text:
Which affects acidity more, in relation to induction? (Distance vs. quantity)

I already know that the answer is III. I'm just looking for further explanation on how the others compare to each other, mainly IV vs V.
I know that iodine is less electronegative than chlorine, making I the least acidic.
Only having one chlorine atom, while the rest have two, puts II next in line.
III is the most acidic, due to the two chlorine atoms being on the same carbon, and closest to the sulfur group.
My question comes in for the comparison of IV and V. While V has two chlorine atoms on the same carbon, they are both 3 carbons away from the sulfur group. Meanwhile, IV has one chlorine atom closer to the sulfur group, but the other isn't on the same carbon atom.
Which one of these is more effected by induction? Proximity to the sulfur group (IV) or being grouped together on a single carbon (V)?

Why do you think structure I is least acidic?
Consider the following facts: In water, HI is more acidic than HCl, and 2-Iodoethanol is more acidic than 2-Chloroethanol.
How does (or does it?) Chlorine being more electronegative than Iodine make the chloro-substituted compounds more acidic?

I'm still pretty new to this so I may be wrong. But I was under the impression that polarizability is only the stronger influence on pKa in binary acids. So HCl is more acidic than HF, because Cl is the larger and more polarizable atom (the larger the atom, the easier it is to stablize the negative charge). However, in more complex structures where the halogens aren't directly attached to a hydrogen, inductive effects are the stronger influence. So, for example, when comparing fluoracetic acid to chloroacetic acid, the former has a pKa of ~2.59, while the latter of ~2.86.

Fluoroacetic acid is more acidic than Chloroacetic acid in water, but not in DMSO, or the gas phase where there're no solvent effects. Now, in terms of chemical bonding, a more electronegative substituent will usually make the bond homolytically much stronger. A calculation of C-H bond enthalpies in Fluoroaceticacid vs Iodoacetic acid would reveal, C-H bond enthalpies in the former are larger than the latter. Yet, the acidity values are different. It is clearly more complex than "polarizability" and "electronegativity", it could be important to think about how these affect acidity.

Inductive effects decrease strongly with increasing distance. Thus, the further away an inductively acting group is, the lower its influence will be. For practical intents and purposes, the otherwise very comprehensive Brodwell $\mathrm{p}K_\mathrm{a}$ database does not even list any β-halo compounds; only the difference between α-hydrogen, carbon and halogen is typically given.
Thus, we can expect compound IV to have practically the same acidity as compound II, and that compound V’s acidity is comparable to the unhalogenated pentan-4-thiol. Considering that all chlorines are very far away from the acidic proton, the effects would be smaller still.


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In a sample of radioactive substance all the atoms don't disintegrate simultaneously. Why?
Radioactivity though spontaneous all the atoms don't disintegrate at a time

Why would they? An atom doesn't have any sort of internal clock. It doesn't "know" what time is it.

Excellent question, have you tried to ask physicists? If they don't know, chemists don't know. We can tell you why this doesn't happen with chemical reactions, but that's a different story I think (though I suspect some of the argumentation is similar)

Crude example would be to take a big hand full of dice. Roll a six and the particular die dies (decays), anything else the die lives. So for each roll the die has same chance of dying. Some die will die on first roll, some will last a large number of rolls.

The basic concept of radioactive decay kinetics is that every single nucleus has the same probability of decaying at any single given time, regardless of the physical-chemical surroundings (it does not know anything about the other atoms). This of course means that they do not disintegrate all at once, because the probability of all of them "deciding to decay" at the exact same time is infinitesimally small.

Spontaneous breakdown has no external cause. To a close approximation, breakdown products do not cause/trigger/contribute to subsequent decay events.   The
other-than-spontaneous breakdown possibilities are slender because nuclei
are small and well-separated inside their electron shells.
Neutron emission (in fissile materials) is the notable exception, because
neutral particles don't interact with electron charge.
The usual neutron energies after a uranium-235 decay are too high to
hit many targets, but with some purification of isotopes
and a bunch of moderator material that slows neutrons down (a 'core', usually
graphite), you can get a chain reaction (analogous to kindling a fire).
In the Oklo region, isotopic abundances indicate that chain reactions were
kindled naturally in the distant past, with water as the moderator.

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What is meant by fractional oxidation state?
What is the real concept behind oxidation state?

Title and body don't cooperate here well... You should elaborate.

https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

Nothing too much, really. When you have a compound like sodium tetrathionate $\ce{Na_2S_4O_6}$, if you calculate the oxidation state of sulfur, you get $\frac{5}{2}$. This means that the AVERAGE oxigation state of sulfur is $\frac{5}{2}$. In fact, the sulfurs in the compound are not equal, you have two of them with an oxidation state of $+5$, and two of them with $0$. But when you balance redox equations, you do not really care about this, and you can use the average value of $2.5$ to find the coefficients.

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why is tertiary carbocation more stable than primary allylic carbocation?
According to general organic chemistry the order of stability is as follows:
aromaticity > resonance > hyperconjugation > inductive effect
Since primary allylic carbocation shows resonance as well as hyperconjugation, moreover resonance is greater than any other effect then why is primary allylic carbocation less stable than tertiary carbocation.

I am not quite sure about the numbers myself. I can believe that the relative energy of a tertiary carbocation is higher than the relative energy of a primary allylic cation (therefore making the tertiary cation less stable). You also need to consider that in a chemical reaction, stability does not only depend on the energy, but also on the reaction speed. Since primary allylic cation is sterically unhindered, it may react faster, making it less stable. Again, it is only a first guess, I do not have numbers for this.

Because simple concepts only get you so far.

The decreasing order of stablity of carbocations is:
aromatic (tropylium, cyclopropenyl) > benzyl > allyl > tertiary > secondary > primary
Allylic carbocations are resonance stabilized whereas tertiary alkyl groups are stabilized by hyperconjugation (and inductive effect). The hyperconjugation (and inductive effect), in general, is quite weaker than resonance (and hyperconjugation effect). The only exceptions to the previous statement I am aware of are halogens (inductive effect is stronger than resonance).

I would put aromatic ring ions (tropylium, cyclopropenyl) first before benzyl.

Hyperconjugation effect on a 3 degree carbocation often overpowers the resonance effect in a primary-allyl carbocation.
If we consider the following compounds,

And,

Solvolysis (a reaction which occurs via SN1 mechanism, causing formation of an intermediate carbocation) will occur at a faster rate in the second compound, even though the carbocation intermediate in the first compound will be resonance-stabilized.
Again, this is because 3 degree carbocations are very stable due to Hyperconjugation effect.

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Lugol's iodine 5 % aqueous solution
There is Lugol's 5 % solution with 5 g of pure iodine and 7.5 g of potassium iodide.
In one ml there should be 12.5 mg of iodine in sum.
But there are suggestions telling that two drops of this solution contains 12.5 mg of iodine and the drops I seen used there are no way 1 ml.
Is it a mistake? Why they don't use measurement in ml rather than drops to be precise?

It's true, 2 drops of Lugol's 5% contain 12.5 mg of iodine/iodide, see this link for the calculation. You cannot just add up the amounts of iodine and iodide, they are not chemically identical. Also, a well-calibrated dropper is much more precise than measuring 1 ml in a syringe.

2 drops would be about 0.1 mL, 20 drops would be about 1mL. While this can be a surprisingly accurate means of measuring aqueous solutions, I disagree with the previous comment; a properly calibrated 1 mL syringe will give you a more accurate and precise delivery of 1 mL than the dropper technique. Depending on your needs though, the dropper method might be sufficient.

@Don_S So are you saying in 1 ml there is no 12.5 mg of iodine in sum? 5 % solution in this case means 5 g pure iodine and 7.5 g of potassium iodide so logically in 1 ml of this solution there should be 5 mg of pure iodine and 7.5 mg of potassium iodide? Is not it correct?

@airhuff there is a big difference between two drops and twenty drops. This solution is intended to be used internally and droppers are comfortable for it but thought it was imprecise.

@airhuff, my intention was that a regular scaled syringe of 5 or 10 mL (even if scaled very precisely) would still not be a match to a dropper, such as the opening in liquid iron supplements for babies, where the opening is designed to give the same volume in every drop - sure, they also fail sometimes for a bunch of reasons, but compared to the syringe, where you have to pull and push manually and you depend on bringing the end of the piston exactly to the scale by sight alone, the error rate of a dropper is probably lower.

@Don_S , OK, we were picturing two very different types of syringes. For what you described, particularly the size (5-10 mL where I was thinking 1 mL) of the syringe, dropwise is probably more accurate.

@Sofiko, yes, that's what I'm saying. Since 5% solution means that there are 5 grams of iodine in 100 mL, when dividing it by 100 you get 0.05 g in 1 mL, which is equal to 50 mg in 1 mL. Since a drop is 1/20 of a mL, dividing 50 mg by 20 gives 2.5 mg of iodine per drop. Do the same math for iodide and you will get 3.75 mg of iodide in 1 drop. You can add up these two amounts and multiply by 2 (for 2 drops) and you WILL get 12.5 mg, but it's not 12.5 mg of IODINE, but of IODINE (5 mg) and IODIDE (7.5 mg) together.

It doesn’t really make sense a priori to have a $5~%$ solution of two solutes … I need more information to tackle this.

@Don_S Please, make it an answer so I can give credit. It is clarified for me now. I was making mistake in calculation. In 1 ml there will be 125 mg of the both and if 1 ml is 20 drops in two drops there will be 12.5 mg.

A Lugol's iodine solution is an aqueous solution of elemental iodine and potassium iodide. This solution may be prepared in different strengths, typically $2{-}15~\%$, while the strength is determined by the content of elemental iodine (see this table from the Wikipedia article, but note that the first column's title is a bit misleading — the leftmost column lists the different strengths of Lugol's solution, and the other columns list the amount of the each constituent (iodine, iodide, total) in a solution of that strength).
Now, this is in essence an iodine solution, but it is prepared from iodine and potassium iodide due to solubility issues of elemental iodine. See this answer on Chemistry.SE to the question Preparation of iodine solution for a simple and concise explanation.
For the calculation of the amount of iodine and potassium iodide in solution, let us take $5~\%$ Lugol's solution, which is prepared from 5 grams of iodine and 7.5 grams of potassium iodide in $\pu{100 mL}$.
If we have $\pu{5 g}\ (\pu{5000 mg})$ of iodine and $\pu{7.5 g}\ (\pu{7500 mg})$ of potassium iodide in $\pu{100 mL}$, then after dividing by $100$, we have $\pu{50 mg}$ of iodine and $\pu{75 mg}$ of potassium iodide in $\pu{1 mL}$.
Since the commonly accepted volume of one drop is $1/20\ (5~\%)$ of $\pu{1 mL}$, then by multiplying $\pu{50 mg}$ and $\pu{75 mg}$ by $0.05$, we get $\pu{2.5 mg}$ of iodine and $\pu{3.75 mg}$ of potassium iodide in one drop of the solution.
For better clarity, here is the calculation again for I2 content:
$$\frac{\pu{5 g}}{\pu{100 ml}}\cdot \frac{\pu{1000 mg}}{\pu{1 g}} = \frac{\pu{50 mg}}{\pu{1 mL}}  \\ \Longrightarrow  \frac{\pu{50 mg}}{\pu{1 mL}}\cdot \frac{\pu{1  mL}}{\pu{20 drops}} = \frac{\pu{2.5  mg}}{\pu{1 drop}}$$
Do the same for $\pu{7.5 g}$ of potassium iodide to get $\pu{3.75 mg/drop}$.
A word about the actual iodine content of the solution following Jan's comment:
In principle, masses can always be added up (unlike volumes, which should not be added up, due to possible interactions between certain solvents — see here).
Therefore, we can say that the total mass of solutes in one drop is $$\pu{2.5mg}+\pu{3.75 mg} = \pu{6.25 mg}$$ ($\pu{12.5 mg}$ in two drops, per OP question). However, we cannot say that two drops contain $\pu{12.5 mg}$ of iodine, because the solution contains both elemental iodine (I2) and potassium iodide (KI), which are two different constituents. Since KI's role in the solution is to facilitate dissolution of elemental iodine molecules (see description of chemical properties, including equations, here), iodide content is irrelevant and the solution's strength reflects only iodine's content.
In other words, $\pu{12.5 mg}$ is the total quantity of solutes in two drops of Lugol's solution, but this quantity has no practical meaning, since the active ingredient is iodine, and its amount in 2 drops of $5~\%$ Lugol's solution is $\pu{5 mg}$.

