text:
Why is the product of sodium and water, sodium hydroxide and not sodium oxide?
In the reaction $\ce{Na(s) + H2O}$ the products are listed as $\ce{NaOH(aq) + H2 (g)}$, but why would $\ce{Na2O}$ not be a possible product?
I searched and found that the reaction $\ce{Na2O + 2H2O -> 2NaOH}$ occurs as well. Does it have to do with the stability of oxides?

As a way of remembering how elements react with water, I often think of the oxide intermediate. First write out the reaction producing oxide and hydrogen. Then consider the oxide reacting with water or forming hydrates. This does not work for halogens, though, and is still just a trick to systematise inorganic reactions. Still gets you further than you might think. As another example: $$\ce{Si + 2H2O -> SiO2 + 2H2}$$ (high temperature). (No further "reaction" here but hydrates and polymers still form when conditions and the state of $\ce{SiO2}$ are right.

Related if not dupe: http://chemistry.stackexchange.com/q/50912

In the usual case every student learns, the reaction of sodium with water is the following — I made note to include the states of matter here as they matter (pun intended).
$$\ce{2 Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$
In the typical setup, you have lots of water and little sodium. Therefore, whatever you get will be a solution or a precipitated salt. Alkali metals, however, often form very soluble salts and indeed all alkali hydroxides are well soluble. Thus, we will never get to see any solid by-products; all will happen in the solution phase.
Technically, it is possible for an oxide anion to be formed. However, this is when Brønsted and Lowry’s acid-base theory kicks in. Remember that water is amphoteric and can react as a base to form $\ce{H3O+}$ or as an acid to form $\ce{OH-}$. Well, equation $(2)$ expands that scheme.
$$\ce{H3O+ <=>[$K_\mathrm{a,1}$][H+] H2O <=>[$K_\mathrm{a,2}$][H+] OH- <=>[$K_\mathrm{a,3}$][H+] O^2-}\tag{2}$$
The oxide anion is connected to hydroxide and water by a simple acid-base reaction. From the definition of $\mathrm{p}K_\mathrm{a}$ values and the ionic product of water, we know that
$$\mathrm{p}K_\mathrm{a} (\text{acid}) + \mathrm{p}K_\mathrm{b} (\text{conj. base}) = 14\tag{3}$$
Since $K_\mathrm{w}$ doubles as the acidity constant of water, $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 14$. Likewise it can be shown, that $\mathrm{p}K_\mathrm{a} (\ce{H3O+}) = 0$. This is a difference of $14$ logarithmic units between two subsequent deprotonation steps. The acidity constants of sulfuric acid or phosphoric acid show that while this difference is large, it is not unusual for the conjugate base (if it is acidic) to have an acidity constand orders of magnitude lower than the original acid).
Condensing that into a conclusion, the acidity constant of hydroxide must be even lower still, to the point where we can say practically no oxide anions can be formed in equilibrium. Likewise, considering that there are many more water molecules than hydroxide anions, any oxide accidentally generated will get protonated immediately.
$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)}\tag{4}$$
Therefore, it is not feasible for oxide anions to exist in water. This is known as the nivelling effect: no acid more acidic than $\ce{H3O+}$ can survive in aquaeous solution for extended time and no base more basic than $\ce{OH-}$.
Since the entire reaction produces products in solution, sodium oxide is not a possibility.

What compounds are formed in a system that contains sodium, hydrogen and oxygen depends on the pressure, temperature, and the number of atoms of the respective type.
If sodium reacts with an excess of water at standard conditions, dissolved sodium hydroxide and hydrogen are formed.
$$\ce{2 Na + 2 H2O-> 2 Na+_{(aq)} + 2 OH-_{(aq)} + H2}$$
Sodium oxide can be synthesized by reaction of sodium hydroxide with sodium:
$$\ce{2 NaOH + 2 Na -> 2 Na2O + H2}$$
Additionally sodium oxide can react with hydrogen to form sodium hydroxide and sodium hydride under certain conditions. The reaction is reversible:
$$\ce{Na2O + H2 <=> NaOH + NaH}$$
From the first two reaction equations above we can conclude that we also could obtain sodium oxide by reaction of sodium with water.

Sodium oxide dissolves in water to form sodium hydroxide:
$$\ce{Na2O(aq) + H2O(l) -> 2NaOH(aq)}$$
So even if sodium oxide is formed, it would be immediately converted to sodium hydroxide.


text:
Determine chemical reaction stoichiometry?
Unfortunately my background on chemistry is very weak but I have to do some experiments in that field as a part of my project.
Here is my problem: I am doing some leaching experiments on realgar (one of arsenic bearing minerals: $\ce{As4S4}$) I wanted to see how this mineral can be oxidized in water and arsenic can be leached out in different leaching conditions. However, I got confused about how I should write the chemical reactions related to my mineral oxidation and after that how I can derive stoichiometry coefficients related to my real conditions. Let say I have $\ce{AsS}$ ($\ce{As4S4}$) and based on my leaching conditions and Eh-pH diagram, I guess I have the following reaction:
$$\ce{4 AsS + 9 O2 + 8 H2O -> 4 H3AsO3 + 4 SO4^2- + 8 H+}$$
After getting some samples I did ICP analysis and found that I have $2\ \mathrm{ppm}\ \ce{As}$ in my solution (which I think should be present in the form of $\ce{H3AsO3}$ based on the reaction). and saw that after a while this $2\ \mathrm{ppm}$ did not change significantly and we can say the reaction reached the equilibrium (roughly speaking) So, now can I derive stoichiometry coefficients based on these data or I should do something else?

