochloric acid. In this case, the $\ce{H+}$ from $\ce{HCl}$ protonates the amine, making the hydrochloride salt. This is still an example of solubility however, as once you remove the solvent, you are left with the original compound, in this case the amine.

Thank you for your answer!
Could you give me an example of a compound which reacts with water but does not dissolve in it? And why is the salt obtained from protonating Ethylamine soluble in water? Lastly, would the results change if we used liquid HCl rather than aqueous?

Metallic sodium is not water soluble, but undergoes a chemical reaction with water to form a water soluble species ($\ce{NaOH}$) and a pretty insoluble gas, $\ce{H2}$. The result is that it appears that the sodium has dissolved, when in fact it has undergone "a process that results in the interconversion of chemical species" (from definition in answer). Regarding ethylamine: amines have a basic lone pair on the amine nitrogen that readily accepts a proton from a strong acid. This gives you a charged species, which is much more polar and generally water soluble than neutral species. Continued..

I hope the amine protination explanation made sense. Again, amines themselves are somewhat polar, and a small compound like ethylamine is polar enough to be water soluble. I just extended that to the example of large hydrocarbons where the amine moiety itself isn't polar enough to make the compound water soluble. In this case, ionic species are so polar that protonating even an otherwise large, aliphatic amine can make it soluble. Regarding pure HCl: it boils at about -85C, so it is very much a gas at room temperature. The gas readily dissolves in water, and the strongest HCl soln is ~38%.

Thanks, it's mostly clear now. But I meant to ask would Ethylamine, Ethanol, Ethyl Methanoate be soluble in a pure (liquid) acid, not if HCl would be soluble in a water.

Unfortunately, every acid has it's own very distinctive properties, so that such a question (is compound X soluble in pure acid) is impossible to fully answer. It's almost like asking if compound X is soluble in liquids. But for the sake of argument, H2SO4 can be almost a pure solution. Ethylamine would be very soluble in H2SO4. Ethanol would undergo a dehydration reaction to ethene gas and water. I suspect that ethyl methanoate would be fairly soluble, though probably less so than in water (~10% soluble in water). The situation would be very different though in, say, HNO3, or other acid.

These questions of mine arose because I couldn't answer a different (but related) question. Though my confusions have been cleared, now I cannot solve this question since I know my logic was incorrect, if that makes sense. Could you tell me how to approach Q6 biii, or would it be better if I asked this as a separate question? http://pastpapers.papacambridge.com/view.php?id=Cambridge%20International%20Examinations%20%28CIE%29/AS%20and%20A%20Level/Chemistry%20%289701%29/2015%20Jun/9701_s15_qp_42.pdf

I think our conversation has already gotten pretty far off from the original question. The best thing to do is to accept this answer (green checkbox to left of answer) if you feel it answered the original question, then, regardless of that, start a completely new question. I hope I was of help to you here, and I'll keep my eye out for your next question.


text:
pH of 0.01 M glycine solution
The question is to find out the pH of $\pu{0.01M}$ glycine solution. It is given that $K_\mathrm{a1} = \pu{4.5*10^{-3}}$ and $K_\mathrm{a2} = \pu{1.7*10^{-7}}$.
Glycine is $\ce{NH2CH2COOH}$. The two $K_\mathrm{a}$ values correspond to the ionization of $\ce{COOH}$ and  $\ce{NH2}$. My idea was to average the two $K_\mathrm{a}$ values and take log to get pH. However this is not yielding the correct answer.

You've actually calculated the isoelectric point of glycine.

And $K_{a2}$ is for Ionization of $\ce{-NH3+}$, not for $\ce{NH2}$

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text:
Is to possible to prepare gallium oxide from gallium sulfate against gallium nitrate or ZnO from ZnSO4 against Zn(NO3)2
Is it possible to synthesize Gallium oxide from Gallium sulfate? Usually I find gallium nitrate to be recommended. Similarly for ZnO synthesis, generally nitrate salt of zinc is preferred to sulfate salt. Why?

It is possible to synthesize everything from anything. What is your question, really?

Generally gallium oxide is formed by decomposition of gallium salts. The choice of your gallium salt depends on decomposition temperature and cost of the salt. Gallium nitrate can be used for this purpose over gallium sulfate due of its low decomposition temperature.
\begin{array}{c|c}
\mathbf{Salt} & \mathbf{Decomposition~temperature}  & \mathbf{Cost (Sigma-Aldrich)}\\\hline
\text{Gallium sulfate} & \text{$\ce{>749°C}$} & \text{110 USD (5g)} \\
\text{Gallium nitrate} & \text{$\ce{200-250°C}$} & \text{$\ce{114 USD (5g)}$}\\\end{array}
Similarly goes for zinc where we would prefer the nitrate salt over sulfate.
\begin{array}{c|c}
\mathbf{Salt} & \mathbf{Decomposition~temperature}  & \mathbf{Cost (Sigma-Aldrich)}\\\hline
\text{Zinc sulfate} & \text{$\ce{>689°C}$} & \text{162.5 USD (1g)} \\
\text{Zinc nitrate} & \text{$\ce{>300°C}$} & \text{$\ce{26.2 USD (100g)}$}\\\end{array}
Note: although the decomposition side products are gaseous, they are both harmful to humans.

I think SO42- salts have higher disassociation temp because it has SO4 2- is more ionic than NO3-

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text:
Why is Cyanidine soluble in water?
Cyanidine is not polar due to its make-up of only hydrogen atoms and carbon atoms, correct? But "like dissolve like", which means it would HAVE to be polar, or no? Is it then soluble solely because of its OH-groups being able to create hydrogen bonds with the water molecules, thereby making it miscible?

And what is it this "polarity" your asking about? I wonder if you know what you're talking about.

Well, I read it was soluble in water. It'd have to be polar, then? @Mithoron

Then ask why it is soluble in water...

@Mithoron Good point. I'll edit.

The very word "polar" seems to be confusing; it has two altogether different meanings.

@IvanNeretin how so?

One meaning is "a molecule which has a lot of highly polar bonds and hence is soluble in water and stuff"; another is "a molecule which has zero dipole moment". These are by no means equivalent.

@IvanNeretin Well, I'm referring to the first meaning in this case.

Then your molecule is highly polar and should be soluble in water.

But the OH-groups are scattered around, wouldn't that affect the polarity of the molecule as a whole in a negative way? @IvanNeretin

You say cyanidine is not polar due to being made only of hydrogen and carbon, but then talk about it having OH groups. It has oxygen. Not only that but it is a charged species (also clearly shown on your structure). So that is two reasons why your initial suspicion of non-polarity because it is s simple hydrocarbon are wrong.

@matt_black Fair point. The oxygen ion has a positive charge, how does this differ from the negative charge in terms of electronegativity? Is it then weaker than an oxygen atom?

@Mithoron Also, I suppose the molecule is asymmetrical as well, making it polar along with the OH-groups?

Don't mention this dipole idiocy, it's irrelevant for solubility.

Cyanidine actually looks to me to be a rather highly polar compound. It's essentially a $\ce{C15}$ hydrocarbon having 5 hydroxyl groups and a formal positive charge.
The ratio of hydroxyl groups to the number of carbons is the same as that of water-miscible isopropyl alcohol. (Of course smaller molecules like isopropyl alcohol also tend to be more soluble than larger molecules.)
The most important feature indicative of it's rather high polarity and water solubility is it's formal positive charge. This is analogous to placing a positive charge on a large amine by protination in order to turn the amine into a polar, water soluble compound.

So the positive charge makes it polar? I suppose it is irrelevant that it is placed practically in the middle of the molecule?

Yes and yes. Usually when we talk about a molecule being polar, we're talking about regions of somewhat higher and lower electron density due to inductive or other effects. We usually call this "delta +" or "delta -" indicating a partial charge. These partial charges, or regions of higher or lower electron density, are far less pronounced than the case of having a full formal charge. The location of the charge can make some difference if it's very sterically "hidden", but that's not the case here.


text:
Why are hexamonodentate ligands optically inactive?
After encountering a lot of problems on optical isomerism of coordination compounds I have come to an idea that only complexes with polydentate or tridentate ligands shows optical isomerism. For example $\ce{[Co(en)3]^{3+}}$ shows optical isomerism but complexes such as those having platinum as the central atom and having six different monodentate ligands don't show optical isomerism. I have no trouble understanding why $\ce{[Co(en)3]^{3+}}$ shows optical isomerism but I am facing trouble finding out a reason for the second case.
The main question is why do complexes having monodentate ligands don't show optical isomerism?Any ideas?

What is the exact species you are looking at? It's not really possible to say anything without knowing what you're talking about.

@orthocresol I just have the doubt that why coordinate compounds having monodentate ligands don't show optical isomerism.

Do you have an example of platinum coordinated to 6 different ligands? It would seem that such a molecule chiral.

@Tyberius the example given in my book is $\ce{[Pt(NH3)(Br)(Cl)(I)(NO2)(py)]}$

@orthocresol now is my question clear?

The compound you stated does show optical isomerism. In fact, there will be 30 isomers! https://en.wikipedia.org/wiki/Octahedral_molecular_geometry


text:
Ordering the reduction potentials for Cr3+, Mn3+ and Fe3+?
I have been asked to order, with explanation, the standard reduction potentials for the following:

$\ce{Cr^3+ + e- -> Cr^2+}$
$\ce{Mn^3+ + e- -> Mn^2+}$
$\ce{Fe^3+ + e- -> Fe^2+}$

(all ions in aqueous conditions)
I am really uncertain what factors to consider; I am even struggling to make this a general born-haber process. I know that $\mathrm{3^{rd}}$ ionization enthalpies will come into play, and that $\ce{Fe}$ will be expected to have the highest one. Will hydration enthalpies also be relevant? Will there be a ligand exchange component (with the solvent, water) that I have to consider also?

Firstly, I'd like to say the explanations we propose here are merely our efforts trying to explain what we observed experimentally in the first place. Let's get some data:
$${\begin{array}{|c|c|} \hline \ce{Cr^3+/ Cr^2+} & \ce{Mn^3+ / Mn^2+} & \ce{Fe^3+ / Fe^2+} \\ \hline \pu{-0.41 V} & \pu{+1.57 V} & \pu{+1.97 V} \\ \hline \end{array}}$$
And while we're at it, let's ponder on electronic configurations as well:
$${\begin{array}{|c|c|} \hline \ce{Mn^2+} & \pu{[Ar] 3d5 4s0} \\ \hline \ce{Fe^3+} & \pu{[Ar] 3d5 4s0} \\ \hline \end{array}}$$
You were correct in thinking that $\ce{Fe^3+}$ will have the highest reduction potential due to the third ionization enthalpy coming into play, all we've got to do now is to think for manganese and chromium. Why don't we put it differently:

Why is $\ce{Cr^2+}$ reducing and $\ce{Mn^3+}$ oxidizing when they both have the same $\pu{d4}$ configuration?