My comment was actually ‘We have three constituents (two solutes, one solvent), so $5~%$ does not carry any meaning a priori.’ However, thank you for clarifying that a percentage of Lugol’s solution is always an iodide percentage. However, the correct way to calculate solution percentages (not v/v; m/m) would be $\text{mass solute}/\text{mass solution}$ — i.e. we would need to have $m(\ce{I2})/(m(\ce{I2}) + m(\ce{KI}) + m(\ce{H2O}))$. But probably simplicity ruled …

My understanding of the historical formula is that the above amounts are not correct. The 5% Lugol's takes it's name from the approximate percentage of the principle solute which is Iodine. The Potassium Iodide is an important part of the formula. Aside from assisting in the dissolving or the Iodine the Iodide part is an important nutritional factor. Some parts of the body need Iodine and others Iodide.
The formula calls for:
10g of Potassium Iodide
5g of Iodine
and Have available 100 ml of distilled water
"Dissolve KI in about 20-30 ml of distilled water. Add iodine and heat gently with constant mixing until iodine is dissolved. Dilute to 100 ml with distilled water. Store in amber glass-stoppered bottle in the dark."
https://www.fda.gov/Food/FoodScienceResearch/LaboratoryMethods/ucm062245.htm
The confusion in the amounts above is because the Iodide content of 10g of potassium Iodide is about 7.5g
The Iodine content of one metric drop ( =1/20ml) is 2.5mg
The Iodide content of one metric drop is 3.35mg
The total dose of Iodine plus Iodide in one drop is 6.25mg


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VSEPR geometry of Bromate Ion
After doing the Lewis structure of Bromate Ion and in order to find the molecular geometry using VESPR method, we have:
central atom: Br
Electrons of the central atom: 7
Electrons that contribute the 3 Oxygens : 3
Electrons that contribute the central atom for the π bonds: -2
Charge of ion (negative) : 1
Total electrons : 9
But if we divide 9 with 2 in order to find the σ-bonding pairs, we get 4,5. what is wrong? Because after searching the molecular geometry of BrO3- it says that it is trigonal pyramid.

I'm not sure what is your difficulty exactly. If you've already done the Lewis structure, then you know the Bromate ion looks like this.  Three terminal atoms + a lone pair indicate that this polyatomic ion is of the type AX3E1. According to this table, this is a trigonal pyramid. So what exactly is the problem? What purpose do these calculations serve?

@Jan I am a bit hesitant to agree that this is a duplicate. The reason for this is that the target question is convoluted with needless information of a pupil-teacher e-mail back and forth and it is hard to find the actual question. I like your answer there and would prefer to see it here.

@Martin-マーチン Fairy ’nuff ;)

There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

How many valence electrons do we have? (add up)
For bromate, the calculation is:
$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?
This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.
$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$
$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$
$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{Br-O}$ bonds are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:

(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image. Originally taken from an answer about iodate hence the central iodine. Replace it with bromine to get bromate.)

Thank you, that was very helpful! According to this, the molecular geometry of ClO2- will be linear? valence electrons=20, electrons we need to form octets: 3x8=24 , electrons we're missing : 24-20=4 or 2 pairs and electrons that will form lone pairs: 20-4=16 or 8 pairs

@MariaP No, it will be bent; you will have two lone pairs on chlorine (and three per oxygen). Linear compounds are very rare for purely single-bonded atoms; to the best of my knowledge, they only occur with beryllium. Multiple bonds are a different story.


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What effect does Baking Soda and Citric Acid have on a soft drink?
In chemistry, we are doing a project where we have to make our own carbonated soda. The ingredients our group is using are:

Baking Soda ($\ce{ NaHCO3}$)
Citric Acid ($\ce{C6H8O7}$)
Flavoring powder (ex. KoolAid flavoring
    powder)
Sugar
Water

Using stoichiometry, our group has already calculated how much citric acid we need based on the 1 gram of baking soda that we are deciding to putting in: we have to use 0.76 g citric acid based on our calculations.
My question is, what effect does baking soda have and what effect does citric acid have on the soda? Does baking soda make it "bubbly"? Does citric acid make it more concentrated?

Hint: the "bubbly" part of a carbonated beverage is the carbonation - dissolved $\ce{CO2}$. What happens when baking soda reacts with acid? What is your acid?

Actually if we care just about the reaction between Baking Soda ($\ce{NaHCO3}$) and Citric Acid ($\ce{C6H8O7}$) the following reaction should ensure :
$$\ce{3NaHCO3(aq) + C6H8O7(aq) -> 3 H2O (l) + 3 CO2(g) + Na3C6H5O7(aq)}$$
So basically the reaction should produce water, carbon dioxide and trisodium citrate (as known as E331 : a food additive)


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How does the degree of saturation influence the colour of the flame?
If the colour of flame produced by the atom of an element is the property of that element then why do unsaturated and saturated carbon compounds give different colours of flame even though both contain the element carbon and hydrogen?

The flames that you see when you burn carbon-containing compounds have colour for a completely different reason that if you add ionic compounds (e.g. sodium chloride) to a flame. Ionic compounds won’t combust, their bonds will merely be broken and the metallic atom then has an electron that can be excited and that can re-relax under photon emission.
For carbon, we are actually talking about proper combustion, i.e. a reaction of the carbon compound with oxygen to give carbon dioxide and water. This comes in two flavours: complete and incomplete.

In complete combustion, all carbon atoms are fully oxidised to carbon dioxide due to sufficient oxygen amounts. We can write a nice equation e.g. for methane:
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}\tag{1}$$
Complete combustion is often shown with methane as it often tends to combust completely.
Incomplete combustion, as the name says, is incomplete, i.e. some carbon atoms won’t make it all the way to carbon dioxide. It is impossible to give an exact equation for incomplete combustion, but one possible one could be the following:
$$\ce{2 CH4 + 3 O2 -> CO2 + C + 4 H2O}\tag{2}$$
Incomplete combustion is commonly shown with unsaturated compounds because they have a greater tendency to display it.

The key factor in incomplete combustion is the fact that elemental carbon (soot) is generated; it is that soot whose electronic transitions cause the yellow colour of a flame. This is not a feature of unsaturated versus saturated hydrocarbons: as the equation above shows, methane can burn both in a complete and in an incomplete manner depending on how much oxygen is added.
Likewise, the most common example for incomplete combustion, ethyne, can be combusted completely if only enough oxygen is added, see this question.

Thanks Jan for catching my mistake. I completely coasted and focused on the "different colours" wording in the question and my mind went to ionic compounds.


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Is there a difference between ‘ammonium hydroxide’ and ‘aqueous ammonia’?
Water ($\ce{H2O}$) can be added to ammonia ($\ce{NH3}$) in order to form ammonium hydroxide ($\ce{NH4OH}$):
$$\ce{NH3 + H2O -> NH4OH}$$
But we know that ammonia is gaseous at STP so writing $\ce{NH3 + H2O}$ equals writing $\ce{NH3(aq)}$ right?
If yes, then the equation will look like:
$$\ce{NH3(aq) -> NH4OH}$$
So my question is: is ammonium hydroxide $\ce{NH4OH}$ simply aqueous ammonia $\ce{NH3(aq)}$? Or am I missing something?
Can I resume the first equation like this?
$$\ce{NH3(g) + H2O(l) -> NH3(aq)}$$

Yes that is the gist of it. https://en.wikipedia.org/wiki/Ammonium_hydroxide In pure water NH3 is like CO2. The most prevalent species is the dissolved gas, not the ionic forms.

That's right. Aqueous ammonia contains a little bit of ammonium hydroxide. Pure ammonium hydroxide doesn't exist at all.

Very related: http://chemistry.stackexchange.com/questions/48373/ammonium-hydroxide-name

Ammonia is a weak base, ammonium is a weak acid. Water is a very weak acid and hydroxide is a very strong base. Taken together, this tells us that equilibrium $(1)$ is pushed towards the reactant side.
$$\ce{NH3 (aq) + H2O (l) <<=> NH4+ (aq) + OH- (aq)}\tag{1}$$
Thus, it is practically impossible to form stoichiometric ammonium hydroxide; the $\mathrm{p}K_\mathrm{a}$ values tell us that ammonium ions and hydroxide ions put together will generate water and ammonia.
However, it is also true that equation $(1)$ is an equilibrium — everybody has seen a solution of ammonia in water display basic properties so hydroxide ions (and thus ammonium ions) are unquestionably there. The predominant species is dissolved ammonia, though.
$$\ce{NH3 (g) + H2O (l) <=>> NH3 (aq) + H2O (l)}\tag{2}$$
It is correct that the most important process when dissolving ammonia in water is $(2)$ — $(1)$ plays a more minor role. Thus, it is also more correct to speak of solutions of ammonia — but at the same time ammonium hydroxide is commonly understood even if not formally correct.
By the way beware: occasionally students are sent off to fetch ‘solid ammonium hydroxide’ as a joke — similar to students getting sent off to fetch ‘vacuum gas’, ‘three phenol rings’ etc.


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How can we predict the colour of transition metal complexes?
I am facing problems in indentifying colour of complex compunds just by seeing the their molecular formula. Is there any method or concept to predict the colour of complex compunds?

See presence of unpaired electrons, complexes having unpaired elctrons absorb visible radiations for d d transitions.

I am asking if a compound is given to us and we need to determine whether it is yellow coloured or not so can we predict that?

The coordination geometry + central metal ion + spin state often gives a good guess what kind of color you will see. However, it is not exact science (if you don't do any calculations), especially with charge transfer bands.