First of all welcome!  Writing a chemical equation is definitely a good place to start when calculating stoichiometry.  Have you tried that?  Showing what you have tried, successfully or not, will help people to help you and they will be more motivated to do so when they see you have already made efforts to solve the problem yourself.

Thanks for the comment. I have edited my question :) Hopefully, it is now more clear :)

This is really a long comment....
You're mingling several concepts in such a way that it is impossible to help you.
Stoichiometry is about balancing a chemical equation.
[Chemical] kinetics is about how fast a reaction occurs. There is scant correlation with stoichiometry.  Kinetics is in general measuring how fast a reaction is moving towards equilibrium.
[Chemical] "Recovery," or extraction deals with what fraction of a particular element (or perhaps mineral) is recovered using a particular process. This again depends somewhat on kinetics but kinetics has a large number of factors. (So for Cu this would be what % of the Cu was extracted...)
Chemical equilibrium does depend on stoichiometry, but for an extration reaction a true equilibrium is rarely reached.

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text:
Solubility of urea in water
I'm hoping that this isn't too obvious of a question (it's been awhile since I've taken chemistry), but I was looking at the solubility table on Wikipedia, which gives the solubility as grams per 100 mL. My understanding is that 100 mL of water weighs 100 grams.
So here's what I was having trouble wrapping my head around: it lists the solubility as 108 (at least at 20 °C). If I understand that right, that would mean that I can absorb 108 grams of urea in only 100 grams of water, right? Can someone explain why it's possible to have more urea than water in this case?

Why would it be a problem to have more of urea than of water by mass?

@Jan It just seems really weird to me that that would be the case, I'm trying to wrap my mind around why that's possible.

Well, that is just the way that it is… You should also look at this a bit differently. Urea has molar mass of $\pu{60 g mol^-1}$ and water has molar mass of $\pu{18 g mol^-1}:$
$$n(\ce{CO(NH2)2}) = \frac{m(\ce{CO(NH2)2})}{M(\ce{CO(NH2)2}) } = \frac{\pu{108 g}}{\pu{60 g mol^-1}} = \pu{1.8 mol}\tag{1}$$
$$n(\ce{H2O}) = \frac{m(\ce{H2O})}{M(\ce{H2O})} = \frac{\pu{100 g}}{\pu{18 g mol^-1}} = \pu{5.6 mol}\tag{2}$$
There are $\pu{5.6 mol}/\pu{1.8 mol} = 3.1$ molecules of water per every molecule of urea.

Thanks, that makes a lot of sense, so the thing that makes that possible is the fact that there are strictly more water molecules than urea molecules (i.e. the total number of molecules in each case is more relevant than the total weight)?

In general the species with the most molecules is the solvent and the species with the fewer molecules is the solute. However solvent can also mean whatever is the liquid and solute something that is a gas (e.g. oxygen in water) or solid (a solid salt that dissolves in water).  So truthfully it is hard to come up with absolute definitions. You have to look at the context in which the words are being used.

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text:
How DFT-D3 incorporates coordination number (CN) into C6
I'm reading Grimme's DFT-D3 paper
but really puzzled by how the $\mathrm{CN}$ is considered in the calculation of $C_\mathrm{6}$ coefficients. (Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg. J. Chem. Phys. 2010, 132 (15), 154104.)
According to Fig. 5 in the paper, it seems the $\ce{AB}$ pair is put in some reference system and then based on the precomputed $C_\mathrm{6}$ values in the reference system we could get the $C_\mathrm{6}$ in the environment.
Especially, I'm confused by the fact in the figure that $\mathrm{CN}^\ce{A}$ and $\mathrm{CN}^\ce{B}$ are in different dimension, what does it mean?

I will quote the paper quite a bit, but I'll try and summarize a bit after the quotes and equations. You might want to start at the bottom and work backward, a lot of this is just for later reference.
First, some background. Equation 3:
$$
E^{(2)} = \sum_{AB} \sum_{n=6,8,10,\dots} s_{n} \frac{C_{n}^{AB}}{r_{AB}^{n}} f_{d,n}(r_{AB})
$$

Here, the first sum is over all atom pairs in the system, $C_{n}^{AB}$ denotes the averaged (isotropic) $n$th order dispersion coefficient (orders $n=6,8,10,\dots$) for atom pair $AB$ (...)

So $A$ and $B$ are nuclear indices, and together specify an arbitrary pair.

Although the $C_{6}^{AB}$ values can be compute easily for any pair of free atoms by using Eq. (5) [the Casimir-Polder integral] in principle, this would lead to a rather inconsistent treatment of dispersion in and between molecules.

For reference, here is equation 5:
$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} \alpha^{A}(i\omega) \alpha^{B}(i\omega) \, d\omega
$$
Instead they calculate the integrals over pairs of hydrides, subtracting the effect of molecular hydrogen (equation 10), in order to capture the effect of being bonded:
$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} d\omega \frac{1}{m} \left[ \alpha^{A_{m}H_{n}}(i\omega) - \frac{n}{2} \alpha^{H_{2}}(i\omega) \right] \times \frac{1}{k} \left[ \alpha^{A_{k}H_{l}}(i\omega) - \frac{l}{2} \alpha^{H_{2}}(i\omega) \right]
$$

At first sight this new approach seems to be a disadvantage because it leads to reference molecule dependent (ambiguous) coefficients. However, as will be discussed in detail below (...), it opens a route to system [coordination number (CN)] dependent "atomic" $C_6$(CN) coefficients. (...) We propose here a radically new "geometric" approach that is based on the concept of a fractional coordination number. (...) We base our ansatz on a fractional CN for an atom $A$, that is a sum of a "counting" function over all atoms in the system,

Equation 15:
$$
\text{CN}^{A} = \sum_{B\neq A}^{N_{at}} \frac{1}{1 + e^{ -k_1 ( k_2 (R_{A,\text{cov}} + R_{B,\text{cov}}) / r_{AB} - 1) }}
$$

where $R_{A,\text{cov}}$ is a scaled covalent (single-bond) radius of atom $A$ (and analogously for $R_{B,\text{cov}}$).