It's because if manganese acts as an oxidizer, it'll get converted into $\ce{Mn^2+}$, which is stable owing to the $\pu{d5}$ configuration. Chromium, on the other hand is notorious for showing the $+3$ oxidation state in its compounds. So, that's a win-win kind of situation we're dealing with.

The subtle role played by the solvent molecules when they're acting as ligands comes into limelight on asking yet another relevant question:

Which is a stronger reducing agent: $\ce{Cr^2+}$ or $\ce{Fe^2+}$?

Answer: $\ce{Cr^2+}$. This is because of more CFSE stabilization of $\ce{Cr^3+}$ as compared to $\ce{Fe^3+}$ when water molecules behave as a ligand for our complex.

this has really helped with pointing me in the right direction-thank you so much!!!


text:
What does "heat is evolved at standard-state conditions" mean?

When $\pu{0.539 g}$ of $\ce{Na(s)}$ reacts with excess $\ce{F2(g)}$ to form $\ce{NaF(s)}$, $\pu{13.3 kJ}$ of heat is evolved at standard-state conditions. What is the standard enthalpy of formation ($\Delta_\mathrm{f}H^\circ$) of $\ce{NaF(s)}$?

I am not asking for the answer. What I would like to know is what "$\pu{13.3 kJ}$ of heat is evolved at standard-state conditions" means. Does anyone know? I am unfamiliar with this terminology.

That is incorrect. Standard state does not include any stipulation of temperature. See https://chemistry.stackexchange.com/q/41862/16683

It simply means that the experiment was carried out under a standard pressure.
This standard pressure can be defined by you to be anything. You may define it to be 1 atm, or 5 atm, or 3 Pa. However, the established convention is to define the standard pressure as $p^\circ = \pu{1 bar}$. See: standard state in the IUPAC Gold Book
What implications does it have for your question? Nothing, really. It just means that the value of $\Delta H$ you derive can be labelled with the word "standard". If you run your reaction at a different pressure $p \neq p^\circ$, then the enthalpy of formation has to be labelled $\Delta_\mathrm{f}H$ instead of $\Delta_\mathrm{f}H^\circ$.
Note that there is no stipulation of temperature. Therefore, the reaction may have been carried out at 10 K, or 1000 K, and it does not have to be 298 K. This simply means that every different temperature will have a different value of $\Delta_\mathrm{f}H^\circ$.


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Why are some molecules unable to absorb infrared radiation?
Methane $\ce{CH4}$ is symmetrical and non polar, and so is oxygen $\ce{O2}$. But $\ce{O2}$ doesn't absorb IR. Why is that? People say that if a molecule can change its polarity then it will absorb IR, but I don't understand how a molecule changes its polarity while absorbing IR.
Please try to keep your answer simple — I don't have a strong background in infrared spectroscopy.

In basic terms, for a molecule to absorb radiation there has to be an oscillating dipole being produced. This can occur by nuclear motion (vibrations, rotations) or electronic motion to produce electronically excited states. You ask about IR radiation, which does not have sufficient energy to produce electronic excitation, so this is ignored in the text below.
The restriction on producing a moving dipole exists because radiation consists of oscillating electric and magnetic fields and to allow energy to go from the radiation into the molecule, there has to be an interaction between radiation and the molecule and the origin of this is the oscillating molecule's dipole. (Of course the radiation also has to have the correct energy to bridge the gap between any two energy levels, but this is just another way of saying that the molecular dipole and the radiation oscillate at the same frequency, i.e. are in resonance.)
Thus in any homonuclear diatomic there is no IR absorption because these molecules have no  intrinsic dipole or one caused by vibrations. Also as they have no permanent dipole there is obviously no dipole generated when molecules rotate. Heteronuclear diatomics, HCl for example, all have a permanent dipole and one that is changed  by vibrational motion, thus there is a vibrational and rotational spectrum in all heteronuclear diatomics.
All homo-nuclear diatomics, $\ce{N2}$, $\ce{O2}$ for example, are transparent to IR radiation and this is of prime importance in global warming, since if these molecules obtain energy, say by collision with water or carbon dioxide that has been excited by IR radiation, they cannot radiate the energy away and so must heat the atmosphere by collisions with other molecules.
In a molecule such as $\ce{CO2}$ although it has no overall dipole it does have moving (oscillating) dipoles due to the way it vibrates. One mode is to stretch one bond at the same time as the other contacts (an asymmetric stretch), and so an oscillating dipole is formed and this interacts with radiation and $\ce{CO2}$ has an IR vibrational-rotational spectrum. The bending vibration is also IR active.
A similar effect occurs in methane, the way the molecule vibrates develops oscillating dipoles and these interact with the radiation and so absorb energy. Molecular group theory is usually used to sort out which of the $3N-6$ ways molecules can vibrate can have IR spectra.
There are complicated higher order effect even in homonuclear diatomics which may generate absorption but these effect are absolutely tiny and can be ignored for most every purpose. The relative magnitudes of electric dipole, quadrupole and magnetic dipole intensities are $I_\mathrm{dip}:I_\mathrm{quad}:I_\mathrm{mag} \approx 1:10^{-5}:10^{-5}.$
We can measure the vibrational frequency of homonuclear as well as heteronuclear diatomics by using Raman spectroscopy, which is a scattering effect (no photons are absorbed), and depends on the shape of the 'electron cloud' in the molecule. This is more properly called its polarisability.

Could you please clarify some physics - why to resonate the frequencies of the dipole and radiation must be the same? They can't resonate if the frequencies differ a bit?

This is normally explained as the Bohr condition; energy of photon matches transition energy gap. As $E=h\nu$ you can think of this also as frequency match. If the energy is not exact the chance of absorption rapidly diminishes in accordance with the line shape of the transition. A typically homogenous linewidth (full width half max) is $ 10^{-3}\pu{ cm^{-1}}$ but can be wider due to Doppler effect, typically $\approx 0.1 \pu { cm^{-1}}$ still very narrow if the transition frequency is $1500 \pu{ cm^{-1}}$

Infrared radiation is absorbed by a molecule when the frequency of the radiation matches that of one of the vibrational modes of the molecule. It is also necessary that the molecule have a vibrational mode for which there is a change in dipole moment. If the energy of a photon does not meet both these criteria, then it will be transmitted rather than absorbed.
According to this Wikipedia article:

A molecule can vibrate in many ways, and each way is called a vibrational mode. For molecules with $N$ number of atoms, linear molecules have $3N - 5$ degrees of vibrational modes, whereas nonlinear molecules have $3N - 6$ degrees of vibrational modes (also called vibrational degrees of freedom). As an example $\ce{H2O},$ a non-linear molecule, will have $3 × 3 - 6 = 3$ degrees of vibrational freedom, or modes.
  Simple diatomic molecules have only one bond and only one vibrational band. If the molecule is symmetrical, e.g. $\ce{N2},$ the band is not observed in the IR spectrum.

So $\ce{N2}$ (or $\ce{O2}$) only has a single vibrational mode $(3 × 2 - 5 = 1),$ no permanent dipole, and can only have a weak induced dipole, so it's absorption in the IR is very small. $\ce{CO}$ also has only a single vibrational mode, but there is a permanent dipole moment and therefore the molecule has the ability to absorb in the IR spectrum.
I've tried to balance "keeping it simple" with giving a meaningful explanation. Don't hesitate to ask for clarifications in the comments.

Probably all molecules with two or more atoms find a way to absorb at least some IR radiation.  A molecule with only nonpolar bonds, such as a nitrogen molecule, may be nonpolar if the electrons are perfectly centered in their orbitals, but if the electrons move there could be a weak temporary dipole.
Still that is much weaker absorption than what you would get with a permanent dipole like water or the individual bonds in carbon dioxide.  Lucky us, otherwise an atmosphere that is 78% nitrogen would launch a huge greenhouse effect.

Very simple answer: IR radiation (in very first approximation) interacts with the individual bond. While methane is unpolar, the bonds in it are polar.
Not terribly polar, but still.

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text:
The effect of baking soda on water electrolysis
Ok, so now I have a new question, what type of electrode should I use. I am intending to use a simple stripped wire as my electrode, but am concerned whether the average copper wire would react and create unwanted byproducts. Is it ok if I use copper, or is some other metal better suited to this experiment.
EDIT:
I believe more baking soda would result in more gases because more free floating ions means more conductivity, which would probably allow more redox reactions to occur, and thus allow more gases to be produced.

Hello and welcome to Chemistry.SE. As this is a homework type of question, you should read through this post on asking good HW questions so that you can get a good answer. For your question, what you really need to do is discuss your thoughts on what would happen and why. Best of luck!

@airhuff yup, just made the change

Yea, you got it. The more current that flows (up to the limits of your system) the more product you will get.

Definitely do not used stripped copper wire, unless you fancy a nice blue solution and electrodes that need constant replacement. You see, copper oxidizes quite readily, so you will end up with lots of $\ce{CuO}$, and a loss in the yield of $\ce{H2}$ and $\ce{O2}$. The best things would be noble metals like platinum, but stainless steel should do fine for your purposes/likely budget.


text:
Is it possible to prepare a solution of hydrogen sulfate in a lab? What is the procedure to do so?
I don't want a buffer solution with it, I want to know if its possible to have a solution with only $\ce{HSO4^{-}}$ ions in it.

Well, you'll at least need a cation to maintain charge balance. Also it will be in some equilibrium with sulfate.

You can prepare such a solution by dissolving a soluble hydrogen sulfate such as sodium hydrogen sulfate in water.
However, since the hydrogen sulfate anion is acidic, there will always be some amount of sulfate ions in the sulution. You can calculate the proportion from its dissociation constant.