Can you provide some examples using the three properties just mentioned ?

The basic principle of coloured compounds is that there is some kind of electronic transition whose energy difference corresponds to a photon whose wavelength is in the visible region ($\pu{400nm}<\lambda<\pu{700nm}$). Thus, to determine the colour of a compound we should always be looking at the molecular orbital scheme; for reference, I have attached the MO scheme of a typical octahedral $\ce{[ML6]^n+}$ compound in figure 1.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
The most common electronic transition corresponding to visible wavelengths is the transition between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$, marked with $\pu{10Dq}$ in figure 1. In simple terms, these five orbitals correspond to the central metal’s $\mathrm{d}$ orbitals meaning that the metal’s electronic configuration plus the strength of the $\ce{L\bond{->}M}$ interaction will determine the colour.
The problem with attempting to predict colours ab initio is that you cannot really predict the difference in energy levels between the ligand orbitals (right-hand side) and metal orbitals (left hand side) — it is mostly the metal and its oxidation state that determines this, a higher oxidation state corresponding to lower-lying metal orbitals, therefore a greater value of $\pu{10Dq}$ and therefore a blue-shift in colour.
However, there are still a few shortcuts that you can use assuming you have memorised the ‘general colour’ of a certain metal in a certain oxidation state. For example, you may consider stronger and weaker field ligands. When going from a slightly weaker ligand to a slightly stronger ligand — e.g. going from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ — $\pu{10Dq}$ is predicted to increase and thus the absorbed wavelength predicted to decrease.
Unfortunately, the generality of this scheme is strongly limited. It fails as soon as:

the complex transitions from a high-spin state to a low-spin state (e.g. $\ce{[Fe(H2O)6]^3+}$ to $\ce{[Fe(CN)6]^3-}$)
the geometry of the complex changes from (distorted) octahedral to something else like tetrahedral (e.g. $\ce{[Co(H2O)6]^3+}$ to $\ce{[CoCl4]-}$).

So you might just be better off memorising the colours of certain complexes rather than attempting to learn the pattern and remember the exceptions.


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Does Nucleophilic substitution require water to happen?
I am under the impression that in a mixture of hydroxide ions, haloalkanes, water, and ethanol, under low temperatures, nucleophilic substitution will occur predominantly, with small amounts of elimination.
When no water is present and the solution is heated, elimination will occur to a greater degree. Is it possible for nucleophilic substitution to occur when no water is present as all?
Thanks a lot.

Keeping other parameters fixed, an elevated temperature may enhance the rate of elimination such the product of elimination is more dominant than the one by substitution.  But there is not a "critical temperature" above which only elimination will occur every time, rather than the structure of the substrate and of the reagent equally influence this.
For the other part, transesterifications as one example work better in absence of water.  See, the oxygen present in the water molecule, with highly concentrated electron density, it may act as a nucleophile, too -- a rather hard one (HASB princible) -- leading to parasitic side reactions.

So is it possible for substitution reactions, say with hydroxide ions, to occur in the complete absence of water?

@BenjaminRoger-Newsome Methoxymethylations, as one example to form ethers (cf. Organic Synthesis, 2007, 84, 102, DOI: 10.15227/orgsyn.084.0102) are performed in absence of water.  The "standard" Williamson ether synthesis is the reaction of an alkylbromide on an alcoholate, the later prepared in situ by the action of metallic sodium on the alcohol (sometimes the latter is used at the same time as solvent of reaction) -- the presence of water would dangerously rapid yield NaOH, hydrogen gas and liberation of a lot of heat.

Ok, thank you. In the case of hydroxide ions acting as nucleophiles or as bases, is it possible for the hydroxide ion to act as a nucleophile in the absence of water?

@BenjaminRogers-Newsome An other classic is the Finkelsteijn exchange of halogenated alkanes in acetone; sometimes "Finkelstein like" exchanges (mesylates, tosylates, etc.) are added to this type of reaction, too.

Nucleophilic / basic ... is a question of definition. And each time, ask the question by stating the potential substrate and its structure, too.  Lewis bases posses by definition a free electron pair and may indeed react as nucleophile, too.   On the other side, 2,6-lutidine, to mention just one organic base frequently used in the (small scale) research lab is, by means of its sterical hinderance, no longer a nucleophile.

In the context of a secondary haloalkane, undergoing elimination, using ethanol as a solvent, where the hydroxide ion is acting as a base, is it at all possible for nucleophilic substitution to occur even a little bit?

Let us continue this discussion in chat.

Whether nucleophilic substitution occurs depends primarily on the nature of the potential nucleophile, the potential leaving group and the molecule potentially attacked nucleophilicly. Secondary effects are caused by the solvend in question.
Most importantly to answer your question, many nucleophilic substitutions are performed in the absence of water since water itself is a nucleophile and would interfere. Here are just a few examples:

The Finkelstein reaction is a nucleophilic substitution performed in acetone
Meerwein methylation can be considered a nucleophilic substitution and is typically performed in dry dichloromethane
Mitsunobu reactions are nucleophilic substitutions and also typically performed in the absence of water in dry organic solvents
The Michaelis-Arbuzov reaction is a double nucleophilic substitution typically performed in triethyl phosphite. Whether its second step is actually a nucleophilic substitution (giving bromoethane) or an elimination (giving ethene and hydrogen bromide) doesn’t even matter.


text:
Mutual solubility of water and hexane
The question says that

the value of mutual solubility of water and hexane is small. Consult bibliography to find it out.

I can't find it and I'm also concerned as to why the question doesn't include a temperature.

Obviously, because the said solubility is small at any temperature.

Sorry, what's the question?

A Google search for hexane solubility in water yields this link as the third result...

I have found this but my problem was that it said mutual solubility and as fas as I know, mutual solubility is the point where the two liquids are miscible. Anyway, thank you.


text:
Hybridization of nitrogen in trisilylamine, (SiH3)3N?
I want to know the hybridization of the central atom in $\ce{(SiH3)3N}$.
I think it should be $\mathrm{sp^3}$, because $\ce{N}$ is attached to three silicon atoms and one lone pair. But actually it is supposedly $\mathrm{sp^2}$.
How is this so?

Central atom in the said molecule is N, to begin with.

@IvanNeretin sorry, I have edited it

Can you add the source of the 'correct' sp2 answer?

Strongly related: http://chemistry.stackexchange.com/q/38599

It is written as backbonding occurs in the molecule , I could not understand it . @Jan

I don't see how's that related. Probably in this compound N is more planar than usual because of steric reasons, hence the higher s character of the lone pair.

@IvanNeretin ‘More planar’ would mean ‘lower s character of the lone pair’. I think the question is related (but far from a dupe) because of the inversion mechanism. However, after reading up on the structure of $\ce{N(SiH3)3}$, I realise that it is far more interesting. I like that molecule!

@Jan Sorry, of course you're right: it is higher p character of the lone pair.

Also @IvanNeretin I don’t think steric reasons are the key; see my answer ;)

Yes, I see. Indeed, the molecule turns out to be more interesting than it seemed.

@TAR86 The planar structure of trisilylamine and its unusually low reactivity for an amine has been known for quite a while; see JACS 1955, 77 (24), 6491–6492 for example. Of course, back then, we rationalised it as pi bonding, but now we know better.

Related?

Ordinarily and according to Bent’s rule, we would expect nitrogen’s lone pair to be in an $\mathrm s$ orbital and nitrogen using its three $\mathrm p$ orbitals to form three bonds to the three silicon atoms. This configuration would allow for the greatest stabilisation.
However, due to nitrogen’s small size, this perfect world already falls apart for ammonia ($\ce{NH3}$), where nitrogen is bound to three otherwise tiny hydrogen atoms. Because an electronically perfect angle of $90^\circ$ would generate much too much steric strain between the hydrogen atoms, $\mathrm s$ contribution is mixed into the bonding $\mathrm p$ orbitals to a certain extent; for ammonia, this extent happens to be almost perfect $\mathrm{sp^3}$ — results for other amines will vary. This electronic situation is not ideal, however it is clearly better than having $\mathrm{sp^2}$ hybridisation and the lone pair in a $\mathrm p$ type orbital. An $\mathrm{sp^2}$ hybridisation of nitrogen in ammonia can be reached, but only as the transition state of nitrogen inversion.
Carrying on to the compound $\ce{N(SiH3)3}$, we would be inclined to again assume a hybridisation of $\mathrm{sp^3}$ in line with the previous paragraph. However, Beagley and Conrad performed electron diffraction studies on $\ce{N(SiH3)3}$ and found the molecule to be practically planar within experimental error.[1,2] A planar molecule without doubt means that nitrogen is $\mathrm{sp^2}$-configured in $\ce{N(SiH3)3}$.
The question remains why. There must be some kind of stabilising interaction of nitrogen’s remaining $\mathrm p$ orbital with something else to keep that molecule planar. Beagley and Conrad suggest — in line with what was thought at the time — that this be due to π bonds with silicon’s remote $\mathrm d$ orbitals.[1]. Numerous evidence, much of which is collected on this site, speaks the opposite (namely that $\mathrm d$ orbitals do not play any role in the bonding situation of main group metals).
Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:
$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$
In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Note and reference:
[1]: B. Beagley, A. R. Conrad, Trans. Faraday Soc. 1970, 66, 2740–2744. DOI: 10.1039/TF9706602740.
[2]: Actually, $\angle(\ce{Si-N-Si}) \approx 119.5^\circ < 120^\circ$. The authors state:[1]

The apparant slight deviation from planarity is associated with a shrinkage effect11 on the $\ce{Si\dots Si}$ distance of about $\require{mediawiki-texvc}\pu{0.007 \AA}$ (see [table]). Spectroscopic results12 are entirely in agreement that the molecule is planar.
11 A. Allmenningen, O. Bastiansen and T. Munthe-Kaas, Acta Chem. Scand., 1956, 10, 261. [sic!]
12 E. A. V. Ebsworth, J. R. Hall, M. J. Mackillop, D. C. McKean, N. Sheppard and L. A. Woodward, Spectrochim Acta, 1958, 13, 202. [sic!]

I still frequently hear d orbitals being implicated in main group bonding situations in books and classes. Can you point me to a good example of evidence against this?

@gannex I can forward you an answer of Ortho’s, can’t I? ;)

To test Jan's argument, I did an NBO analysis of your structure (optimized at PBE-D3/def2-SVP with NWChem 6.6 using a conformational search with MMFF94s and Avogadro as the starting point; a frequency calculation determined it was a true minimum). I used NBO version 5.9.