So the CN$^A$ is some bond distance-dependent scaling factor for the specific atom $A$ under the influence of all other atoms $\{B\}$. Here is how they incorporate their coordination number scaling into the dispersion coefficients. Following emphasis mine:

Turning back to the dispersion coefficients we propose the following general approach. For each element in the Periodic Table, at least one reference molecule is used to compute $\alpha(i\omega)$ values in Eq. (10). (...) If an element is commonly investigated, and/or often found in different chemical environments, we suggest to use several representatives for which values are computed. In the case of carbon for example, ethyne, ethene, and ethane can be used for CNs between 2 and 4. Values computed for $\ce{C-H}$ and the carbon atom complete the set. For the atom of interest in any of the reference systems also the CN is calculated and stored. Using Eq. (10) the $C_{6,\text{ref}}^{AB}$(CN$^A$,CN$^B$) values are computed for this pair with the two atoms in their specific environments. These serve as supporting points in an interpolation procedure to derive the coefficient for any combination of fractional CN values. Because we base our approach on pair-specific coefficients $C_{6}^{AB}$, a two-dimensional (2D) interpolation scheme must be used. After extensive testing we propose a simple Gaussian-distance ($L$) weighted average,

Equation 16:
$$
C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B}) = \frac{ \sum_{i}^{N_A} \sum_{j}^{N_B} C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B}) \times L_{ij} }{ \sum_{i}^{N_A} \sum_{j}^{N_B} L_{ij} }, \\
L_{ij} = e^{ -k_3 [ (\text{CN}^A - \text{CN}_i^A)^2 + (\text{CN}^B - \text{CN}_j^B)^2 ] },
$$

where $N_A$ and $N_B$ are the number of supporting points (= number of reference molecules) for atoms $A$ and $B$, respectively. The CN$^A$ and CN$^B$ are coordination numbers for the atom pair $AB$ in the system of interest, and the CN$_i^A$ and CN$_j^B$ are those for the two reference systems $i$ and $j$, for which $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ is the precomputed value. (...) For most of the elements only two reference values are necessary to cover the typical bonding situations. For "important" elements such as carbon, however, more values (5 in this case) seem appropriate. Currently we have computed in total 227 reference systems for the elements up to $Z=94$ which leads to about $2.6\times10^4$ difference values for $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$. They are computed once and read at the beginning of any computation.

Conceptually, here is how a calculation would work:

Your SCF has converged and it's time to calculate the -D3 correction. This table is read in from a file into some data structure so that all the $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ can be looked up easily.
Loop over the atom pairs $AB$ of your system as part of equation 3. For atom $A$, calculate its coordination number CN$^A$ using equation 15. Do the same thing for atom $B$. In the equation, $B\neq A$ means all atoms other than $A$, not atom $B$ in the pair!
For this pair, the $C_{n}^{AB}$ in equation 3 is really $C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B})$, which requires evaluating equation 16. What equation 16 does is interpolate between those precomputed $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ for a series of coordination numbers. Because there are two CNs, the interpolation is on a 2D grid, rather than a line or some higher dimension. The interpolation function is chosen to be the $L_{ij}$. From the above quote, each $i,j$ sum is only about 2-5 points, so this is very fast; even the $\ce{C-C}$ grid will only total $5\times5=25$ points.

So, to answer your question,

According to Fig. 5 in the paper, it seems the AB pair is put in some reference system and then based on the precomputed C6 values in the reference system we could get the C6 in the environment.

This is exactly correct. It is done by interpolation on a grid of reference values, where the coordinates/points on the grid are the coordination numbers, and there is a grid for every possible $AB$ pair.
$\tiny{\text{Phew.}}$

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text:
How to determine parameters range?
I am into running some leaching experiments and wanted to study the effect of different parameters on my dissolution results. But I am confused with determining the parameters range. For example, I wanted to investigate the effect of temperature on my results. But I am not sure in which temperature range I should run my experiments. Should I test different temperatures (or other parameters as well) and see the change in the results and if by changing more in the level of that parameter, no change happened, so I can say for example that level is one of my upper of lower range?
The system I am going to study is: Oxidation of arsenic bearing sulfide (AsS) in aquatic leaching medium. I want to see the effect of different parameters; pH, Eh, temperature, oxidant agent on the As dissolution behavior. However, choosing the right range for each of these parameters are my main concern.
I am going to derive the equilibrium state for my mineral dissolution (so I can derive chemical stoichiometry) and then I should study the dissolution rate of my goal element (As).  I am going to study both synthesic (pure) ore and real ore from mine location, therefore I asked my question as a general one.

Can you provide more information, such as what the actual chemical system is? As it stands, your question is very broad.

Your question is sort of a mess. (A) Are you trying to study the kinetics or the equilibrium for the reactions? (B) Is this for some sort of leaching from ore or "pure" AsS? (C) You need to first design an experiment to see what factors are important (D) Given (C) you vary each factor of interest over a range to try to fit appropriate constants to the data.