Do you want to prepare a $\ce{HSO4-}$ solution? Let's take the sodium salt as example.
One method involves mixing stoichiometric quantities of sodium hydroxide and sulfuric acid which react to form sodium bisulfate and water.
$$\ce{NaOH + H2SO4 → NaHSO4 + H2O}$$
Note, the reaction is exothermic. Also, as mentioned by @airhuff and @aventurin, the solution would contain some amount of sulfate ions in the solution which are in equilibrium.
Second method is to retrieve as a byproduct from the reaction of excess sulfuric acid with any of the sodium salts ($\ce{X- = CN−, NO3−, ClO4−}$):
$$\ce{NaX + H2SO4 → NaHSO4 + HX}$$
The advantage of this process is that the acids produced have a lower boiling point than the reactants and are separated from the reaction mixture by distillation. However I would not recommend using cyanide or perchlorate as sodium cyanide is a potent poison and perchlorate is strong oxidizer. Do not attempt these reaction if you are not an experienced chemist.
You can also try making it from sodium chloride. You can check this video.
$$\ce{NaCl + H2SO4 → NaHSO4 + HCl}$$


text:
Is lithium bicarbonate an aqueous solution of lithium carbonate?
This question is related to:

Are Lithium bicarbonate and Magnesium bicarbonate both soluble in water?
What type of chemical reaction is this?
How can bromine water be called a solution?

In the first question, the page from JD Lee (Concise Inorganic Chemistry 5 ed) states:

Lithium is exceptional that it does not form that it does not form a solid bicarbonate, though $\ce{LiHCO3}$ exists in solution. All the carbonates and bicarbonate are soluble in water.

and then in the sideline(the most important line):

If it exists in water, it must definitely be soluble in water.

Now, Lithium carbonate is prepared by bubbling $\ce{CO2}$ gas to a solution of lithium hydroxide, $\ce{LiOH}$.(source)
$$\ce{2LiOH + CO2 → (Li2CO3 + H2O)}$$
But in another source (endmemo), it shows that lithium hydroxide and carbon dioxide directly combines to form lithium bicarbonate.
$$\ce{LiOH + CO2 -> LiHCO3}$$
So, basically lithium bicarbonate is aqueous solution of lithium carbonate. Isn't it?
Now, what I gather from the fact that lithium carbonate is soluble in water and form a solution where the components do not react with each other. However, when the two reaction are compared, we observe that lithium carbonate reacts with water to form lithium bicarbonate. It seems contradictory with the above statement. How is this possible?
In the 3rd question, I said that if bromine reacts with water to form a mixture of acids, how can it be called a solution? Components of solution do not react with each other. Same case goes on with this reaction. If lithium carbonate is in aqueous solution, it should dissociate into component ions i.e $\ce{Li+}$ and $\ce{CO3^2-}$ but instead it goes on form $\ce{LiHCO3}$ and thus the solution has $\ce{Li+}$ and $\ce{HCO3-}$ ions. Why?

Lithium bicarbonate solution is something like a lithium carbonate solution in the sense that precipitation would produce the carbonate instead of the bicarbonate.  But what is in solution is mostly lithium and bicarbonate ions.  Precipitation is accompanied by a reaction in which the highly polarizing lithium ions react with bicarbonate:
$2\ce {Li+}+2\ce {HCO_3-}\rightarrow \ce {Li2CO_3}(s)+\ce {H2O}+\ce {CO2}$
Other alkali metals give less polarizing ions and do not react in the same way; they just precipitate the bicarbonate salt.

In a nutshell:

Depends how you define the solubility. The solubility product constants for lithium carbonate $\ce{Li2CO3}$ and magnesium carbonate $\ce{MgCO3}$ are $\ce{8.15×10^{-4}}$ and $\ce{6.82×10^{-6}}$ respectively. Thus, the solubility of magnesium carbonate in water at RT two orders of magnitude lower comparing to the solubility of lithium carbonate... approximately $\ce{0.014g/100ml}$ vs. $\ce{1.29g/100 mL}$.
It is a combination or synthesis reaction.

... basically lithium bicarbonate is aqueous solution of lithium carbonate.

Not really. It will be a mixture of lithium cations, carbonate and bicarbonate anions.

However, when the two reaction are compared, we observe that lithium carbonate reacts with water to form lithium bicarbonate. It seems contradictory with the above statement. How is this possible?

All salts of either weak base and/or weak acid will hydrolyze (react with water). For most cases its not important, thus no needs to complicate it. Where it is important, there is no need to name it, it is clear as it is. BTW what would be the proper name in your opinion?

The same logic - simplicity. That's why we have bromine water (: Can you come up with alternative names? Would be interesting to find out what do you have on your mind on this subject.


text:
Cyclohexene reduction reaction with hydrogen gas and ethanol
I have a following reaction (see below). I found the section of reduction and oxidation states in my book but there is nothing about cyclohexenes. I know that the double bond will be reduced in this reaction but what I have no idea what happens to those substituents. Why do they move around and what are their orientations?

You know how reaction works, but not rotated molecule... Wow

Note: This is a homework question, so I will not give a complete answer.
$\ce{H2/Pd}$ is a reducing agent, and it acts so by adding hydrogen to the double bonds. Alkenes are converted to alkanes. This proceeds via a concerted pathway, so rearrangements due to carbonation shifts do not occur. Now I think you can proceed from here.
As far as those substituents are concerned, think carefully on what would happen.

The reaction with $H_2/Pd$ is one of the ways to reduce alkenes to alkanes. This takes place over the Palladium surface which has $H$ adsorbed on it. Hence we can think of cis addition in this case.
It is clear through the mechanism that there is no formation of carbocations. So we need not worry of rearrangements. Now from the options up there it is quite clear I suppose. Hope it was helpful.


text:
Which of these functional groups is soluble in aqueous HCl and/or NaOH?
http://pastpapers.papacambridge.com/view.php?id=Cambridge%20International%20Examinations%20%28CIE%29/AS%20and%20A%20Level/Chemistry%20%289701%29/2015%20Jun/9701_s15_qp_42.pdf

I need some help to solve Q6 biii) of this paper, asking about which functional groups are present in E, F, G and H. I got the correct answers thinking that if something reacts with an acid or base, then it dissolves in them. But then I realized that amides and esters also react with both ( though heat is required, so that may be why they aren't said to be soluble ), and I also realized that in general my 'dissolves = reacts' notion is very flawed. So I discarded this concept, and tried to work on it fresh, but I am now very much stuck.
My approach before was this :
Esters do not dissolve in cold aqueous bases or acids, nor do amides or alcohols.
Amines dissolve in aqueous acids. Phenols and Carboxylic acids dissolve in aqueous bases.
From the molecular formula, we know that each compound contains either an Amine or an Amide functional group, so the rest of it is simply working out which combination fits which functional groups. Thanks!

Which of these functional groups is soluble in aqueous HCl and/or NaOH?

Functional groups don't have solubility. Their presence in a compound, however, can have an effect on solubility.

My approach before was this : Esters do not dissolve in cold aqueous bases or acids, nor do amides or alcohols. Amines dissolve in aqueous acids. Phenols and Carboxylic acids dissolve in aqueous bases.

That is the correct approach. Amines dissolve in aqueous acids because they are protonated to form a positively charged ion (unless other charged groups are present). Conversely, phenols and carboxylic acids are deprotonated to form phenolate and carboxylate anions.

From the molecular formula, we know that each compound contains either an Amine or an Amide functional group, so the rest of it is simply working out which combination fits which functional groups.

F and G are soluble in acidic solution, so they might contain an amine plus two oxygens from another functional group. The choices are carboxylic acid or ester, and the solubility in basic solutions shows that F has the former and G the latter.
E and H are insoluble in acidic solution, so they might contain an amide instead, plus one oxygen from another functional group. The choices are phenol and alcohol, and the solubility in basic solutions shows that H has the former and E the latter.


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Stability of compounds
Which is more more stable, cis-1-ethyl-3-methylcyclohexane  or trans-1-ethyl-3-methylcyclohexane?
I think trans- is more stable than cis- but in my textbook the answer is different.

With cis both substituents can be equatorial.

Note: This is a homework question, so I will not provide a complete answer (site policy).
Cyclohexane usually exists in the chair-conformation. See the picture below.

The different substituents on a cyclohexane ring (here they are all hydrogen) can be either axial or equatorial. Axial substituents point along the axis of the cyclohexane ring, while the equatorial substituents point outward the equator of the cyclohexane ring. I have marked the axial and equatorial positions in red and blue respectively:

Adjacent axial or adjacent equatorial positions are trans with respect to each other. Equatorial positions are considered more stable compared to the axial positions, due to steric factors.
With the knowledge above, you should be able to correctly decide on the answer.


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How do we know which compounds have higher solubility?
Is there any way we can look at the chemical formula of various compounds, do some calculations or something and figure out which compounds will have higher solubility as compared to the others.

Your question needs more clarification. Are you really asking if you can predict solubility from the formula, or from the chemical structure? These are two different things.

It depends on the solution and the solute. Usually, the rule is "like" dissolves "like". A polar solution will dissolve polar compounds and a non-polar solution will dissolve non-polar compounds.
https://www.youtube.com/watch?v=PVL24HAesnc
As far as I know, to find the solubility you have to determine it experimentally. This can be done by "feel" first. If you have a compound that doesn't have many polar groups on it, chances are it can be dissolved by a non-polar solution. If you know what the structure looks like try to match it with a solvent that looks similar to it.

As a partial answer to this broad question, one may provide a prediction about the partition of an organic compound between water and octanol, $\log p$.  This often serves as one of the preliminary checks (prior to chemical synthesis) of compounds foreseen for a pharmaceutical application among the Lipinski rules. An example like this or this may provide a first experience; some molecular editors offer such an estimate, too.
As usual, as such a prediction is based on a model, there are shortcomings; like i) how to treat naturally ionic compounds, or ii) how to account for amino acids, for example. While it may guide between more or less likely water/octanol soluble compounds, an experimentally determined value is preferable.


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Why don't effervescent vitamin C tablets react with the vitamin C?
I have a packet of effervescent Vitamin C tablets. When placed in a glass of water, a tablet will fizz and dissolve. They have the following ingredients (all included for completeness):

citric acid
ascorbic acid
sodium hydrogen carbonate
sorbitol
sodium carbonate
inulin
potato starch
zinc citrate
tricalcium phosphate
maltodextrin
riboflavin phosphate sodium
sweetener & flavours
tocopherol

I'm guessing that the citric acid is intended to react with the sodium hydrogen carbonate to effervesce.
Given that vitamin C (ascorbic acid) is also an acid, surely it could also react with the sodium hydrogen carbonate, removing the vitamin C from the finished drink?
Does this happen to any extent? If so, in what proportions? If not, why not?
The ingredients state how much vitamin C is present in the tablet, not in the finished product. This is also specified as a proportion of government recommended recommended daily amounts, which suggests that one would receive that amount if one took a dose as directed. If the vitamin C does react, does this mean that the label doesn't indicate how much vitamin C is available?
(I last studied chemistry at GCSE level in [high] school, so please excuse the lay question. A simple answer would be most useful!)