Figure 1: optimized geometry (angles in degrees and distances in angstrom)
The obtained geometry is in perfect agreement with Jan's answer, showing a $\ce{Si-N-Si}$ angle of 120°.
The five most significative NBO second order stabilisation energies are:
              |                  |   E(2)   | E(j)-E(i) | F(i,j)
Donor NBO (i) | Acceptor NBO (j) | kcal/mol |    a.u.   |  a.u.
==============|==================|==========|===========|========
LP ( 1) N 1   | BD*( 1)Si 2- H 6 |   5.08   |    0.43   |  0.043
LP ( 1) N 1   | BD*( 1)Si 2- H 7 |   5.08   |    0.43   |  0.043
LP ( 1) N 1   | BD*( 1)Si 3- H 8 |   5.08   |    0.43   |  0.043
LP ( 1) N 1   | BD*( 1)Si 3- H 9 |   5.08   |    0.43   |  0.043
LP ( 1) N 1   | BD*( 1)Si 4- H12 |   5.08   |    0.43   |  0.043
LP ( 1) N 1   | BD*( 1)Si 4- H13 |   5.08   |    0.43   |  0.043

That is, a six-fold $\ce{n_\ce{N}} \rightarrow \sigma^*(\ce{Si-H})$ donation, worth of 5.08 kcal/mol each, seems to be the most significant delocalisation.

Figure 2: $\ce{n_\ce{N}} \rightarrow \sigma^*(\ce{Si-H})$ delocalisation scheme
On the other hand, the natural electron configurations are as follows:
Atom  |  Natural Electron Configuration
------|----------------------------------
 N    | [core]2s( 1.53)2p( 5.12)
 Si   | [core]3s( 1.01)3p( 1.96)3d( 0.02)
 H    |       1s( 1.15)

Thus, the electron configuration of $\ce{N}$, according to NBO analysis, is $\ce{1s^{2} 2s^{1.53} 2p^{5.12}}$.
Furthermore, the nitrogen lone pair is of pure $\pi$ character.
Looking closer to the $\ce{N-Si}$ bond we see:
    (Occupancy)   Bond orbital/ Coefficients/ Hybrids
-------------------------------------------------------------------------------
  1. (1.97614) BD ( 1) N  1-Si  2
              ( 81.00%)   0.9000* N  1 s( 33.32%)p 2.00( 66.66%)d 0.00(  0.02%)
              ( 19.00%)   0.4359*Si  2 s( 23.84%)p 3.16( 75.42%)d 0.03(  0.74%)

Thus,

The $\ce{N-Si}$ bond is polarised towards $\ce{N}$ ($0.9000 h_\ce{N} + 0.4359 h_\ce{Si}$), as expected due to electronegativity;
The $\ce{Si}$ hybrid ($h_\ce{Si}$) is of $\ce{sp^{3.16}}$ character, as expected; and
The $\ce{N}$ hybrid ($h_\ce{N}$) is of $\ce{sp^{2.00}}$ character.

In agreement with Jan's answer.

I will now study this answer carefully so as to shamelessly rip off its technique. Very nice! Can you suggest any good reading for learning more about how to best carry out and/or interpret NBO analyses?

I really like Discovering Chemistry With Natural Bond Orbitals, a book by NBO author Frank Weinhold

@FelipeSchneider I've also been wanting to learn about techniques like this. Thanks for the reference!!

By any chance, do you still know which NBO version you have used? I don't think it'll make a huge difference as major improvements in the theory were towards the use with transition metals, but in general it would be nice to know.

@Martin-マーチン Excellent suggestion, it was version 5.9. I added it to the answer :)

We call this weird thing as back bonding. The lone pair kinda delocalises or seeks refuge in the empty d orbital of Si, basically providing each N-Si, bond, on an average, a one third of an extra bond.

No d-orbitals, please...

It's actually due to the back bonding basically silicon is an element in the 3rd period and nitrogen is an element in the 2nd period so here silicon will contain empty orbitals of 3d and whereas nitrogen has lone pair of electrons so therefore back bonding occurs so while calculating the steric number in order to calculate the hybridization the lone pair on central atom nitrogen isn't considered which makes that the steric number is 3 and the hybridization is sp2.

Did you not read the answer given by @TheScholarAM and the comment about the lack of d-orbital participation?

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


text:
Will cinnamaldehyde undergo self-aldol condensation?
Will cinnamaldehyde undergo self-aldol condensation?

My book says it won't, but since it has an α-hydrogen, I'm confused.

Not every α-hydrogen automatically means that a ketone is enoliseable. It would need to be an α-hydrogen capable of aligning with the π system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the phenyl ring, the double bond and the aldehyde meaning that this compound is entirely planar. Thus, there is no way for the α-hydrogen to align favourably for deprotonation to occur. If it did, it would generate an allene-olate $\ce{Ph-CH=C=CH-O-}$ which is not particularly favourable.
In general, α,β-unsaturated ketones and aldehydes are deprotonated vinylogously, meaning that the double bond shifts away upon deprotonation. For example crotonaldehyde $\ce{H3C-CH=CH-CHO}$ is deprotonated in the following way:
$$\ce{H3C-CH=CH-CHO ->[][- H+] H2C=CH-CH=CH-O-}\tag{1}$$
I.e. the γ-hydrogen is abstracted in place of the α-hydrogen.

No aldol condensation is not possible here, as you would have to extricate an alpha hydrogen(alpha to the carbonyl group), but if you do extricate a hydrogen atom from the alpha position it would result in the formation of Vinylic Carbanion which is very very unstable.
Thus the molecule would not want the alpha hydrogen to get extricated, thus aldol reaction would not proceed as you need to remove the alpha hydrogen in order to proceed forward with the reaction.

You might want to read the question you linked; it actually does not say that vinylic carbanions are rather unstable. They are actually more stable than alkyl carbanions.

(Side note: if you remove that wrong bit from your answer I’ll upvote it because the remainder is pretty good.)

You mean the comment?

Everything that calls the vinylic carbanion very unstable. And probably everything that talks about vinylic at all because the anion wouldn’t be a vinylic carbanion.


text:
How to move forward in balancing this equation?
I want to get the molecular formula for this reaction:
$$\ce{{$trans$-stilbene} + {$m$-chloroperoxybenzoic acid} -> {$trans$-1,2-diphenyloxirane}}$$
and if you do you get this:
$$\ce{C14H12 + C7H5ClO3 -> C14H12O}$$
but what do I do about the chlorine not being present in the products, is it released as $\ce{Cl2}$?

The reason you cannot balance the reaction is because you did not consider everything that happens. In the Prilezhaev epoxidation, mCPBA is a stoichiometric reactant but only one single atom of mCPBA is transferred to the product. Therefore, these compounds are often written on the reaction arrow rather than as reactants and products:
$$\ce{C14H12 ->[C7H15ClO3] C14H12O}\tag{1}$$
However, that still does not tell you what exactly happened. If you take a close look at the mechanism, you will see that mCPBA remains essentially unchanged except that it loses its peroxy function; the second product is mCBA or meta-chlorobenzoic acid. Thus, a full reaction equation would be:
$$\ce{C14H12 + C7H15ClO3 -> C14H12O + C7H15ClO2}\tag{2}$$
Notice how this balances automatically.

THANK YOU! @Jan! Can  you help me further to reach the answer I was looking for, which was what the theoretical yield of the C14H12O would be when I used 1,8g of trans-Stilbene and 3,45g of mCPBA with an obtained product of 0,247g. I get the theoretical yield to be 33g but that must be wrong!

@O.kth The typical steps are: 1) Calculate the amount of substance of each reactant 2) determine the limiting reagent 3) using that and the equation to determine the expected amount of product 4) calculating the expected mass of product. Using your numbers, the theoretical yield should be $\pu{1.96g}$.


text:
how to separate aluminum from aluminum alloy
In an attempt to produce aluminum oxide ($\ce{Al2O3}$) also known by alumina I try it out to dissolve two old hard drive frame in brine solution one as the cathode and the second as anode, in the first time the solution turn white and the reaction stopped,At the second time with a new solution very saturated with $\ce{NaCl}$ the solution turned dark grey and did dissolved a part of one of the electrodes which are hard drives and the reaction stopped again maybe the dissolved part shortcut.
Finally to save the situation I used 150g of sodium hydroxide but after a night in solution $\ce{NaOH}$ not much happened and then I figured out that the hard drives are an aluminum alloy.
in the electrolysis part I used an old dell PSY 12v. my question is how to get the aluminium from the aluminium alloy?
EDIT:
die-cast from A380 alloy (other names ADC10, SC84A, LM24, A03800, A13800, AlSi8Cu3Fe) composition:
Melting Point              1,000°F      537.778°C
<0.35% =               231.9°C  =Tin
<3.0%   =         419.5°C  = Zinc
660.3°C  =aluminum
3.5% +/- 0.5% =       1,085°C =  Copper
<0.1% =        1,091°C  = Magnesium
8.5% +/- 1.0% =       1,414°C  = Silicon
<0.5%          =      1,455°C  = Nickel
<1.3%      =          1,538°C  = Iron
<0.5%  =     2,061°C  = Manganese
Trace = <0.5%

What do you want to achieve? If you just want pure Alumina, I'd buy it.

My idea is to first react all your alumina with NaOH to create $\ce{AlOH3}$. Put that resultant in a calcium solution and a single replacement reaction should occur.

How are you going to get a calcium colution? It does not exist in water and molten calcium may produce weird things like ternary oxides and intermetallic compounds. YOU CAN'T SMELT ACTIVE METALS AT HOME!

Do a process of anodic oxidation to your hard drive frame, and using acid to remove other metal.
Peel off the aluminum oxide.
Repeat step 1 and 2 until bulk aluminum oxidized fully.
Electrolyze Al2O3.

Do you think this is necessary?

the solution that passes in my mind is to melt the alloy since it has a low temperature 537.778°C and during the process  add carbon to precipitate the copper, Nickel, iron, manganese, Silicon and magnesium since they have a higher melting point and will not melt.
The remain product is aluminum and Tin Zinc .
EDIT
The second solution is the electroplating process on steel or iron since aluminum doesn't depose on any metal it gonna precipitate at the bottom and other metals will depose on steel sheet

Won't work. The components in alloys are usually soluble in each other in the melt. What's that trick with the carbon?

@Karl I have found that the first solution is called " Liquation "

Shouldn't that be an edit to the question?

If you know your alloy components, you can try to use an acid that will selectively dissolve one of your components. If there is not a big difference in solubility of the components, you can try to dissolve the entire sample and try to electroplate the desired metal or perform some other reaction(s) that will isolate aluminium.
Do this activity, even just lab scale can be dangerous, so take caution.

A surface etch won't change the bulk composition. "Dissolve" typically means water and you can't plate Al from an aqueous solution.

@j.ari first search i found that the hard drive disk frame are made from A380 alloy, die-cast from A380 (other names ADC10, SC84A, LM24, A03800, A13800, AlSi8Cu3Fe) composition.

@MaxW I'm imaging ore refining plants that put raw ore into batch reactors full of conc. acid that dissolves the solid ore, then the solution is processed further to isolate the desired metal(s) through other reactions.