Thanks for the comment, I am going to derive the equilibrium state for my mineral dissolution (So I can derive chemical stoichiometry) and then I should study the dissolution rate of my goal element (As). You are right about the purity, but as I am going to study both synthesis (pure) ore and real ore from mine location, I asked my question as a general one.

Well a great method used is to do a Design of experiment


text:
Which molecule is more acidic: CHF3 or CHBr3?
I know that fluorine is more electronegative than bromine. However, because of the size of bromine, it is more stable with a negative charge. In the case of $\ce{HF}$ vs. $\ce{HBr}$, to me, $\ce{HBr}$ is without a doubt the stronger acid. When it comes to comparing $\ce{CHF3}$ with $\ce{CHBr3}$, we are supposed to compare the stability of their conjugate bases.
In $\ce{CF3^-}$, the fluorine can hold the electronegative charge quite well. It also exerts a strong inductive effect.
In $\ce{CBr3^-}$, the bromine is less electronegative than the fluorine but is more polarizable, so I think it can hold the negative charge more efficiently.
With this understanding, I am concluding that $\ce{CBr3^-}$ is more stable which would mean $\ce{CHBr3}$ is the stronger acid. However, I've been told otherwise.
I'm aware that the $\ce{C-Br}$ bond length is longer than the $\ce{C-F}$ bond length. I'm wondering if that short bond length causes $\ce{CF3^-}$ to be more stable than $\ce{CBr3^-}$. If this is true, is it okay for me to think about $\ce{CF3^-}$ dissociating vs. $\ce{CBr3^-}$ dissociating? In that scenario, I believe $\ce{CBr3^-}$ dissociates more easily than $\ce{CF3^-}$, meaning $\ce{CF3-}$ is more stable.
Note: This logic wouldn't apply to the instance of $\ce{HO-}$ vs. $\ce{HS-}$ because $\ce{HS-}$ is more stable than $\ce{HO-}$ (even though $\ce{H-O}$ bond length is shorter than $\ce{H-S}$ bond length).

related How to prove that CHCl3 is more acidic than CHF3?

Wikipedia list the pKa of Fluroform as $25$-$28$, while Bromoform has a pKa of $13.7$, so interestingly Bromoform is much more acidic then Fluoroform. However, I'm not certain I have a satisfying explaination of why this is the case.

The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen.
\begin{array}{|c|c|c|} \hline
\text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline
\ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline
\ce{CHCl3} & 15.7 & 3.16\\ \hline
\ce{CHBr3} & 13.7 & 2.96\\ \hline
\ce{CHI3} & \sim 25 & 2.66\\ \hline
\end{array}
Looking at the $\mathrm{p}K_\mathrm{a}$'s, the first thing we notice is that the two haloforms at the beginning and end of the series are weaker acids than the two haloforms in the middle of the series. This behavior is often seen when there are at least two factors affecting acidity and they are in opposition to each other.
Let's consider inductive, steric and resonance effects and examine their expected effect upon haloform acidity.

Inductive effects should correlate with electronegativity.  Based only on inductive effects we would expect fluoroform to ionize and produce the most inductively stabilized cabanion ($\ce{CF3^{-}}$), while iodoform would ionize to produce the least inductively stabilized carbanion ($\ce{CI3^{-}}$).
Steric effects are often discussed in terms of bulky groups coming too close to one another, but what we are really saying is that the electron clouds around the groups are coming too close to one another causing electron-electron repulsion. Iodoform is a very bulky molecule.  The iodine substituents are very large and there are many lone pair-lone pair repulsions.  Upon ionizing iodoform can relax from a crowded $\mathrm{sp^3}$-like geometry to a more open and spread out $\mathrm{sp^2}$-like geometry.  As the halogen gets smaller there is less of a steric \ electron repulsion driving force for haloform ionization.
Resonance effects are not a significant stabilizing factor in the haloform series. Significant low energy resonance structures cannot be drawn. Thanks to TanYongBoon for noting this.

Conclusion: Inductive effects and steric effects (steric meaning electron-electron repulsion) control acidity in the haloform series. The two effects are in opposition to one another.  The inductive effect strongly stabilizes the fluoroform anion and plays a lesser role as we move down the series.  Steric \ electron repulsion effects strongly stabilize the iodoform anion and plays a lesser role as we move up the series.  Said differently, only one effect stabilizes the fluoroform (inductive) and iododform (steric) anions, but both inductive and steric effects play a role in stabilizing the chloroform and bromoform anions making chloroform and bromoform the most acidic molecules in the series.

How would the F2C=F- resonance structure make sense? The fluorine atom in that structure has 10 electrons around it. That is not allowed for Period 2 elements. Resonance would not be able to stabilise the carbanions.

@TanYongBoon Thanks for noticing that, I've edited the resonance section; the conclusion remains unchanged.

Why does the geometry change from tetrahedral in the haloform to trigonal planar in the carbanion? My friend is saying that it should change to a trigonal pyramidal-like geometry instead.

@TanYongBoon   Not all of the haloforms form planar carbanions, only the larger ones where relief of steric strain becomes a driving force.

Is it possible that some sort of p-d interaction occurs between the p orbital of the carbon atom with the d orbitals of chlorine/bromine/iodine? I understand that d orbitals are energetically inaccessible for bonding when describing bonding in most molecules. But can it happen here? This is so as to delocalise the negative charge on the carbon. I mean certainly there was be some sort of delocalisation of charge right? Otherwise, the carbanion would be too unstable.

Another important consideration which many credible sources I have read have mentioned is the repulsion between the lone pairs on the halogen atoms and the lone pair on the carbon. This repulsion destabilises the carbanion. The strength of this repulsion decreases from F to I.