Citric and ascorbic acids aren't removed, just partially neutralised.

Maybe it does, but nobody cares.

I care! I've edited the question to add detail about the amount of vitamin C that is indicated on the packaging.

Acid-base reactions are fast and reversible (barring any additional reaction that occurs). That is, once you move the conjugate base of ascorbic acid to a low-pH environment (e.g. your stomach), it rapidly reverts to the protonated form. (Ascorbic acid does not undergo any pH-dependent further reactions, as opposed to carbonate, which when protonated to carbonic acid can decompose to carbon dioxide and water.)
Actually, as the pKa (roughly the "halfway pH" of acid/base reactions) of ascorbic acid is 4.2, the moment the protonated ascorbic acid enters your bloodstream (roughly pH 7.3), it immediately loses that proton again, reverting to the deprotonated ascorbate ion.
So in a practical sense, it really doesn't matter which form you take your Vitamin C in - either (protonated) ascorbic acid or (deprotonated) ascorbate acts in you body the same way, because they can rapidly interconvert based on the pH of the solution they happen to be in at the time.
(Because they typically work in buffered aqueous environments, biochemists typically play fast-and-loose with the difference between the protonation states of molecules. Unless you're explicitly dealing with pH-dependance, they're typically lumped all together, with the implicit understanding that at a given pH they'll adjust to the appropriate relative concentration. This can get confusing if you're a stickler for nomenclature.)


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How can the standard deviation for a set of experimental value be interpreted in terms of precision?
I have read from many websites that the sample standard deviation can be used to measure the precision of experimental values. I understand that the lower the standard deviation the better, as it would mean that the values are close together (precise). But what standard deviation is small enough to be considered precise?
As an example, in a redox titration experiment I obtained titrant values of $16.3$, $16.4$, $16.5$ and $\pu{17.1 mL}$.
From these four measurements, I thus calculate an average of $\pu{16.58 mL}$ and a standard deviation of $\pu{0.35 mL}$.  The uncertainty of the burette is $\pm\,\pu{0.05 mL}$.
From this standard deviation, would this experiment's results be considered precise?
If not, why not, and how would I interpret it?
Is there a better method for quantifying precision?

No, it's not precise, 'cause 4th value is outlier.

One std dev corresponds to obtaining data that is within $\approx 68 $ % of the mean, ($\pm34$ %) this means that $32$ %of the time results will be outside this range by random chance and you seem to have observed one of these.  If you calc 2 std dev then your values should be outside this range only $\approx 5$ % of the time. Of course your results may not be accurate but that is a different matter.

But what standard deviation is small enough to be considered precise?

"Sufficiently precise," ironically enough, is something that can't really be defined precisely in general terms.
In other words, a level of precision that is perfectly fine in one context might be unacceptably low in others.
For example, if I'm trying to estimate how much table salt I need to buy to supply my catering business for the next six months, I might be perfectly happy with a not-very-precise calculation at $\pm\,30\%$, since salt is cheap and will keep just fine if I buy too much.
On the other hand, if I'm a pharmacist compounding a prescription containing a drug where the level in the body leading to nasty side effects is not all that much higher than the level that provides the desired pharmacological activity, I probably want to use (and am probably required by law to use!) methods and apparatus that'll give me, say, $\pm\,3\%$ precision on my measurements.

Is there a better method for quantifying precision?

Not really.  Pretty much every metric I've seen for quantifying precision involves the standard deviation in some way.  Depending on the situation, one might use:

$s$, the standard deviation by itself

$s/\bar x$, the standard deviation divided by the mean, also called the relative standard deviation or coefficient of variation (CV)

$s/\!\left(\bar x \sqrt n\right)$, the relative standard deviation divided by the square root of the number of measurements

This is equal to the standard error $\left(s/\!\sqrt n\right)$ divided by the mean

The "Comparison to standard deviation" subsection of the Wikipedia article on the coefficient of variation lists some situations where these different quantities would be useful relative to others.

From this standard deviation, would this experiment's results be considered precise?
If not, why not, and how would I interpret it?

In my opinion: Yes, these results are precise. In this case, since the mean value is far from zero and there's only one dataset being examined, I would use the coefficient of variation (CV) as my metric of precision.  With these values, I get a CV of $1.88\%$.  On the whole, I think this is an entirely reasonable CV for a titration experiment. The data could exhibit more precision, obviously, but that low of a CV implies very good repeatability to me.
As porphyrin noted, that $\pu{17.1 mL}$ measurement you obtained is perfectly reasonable in a statistical sense, as it's $25\%$ of a small dataset, residing $1.7$ standard deviations away from the mean. This puts it well line with the expected $32\%$ occurrence of values $x$ where $|x-\bar x|/s > 1.0$.
In other words, I disagree with Mithoron: I don't think that value is an outlier at all.  The usual metric I'm familiar with for identifying outliers is where $|x-\bar x|/s > 3.0$, which should occur only $\approx 0.34\%$ of the time for normally-distributed data.

The uncertainty of the burette is $\pm\,\pu{0.05 mL}$.

A brief note: I believe the small uncertainty on the burette as compared to the standard deviation in your data $(0.05 / 0.38 \approx 13\%)$ indicates that the majority of the spread in your measurements comes from elements of the experimental apparatus/procedure other than the burette.

Further reading: This article at Inorganic Ventures is a good writeup of the mean and standard deviation, and how they relate (or not) to accuracy and precision.


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What compound is this?
What compound is this please? A painkiller?

Please try to include full details when asking a question, otherwise it may sound you just want to get done with your homework. Include details that will be beneficial to other users as well.

It is diclofenac, which is described there as "applied to reduce inflammation and as an analgesic reducing pain in certain conditions".
By drawing the structure in emolecules (search mask here), I selected the entry as the free acid (and not the sodium salt, for example). A click on "LSKB" below the structure representation linked to the next database in which the restraint "where to find" was lessened to "chemicals in literature". Eventually I was guided towards this page, which a first entry name not following systematic nomenclature; an often seen pattern for pharmaceutical products:

To round the picture, if looking for physical chemical data, I would repeat the query, then including its salts (like the sodium salt).

This is also a good structure search engine that works in android and iphones as well:http://www.organic-chemistry.org/chemicals/structuresearch.htm


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What does the first $1$ mean in $\ce{Li}_2 1 ^1\Sigma_g^+: (1s\sigma_g)^2(1s \sigma_u)^2(2s \sigma_g)^2$?
I am trying to learn the nomenclature behind molecular states and in (Demtröder, 2010) I came across this expression:
 $$\ce{Li}_2 \color{red}{1} ^1\Sigma_g^+: (1s\sigma_g)^2(1s \sigma_u)^2(2s \sigma_g)^2$$
now I can explain the origin of every term except the $1$ which I have put in red. What does this 1 mean?

It probably denotes the state of the inner electrons.

@aventurin are you sure, this would seem like a very redundant piece of information  since I can already see what state the inner electrons are in - and in any case surly in most cases this will be $1$.

No, I'm not sure.

I could not find your example in Demtroder's Laser Spectroscopy book, but found on p575-576 $\ce{Na}3~ ^2P_{3/2}$ in which the $3$ refers to the Rydberg series starting at the 3P level so $n=3$ . He also uses notation such as $3\mathrm {p}^{3\mathrm P}$ to indicate a 3P state arising from a 3p level.

@porphyrin I think you are right, this pdf describes the classification on pg15-16 and that it has replaced the conventional, $X$, $A$, $B$... and $a$, $b$,... notations. If you want to write up an answer summarizing this I will accept it.

The conventional labelling of molecular states starts at X for the lowest state and A, B, C for higher ones and a, b, c, for equivalent states of the same symmetry but different multiplicity; see Herzberg spectroscopy books for many examples. These labels, however, reflect the order in which the states were discovered and so do not reflect their relative energies. It is becoming more common to precede a molecular state's symmetry label by an integer e.g. $1~^1\Sigma _g^+,~ 2~^1\Sigma _g^+,~3~^1\Sigma _g^+$ to reflect the energy ordering of states with the same symmetry. This notation reflects the way atomic levels of the same symmetry are labelled as $1$s, $2$s, $3$s, etc.
As an example, the conventional notation for the first four electronic states of  $\ce{C2}$ in order of energy is
X $^3\Pi_u$ ;  a $^1\Sigma_g^+$ ; b $^1\Pi_u$ ; A $^3\Pi_g$
which becomes
$1~ ^3\Pi_u$ ;  $2 ~^1\Sigma_g^+$ ; $1 ^1\Pi_u$ ; $3~ ^3\Pi_g$
In atoms the notation for a set of transitions starting from the 3p level in sodium is  Na $3~ ^2P_{3/2}$ .


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Tagging functional group
Is it a standard practice to name functional group by the greek alphabet of the carbon it is attached to?
Recently read a question in which the order of translation (protein synthesis) was asked and there were a couple of options with the $\ce{NH2}$ was labelled as $\ce{\alpha-NH2}$ and $\ce{COOH}$ as $\ce{\alpha-COOH}$.
I read wiki, it says H atoms are labeled in that way but are functional groups too labeled in the same way?

As you can see, they are.

Definitely not standard though. Standard is nothing but the IUPAC name. These are simply useful notations chemists use in certain chemicals as they have useful properties being that way.