@MaxW smelting aluminum is done with electricity up to 12 Kw to produce 1 Kg of pure aluminum no acids its an expensive process.


text:
Quantum mechanical (or QFT) description of mesomeric effect
This is rather a question about finding the right literature, since the answer would probably exceed a typical forum post.
In organic reactions, mesomeric and inductive effects are of tremendous importance for the prediction of reactivity. I can work with Lewis structures and electronegativity, but I still wonder if there are more accurate descriptions arguing in terms of symmetries of the potential that is generated by the nuclei; how it allows for different coexisting standing waves; and that also consider 3D geometry dependences of the inductive effect.
Most texts argue in terms of atomic orbitals that are somehow bent and combined and hybridized, but electrons don't have a memory so they don't mourn after the times when they belonged to single atoms. Is there any readable description that looks at the whole molecule from the start, possibly written by physicists? Or is this just a thing for computers?
I searched the local university library and google scholar, to no avail. I have Clayden and Anslyn/Dougherty, but those don't really answer this question.

Maybe you should look into Valence Bond theory, the most updated book is by Sason Shaik A chemist's guide to valence bond theory. I found a theses by Avital Shurki on the subject named: VB theory as a tool for understanding structure and reactivity in chemistry. Another option is to look into Molecular Orbital theory, I found this book that looks promising: quantum chemistry of organic compounds: mechanisms of reactions. Molecular Orbital is mostly used since it is linear and more easily computed. But Valence bond is a standing theory, both have faults. Choose what suits you better.

If you are looking for quantum chemical description of organic molecules, I'd suggest you take a look at Computational Organic Chemistry by S. M. Bachrach.

Thanks for your recommendations! Though I was partly (naively?) hoping to find a concise (review?) text treating just either or both mesomeric and inductive effects on a quantum level, I will definitely try these! I understand that any theory is just a descriptive approximation and neither ist identical to reality. I avoided computational chemistry books for fear of overwhelming empirical mathematical descriptions, but I'll try this one, too.

As far as I understand you search a text which provides you with the information about how to quantify the mesomeric and inductive properties of parts of an three dimensional organic molecule. The quantification should account for symmetries in the underlying potential.
From my point of view, any modern quantum chemical program provides you exactly this. So a good way to go are the usual books from Szabo and Ostlund or (a bit more modern and complete) from Helgaker, Jorgensen and Olsen. Both start from a quantum description of fermionic systems and later on introduce the potentials arising due to the physical nature of the system. However, geometrical symmetries are not covered in these books (maybe apart from solving the Schrödinger equation for a particle in spherically symmetric field - $\ce{H}$-like systems). If you're interested in general in molecular symmetries, then the book of Bunker and Jensen might be of interest.
However, those texts usually use the linear combination of atomic orbitals for molecular orbitals (LCAO-MO) approach, i.e., a certain number of atomic orbitals is combined to form molecular orbitals (MOs). Effects like "mesomerization" are obtained directly as a result because MOs are "spread over the whole molecule", i.e., the probability density to find an electron at position $\mathbf{r}$ is roughly constant along the bonds of the mesomeric system.
Your question states that

[m]ost texts argue in terms of atomic orbitals that are somehow bent and
  combined and hybridized [...]

I agree that the usual chemistry starter books do this. And as you state, the bendedness of such an orbital (like the $sp^3$) hybridization arises from the symmetry of the underlying potential (in the $sp^3$-case the tetrahedral [$T_d$] potential of the four $\ce{H}$ atoms around $\ce{C}$ in methane). In LCAO-MO theory, however, symmetry is not necessary, i.e., you can place your atomic orbitals with their angular momenta at the atoms, and you would still get the correct answer. But by symmetry adaption of your basis (basis, because the atomic orbitals form the basis within which the MOs are spanned), you can obtain basis functions that already have a symmetry corresponding to the molecular geometry. This step is not entirely complicated but requires a bit of knowledge about group theory. If you are interested particularly in this part, I would recommend a book about Hückel MO (HMO) theory, like, e.g., Elektronenstruktur organischer Moleküle by Martin Klessinger or the book of Heilbronner and Bock. HMO theory is a simplified LCAO-MO theory in order to treat organic molecules that consist mainly of $\pi$-conjugated systems. In these books they explain how to symmetry-adapt a basis of $p_z$ orbitals e.g. for the calculation of the MOs of $D_{6h}$ symmetric molecules like benzene. While for the non-symmetry adapted basis you need to solve a $6\times6$ eigensystem, it simplifies very much in a symmetry adapted basis.
Lastly, I would like to mention that inductive as well as mesomeric a mere qualitative models in order to quickly come up with an idea how a system changes its properties under substitution of specific parts. None of those quantities is are actual calculable properties (there is no operator that returns the expectation value for this property when applied on a wave function). In this way a quantum chemical theory cannot teach you directly about

3d dependences of the inductive effect[.]

But you can calculate your system and its properties (absorption spectra, electron distribution, electronic/charge density, electric dipole moents, ...) and do the same for a substituted system. From both data you can come up with a hypothesis about the source and change of mesomeric and inductive effects. The same arguments hold, of course, for aromaticity. There, it suits to cite a professor:

Aromaticity is like a unicorn - something everybody knows, but nobody has ever seen.


text:
Dimethylene Cyclohexane to Octalin-1,4-Dione | Hydroboration vs. Ozonolysis Pros Cons
With regards to the following synthesis, what are some pros and cons of using hydroboration oxidation followed by oxidation of the alcohols compared to ozonolysis.
More specifically, how do the two compare as to yields of the final product, cost / amount of reagents, speed of reactions, unwanted side reactions?
Here is the synthesis, along with the two ways to prepare the cyclohexadione:

You lost some carbon atoms in your hydroboration.

What Zhe said. Of your two approaches, only one works. Now, there are other oxidative cleavage methods...

It's not part of your question, but your last line won't work. Alpha bominate carbonyl compounds using $\ce{Br2}$ in acetic acid. You need a base to eliminate the halide, not sulfuric acid.

Comparison of the two approaches as far as yield/cost/etc. is void because one of them does not provide the same product as the other. Hydroboration-oxidation of 1,4-dimethylenecyclohexane yields 1,4-bis(hydroxymethyl)cyclohexane with undergoes oxidation to yield 1,4-cyclohexanedicarbaldehyde or 1,4-cyclohexanedicarboxylic acid.

Now, you may be able to take the carbocylic acid, covert to acid halide and then react with $\ce{LiCuMe2}$ to get the ketone, which should undergo Baeyer-Villager oxidation followed by hydrolysis to give the alcohol you want. Or you could covert the diacid to the diamide and subject it to Hofmann degradation to get the amines, which you might be able to convert to the diazo salt with HONO, which may hydrolyze to give the desired diol. Both routes ae significantly more convoluted than ozonolysis, which decreases oveall yield and increases cost.
Now, there are other oxidative cleavage reactions which might be cheaper.

As others have said hydroboration does not give you the product you want.
The major drawback with ozonolysis is having access to the kit and an oxygen supply. A good alternative is the Johnson-Lemieux procedure with Osmium Tetroxide (available encapsulated and much safer than it used to be) and Sodium Periodate. This should give you decent yields of the diketone.


text:
Why does copper(II) carbonate decompose into copper(II) oxide rather than copper(I) oxide when heated?
I recently did a lab at school surrounding the decomposition of copper(II) carbonate into copper(II) oxide, and subsequently into pure copper, when sufficient heat is added. What I don't understand is why the copper(II) carbonate can't decompose into copper(I) oxide, rather than copper(II) oxide.
For example, copper(II) carbonate decomposing into copper(II) oxide would be:
$$\ce{CuCO3 ->CuO + CO2}$$
While copper(II) carbonate decomposing into copper(I) oxide would be:
$$\ce{4CuCO3 -> 2Cu2O + 4CO2 + O2}$$
If anyone could help me understand why copper(II) carbonate decomposes into copper(II) oxide rather than copper(I) oxide, I would greatly appreciate it.
Edit: I've been nosing around Wikipedia a bit and discovered that copper(I) oxide reacts with atmospheric moisture to become copper(II) oxide. This would explain why copper (I) oxide is not present; it has degraded into copper(II) oxide. I'm still interested in how copper(I) oxide and water react with respect to reactants and products, so if anyone knows what the equation between the products and reactants is, I would like to know.

Hello and welcome to Chemistry.SE. A good way to familiarize yourself with the way this site works is by taking the short [tour]. Also, as this is a homework-type of question, you can find a good discussion of the homework policy here. Regarding your question, do you have any ideas about why copper(I) was not formed in the first step? How hot did you have to heat the CuO to form pure Cu?

It comes to stability of products, in this case interaction of metal ion with ligands (here oxygen). In water solutions $\ce{Cu^2+}$ is usually more stable than $\ce{Cu^1+}$, it's quite different story in non-aqueous solvents like acetonitrile. In solid: $\ce{CuCl2}$ is more stable than $\ce{CuCl}$, but iodide $\ce{CuI}$ is more stable than $\ce{CuI2}$.
The simplest explanation would be hard-soft acid-base theory: $\ce{Cu^2+}$ is a hard acid, and $\ce{Cu+}$ is a soft one.  Oxygen in copper oxide or $\ce{Cl-}$ in chloride are hard bases and $\ce{I-}$ is a soft one. The same to the same: soft acids goes to soft bases, and hard acids likes the hard bases.

Since @Kris_R already answered that Cu(II) is much more stable than Cu(I) according to HSAB theory and thus form $\ce{CuO}$ rather than $\ce{Cu2O}$ on decomposition but it is further noted that copper(II) oxide can be further decomposed to copper(I) oxide on further application of heat.
$$\ce{4CuO ->[1026-1100°C] 2Cu2O + O2}$$
(reaction source)
So, complete reaction would be:
$$\ce{CuCO3 ⟶[\Delta 1] CuO + CO2}$$
$$\ce{4CuO ->[\Delta 2 ] 2Cu2O + O2}$$
Note that you can't write both reactions in one single equation(as mentioned in the question) as the two reactions proceed at different temperatures.


text:
What causes the deshielding in carbon - carbon bonds?
I thought of this question while reading through a chapter on Carbon 13 NMR. It's easy to see that in a bond with two different atoms, such as in a bond involving $ \mathrm C$ and $\mathrm O$, that the difference in electronegativity causes a deshielding effect. It's also easy to imagine that the deshielding effect is proportional to the number of electrons involved in the bond. These two facts are consistent with the Carbon 13 NMR spectra for single and double bonds between oxygen and carbon atoms. A single bond between carbon and oxygen generally causes a chemical shift of $ 50 - 100 $ p.p.m. A double bond between these two atoms is close to the $150 - 200$ p.p.m range.
However, there is a discrepancy with this approach when we look at carbon carbon bonds. Firstly, there is no difference in electronegativity between these two atoms, so we would not expect a significant chemical shift, however there is a significant observed chemical shift in the NMR spectra. Let's ignore this for a moment. We have for atoms involved in carbon carbon single bonds a chemical shift in the $0 - 50$ p.p.m. range. In carbon carbon double bonds, a shift of $100 - 150$ p.p.m. is observed. However, we have for carbon carbon triple bonds, shifts of $50 - 100$ p.p.m. observed.
Clearly, electronegativity is not the only factor affecting deshielding. Also, we cannot rely on the assumption that the chemical shift is directly proportional to the amount of bonds between the atoms. So, my question is this: What causes the deshielding between atoms involved in bonding? How can we explain the observed spectra theoretically? Our current theory is clearly inadequate.