Also, I don't mean to pick on your answer. I am just very confused with some bits of this issue and I am feeling very depressed over it right now. Are you sure that the data from Wikipedia is credible?

@TanYongBoon  1) p-d interactions play an extremely minor role, maybe something on the order of 1% or so of the wavefunction  2) as I mention in my answer, "steric effects" are really just electron-electron repulsions, so the lone pair repulsions are included in this effect and it increases from F to I  3) I assume the Wikipedia data is credible, but I didn't collect it so who really knows  4) confusion and struggle are fine, you emerge knowing more than when you started, but no need to get depressed.  Maybe work on other problems for a while and come back to this in a month or so. Good luck!

Experimentally, according to The Relative Rates of Formation of Carbanions by Haloforms J. Am. Chem. Soc., 1957, 79 (6), pp 1406–1412 the rates of formation of the corresponding anions from the following haloforms was as follows:
$$
\begin{array}{|c|c|}\hline
\text{Haloform}&\text{Rate}\\\hline
\ce{CDCl2F}&0.89\\\hline
\ce{CDBrClF}&21\\\hline
\ce{CDCl3}&47\\\hline
\ce{CDBr2F}&180\\\hline
\ce{CDCl2I}&275\\\hline
\ce{CDBrCl2}&290\\\hline
\ce{CDBr2Cl}&1435\\\hline
\ce{CDBr3}&5790\\\hline
\ce{CDI3}&6010\\\hline
\end{array}
$$
The article has the following explanation:

While the ease of carbanion formation of chloroform
  has been attributed to the electronegativity
  of the chlorine atoms, our data make it clear that
  other factors must also be of at least comparable
  importance. There appear to be three factors that
  might be considered in explaining why the effect of
  the a-halogens on rates of carbanion formation is
  almost the reverse of that expected from the inductive
  effect. One factor is B-strain.[reference 17] That is,
  there may be repulsions between the three halogen
  atoms of the tetrahedral haloform that are somewhat
  relieved upon formation of the more nearly
  planar carbanion. Steric effects should increase in
  size quite rapidly once they have become at all important.
  However the replacement of chlorine by
  bromine in our compounds produces a significant
  increase in reactivity. If this is due to B-strain
  counteracting the decrease in electronegativity
  (0.2 unit), then the replacement of bromine by iodine
  should increase the reactivity at least as much
  since the decrease in electronegativity (0.3 unit) is
  not much larger. Another argument against the
  importance of B-strain in the present instance relates
  to the effect of iodine compared to bromine in
  two cases If the comparable reactivities of CDCl2I
  and CDBrCl2 are due to an approximate equality
  of B-strain and the inductive effect, then the B-strain
  factor should be much more important in a
  comparison of CDI3 and CDBr3. The observation
  that the latter two haloforms are equally reactive,
  within experimental error, shows that B-
  strain should not be contributing more than about 15% to the reactivity in this case. Its contribution
  should be less in all other cases.

The article goes on to discuss resonance and polarizablity effects.
See also page 109 of Fluorine in Organic Chemistry which cites to the above article and says for $\ce{CHF3}$, pKa=31, and that $\ce{CHF3}$ in experiments as above forms anions at an immeasurably low rate.

To compare the acidic strengths of the haloforms, we need to check the relative stabilities of their conjugate bases.
CF3- , CCl3-, CBr3-.
Here, only CCl3- has effective backbonding and hence the negative charge partially gets stabilised by back donation to the vacant 3dπ orbitals of Cl. Thus, CHCl3 is a stronger acid among them.
The acidic strengths of the other two csn be compared their inductive effects.
F is very electronegative and hence stabilises the negative charge on the C atom. So, CHF3 is a better acid than CHBr3.

3d orbitals of chlorine do not play any significant role in bonding. This answer follows an outdated and proven wrong concept. –1.


text:
Help with derivation on why antibonding orbitals are more antibonding
I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words,
$$-1<\int\psi^*_i\psi_j\,dx<1$$
He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,
$$S_{ij}<0\Rightarrow\text{antibonding}$$
$$S_{ij}>0\Rightarrow\text{bonding}$$
$$S_{ij}=0\Rightarrow\text{nonbonding}$$
We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,
$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$
Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom.
The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm
In the end he got two wavefunctions with different coefficients
$$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$
$$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$
He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words,
$$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$
And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.
But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

Possible duplicate of How can antibonding orbitals be more antibonding than bonding orbitals are bonding?

@Todd Minehardt The question only addresses the energy. It also doesn't consider the overlap being negative. If the overlap integral can be negative, then the antibonding energy can be lower.

OK - retracting VTC. One thing, though: you state that "He mentioned that since $S_{ij}$ is always positive..." and then go on to say "...if $S_{ij}$ can be negative (like he mentioned)...", which are contradictory statements. Can you please clarify?

I don't think I can clarify further since I didn't understand that part very well. One statement should make the other statement untrue. But it doesn't for some reason? And I am trying to figure out why. If  $S_{ij}$ can be negative,  the term in front of $\psi_{1s}+ \psi_{1s'}$ will be smaller than the term in front of $\psi_{1s} - \psi_{1s'}$. This will be true for the energy as well. Thus, we end up in a situation where the bonding orbital is more bonding than the antibonding orbital is antibonding. Which should be a problem. I am trying to figure out if one statement is false.