In the study of amino acids (and peptides, proteins etc), it IS standard practice to label the alpha carbon as such. More carbons leading into the R-group may also be labeled with Greek letters, but this is not as necessary as having a label for the stereogenic center, the alpha carbon. Personally, I've found Greek letters helpful in remembering the structures of the different amino acids, for example I always think of Lysine as "the one with the epsilon-amino side chain". Note that in chemistry in general, whose nomenclature follows IUPAC guidelines, it is NOT "standard" to use Greek lettering, however it is still generally recognized. The practice is probably most prominent in the study of fatty acids (another biochemistry subfield), which results in such Greek -rich categorizations as "omega-3 unsaturated fats".


text:
Vapor Pressure Paradox
This is quite an interesting situation I had thought of. Consider two beakers, each containing $\pu{180 g}$ of water. Also, suppose one of the two has $\pu{1.8 g}$ of glucose dissolved in it. Let's call the one without glucose as A, and the one with glucose as B. I keep a small bell-jar that is large enough to just cover both of the two vessels. Now I will try to measure the vapor pressure inside the jar. Here's a little diagram to show what I mean:

I do not have access to bell-jars or manometers so I'll try to use Raoult's law to find the vapor pressure inside the vessel. The law in a simplified way would be:
$$P_{\text{solution}}=P_{\text{solvent}}x_{\text{solvent}}$$
Where $x$ denotes mole fraction, and solute is considered non-volatile. Keeping vessel A in mind, I get the vapor pressure as 23.8 mmHg, while applying Raoult's law for vessel B, I get the vapor pressure as:
$$P=23.8\left(\frac{10}{10.01}\right)=23.7 \text{mmHg}$$
Now, both of the values can't be true at the same time, and that's what I believe is the paradox.
How can this paradox be resolved?
My thoughts are that perhaps water from vessel A condenses back onto B, thus slowly increasing water content in B, increasing the vapor pressure. However, since theres solute already in B, the vapor pressure can never reach the value provided by the pure water.
What are your thoughts? Does the paradox ever resolve or does it remain in an unbalanced situation?

It is not a paradox. It only means those two are not in equilibrium and the solvent slowly moves between the two.

... and will reach equilibrium only when all the solvent in A has become vapour and solvent in B.

Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.
Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.
An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.

wait...I thought vapor pressure only depended on the temperature? why do small droplets have a different vapor pressure than large droplets?

Your post suggests (in fact, requires) you believe that Raoult's Law is accurate to 0.5%. It is not. Quantum Mechanics, the foundation of modern chemical theory, was created/discovered in the 1920s, Raoult's Law in the 19th Century. That's pretty good evidence that the law will be, at best, an approximation to the actual state of the system. As greg commented, you assume equilibrium - or at least, you should have understood that Raoult's Law requires equilibrium or near equilibrium. You are correct that since the two liquids have different vapor pressures, then equilibrium can not (yet) exist. The outcome would be (given an infinite amount of time, given a isolated system, given no bacteria nor oxidants present to consume the glucose, that the pure water (pure, right? no dissolved salts, but what about the dissolved air?) that the glucose solution would increase in mass & volume eventually overflowing its container. There are other considerations, which we ignore: surface tension and gravitational potential energy (adding more volume raises the height). One thing I learned fairly recently is that if you picture a cylinder of gas, the thermodynamics I learned in chemistry classes (including graduate level) and the thermodynamics I learned in freshman Physics, says that that cylinder (as a well insulated and closed system) will eventually attain thermal equilibrium. That is wrong. The temperature and density of the gas in the cylinder will depend on height. Thermal equilibrium will "never" be achieved.(although for a sufficiently "short" cylinder, the temperature and density gradient might be so small as to be impossible to measure.) ("never" meaning as long as the planet and cylinder exist. In the long run, the planet will most likely (we don't know for sure) be pulled into the Sun... but I digress...). Chemistry talks about many things, it is much more of a pragmatic science that Physics - which is more theoretical. (Although there may be more pragmatic physics used day-to-day than theoretical chemistry...) One way to divide chemical theory (there are many ways) is between Thermodynamics (equilibrium or near equilibrium) and Kinetics. I advise you always keep in mind whether a particular relationship or process or theory is thermodynamic (usually about energy, about path-independent initial and final states) or kinetics (about speed, shape, and specific process (path)).

You suggest that quantum mechanics trumps Raoult's law, but then you don't explain what you mean (or I missed it). Could you explain how quantum mechanics negates Raoult's law?


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If you add AlCl3 to water, why does the pH decrease?
You are not adding any hydrogen ions, so why does the pH of the solution decrease?
Thank you in advance.
EDIT: I will add the context of my asking this question to clarify some details. This is not a homework question. I am preparing for an upcoming examination in Chemistry, and I found this statement in the mark scheme of a practice exam I was doing, and yet I don't understand why it is true.
I have learnt pH as simply being the negative log of the concentration of H+ ions. I fail to see how adding the above salt changes the pH, especially since HCl is a strong acid (hence it completely dissociates in solution).
Thought process: Is the pH change a result of Aluminium forming Aluminium Oxide?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

@M.A.R. Hi, thanks for your comment. Please see my edit, marked "EDIT"

https://chemistry.stackexchange.com/questions/41675/mgcl2-acidic-or-neutral-in-water

Your thought process shouldn't be another question!

@PrittBalagopal Well, answering either question would in the process answer the other, i.e. they are linked. And as it is, my thought process is only questions, no answers!

A more clarified reaction in addition to Berry Holmes' answer:
$$\ce{AlCl3 + 4H2O <=> [Al(OH)4]- + 3Cl- + 4H+}$$
The reason this happens is the Lewis-Acidity of $\ce{Al^3+}$ as well as the higher affinity for $\ce{OH-}$ ions when compared with $\ce{Cl-}$.

Equilibrium of dissociation of complex won't get that far here, remember it's acidic solution Daniel's answer gets this right.

Hi tillyboy, thanks for your answer! Right now, I have 2 answers and I am not sure if they conflict. Could you thus clarify the following thought: Does Aluminium directly form the Al(OH)4 complex or does it first form Al(H2O)4 which then releases the H+ to form Al(H2O)3(OH) and so on until 4 protons are released? Thank you very much again.

@Mithoron: Well, that is the general equation of the complete reaction. Of course that reaction consists of many individual steps with their own equilibrium constant. You could determine through thermodynamical calculations where the equilibria of the individual constituent reactions are, although I am way to lazy to conduct said calculations :P (Actually after re-reading the question two times, where is it said, that the solution is acidic?)

@K.T. I am not quite sure about the mechanism here, but my best guess is actually that it is a mixture of both: The hydroxyl ions from dissociated $\ce{H2O}$ molecules attach to the aluminium, shifting the equilibrium of auto-dissociation, while other water molecules attach to the aluminium, gradually throwing away their protons. Note however that the equation above is a equation that sums up a lot of reactions but doesn't give any information about the equilibrium state. In the end you will be left with a solution of $\ce{[Al(H2O)4]^{3+}}$, $\ce{[Al(H2O)3OH]^{2+}}$

$\ce{[Al(H2O)2(OH)2]^+}$, $\ce{[Al(H2O)(OH)3]}$ and $\ce{[Al(OH)4]-}$, as well as your $\ce{Cl-}$, $\ce{H+}$ ions.

When you add $\ce{AlCl3}$ to water a reaction takes place:
$$ \ce{AlCl3 + H2O <=> Al(OH)3 + HCl} $$
But wait, now I've an acid, i.e. hydrochloric acid ($\ce{HCl}$) and probably a base ($\ce{Al(OH)3}$). Close enough, but $\ce{Al(OH)3}$ is actually amphoteric in nature, which means it can behave both as an acid and a base.
Think of it as a tennis match, on one side you've a strong player (Novak Djokovic, Roger Federer or Rafael Nadal – you name it) and on the other side you've a teen who has barely known tennis for a month. For sure, the strong player is going to win. But which of the two is the strong player in our case?
If you're guessing it to be $\ce{HCl}$, I'll give you points. I think $\ce{Al(OH)3}$ is double minded, it doesn't know for sure if it's going to act as an acid or a base. $\ce{HCl}$ on the other hand knows that it is a strong acid. $\ce{HCl}$ wins the game, making the solution acidic and so, the pH decreases.

Hi Berry, thank you for your very fun to read and pedagogical answer! In fact, my thought process was similar (although I did not state this in the post): Adding AlCl3 to water forms HCl, which is an acid.
However, since it is a strong acid, it is completely dissociated in solution, so I don't understand why the presence of Cl makes a difference to the pH?
In other words, before adding AlCl3, there were a certain number of H+ ions, and after there are still these many ions left. Why does the pH change?

@berryholmes Indeed great answer, any journal reference to this behaviour of AlCl3? What's your opinion about LaCl3 in water (aq) and its pH? Can you comment on the pH of LaCl3 in water and pKa of La(OH)3? Because LaCl3 (aq) rapidly forms La(OH)3 percipient.

First of all, $\ce{Al^3+}$ will separate itself from $\ce{Cl^-}$ in water, due to the nature of this salt (don't know why, the other answers have included this I believe).
Because in water, the $\ce{Al^3+}$ ion surrounds itself with $\ce{H2O}$ molecules, with the partially positively charged H-atoms turned towards the $\ce{Al^3+}$.
With enough electric force pulling all of the $\ce{H2O}$ atoms in there will be so many H2O molecules surrounding the $\ce{Al^3+}$, all being pulled in by the positive charge, that the molecules are too close together, and eventually one of the $\ce{H^+}$ ions is whipped out of this "$\ce{H2O}$ circle".
So $\ce{[Al(H2O)6]^3+}$ will become $\ce{[Al(H2O)5(OH)]^2+}$ + $\ce{H^+}$
Not all $\ce{[Al(H2O)6]^3+}$ 's will perform this action, and therefore it, like all metal aquo complexes, is a weak acid. For example, it works with $\ce{Fe^3+}$ as well.

Hi, thank you for your answer. However, I have some queries:

How does Al forming a complex with H2O ligands alter the pH?
I don't understand what you mean by "metals are all weak acids".
Thanks for your answer, my knowledge is roughly secondary school level as well.

@K.T. 1) because there are so many H2O molecules surrounding the Al3+, all being pulled in by the positive charge, that the molecules are too close together, and eventually one of the H+ ions is whipped out (will edit this bit in) 2) all positive metal-ions are weak acids, it's just a fact, and it's an explanation to the "do sometimes" bit before the brackets.

@airhuff ty for the edit, haven't got a clue how to do subscript and superscript.

No problem. See this for a discussion/tutorial on how to use MathJax formatting for chemical and mathematical equations and such. It's a pretty small learning curve ;)

"don't know why, the other answers have included this I believe" Do you answer based on other answers to the question? See this for why ions split: https://en.m.wikipedia.org/wiki/Dissociation_(chemistry)


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Hybridization of heteroatom
Why is the hybridization of oxygen in the seven membered ring pictured below $sp^3$, while the oxygen has a hybridization of $sp^2$ in the 5 membered ring? This also may be an error in the answer key as well.

Count the number of pi electrons - what would happen if that oxygen were to be sp2 instead of sp3?