Carbon-13 shifts are determined primarily by paramagnetic shielding $\sigma_\mathrm{p}$ - whereas electronegativity affects diamagnetic shielding $\sigma_\mathrm{d}$ (and is a relatively small effect for C-13). $\sigma_\mathrm{p}$ is inversely proportional to $\Delta E$, an averaged excitation energy, and hence the shifts can be related to the energies of $n \to \pi^, \sigma \to \pi^, \cdots$ transitions. (Note that $\pi \to \pi^*$ transitions don't count due to symmetry considerations.) More information - Gunther NMR Spectroscopy 3rd ed.; pp 409 onwards


text:
What is the size of a Sulfate ion?
In terms of nanometers, what would the size be of an individual sulfate ion in solution?
Would its size exceed 0.5 nm? Enough to displace something that is held together with a gap of 0.35 nm?

Hello and welcome to Chemistry.SE. If you want to learn more about how this site works, the short [tour] is a good place to start. Best of luck with your question!

The radius of a sulfate ion is  0.242 +/- 0.007 nm. (Yirhak Marcus. Chem. Rev. 1988, 88, 1475-1498.)
In order to know if it would displace something could you tell us what exactly you are trying to displace?


text:
Correct stability order of resonance structures
Question no 9 as far as I know none of the rules of relative order of stability apply to this( completion of octate,charge seperation etc.) So they all must be equally stable but the answer is (D)

Compound II and III have 2 aromatic rings whereas I and IV have only 1 aromatic ring, and more the number of aromatic rings, the more the compound is stable.

Aromatic molecules are very stable, and do not break apart easily to
  react with other substances. Organic compounds that are not aromatic
  are classified as aliphatic compounds—they might be cyclic, but only
  aromatic rings have special stability (low reactivity).

Thus stability of II= stability of III and stability of I = stability of IV.
And since II and III have greater number of aromatic rings than I and IV,
Stability(II,III) > Stability(I,IV)
Thus the answer would be [Option D] and not [Option C].


text:
Why do the two cycloheptadiene isomers both show four distinct shifts in carbon-13 NMR?
In the following question I have filled in the correct answer for i:

For ii I thought that should be three for Q and two for P,
For Q there are three distinct enviroments for the C atoms:

$\ce{-\mathbf{C}H2-{}}$
$\ce{-\mathbf{C}H2-C=C-{}}$
$\ce{-\mathbf{C}=C<}$

and for P there are two:

$\ce{-\mathbf{C}H2-C=C-{}}$
$\ce{-\mathbf{C}=C<}$

Howevere the correct answer is 4 for both P and Q.
So then I thought maybe they want me to include the splitting of the peaks but in both P and Q there is too much splitting for this to be the case...
Why is the correct answer 4?

What you identified by determining hybridisation and bonding partners weren’t carbons with identical signals but carbons whose signals would appear in a similar range of the NMR spectrum. Two atoms will only give identical signals if they are homotopic or enantiotopic (the latter only in the absence of chiral compounds).
To determine topicity, you first need to determine which elements of symmetry exist in a molecule. Both your cycloheptadienes have the point group $C_\mathrm{2v}$, meaning that there is:

1 axis of rotation that transforms the molecule onto itself by rotating $180^\circ$
2 perpendicular planes of symmetry
(and identity, which is always a symmetry operation)

One of the two planes is the paper plane. The second plane interacts with the paper plane where Long’s answer drew a dashed line. The rotation axis is where the two planes intersect, i.e. the dashed line.
With that in hand you can determine topicity. Homotopic atoms are those:

that are transformed onto each other by a rotation which is a symmetry operation for the entire molecule; or
that can be transformed onto each other by rotation of (a series of) bonds in such a manner that the final shape cannot be distinguished from he original without labelling.

It is the axis of rotation that transforms each carbon of a certain colour onto its counterpart of the same colour. Thus, all carbons of a single colour are homotopic. Homotopic atoms are always chemically identical and thus will give a signal with the same chemical shift. Chemically non-identical atoms in similar environments will give similar but not identical shifts.
Enantiotopic atoms are those that are transformed onto each other by a plane of symmetry inherent to the molecule. These are only chemically equivalent in achiral environments.

There are indeed 4 carbon environments in both of these molecules, shown below;
A plane of symmetry can be drawn through each, with one environment having only one carbon nucleus. All other sites have 2 carbons.
I don't know that any words will make it any clearer than the picture, but if you aren't sure after looking at this, I can elaborate.

I can see how they are different, but I don't understand how it makes the four peaks.

Each unique generates a different environment. There are four different environments (red, blue, green, purple). Each will have there own signal. The intensities will be 1:2:2:2 (red:blue:green:purple)


text:
Calculating ratio of bicarbonate ion to carbonic acid in blood
For the following reaction, $\ce{K_a} = \pu{4.2e-7}$ :
$\ce{H2CO3(aq) + H2O(l) <=> H3O+(aq) + HCO3-(aq)}$
We're given the pH of blood, which is $7.4$. We need to calculate the ratio ("$X$") of bicarbonate ion to carbonic acid in blood.
I know I need to convert the $\ce{K_a}$ to $\ce{pK_a}$, and did so as follows:
$\ce{pK_a} = -\log(\ce{K_a}) = -\log(\pu{4.2e-7}) = 6.4$
But I'm not sure how to proceed from here. Any help would be greatly appreciated.

Are you familiar with the Henderson–Hasselbalch equation?

For these types of problems, the general approach would be to use the Henderson-Hasselbalch equation. For a general weak acid/conjugate base system, this has the form: $$\mathrm{pH}=\mathrm pK_\mathrm a+\lg\left(\frac{\ce{A-}}{\ce{HA}}\right)$$ Since you are given $\mathrm{pH}$ and have found $\mathrm pK_\mathrm a$, it won't require much more work to obtain the ratio.

Using the Henderson-Hasselbalch equation
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10}\left({[\ce{A-}]\over[\ce{HA}]}\right)$$
and substituting bicarbonate and carbonic acid for $\ce{A-}$ and $\ce{HA}$, respectively, we arrive at:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10}\left({[\ce{HCO3-}]\over[\ce{H2CO3}]}\right)$$
Given $\mathrm{pH}$ for blood and computing $\mathrm{p}K_{\mathrm{a}} = 6.376$ from the information given, we have
$$7.4 = 6.376 + \log_{10}\left({[\ce{HCO3-}]\over[\ce{H2CO3}]}\right)$$
or
$$1.024 = \log_{10}\left({[\ce{HCO3-}]\over[\ce{H2CO3}]}\right)$$
Finally, the ratio of bicarbonate to carbonic acid is solved for as
$$10^{1.024} = \left({[\ce{HCO3-}]\over[\ce{H2CO3}]}\right) = 10.57$$
which makes sense at given that the $\mathrm{pH}$ for blood is almost exactly one unit higher than the $\mathrm{p}K_{\mathrm{a}}$ for carbonic acid.


text:
Biphenyl symmetry group dependence on torsional angle
What is the symmetry group of the biphenyl when the torsional angle between the phenyls is 20 degrees? What about when it is 90 degrees?

The $20$ degree structure exhibits the $D_2$ point group and the $90$ degree $D_{2d}$.
The figure below shows an $D_{2d}$ structure (a spiropentadiene) but the similarity in symmetry should be clear (I hope ). (From molecule-viewer.com)

The biphenyl at an angle not $90$ degrees is shown below


text:
Relation between molecule symmetry and crystal symmetry
How much is crystal symmetry dependent on molecule symmetry?
Can one make any kind of inferences on the symmetry of the molecule looking at the symmetry of its crystal or the other way around?

Not much. If the molecule is chiral (and you don't have a racemic mixture), then so is its crystal, and vice versa. That's about it.

But I read somewhere that the reason that snow crystals have a triad axis of symmetry passing through has something to do with the shape of the water molecule.

If the shape of the molecule were different, then it would be some other molecule, and consequently the crystal would be different, too. It is not going to be any more straightforward than that. True, the snowflakes have sixfold (or threefold, for that matter) symmetry axes. Do you think water molecule has such axes? It has none.

There are plenty of inorganic materials that essentially built up from "balls" (ions made of single atom), yet you have a wide variety of crystal structure even of those.

I have a subtle thought in my mind from my crystallography courses I took at my masters program. I remember something along the lines of: you can't have a crystal with a higher symmetry group than the molecules building it. This is why most organic molecules have really asymmetric crystal structures, while inorganic molecules (usually much more symmetric) can build up many different space groups. I do not have a reference for this - it may not even be true. But this is what I feel is almost right.

@Ezze That's wrong, you can. Water ice, if anything, is an example of a crystal which is more symmetric than the molecule it is built of.

There is only loose connection.
If you have molecule that shows C3 symmetry like Ph3P->BCl3 there is a good chance that crystal will have either hexagonal or trigonal cell. On the other side I've seen structure of C2 symmetrical molecule (small metalo-organic) where three molecules built up a helix which led to P3(1) space group.
The only "hard" rule is, as Ivan wrote, is that if your compound is chiral and you are working during crystallization with pure enantiomers (not racemic mixtures) then the space group of crystal must also be chiral.
On the other side racemates can give crystals both in chiral (pure or racemic twins) or in centrosymmetric space groups. Eventually you will see quite often, that molecules that in solution have high(er) symmetry are not placed on special but on general position in crystal, what eventually leads to "not quite symmetric" geometry.

To be honest, I don't think you can generally infer any relation between the symmetry of your molecule and the symmetry of the crystal.
Remember that crystallographic symmetry only describe how the asymmetric units populate the unit cell. As such if you work with large chiral molecules (ahem proteins/DNA/RNA) there will be no molecular symmetry.
The asymmetric unit may contain symmetry in the form of non crystallographic symmetry (NCS) which can influence the observed diffraction pattern.
The only general rule as stated by others is that if the molecule that have been crystallized are chiral, then the only allowed spacegroups are the 65 chiral spacegroups. On the other hand if the molecule is not chiral, it could crystallize in any of the 230 possible 3-dimensional spacegroups.


text:
Why second ionisation energy greater than the first?
Why is the second ionisation energy for atoms with more than 1 electron in the outermost sublevel greater than that of the first, if you are removing an electron from the same sublevel ? Why is there a greater attraction for an electron in the same sublevel?

The ionisation energy basically measures how tightly an electron is held. That in turn is determined by a couple of factors, namely:

Nucleus-electron attraction. If this increases, the ionisation energy increases.
Electron-electron repulsion. If this increases, the ionisation energy decreases.