For the 1s orbital in hydrogen, the wavefunction is
$$\psi_\mathrm{1s} = N\mathrm{e}^{-r}$$
where $N$ is some normalisation constant. The normalisation constant is not unique: if $N$ is a satisfactory normalisation constant such that $\left<\psi_\mathrm{1s}\middle|\psi_\mathrm{1s}\right> = 1$, then for any value of $\phi$, the normalisation constant $N\mathrm{e}^{\mathrm i\phi}$ is also valid.
However you can make the assumption, without any loss of generality, that the wavefunctions of both 1s orbitals in dihydrogen ($\ce{H2}$) are of the same phase (i.e. both normalisation constants have the same value of $N$ and $\phi$). As a simple example we can just let $N$ be a positive real number. With this, the wavefunction of the 1s orbitals always have a positive value, and the overlap integral
$$S_\mathrm{AB} = \left<\psi_\mathrm{1s,A} \middle| \psi_\mathrm{1s,B} \right> = \int \psi_\mathrm{1s,A}\psi_\mathrm{1s,B}\,\mathrm{d}\tau$$
always has a positive integrand (no imaginary components, so I dropped the complex conjugate). The integral is therefore positive.
(Note that even though both 1s orbitals have the same exponential form, $\psi_\mathrm{1s,A}$ is not equal to $\psi_\mathrm{1s,B}$ since the orbitals are centred at different points in space.)
In fact, this choice of phase is required so that you can express the bonding and antibonding MOs as the specific linear combinations presented. The "in-phase" linear combination, i.e. the bonding MO, only has the form $N(\psi_\mathrm{1s,A} + \psi_\mathrm{1s,B})$ if you let the two orbitals have the same phase. If you had a different choice of normalisation constant, the bonding MO would be represented by a different linear combination $N(\psi_\mathrm{1s,A} + z\psi_\mathrm{1s,B})$ where $z$ is some complex number determined by the relative phases of the 1s orbitals.
As such, by virtue of expressing the MOs in that form, your overlap integral is necessarily positive.
Note that the above discussion only really applies to 1s + 1s overlap. With other orbitals, which have both positive and negative regions, you can end up with a negative overlap integral depending on the relative phases that you choose.
For example in $\ce{BeH2}$ if you consider the middle Be 2p orbital to overlap with both hydrogen 1s orbitals, one "overlap" will be formed with the positive lobe of the 2p orbital and the other "overlap" will be formed with the negative lobe of the 2p orbital. Assuming that you phase both 1s orbitals to be positive, then the first overlap integral will be positive (positive 1s times positive 2p gives a positive integrand) and the second will be negative (positive 1s times negative 2p gives a negative integrand).

In general your choice of phase is arbitrary; it does not affect the final physical result, and is simply chosen to make the calculations easier. That is to say, the eigenvalues (MO energies) that you get out of it will still be the same. The eigenfunctions may take on a different form, but that is physically unimportant.


text:
How or why can carboxylic acids be protonated?
In the wikipedia article "Step-growth Polymerization/Kinetics", it is stated that:

The simple esterification is an acid-catalyzed process in which the
  protonation of the acid is followed by interaction with the alcohol to
  produce an ester and water

My question is: How are acids even protonated? Aren't they the ones supposed to be giving out protons? How are they the ones accepting the protons?

Probably this could help you :      http://www.organic-chemistry.org/namedreactions/fischer-esterification.shtm

Everything can be protonated, given enough effort.

@IvanNeretin Although I would love to see the stuff that protonates superacids — on paper only, that is!

@Jan Naked proton would protonate any neutral molecule, I guess.

Acidic and basic is all relative--we just happen to have arbitrarily used water as the point where we switch from calling it an acid to a base. If you instead think of things as "more acidic" and "less acidic" (or "more basic" and "less basic"), then all that's happened is that you have something in solution more acidic than COOH.

(well, perhaps not all that arbitrarily)

A general acid-base reaction is one of the kind $(1)$.
$$\ce{\underset{\text{acid}}{HA} + \underset{\text{base}}{B} <=> \underset{\text{conj. acid}}{HB+} + \underset{\text{conj. base}}{A-}}\tag{1}$$
All that the acid needs to react as an acid is some hydrogen atom bonded to something, and all that the base needs to react as a base (although it is not shown here) is a lone pair which can accept the acid’s proton. Depending on the relative strength of the acid and the conjugate acid, this equilibrium will be shifted. If the acid is stronger, it will be shifted towards the product; if the conjugate acid is stronger, towards the reactant. If the acidity are within reasonable range of each other, you can always expect a catalytic amount of the stronger acid to be protonated.
The above was the general introduction; now how does this apply to your question?
The ‘acid’ you are talking about is a carboxylic acid. These have the functional group $\ce{-COOH}$ and each of the oxygens in there has two lone pairs. Thus technically, a carboxylic acid could be protonated up to four times.[1] We only need one further protonation, so we can define the carboxylic acid to be compound $\ce{B}$ above — the base in the acid-base reaction.
In fact, acid-catalysed esterification is typically catalysed by the addition of a small drop of sulfuric acid. An average carboxylic acid has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 5$, sulfuric acid $-3$. The hydronium ion $\ce{H3O+}$, that forms if water takes the place of the base in $(1)$, has a $\mathrm{p}K_\mathrm{a}$ of $-1.7$ or $0$, depending on who you ask. Other alcoholic oxonium ions have similar $\mathrm{p}K_\mathrm{a}$ values. The lower the $\mathrm{p}K_\mathrm{a}$, the stronger the acid. Thus, we can assume that most carboxylic acid molecules remain as $\ce{-COOH}$, while the catalytic amounts of sulfuric acid (practically completely deprotonated once) protonated a few solvent molecules at equilibrium. The stronger acid — sulfuric acid — practically governs that none of the weaker acid — the carboxylic acid — can lose its proton.[2]
Even though it is equilibrium, a few of the acid molecules — the ‘next-weakest links in the chain’ — will also be protonated at a given point in time, even though we can expect the $\mathrm{p}K_\mathrm{a}$ value of a protonated carboxylic acid to be rather low. This small proportion is enough to get the reaction going.