Honestly, this is a bad question. Are we assuming the ring is flat? If it's puckered, you could make a case for $sp^{2}$. If it's flat, $sp^{3}$ is probably better...

It relates to the Hückel rule for aromatics. With the electron pair of the oxygen participating in the $\pi$ system of the oxazol (5-membered ring)  we get 6 $\pi$-$\ce{e-}$. Doing the same for the oxazepin would get 8 $\pi$-$\ce{e-}$.
Hückels rule states that planar, circular $\pi$ systems with $4n+2$ electrons are aromatic, resulting in a lower overall energy if the system is conjugated. (with $n$ being a positive integer or zero)
The other statement that Hückels rule does is that $\pi$ systems with $4n$ electrons are anti-aromatic, which avoid conjugation as it would increase the overall energy of the system.
Therefore it is favoured for your oxazol to conjugate, which results in a sp$^2$-hybridization for the oyxgen, while the oyxgen in the oxazepin avoids conjugation, resulting in a sp$^3$-hybridization.
The Hückel rule in turn can be quantized by a Hückel method computation or  graphically utilizing a Frost-Musulin-diagram.


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Chemistry of solids: allowed wavevectors
I'm a complete beginner in the chemistry of solids, and many of the things I came across don't make sense. I have two subquestions, which I think it would be most logical to post as a single question.

The longest wavevector of a commensurate matter wave is $k_{max} = \pi / d$, where d – lattice-plane spacing, so that amplitude minima and maxima of the wave  coincide with adjacent lines / planes of atoms. I don't understand this explanation – why is, for example, $k = 2\pi / d$ inconsistent?
Bloch's equations dictate that the value of the electron wavevector isn't uniquely defined, and the vector is uncertain to modulo G, where G – a general Reciprocal Lattice (RL) vector. This means that wavevectors greater than $\pi / d$ are permitted and can be found in Brillouin's zones of order greater than 1 in the RL. This seems to be in contradiction to the point above – what am I missing?

Could you cite where you are getting these statements?

@QuantumAMERICCINO From my undergraduate course and the textbook "The Physics and Chemistry of Solids" by S. Elliott, although I'm not sure if I'm interpreting them correctly

In a crystal, electron waves are reflected off the periodic lattice and this (Bragg) reflection is the cause of the energy gaps also called a band structure. If there is a weak periodic potential in a 1D crystal then the energy gaps occur at wavevector $k=\pm \pi/d$ where d is the lattice constant.
If $\bar G$ represents the reciprocal lattice vector then the general Bragg condition for diffraction of a wave with wavevector $\vec K$ can be written as
$(\vec k + \vec G)^2=k^2 $
In one dimension this condition becomes
$$k=\pm \frac{1}{2}G = \pm n\pi/d$$
and $G=2\pi n/d$ is a reciprocal lattice vector and $n=1,2,3..$  an integer. This is where your factor of 2 appears.
The reflection at $k=\pm \pi/d$ occurs because one wave is reflected of an atom and interferes constructively with phase difference $2\pi$ with one reflected from the nearest neighbour atom. The Brillouin zone is the $\bar k$ space between $-\pi/d$ and $+\pi/d$.
(The reciprocal lattice vectors have the form $\vec G =h\vec A+k\vec B + l\vec C$ for integers h, k, l. The vectors are defined as
$$\vec A= 2\pi \frac{b~ \times~ c}{a\cdot b~ \times~ c}~~
\vec B= 2\pi \frac{c~ \times~ a}{a\cdot b~ \times~ c}~~
\vec C= 2\pi \frac{a~ \times~ b}{a\cdot b~ \times~ c}$$
for primitive lattice vectors a, b, c)


text:
Explain mechanism for first step in synthesis of acetone and phenol
"The following reaction is the first step in the industrial synthesis of acetone and phenol. AIBN initiates radical reactions by breaking down upon heating to form two isobutyronitrile radicals and nitrogen gas. Using an isobutyronitrile radical to initiate the reaction, write a mechanism for the following process."

I was able to write a mechanism but I have to explain it to the class and I don't know how.

Count your electrons and your hydrogen atoms. Your product isn't quite right.

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

You still need to write a mechanism that explains how the radical leads to the product

AlBN is a sideway reagent (radical initiator). Cumene is the reactant and basicially reacts with radical O2(.) Since you create peroxide is a radicals mechanism (not neutral ALBN product that you wrote).
Essentially is the Hock process mechanism. You should put the isopropyl cyanide radical in place of R, in the following mechanism and you are good.
https://upload.wikimedia.org/wikipedia/commons/thumb/a/af/Hock-Phenol.png/751px-Hock-Phenol.png
Nevertheless, the first step is the creation of isopropyl cyanide radical. The mechanism is: https://upload.wikimedia.org/wikipedia/commons/thumb/5/56/Formation_of_Radicals_from_AIBN.png/1280px-Formation_of_Radicals_from_AIBN.png
DO NOT CONFUSE HALF-ARROW SYMBOL (which indicates movement of one electron) with DOUBLE (2 electron movement)


text:
Carboxylate salt of Iron
I read somewhere that Fe does not react with formic acid. Because carboxylic acids are very weak. But by searching I found an article about carboxylate on wiki. It mentions about carboxylate metal salts. So,  is there carboxylate of Iron? What is its chemical formula and how is it prepared?

This paper has some discussion about iron carboxylate salts (beyond just the below excerpt): several candidate materials, including FeO(OH) from Aldrich (iron(III) oxide, hydrated; catalyst grade,  30–50  mesh;  cat.  #  371254),  did  successfully  dissolve in carboxylic acids such as oleic acid to form iron carboxylate salts..

Yes you are right. Iron does not reacts with formic acid directly to form iron formate. You need iron salt and acetate salt so that the spectator ions combines with each other to form iron formate and another salt. You can observe this in my answer.

The formula of carboxylate ion is $\ce{RCOO-}$ where R=alkyl group or hydrogen. Iron forms two types of ion $\ce{Fe^2+}$ and $\ce{Fe^3+}$. So the general formula for iron carboxylate would be $\ce{Fe(RCOO)2}$ or $\ce{Fe(RCOO)3}$. Here I will discuss iron acetates(R=CH3-)and iron formate(R=H-).

Iron(II) acetate

Iron(II) acetate is an coordination complex with formula $\ce{Fe(CH3COO)2}$. It is a white solid, although impure samples can be slightly colored. A light green tetrahydrate is also known, which is highly soluble in water.
Iron powder reacts with hot acetic acid to give the product.
$$\ce{Fe + 2CH3COOH → Fe(CH3COO)2 + H2}$$

Actually It is not a salt. It adopts a polymeric structure with octahedral Fe(II) centers bridged by acetate ligands.

Iron(III) acetate

Ferric acetate is the coordination compound more commonly known as basic iron acetate. With the formula $\ce{[Fe3O(OAc)6(H2O)3]OAc}$ where $\ce{OAc− = CH3COO−}$, it is a salt, composed of the cation $\ce{[Fe3(μ3-O)(OAc)6(H2O)3]+}$ and an acetate anion. The formation of the red-brown complex has long been used as a test for ferric ions.
Basic iron acetate is produced by treating aqueous solutions of iron(III) salts with acetate salts.Another process is solid iron is mixed with hydrogen peroxide to form iron(II/III) hydroxide, which then react with vinegar/acetic acid or acetate salts to form iron(III) acetate.

Early work showed that it is trinuclear. The Fe centres are equivalent, each being octahedral, being bound to six oxygen ligands, including a triply bridging oxide at the center of the equilateral triangle. The compound was an early example of a molecular compound of iron that features an oxide ligand. Ignoring its 24 hydrogen centres, the cation has D3h symmetry.

Iron formate

Iron(II) formate dihydrate($\ce{Fe(HCOO)2.2H2O}$)

Prepared by the double decomposition reaction between ferrous sulfate and sodium formate in  presence of an acid.

Iron(III) formate dihydrate($\ce{Fe(HCOO)3.2H2O}$)

Iron(III) formate dihydrate is obtained from freshly precipitated
iron(III) hydroxide dissoluted in excess formic acid, and subsequent crystallization by evaporation in vacuum at 35 C.

Excellent answer, very thorough. +1

You put C-Fe bond in formate...

@Mithoron, I couldn't find any picture of ionic form of iron(II) formate

Then delete it or draw a new one, now it may be misleading.


text:
Wittig versus Corey–Chaykovsky reaction
The Wittig and Corey–Chaykovsky reaction both involve an ylid, phosphonium ylid in the former versus a sulfonium ylid in the latter, attacking a carbonyl compound.
In the Wittig reaction, a concerted [2+2] cycloaddition leads to direct formation of a four-membered ring, but this does not happen with the sulfur ylids. Instead, nucleophilic addition to the carbonyl occurs to give a betaine, which then decomposes via intramolecular nucleophilic substitution by the oxyanion. Thus, in the Corey–Chaykovsky reaction, we end up with an epoxide, while the Wittig reaction affords an alkene.
Possibly formation of the strong $\ce{P=O}$ bond is a key driving force for the wittig reaction (the P=O bond strength in $\ce{Ph3PO}$ is $\pu{529 kJ mol-1}$ vs. the $\ce{S=O}$ bond strength in $\ce{Ph2SO}$ which is $\pu{367 kJ mol-1}$).
I am looking for an answer that explains the differences between the outcomes of these two reactions, and their respective intermediates. It would be great if the answer can be supported with references, and/or calculations.

Overview of mechanisms
There have been recent excellent summaries of the reaction profiles of the Wittig1 and the Corey–Chaykovsky2 reactions. For the purposes of answering this question, there's no need to go into exquisite detail about each individual step (a lot of the research has been poured into understanding stereoselectivities which isn't relevant here).
The general mechanisms are shown below, which should suffice; however, the interested reader is encouraged to consult the articles above.

Both reactions are considered to be essentially irreversible. As such, an argument based on the thermodynamic stability of the products cannot be valid.
For a kinetically controlled reaction, one should look at the activation energies of betaine vs. oxethietane or oxaphosphetane formation from the ylid and carbonyl.