More formally these can be quantified by $Z$, the nuclear charge, and $\sigma$, the shielding parameter. The effective nuclear charge, which is hopefully a term you have come across, is defined as
$$Z_\mathrm{eff} = Z - \sigma$$
and the larger $Z_\mathrm{eff}$ is, the larger the ionisation energy. You don't necessarily need a formula to prove this, as it should be pretty intuitive. Larger effective nuclear charge $\implies$ electron is held more tightly $\implies$ more energy needed to ionise. But here is the mathematical relationship:
$$\text{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$
As long as you are staying within the same subshell, i.e. the principal quantum number $n$ is not being changed. So, the ionisation energy is determined by $Z_\mathrm{eff}$.
When you remove an electron, you decrease the electron-electron repulsions; this means that $\sigma$ decreases. As a result $Z_\mathrm{eff}$ decreases and the ionisation energy decreases. (Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus. However, within this inorganic chemistry framework, the nuclear charge $Z$ is simply the number of protons in the nucleus and therefore remains constant.)

Is the additional culonic attraction after first ionisation accounted for in $Z_{\text{eff}}$, i.e. in $\sigma$, or instead just the decrease in repulsion between negative charges? (Or am I just confused and this is the same thing?)

I guess it is two sides of the same coin. Less repulsion means that electrons on average are closer to the nucleus, so experience a stronger Coulombic attraction from the nucleus. I should make that slightly clearer.


text:
How to determine the order of a reaction with not a commensurate rate equation
As an example of such reactions let's take decomposition of $\ce{N2O}$ which I got the equation from a textbook :
$\ce{N2O = N2 + 1/2 O2}$
with the rate equation of $$-r_{\ce{N2O}}=\frac{k_1[\ce{N2O}]^2}{1+k_2[\ce{N2O}]}$$
So how do we determine the overall order of reaction or order of reaction with respect to a reactant/product ?

To be honest with you, the term "order of reaction" is not an exact term. If I want to pull out the rigorous kinetician out of the closet, I'd say that only elementary reactions have orders, and we have first or second (or zero or maybe third) order reactions only, and they are obvious to spot.
What you have is not an elemental reaction, but an overall - somewhat empirical - net reaction rate for a complex mechanism. I personally do not like to talk about orders of non-elementary reactions, but some authors do this. In this context, I'd say your reaction is first order in $\ce{N_2O}$, simply because you have concentration squared divided by approximately concentration. Note that if $\ce{[N_2O]}$ is sufficiently large (so the $1$ in the denominator is negligible), the equation becomes $-r =\ce{\frac{k_1 [N_2O]^2}{k_2 [N_2O]} = k' [N_2O]}$, so in high concentration limit the unimolecular approximation becomes exact.

Also, in the case of sufficiently small concentrations of $N_2O$, we could approximate the rate law to be 2nd order overall.


text:
Why is benzoic acid more acidic than p-chlorobenzoic acid?
We know that in case of halogens, inductive effect is more dominant than resonance effect. So, the -$\ce{Cl}$ on the para position will withdraw electrons by a -I effect and should make p-chlorobenzoic acid more acidic than benzoic acid. But why is it not so? Am I missing something?

Where are you getting your information? So far as I can find, 4-chlorobenzoic acid has a $\mathrm{p}K_\mathrm{a}$ of $3.98$ and benzoic is $4.2$.

No! -COOH is meta directing group and hence it isn't stabilised at the para position. Hence benzoic acid is more acidic than para-chloro benzoic acid(because the resonating structure of benzoic acid is more stable than p-chloro benzoic acid due to the reason mentioned above).

The meta para comes in external electrophilic attack, not in acidity.  For acidity, we have to consider in which case COOH can easily lose its proton. If more electron withdrawing group is attached to COOH its acidity increases

The resonance with Benzene deatabilises conjugate bases if you think properly.Resonance  is less  favoured in P chloro benzoic acid as +ve charge is created near Cl which is unstable due to -I . So electron pair on $ O^-$ can get delocalise very easily in ap chloro benzoic acid  rather than benzoic acid. So acidity of Benzoic acid less than p chloro benzoic acid.


text:
How to calculate the solubility of lead (II) carbonate in water, given that bicarbonate will also form?
Calculate the solubility of lead (II) carbonate, $\ce{PbCO3}$, ($K_{\mathrm{sp}} = 1.5 \times 10^{-15}$) in water, given that carbonate is a stronger base than water, so will also form bicarbonate. When bicarbonate acts as an acid: \begin{align}
\ce{HCO3- &<--> H+ + CO3^2-} & K_{\mathrm{a}} &= 4.7 \times 10^{-11}.
\end{align}
I think the equation is, in light of airhuff's suggestion, $$\ce{PbCO3 -> Pb^2+ + CO3^2-}.$$
What I'm most confused about is how the bicarbonate and $K_\mathrm{a}$ comes into play in this question. It makes a traditional $K_\mathrm{sp}$ question a little trickier.
Possible answer?:
\begin{align}
\ce{PbCO3 &<--> Pb^2+ + CO3^2-}
&\implies K_1 &= K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{CO3^2- + H2O &<--> HCO3- + OH-} &\implies K_2\\
\ce{PbCO3 + H2O &<--> Pb^2+ + HCO3- + OH-} &\implies K_3\\
\ce{HCO3- &<--> H+ + CO3^2-} &\implies K_4 &= K_\mathrm{a} = 4.7\times10^{-11}\\[3ex]
K_3 &= K_1 \cdot K_2\\
K_3 &= \frac{K_\mathrm{w}}{K_4}\\
K_2 &= \frac{1\times10^{-14}}{4.7\times10^{-11}}\\ &= 2.1\times10^{-9}\\
K_3 &= (1.5\times10^{-15})\cdot(2.1\times10^{-4})\\ &= 3.15\times10^{-19}\\
S_{\ce{PbCO3}} &= [\ce{Pb^2+}][\ce{HCO3-}]\\ &=[\ce{OH-}]\\
K_3 &= [\ce{Pb^2+}][\ce{HCO3-}][\ce{OH-}]\\ &= S^3\\
S &= \sqrt[3]{K_3} (3.15\times10^{-19})\\
  &= 6.8\times10^{-7}~\mathrm{M}\\
\end{align}
Does the way I solved it look okay?

I'm pretty sure this will be closed shortly if you don't show some more work. How about writing out the chemical reaction for the solubility of $\ce{PbCO3}$? Take a shot at combining the reactions an calculating the solubility? I've seen you post some stuff before and I think you can at least make a good effort toward solving this one.

I've tried to solve this question and reached a final answer to what I think may be right?

@orthocresol , I think I got this pretty complex formatting right, other than that I couldn't figure out how to do a cubed root so did a power to the 1/3. You've been helpful in getting all this through my head and I just wanted to bounce this one off you...thx for any input. This message will self destruct rather than be left here ;)

@airhuff any comments about the actual question will be most helpful. Thanks.

Greg, I just nominated this question to be reopened based on all the work you've put into it. This doesn't mean it will be reopened, just that I think it should be. Regarding the formatting, it's much easier to read when properly formatted. One of the reasons I did all that work to re-format was to increase the likelihood that it would be re-opened. I don't have the time to look closer at it now, but it looks like you're on the right track anyway.

@airhuff (square)roots: \sqrt[3]{K_3}. I reopened, +1 for effort!

I haven't worked out the answer, but yours isn't correct. You messed up a bit on the nomenclature for carbonic acid. $K_4 = K_{\alpha2}$ and $K_2 = K_{\beta2} = \dfrac{K_w}{K_{\alpha2}} $. The key to the solution is to note that $\ce{[Pb^{2+}] = [CO3^{2-}] + [HCO3^-] }$

List known reactions and equilibrium constants
\begin{align}
\ce{PbCO3 &<--> Pb^2+ + CO3^2-} \quad \quad \quad \quad \quad K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{HCO3- &<--> H+ + CO3^2-} \quad \quad \quad \quad \quad \quad K_\mathrm{a2} = 4.7\times10^{-11}\\
\ce{CO3^2- + H2O &<--> HCO3- + OH-} \quad \quad \quad \quad \quad K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = 2.13 \times 10^{-4}\\
\end{align}
Start off by just evaluating quick and dirty

Assume $\ce{[Pb^{2+}] = [CO3^{2-}]}$, $\ce{[Pb^{2+}] = \sqrt{1.5\times10^{-15}} = 0.387 \times10^{-7}}$

OK, the assumption is no good and this is going to be fairly nasty...

The pH is close to the autoionization of water
$\ce{CO3^{2-}}$ is a strong base so $\ce{HCO3^{-}}$ can't be ignored.

GIVEN:

Starting off with pure water so $\ce{[H+] = [OH-] } = 1\times 10^{-7}$ before $\ce{PbCO3}$ is dissolved.
$\ce{[CO3^{2-}] = 0}$ and $\ce{[HCO3^-] = 0}$ before $\ce{PbCO3}$ is dissolved.

WORK
Know that

The final pH > 7 due to the formation of $\ce{HCO3^{-}}$
$\ce{[Pb^{2+}] = [CO3^{2-}] +[HCO3^-]}$
$\ce{[OH^-]_{final} - [OH^-]_{water} = [HCO3^-] }$ or $\ce{[OH^-]_{final} = 1.00\times 10^{-7} + [HCO3^-] }$

For the ratio $ \dfrac{\ce{[CO3^{2-}]}}{\ce{[HCO3^-]}}$ we can use the expression of the $K_\mathrm{a2}$ equilibrium to find that:

At pH 7 ratio = 0.00047
At pH 8 ratio = 0.0047
At pH 9 ratio = 0.047

So it would seem that $\ce{[HCO3^{2-}] >> [CO3^-]}$ is a reasonable assumption, which also means that $\ce{[Pb^{2+}] \approx [HCO3^-]}$.
Now we have the two equilibrium equations
$K_\mathrm{sp} = 1.5\times10^{-15} = \ce{[Pb^{2+}][CO3^{2-}]}$
$K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = \dfrac{\ce{[HCO3^-][OH-]}}{\ce{[CO3^{2-}]}}$
which we can multiply together to get
$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$
APPROXIMATE CUBIC SOLUTION
if we let $x = \ce{[Pb^{2+}]}$ then since $x = \ce{[Pb^{2+}] \approx [HCO3^-]}$ and $\ce{[OH-] = [HCO3^-] + 1.00\times10^{-7}}$ we have
$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} =(x)(x)(x + 1\times10^{-7}) =x^3 + (1\times10^{-7})x^2$
Now if we multiply both sides by $10^{21}$ we are trying to solve:
$319.149 = x^3 + x^2$
neglecting the square term we get $x=\sqrt[3]{319.149} = 6.834$, or $\ce{[Pb^{2+}] = 6.834\times10^{-7}}$.
But looking at our cubic equation again
$319.149 = x^3 + x^2 \text{ =?= } 6.834^3 + 6.834^2 = 365.821 $
which is too far off since we need values to about 1%. Solving the cubic by iteration we get $x = 6.516$

Using 2 guesses then interpolation
    6.834           365.876
    6.000           252.000
    6.492           315.758
    6.517           319.257
    6.516           319.116
    Check
    $319.149 = x^3 + x^2 \text{ =?= } 6.516^3 + 6.516^2 = 319.116 $

which gives:
$\begin{align}
\ce{[Pb^{2+}]} &= 6.516 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.516 \times10^{-7}\\
\ce{[OH^{-}]} &= 7.516 \times10^{-7}\\
\end{align}$
We can calculate
$\ce{[CO3^{2-}]} = \dfrac{K_\mathrm{sp}}{\ce{[Pb^{2+}]}} = 0.023\times10^{-7} $
$\ce{[H+]} = \dfrac{K_\mathrm{w}}{\ce{[OH-]}} = 0.133\times10^{-7}$
CHECK:
$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.695\times10^{-11}$
So to the correct number of significant figures we have
$\begin{align}
\ce{[Pb^{2+}]} &= 6.5 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.023 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.5 \times10^{-7}\\
\ce{[H^{+}]} &= 0.13 \times10^{-7}\text{,}\quad pH = 7.88\\
\ce{[OH^{-}]} &= 7.5 \times10^{-7}
\end{align}$
EXACT FORMULIC SOLUTION
Starting with the formula
$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$
We let $x = \ce{[Pb^{2+}]}$ as before.
$\ce{[HCO3^-] = [Pb^{2+}] - [CO3^{2-}] =} x - \dfrac{K_\mathrm{sp}}{x} = \dfrac{x^2 - K_\mathrm{sp}}{x}$
$\ce{[OH-] = [HCO3^-]} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 - K_\mathrm{sp}}{x} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x}$
$\begin{align}
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} &= (x)(\dfrac{x^2 - K_\mathrm{sp}}{x})(\dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x})\\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &=(x^2 - K_\mathrm{sp})(x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}) \\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - K_\mathrm{sp}\sqrt{K_\mathrm{w}}x + K^2_\mathrm{sp}\\
0 &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - \dfrac{K_\mathrm{w}K_\mathrm{sp}+ K_\mathrm{sp}K_\mathrm{a2}\sqrt{K_\mathrm{w}}}{K_\mathrm{a2}}x + K^2_\mathrm{sp}\\
\end{align}$
$x = 6.531198\times10^{-07}$, and other values same as Excel also...
EXACT SOLUTION WITH EXCEL
Solved first by using an excel spreadsheet (with too many significant figures, but extra digits help show that the calculations are consistent)...*
$\begin{align}
\ce{[Pb^{2+}]} &= 6.531199 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.022967 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.508231 \times10^{-7}\\
\ce{[H^{+}]} &= 0.133187 \times10^{-7}\text{,}\quad pH = 7.8755376\\
\ce{[OH^{-}]} &= 7.508230 \times10^{-7}
\end{align}$
CHECKS:
$\ce{[H^{+}][OH-] = 0.9999986\times 10^{-14}}$
$\ce{[Pb^{2+}][CO3^{2-}]} = 1.500020\times10^{-15} $
$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.700057\times10^{-11}$


text:
How to draw wedged double bond in ChemDraw?
How can I draw a wedged double bond in ChemDraw? When I try to draw BINAP, making the wedged double bond seems impossible. It is however possible to draw a bold double bond.

I don't remember ever being able to do that. But I also think that that would look weird...

Note that the shown structural diagrams are not in accordance with the section on hindered biaryls in Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006).

As Zhe pointed out in the comments, bolding both bonds looks 'weird', and if you look in textbooks or throughout the chemical literature, BINAP (and related systems) are commonly drawn with only the outer edge bolded (as shown on the left in the image below).

Although the feature you want isn't directly supported in ChemDraw,  one can get both lines bolded by simply selecting the bond in question, right-clicking, and changing the line width to match the bold width (this gives you the structure on the right hand side of the image below).
That said, just because you can do this, doesn't mean you should, it goes against the standard way of drawing these systems, and doesn't add anything in terms of clarity (the first diagram emphasises the stereochemistry perfectly well, the additional bold bond adds nothing).


text:
Is this online fluorophore database reliable?
I need fluorescence spectra for various fluorophores in certain dispersion media for some calculations. There is an online database of University of Arizona: http://www.spectra.arizona.edu/
Is using such online resources reliable (not for formal purposes)?
I'm sorry for not asking a technical chemistry question.

No idea what you're trying to do. // The spectra are self-consistent. On a different instrument with different path lengths, a different grating, a different photodetector and etc etc etc the spectra will be a slightly   different, but they won't be radically different.

That is the correct domain name for the respected and accredited University of Arizona. I would feel confident using their data.
In general, that's the key: being certain that the source is who they say they are, and that they are a source that you have some reason to trust.
Regarding security, it's unlikely someone would spend the time, money and effort required to falsify some fluorescence spectra database.  If it were a database of controversial climate data, or clinical trials for a new medication, etc., you might want to investigate what the site does to secure their data.
I would also be cautious of using open wiki data like Wikipedia. Although I've personally gotten into the habit of trusting those sites more than I should, anybody can edit them. And while they try to police the sites for poorly documented data and such, it's a lot to keep up with. A good wiki article will have sources for their data, and reaching out directly to the source, if possible, is the better way of getting reliable data from a wiki site.

As airhuff already pointed out, the origin of the database is the University of Arizona. The data was compiled by Urs Utzinger, who seemingly still holds a position there in biomedical engineering.
I can't comment on the Flash part, since I do not use Flash.
The SQL dump, which can be used to recreate the database, contains lots of information on the data sources for the different spectra.
In summary, I would trust the data.


text:
What does a higher half cell potential mean?
I was taught in class that a positive large half cell potential generally meant that the reaction was more favorable.  What exactly does this mean?  Doesn't a reaction with potential 4V need a greater input of energy to occur? Meaning that a lower electron cell potential would occur first and is therefore more favorable?
Also how why does Zn(II) +2e- ---> Zn favor the left side of the reaction?  Shouldn't Zinc naturally tend towards being neutral?

And positive cell potentials suggest spontaneity right?

That's not quite correct.
Half-reaction potentials are only really meaningful when you complete the redox pair, that is, you must have an oxidation and reduction reaction. The sum of the potentials is the potential for the full redox reaction at standard concentration and state. In that sense, the more positive either of the potentials is, the more favorable the overall reaction will be.
There is no preference for a neutral species in general. If you put sodium in water, the electron is quickly liberated and the end products of hydrogen and sodium ion are much more favored than the reactants of water and sodium.

In a practice examination that I'm using to study for my upcoming midterm, one of the questions asks to choose three elements from a row of the periodic table and to give the order of the increasing oxidation half cell potential of the neutral element.  Is it correct to assume that a more electronegative element would have the lowest half cell oxidation potential?  Because it would be the most resistant to losing its electrons?

That seems correct. You can also confirm this easily by consulting a table of reduction values. Your statement then becomes: electronegative elements should have more positive reduction potentials.


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Required alkenes and reagents for alcohol products
In the reactions of alkene, which alkene and reagents would be necessary to produce

2,3-dimethyl-2-butanol
3-bromo-2-methylbutan-1-ol

For 2,3-dimethyl-2-butanol, I came to the conclusion that it would be the 2,3-dimethyl-1-butenol, assuming that a rearrangement would happen. I am right about this one?

Welcome to Chemistry.SE! This is a Q&A site, but it is not a do-my-homework service. The members will be glad to help and point you in the right direction, but you have to show us your efforts, i.e. what you've done so far and what is it exactly that you don't understand. Otherwise, your question may be voted down and eventually closed.

Do you mean 3-bromo-2-methylbutan-1-ol for your second product?

Yes, exactly this

Let me try to give you some hints without answering your homework question directly ;-)

Draw your products!
You can still take rearrangements into account on a later stage if nothing else works out, but initially: think simple
Would you agree that the most simple way to convert an alkene to an alkanol is the reaction with water under acidic conditions? You probably remember how this works:

 proton adds first, a cation is formed, then comes water as the nucleophile

Will this work if you need to introduce $\ce{Br}$ and $\ce{OH}$ at adjacent positions?

 At 3., $\ce{H+}$ and (formally) $\ce{OH-}$ were added. The combination of $\ce{Br+}$ and $\ce{OH-}$ doesn't sound like a common reagent, but your teacher might have talked about the addition of halogens to alkenes :) With bromine, the $\ce{Br2}$ adds to the alkene and forms a three-membered ring, a bromonium ion. Then $\ce{Br-}$ comes along and adds too. Formally, this is like adding $\ce{Br+}$ and $\ce{Br-}$ to the adjacent carbon atoms of the $\ce{C=C}$ bond. Now imagine that there is not just bromine, but also a lot of water. The first step, the formation of the bomonium ion, will be the same, but now you have (lots of) water as a nucleophile that competes with $\ce{Br-}$!

Thank you for your feedback! My teacher didn't talked about epoxides yet. What i know is some very basic things that i searched. May i ask you: When you say "Adjacent positions" what do you mean? Sorry if it is a dumb question, but english is not my main language, and this happens sometimes =D.

@terra2322 No worries :) It is not a dumb question at all and reading buteno, I already thought that your first language might be spanish. Adjacent just means being neighbours, side to side. So, carbon 2 and carbon 3 are adjacent positions, while carrbon 1 and carbon 3 are not.

@terra2322 Sorry about the epoxide! I forgot about a much simpler reaction and have updated my answer.

Aaah, i see. I think i understand how this kind of exercises are done.

I'm gonna keep doing it thinking like that. Thanks!


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What is the pH of kvass?
I know that it is less than 7.0, but would be nice to know exact value of pH of natural fermented kvass (not soda-like counterfeit). Another aspect is to know if kvass is more gentle than Cola (pH=2.5) ?

According to this paper$^{[1]}$, the kvas in their study started out at a pH of 6.1 prior to fermentation, then became stable at about a pH of 3.8 after 12 hours.  This is of course significantly less acidic than a cola with a pH of 2.5 (and some colas can have a pH closer to 2).
1) QUALITY CHANGES OF NATURALLY FERMENTED KVASS DURING PRODUCTION STAGES, Department of Food Technology, Faculty of Food Technology, Latvia University of Agriculture, Liela Street2, Jelgava, Latvia

Wow, thanks a lot !!! I couldn't get answer on Quora for two days, and here I got it so fast.

No problem, I'm very glad it was helpful :)


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Periodic trends on half cell potentials
In a practice examination that I'm using to study for my upcoming midterm, one of the questions asks to choose three elements from a row of the periodic table and to give the order of the increasing oxidation half cell potential of the neutral element. Is it correct to assume that a more electronegative element would have the lowest half cell oxidation potential? Because it would be the most resistant to losing its electrons?
Would a correct sequence (highest oxidation half cell potential to lowest) be B, C, N because as the sequence goes from left to right it becomes increasingly electronegative and therefore less likely to be oxidized?

https://www.google.co.in/search?q=electrochemical+series&espv=2&biw=1440&bih=803&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiUjLnF_LrSAhXIPI8KHdkaACEQ_AUIBigB#imgrc=p2M3BgmtGk2ChM:

No, it's much more complicated, in general.