Phosphorus ylids
The mechanism of the Wittig reaction has been studied very thoroughly, and the general consensus is that betaines are not intermediates. See: Which is the currently accepted mechanism of a Wittig reaction?
As described in the answers there, true betaines have never been observed as intermediates in "salt-free" Wittig reactions, and have been rigorously disproven in some cases. Betaines can be obtained by deprotonation of their conjugate acids, called β-hydroxyphosphonium salts. However, the only observed products of this deprotonation are either oxaphosphetanes3 or the alkene + phosphine oxide products4. There is a large body of evidence for a four-membered transition state leading directly to an oxaphosphetane.1
As for computational studies, Harvey and Aggarwal5 found a betaine intermediate at 14.6 kcal/mol higher than the cis oxaphosphetane for the reaction of $\ce{Me3P=CHMe}$ with acetaldehyde, but could not find a corresponding transition state. For the reaction with $\ce{Ph3P=CHMe}$, the betaine was found to be 16.2 kcal/mol higher than the oxaphosphetane, and the TS leading to the betaine 3 kcal/mol higher than the TS leading to the oxaphosphetane.
If the phosphorus betaine is never formed, the next step to form an epoxide (intramolecular nucleophilic substitution) cannot occur. Hence, the Corey–Chaykovsky reaction will not occur with phosphorus ylids.

Sulfur ylids
Very few cases of oxathietanes are known. A rare case was provided by Okazaki in 1996,6 characterised by NMR and crystallography:

However, the oxathietanes possibly formed in a typical Corey–Chaykovsky reaction differ from this in structure. For the model reaction of $\ce{H2S=CH2}$ with formaldehyde, an early computational study found a local minimum corresponding to an oxathietane with S–O bond length of 2.10 Å,7 and postulated that a Wittig outcome (alkene formation) did not occur because of the high activation energy to pseudorotation about sulfur as well as [2+2]-cycloreversion. However, these studies also did not take solvation effects into account, which will stabilise the highly charged betaine.
In a more recent DFT study, Harvey and Aggarwal2 were able to reproduce the molecular geometry of Okazaki's oxathietane, proving that there was no inherent computational bias against such structures. However, attempts to find a stable oxathietane from the addition of $\ce{Me2S=CH2}$ to formaldehyde, or from the addition of $\ce{Me2S=CHPh}$ to benzaldehyde, failed. In particular, the S–O distances in the optimised intermediates were 2.81 and 2.48 Å respectively (the full geometries are in the supporting information). Geometry optimisation starting from an oxathietane structure also yielded nothing but betaines.
Likewise, if an oxathietane cannot be formed, then the decomposition to alkene and sulfoxide cannot occur. Hence, sulfur ylids do not undergo the Wittig reaction.

Why the difference?
Harvey and Aggarwal2 wrote of their inability to find oxathietanes:

Clearly, the energetic gain associated with bonding between these atoms [sulfur and oxygen] is more than offset by the ring strain it would entail.

Evidently, the key to understanding the different behaviour lies in bond strengths: not the double bonds in the product, but the single bonds in the oxetane intermediate. Kutzelnigg8 has found binding energies for $\ce{H2P-OH}$ (relative to $\ce{PH3 + O}$) and $\ce{HS-OH}$ (relative to $\ce{SH2 + O}$) to be 114 and 59 kcal/mol respectively. A qualitative explanation is not easy, and I am not really inclined to try.

Extra: Phosphorus betaines
As mentioned above, phosphorus betaines can be independently generated by deprotonation of β-hydroxyphosphonium salts. If we ignore the immediate cyclisation to form oxaphosphetanes, it is theoretically possible that these betaines could undergo intramolecular nucleophilic substitution to yield epoxides:

In a 2005 study, Aggarwal and Harvey again showed that such a process was much more energetically unfavourable for phosphorus than for sulfur.9 This was mainly rationalised by the poor leaving group ability of $\ce{PR3}$ compared to $\ce{SR2}$. Firstly, the intrinsic leaving group ability of phosphorus is lower (this is a purely kinetic measure of leaving group ability, based on the activation energy of the reaction $\ce{X + CH3X+ <=> X+CH3 + X}$). Secondly, the C–P bond is stronger than the C–S bond, which adds a further thermodynamic cost to epoxide formation. The difference in the energy profiles is striking:

For $\ce{X = PMe3}$, epoxide formation from the betaine is associated with an activation energy of 28.3 kcal/mol. Compared to an essentially barrierless cyclisation to the oxaphosphetane, it is easy to see why such a process cannot compete.
For $\ce{X = SMe2}$, on the other hand, the activation energy is only 5.4 kcal/mol. In fact, as the authors had noted in their earlier paper,2 the major challenge in the Corey–Chaykovsky reaction is getting the betaine into the correct conformation for intramolecular $\mathrm{S_N2}$ (i.e. going from 4-syn to 4-anti in the scheme above)!

References

Byrne, P. A.; Gilheany, D. G. Unequivocal Experimental Evidence for a Unified Lithium Salt-Free Wittig Reaction Mechanism for All Phosphonium Ylide Types: Reactions with β-Heteroatom-Substituted Aldehydes Are Consistently Selective for cis-Oxaphosphetane-Derived Products. J. Am. Chem. Soc. 2012, 134 (22), 9225–9239. DOI: 10.1021/ja300943z.
Aggarwal, V. K.; Harvey, J. N.; Richardson, J. Unraveling the Mechanism of Epoxide Formation from Sulfur Ylides and Aldehydes. J. Am. Chem. Soc. 2002, 124 (20), 5747–5756. DOI: 10.1021/ja025633n.
Vedejs, E.; Marth, C. F. Mechanism of Wittig reaction: evidence against betaine intermediates. J. Am. Chem. Soc. 1990, 112 (10), 3905–3909. DOI: 10.1021/ja00166a026.
Maryanoff, B.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108 (24), 7664–7678. DOI: 10.1021/ja00284a034.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.
Ohno, F.; Kawashima, T.; Okazaki, R. Synthesis, Crystal Structure, and Thermolysis of a Pentacoordinate 1,2λ6-Oxathietane: An Intermediate of the Corey−Chaykovsky Reaction of Oxosulfonium Ylides? J. Am. Chem. Soc. 1996, 118 (3), 697–698. DOI: 10.1021/ja953533u.
Volatron, F.; Eisenstein, O. Wittig versus Corey-Chaykovsky Reaction. Theoretical study of the reactivity of phosphonium methylide and sulfonium methylide with formaldehyde. J. Am. Chem. Soc. 1987, 109 (1), 1–14. DOI: 10.1021/ja00235a001.
Wallmeier, H.; Kutzelnigg, W. Nature of the semipolar XO bond. Comparative ab initio study of H3NO, H2NOH, H3PO, H2POH, H2P(O)F, H2SO, HSOH, HClO, ArO, and related molecules. J. Am. Chem. Soc. 1979, 101 (11), 2804–2814. DOI: 10.1021/ja00505a002.
Aggarwal, V. K.; Harvey, J. N.; Robiette, R. On the Importance of Leaving Group Ability in Reactions of Ammonium, Oxonium, Phosphonium, and Sulfonium Ylides. Angew. Chem. Int. Ed. 2005, 44 (34), 5468–5471. DOI: 10.1002/anie.200501526.


text:
In an octahedral complex, what happens to the electrons donated by the ligand?
I am trying to understand how complexes are coloured. After some reading, I found out this was due to the d-d splitting induced by the coordinate bonds of ligands to the central metal ion. The electrons in the d-orbitals then transition between each other, and absorb light in the process.
However, I do not understand what happens to the electrons donated by the ligand. Don't they form a sigma bond with the d-orbitals of the metal ion? In which case, wouldn't they go into the d-orbitals and 'fill it up', thus disallowing transitions?
Or is it the case that the d-orbitals of the central metal ion are not involved in the bonding at all?

This is a very good question, I must say. It requires the understanding of the very fundamentals.
You're right, if the electrons from the ligand pair up with the electrons of the metal, the electrons cannot undergo $d$-$d$ transitions, and no color would be seen.
But do they pair up? Think carefully. Recall one of the most basic assumptions of the Crystal Field Theory: The attraction between the ligand and the metal is assumed to be purely electrostatic in nature. This very important assumption is the sole reason for the explanation of colors of these complexes. As the ligands approach the metal atom, they create a crystal field. These fields tend to repel the electrons of the metal and increases their energy. A perfectly symmetric spherical  crystal field would increase the energy level of all the orbitals equally and to the same level. But this isn't the case in actual coordination complexes. An octahedral crystal field would approach along the axes of the $d_{x^2+y^2}$ and $d_{z^2}$ and tend to repel them to a greater extent than the other orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$). This difference in energies of the orbitals is known as crystal field splitting. We can now group these orbitals into two groups, three low energy $t_{2g}$, and two high energy $e_g$ orbitals.

Keep in mind that the ligands do not pair with the electrons of the metal. They simply repel the orbitals electrostatically and increase their potential energy in the process.
So, now how do the complexes actually get their color then? When the orbitals split, the difference in their energies is called Crystal Field Stabilization Energy (CFSE) and is denoted by $\Delta_{\text{o}}$. When photons of light are incident on the complex, it absorbs the photons which possess the energy equal to that of the value of $\Delta _{\text{o}}$. From quantum theory of electromagnetic waves, it's known that the energy of a photon is given by:
$$U = \frac{hc}{\lambda}$$
Where $h$ is known as the Planck's Constant, with a value of $\pu{6.626×10^{-34}Js}$, and $\lambda$ is the wavelength of light. If the value of the wavelength lies in the visible light spectrum, you can find the color of the light absorbed by the complex.
How do you find the color emitted out? There's a very easy and fun way to find that out. Take a look at this color wheel:

Find where the color of absorption lies. Then the color of the complex lies on the opposite side of the color of absorption.
This is a brief discussion about the colors of complexes. Let me know if you need more clarification.

Hi, thank you very much for your answer. However, I still do not understand where the ligand's donated electrons go...Do they go into any orbital of the transition metal ion?

@K.T. The electrons remain with the ligand itself. The ligand electrons simply repel the metal orbitals electrostatically.

But doesn't the fact that they form a coordinate bond with the transition ion imply that they don't remain within the ligand? Thanks.

@K.T. No no no.., forget the very idea of coordination.  Crystal field theory makes sure that there's no coordination or donation of electrons, just a simple electrostatic attraction.

@PrittBalagopal, you might want to clarify your use of the word 'ligand field', in CFT (which is what you're describing), the ligands are just point charges, the idea of a ligand field/ligand field theory is a somewhat more complex theory that makes use of molecular orbital theory.

@PrittsaysReinstateMonica Why f or p orbitals don't split?

An alternative to the crystal field theory approach presented in Pritt Balagopal's answer is to make use of ligand field theory. Neither theory is perfect, and crystal field theory can explain many properties of coordination compounds (even if not always for the right reason).
Ligand Field Theory
Ligand field theory (LFT from here on in) is a newer and more in-depth way of considering the bonding/properties of coordination complexes, and was developed by combining crystal field theory (CFT) with molecular orbital theory (MO theory). The major difference is that the ligands are no longer considered as electrostatic point charges.
LFT works by taking the orbitals on the metal centre (s, p, d) and mixing them with a set of (symmetry adapted group) orbitals which represent the ligands (doing this requires some treatment with group theory). The process of doing this is a question in itself (and has been discussed on chem.SE previously), but what we really care about right now is the results obtained from the ligand field approach.
With LFT, the resulting diagram differs depending on the type of ligands involved (in CFT, there is an octahedral crystal field which is largely the same, irrespective of the ligand type, since all ligands are considered as point charges).
What remains the same is that we still end up with an eg MO and a t2g MO, its just that the orbitals used to obtain those are different depending on the ligand.

Fig 1: Ligand field diagrams for three common ligand types, reproduced from Ligand Field Theory and its applications, Figgis.
Once we have our ligand field diagrams, we simply have to fill with electrons (taking into account the metal and the ligands). If you go through and do this, you should end up with a diagram such as that in figure 2.

Fig 2: Filled ligand field diagrams, reproduced from Inorganic Chemistry, Meissler and Tarr.
On the left hand side of the above diagram, is the key portion of the filled ligand field diagram for the CN case, and on the right hand side, the key portion of the filled ligand field diagram for the F case.
From CFT, you still would have gotten to the fact that the t2g orbital ends up filled with 3 electrons in both cases. What differs (and where the power in LFT lies) is that we can now see where these electrons come from (i.e. the contributions).


text:
How does the partial pressure of gas C change with the change in volume of the container?
Suppose I have the equilibrium in a closed container:
$$\ce{3A(s) + 4B(g) -> 4C(g) + D(s)}$$
What happens to the partial pressure of C if the volume of the container is halved?
I was pondering about this question, because by the gas equilibrium constant
$$K_\mathrm{p} = \frac{[C]^4}{[B]^4}$$
 the partial pressure of each should remain equal to each other. But using an ice table, $K_\mathrm{p} = \frac{[C]+x}{[B]-x}$, or you can flip it around. Either way, $x$ should equal 0 by this equation for them to remain the same, which means that the partial pressures don't change, which then means that the gases are turning into the solids. But by the chemical equation 1 mole of one gas will produce 1 mole of the other gas, so how does the gas even turn into the solid? So wouldn't the partial pressures of both increase?
Another way to think about it is to imagine a container with only B and C. That means a reaction cannot even occur, so when volume decreases, the partial pressures increase.
This is my thought on the problem. Can someone please explain to me what really happens?

Partial pressures change, but they change in such a way that the equilibrium constant remains unaffected. Also, a fundamental flaw in your reasoning is that you've written the expression of $K_c$ and not $K_p$.

The equilibrium constant $K_P$ is independent of pressure but does depend on temperature which we assume to be constant. The equilibrium constant $K_P$ is the ratio of partial pressures $p_i$ of the gaseous substances raised to the stoichiometric powers, $K_P= p_C^4/p_B^4$ and is defined at a standard state of $1$ atm. pressure.
Writing the equilibrium constant for the reaction $\ce{mB <=>lC}$ in terms of the mole fraction $x_i=n_i/n$ for n total moles gives $K_x= x_C^l/x_B^m$ where l and m are constants, both four in your question. The mole fraction i of a species behaving as a perfect gas can also be written  $x_i=p_i/P$ where $p_i$ is the partial pressure and P the total pressure.
Rewriting the equilibrium constant in terms of partial pressures of B and C gives
 $$K_x=\frac{p_C^l}{p_B^m}P^{m-l}= K_PP^{\Delta n}$$
where $\Delta n$ is the change in the number of moles of the gases. Thus unless the change in the number of moles is zero, which it is in your example, the mole fractions of the components  will change with change in total pressure even though $K_P$ is constant. If $\Delta n > 0$ the pressure will reduce the mole fraction of products.

If you half the volume then pressure will double. Now by Le Chatelier's principle the equilibrium will shift in order to lessen the pressure, which means the equilibrium must shift to that side where number of moles of gaseous species are less.But in your reaction number of gaseous moles on both sides of the equation are same.Hence there will be no change in equilibrium.


text:
Product of Friedel-Crafts cyclisation of (3-methylpent-4-en-1-yl)benzene
My proposed mechanism and final product is:

But the book says the answer is:

I know that a 6 membered ring is more stable than a 5 membered ring so I made the product with the 6 membered ring (however, the carbocation was a secondary carbocation).
However, the book claims the answer to be the one with the 5 membered ring (the carbocation here was a tertiary carbocation).
Is the answer provided in the book wrong?

Then I suppose the carbocation thus formed rearranges to 3° and then is being attacked by benzene. Later through electrophilic substitution the product given in the book is obtained.

Related: Why are 7-membered rings less likely to form than 5- and 6- membered rings?

The comments seem to underestimate the fact that a tertiary carbocation is more favorable due to inductive effect. So, then it will form pentacarbon ring on the second step. And I don't think there is much strain in a pentacarbon ring.

I think the carbocation will rearrange to a more stable one.

Though the question is old and pretty much answered in the comments, I ran into this question and draw the mechanism, so I thought about summarizing all that was brought up in an answer.
The carbocation is first formed on the secondary carbon. However, molecules are more stable when the charges are delocalized (this page talks about the stability of carbocations). When the positive charge is in the secondary carbon, it can only "share the burden" of having this charge with 2 neighbor carbons. So the tertiary carbon donates an hydrogen through a sigma to atomic orbital interaction (the sp3 hybdrization allows a π-type interaction) resulting in the more stable tertiary carbocation seen in the mechanism below:

Then the reaction follows analogously to the pathway you deduced for the 6 membered ring: First a π to atomic orbital interaction between the aromatic ring and the tertiary carbocation, which I believe to be the Rate Determining Step since it breaks the ring aromacity (bigger energy gap). Then finally the sigma to atomic interaction between the hydrogen and the ring carbocation, that restores aromaticity and releases the hydrogen.


text:
Regioselectivity in Friedel–Crafts acylation of thiophene

In the above reaction, I  understand that $\ce{AlCl3}$ is acting as a Lewis acid to generate an acylium ion, which is a strong electrophile.
Would this electrophile attack at carbon 2 or 3?
I think it would attack at C-3, as the resulting positive charge at C-2 would be stabilized by resonance using the lone pair on sulfur.

Related: Regioselectivity in electrophilic substitution of pyrrole

Friedel-Crafts acylation reactions are well documented on thiophenes, and go with high regioselectivity for the 2-position (or 5-position, which is equivalent in the case of unsubstituted thiophene) rather than the 3-position (or 4-position). Joule and Mills' Heterocyclic Chemistry (5th ed., p 327) gives the following two examples:

The regioselectivity can be explained through consideration of the intermediates involved when the thiophene attacks the electrophilic carbenium species.

One is able to see that the intermediate species produced by attack at the 2-position is able to be drawn in three resonance forms, whilst the intermediate produced by attack at the 3-position is only able to be drawn in two resonance forms. As a general rule of thumb, the more resonance forms one can draw, the more 'stable' and intermediate, and hence the energy required to access it is lower, hence in the case of thiophene we get preferential electrophilic substitution at the 2/5-position.
(When looking at the intermediates, we can also consider the type of conjugation involved. Attack at the 2-position yields a linearly conjugated intermediate, which is lower in energy than the cross conjugated intermediate from attack at the 3-position).


text:
Why does neighbouring group participation increase the rate of reaction?
I don't see any reason why the transition state of an SN2 reaction that happens after an internal nucleophilic group attacks(which is known as NGP) would be more stable. In fact, the nucleophilic group that attacks is often a strong nucleophile that is quite reactive (more than the leaving group which is usually a weak base) which would only destabilise the transition state when the external nucleophile does attack.
My question is: Why is the internal attack and the external attack combined faster than simply the external attack since the rate of such reactions where neighboring groups are involved is greater than normal SN2 reactions?

I suppose these neighboring groups switch on abilities like dipole in the bond and make way for easier substitution.

Think entropy, the neighboring group is in the same molecule and held in close proximity.

Possible somewhat answered here: https://chemistry.stackexchange.com/a/65872/21296

@NotBaran No, that does not answer my question at all. I understand the reason NGP occurs, I just don't understand why it increases the overall rate of the reaction

This is a bit of a tautology, as you would only hear about NGP if it increased the rate of the reaction. If there was a NGP pathway available that didn't increase the rate of the reaction, it would simply not be mentioned.

In neighbouring group participation (NGP), an internal nucleophile first displaces a leaving group in an intramolecular sense, forming a ring (usually 3 or 5 membered) which is then opened by the incoming nucleophile intermolecularly.

General mechanism of neighbouring group participation (taken from March, Advanced Organic Chemistry, 2007

In the scheme above, Z first displaces X to form a 3 membered ring, which is then opened by Y. Your question then becomes why is Z faster than Y at substituting X?
Collision
In order for the substitution to take place, the nucleophile must come into alignment with the leaving group. In a general sense, intramolecular reactions are usually faster than intermolecular since the two reacting groups are tethered together – there is no requirement that they collide in order to react. This is especially evident for intramolecular formation of 3 and 5 reasons, since those ring sizes are especially favourable. (Note that intramolecular reactions isn't always favourable, for example in macrocyclisations of large rings, where high dilution is needed to close the ring rather than getting intermolecular reactivity).
Entropy
The reaction between the substrate and the external nucleophile Y in an intermolecular fashion is accompanied by a decrease in entropy (two molecules are becoming one in the transition state). Whilst this also happens in the intramolecular sense (the molecule is becoming more rigid), it is significantly less than in the bimolecular case.
Its also worth pointing out that its not always 100% clear whether a molecule is reacting via an NGP type pathway, and indeed there are many cases where its been shown that both are operating at the same time, with the intra being faster than the inter, but not sufficiently rapid to prevent any direct intermolecular reaction.

My question was: why is the internal attack+the external attack combined faster than simply the external attack since the rate of such reactions where neighboring groups are involved is greater than normal Sn2 reactions

You may wish to edit your post - that comment is a lot clearer than your original question