Notes:
[1]: However, each successive protonation introduces a positive charge which always makes further protonation less favourable. A positively charged compound is, generally, also more inclined to transfer hydrogen to something else, i.e. is more acidic.
[2]: This can actually be extended. There are a number of superacids with even lower $\mathrm{p}K_\mathrm{a}$ values — in fact, those values are so low that they can no longer accurately be calculated and instead a different value, the Hammett acidity is used. Some of the strongest superacids are even able to protonate methane which is unsuspiscious of providing a lone pair an acidic proton could interact with. Hence Ivan’s comment of:

Everything can be protonated, given enough effort.

Except the strongest superacid known?

@hBy2Py Until we find a stronger one.

I got lost in this. I think the OP is looking for the mechanism for Fischer Esterification. http://www.organic-chemistry.org/namedreactions/fischer-esterification.shtm

@MaxW I think, OP gets the idea of a Fischer esterification but wants to know why the thing that they learnt gives protons (‘acid’) is suddenly acting like the thing that they learnt accepts protons (‘base’). Tbh, I feel like I wrote better answers, too. (I got lost in my thoughts writing it …)

The point is that the mechanism clearly shows the carboxylic acid accepting another proton to form $\ce{R-C-(OH)2^+}$. So it is, as you stated, acting as a base. I'd also point out that this reaction happens in "relatively" anhydrous solutions.


text:
Why is 1H-pyrrol-1-ium non-aromatic?

It has 4n electrons and is completely conjugated. It should be anti aromatic.
Why is this compound non-aromatic?

Pyrrole is aromatic because it has six electrons in its conjugated $\pi$ system. This thing has four.

@IvanNeretin This thing has 4 electrons, so it must be anti aromatic. Why it is non aromatic?

Why should it be antiaromatic? It does not even have a cyclic conjugated $\pi$ system anymore.

Not fully conjugated.  There is a saturated atom in the cycle.

It is non-aromatic even though it has 4$\pi$ electrons because nitrogen cannot form five bonds. Hence delocalisation is not possible and so it is non aromatic.


text:
The intuition behind carbon's covalent bonds
As an introduction to hybridization, my textbook discusses that "The formation of covalent bonds often starts with the excitation of the atoms." Taking carbon as an example, the author says:

Carbon  forms four covalent bonds. Yet if we consider the elctron configuration in the carbon atom, we would not predict this as it has only two singly occupied orbitals available for bonding. ($1s^2\, 2s^2\, 2p_x^1\, 2p_y^1$)

I don't understand why carbon's bonding is counterintuitive. Why don't the 4 added electrons just go into the $p_x$ and $p_y$ orbitals then into some newly formed $p_z$ orbital? Why does the existence of "only two singly occupied orbitals available for bonding" complicate things?
As an extension, why is the excitation of the s-orbital electron to $p_z$ needed?
Thank you for your answers and please try to respond keeping in mind that I am a high schooler (more specifically, part of the IB Diploma program).

The most naive answer would guess that since every orbital wants two electrons and there are two orbitals with only 1 electron in each, then 2 more electrons are needed.
However carbon typically combines the one $2s$ orbital and three $2p$ orbitals to form four equivalent $2sp^3$ orbitals each with one electron. The four hybrid orbitals have a tetrahedral shape. So after hybridization, the carbon atom wants to create four bonds to fill the four electron shell vacancies.

My answer is very similar to MaxW's answer, but maybe it can still provide a
bit of insight.
Consider methane as our current molecule:
1 carbon atom reacts with 4 hydrogen atoms to form 1 methane molecule.
Most/Many textbooks explain this by hybridization of three p orbitals
and one s orbital to form four energetically equivalent (degenerate) sp3
orbitals. This is the valence bond model.
"[Carbon] has only two singly occupied orbitals available for bonding."
The author suggests that since only two electrons are available carbon should
form two bonds. This would leave the last carbon p
orbital completely empty, though, and, tbh, this would to me be completely
counterintuitive. To say it in a sloppy way: Chemical bonds/orbitals are
all about being either completly full or completly empty (half-filled is also
possible sometimes). Leaving one of the three p orbitals completly empty is
absolutely not what any chemist would expect in such a situation. To suggest
that carbon would only form two bonds is actually the really counterintuitive
thing here (at least for me). It would even make more sense that carbon loses
its two p electrons in order to have noble gas configuration, but ok... let's
leave that aside
"Why don't the 4 added electrons just go into the px and py and some new pz?"
The thing is that every bonding orbital can contain maximum two electrons. A bond between two atoms is only established, if every atom provides one electron each. If one carbon p orbital is empty then it would not make sense to put in two electrons from hydrogen. If two hydrogen electrons are put in the same bonding orbital it only means that the two hydrogen atoms are now bonded and form an H2 molecule.
So since every electron from the 4 hydrogen atoms needs an electron from
carbon, the best way is to take the two electrons from the 2s
shell. In that way carbon can provide its two 2s electrons for two bonds with
two hydrogen atoms and the 2 p electrons can bind with another two hydrogen
atoms. This will give four bonding orbitals, which are all completely full.


text:
Thermodynamics: Relation between different types of energy: My Analogy
I have always been confused between the different kind of energies in thermodynamics so today I thought of finally getting it right. Here's my analogy(which is wrong and I need help in improving it, please read it once):
Let us suppose we take a gas at temperature $T_1<T_{room}$ in a beaker fitted with a piston and conducting walls.
Its present energy can be represented by the following diagram:

When the temperature increases both the pressure volume energy and internal enrgy increase by fixed amounts, independent of path(being state variables), according to the equation $\Delta(PV)=nR\Delta T$ and $\Delta U=3/2\ nR \Delta T$
As enthalpy is the sum of both these energies, even it is a state variable, and its change is independent of the path followed.
Now event though $\Delta (PV)$ is a state function and there is a fixed change in its value, there are 2 ways of changing (increasing in this case) it. One by increasing volume and the other by increasing pressure.
The problem is that in the process of increasing pressure-volume energy by increasing volume, the system has to do some work, which means it has to throw out some energy. But we can't let that happen. So we need to provide the system with some extra energy that it can throw out to do work. We haven't increased the energy of the system by doing that as it has come back to the surroundings in the form of work.
There is this fundamental change in energy of the system that is absorbed by the system from the surroundings and there is this extra energy $P\Delta V$ that is also absorbed but then thrown out in the form of work. The total energy absorbed is termed as heat $q$ and the energy thrown out is called work $(-w)$. They depend on the path as it depends on the way we increase the $PV$ energy.
Problem:
The biggest problem with this analogy is it cannot explain $\Delta U=q+w$. According to my analogy $\Delta U+\Delta (PV)=q+w$. This means pressure-volume energy is a part of internal energy and not something different. But then what would be the significance of enthalpy $H=U+PV$?
Moreover I don't know where to put:
1. gibbs free energy as I don't have an intuitive understanding of it, I only know the mathematical formula and the definition: maximum useful work that the system can do and
2. energy lost due to entropy ($T\Delta S$).
It would be great if someone could help me improve my understanding of thermodynamics as I am tired of getting confused each time I open this chapter. An approach coinciding with this one (something like a venn diagram) would be appreciated but anything is welcome. Please help, I really need it!

This is not an attempt at a complete answer but I think its relevant to share this great explanation from "An Introduction to Thermal Physics" by Schroeder:

So basically the whole of H is required to summon up a rabbit but we need to only provide the differences as work. Then the energy required to summon up the rabbit is not just U but U+PV. Hence pressure volume energy doesn't count under internal energy. Am I right? Btw thank you this image is pretty useful.

Yes. The internal energy of the rabbit would remain same if he were in vacuum. The energy TS flows in spontaneously because of the difference in temperature. If you were creating him in Zero Kelvin, you'll have to do the full work.

Okay, but what about the proof of $\Delta U+\Delta(PV)=q+w$ in my question? If we consider pressure volume energy not to be a part of   internal energy to be different, then the 1st law is violated

At first glance I could not really follow your argument. I'll have a look again in the evening.

Sure, just please do.

Basically, the total change in energy is $q+w$ but that accounts for both, change in internal energy as well as change in pressure volume energy. How can that happen as $\Delta U=q+w$

Let us talk in terms of changes. $\Delta U$ would be the total measure of change in energy of the system. This change can be manifested either by work done or heat exchanges. Hence, $\Delta U = W + \Delta Q$. (You may be using a different sign convention for work but it doesn't matter.) $\Delta H$ is the measure of heat changes at constant pressure. It is a very useful parameter when the system is in contact with a constant pressure reservior such as the atmosphere and hence it is the measure used for heat of reactions.

Then to create a rabbit, why do we need to provide the whole of H and not just U? And what about the change in pv energy?

Now, $\Delta H = \Delta U - \Delta (PV) $ but $\Delta (PV) = P\Delta V $ at constant pressure and $ P\Delta V $ is the work done so we get enthalpy change is the change in internal energy minus the work done ie the heat change as was expected.

Relating to this diagram, first of all, we cannot use $\Delta U = W + \Delta Q$ because there are more terms involved for the chemical potential etc. So, don't think of creating a rabbit but rather think of creating a "balloon of heat" because then it won't require any chemical changes.

Now to create such a balloon in vacuum we only need to expand (no work done) and then supply the heat. This is heat supplied at constant volume which equals change in internal energy. But to bring the balloon to atmospheric pressure you need to do extra work ($=P\Delta V = PV$ as we are starting from zero volume). Or you could do the same thing by supplying heat to balloon at constant pressure. Both ways you get enthalpy change.

Let us continue this discussion in chat.

I agree that many chemistry (and physics) texts present these terms in a very confusing way.
Overall, the ways these are used are:

Internal energy is useful for incompressible things because they're at constant volume. Conservation of energy for liquids, ionic stuff, and solutions can be handled with internal energy. A classic example is analyzing a battery.
Enthalpy is useful for compressible things because those processes are often not at constant volume. Conservation of energy for phase changes and gas phase processes usually require enthalpy. A classic example is a gas expanding through a valve or turbine.
Gibbs energy is useful for determining which version of a system will exist. A classic example is determining which direction a reaction will go.

I didn't really understand them until I had to use them, and specifically I needed to do engineering problems with enthalpy. I would recommend working problems with each of these terms.
It sounds like you're a chemistry student, so your textbook can probably supply internal energy problems. Second, skip to later in the book to the chapter on spontaneous reactions and work some of the simple problems that require you to use Gibbs energy (side note: IUPAC prefers no 'free') to determine whether a process is spontaneous. Third, find an engineering thermodynamics textbook and work some of the problems involving isenthalpic processes (isenthalpic being constant enthalpy, although an engineering text will call these adiabatic which is only slightly different).
To answer your question specifically, I think the trouble is that if you let your system change volume, then you by definition no longer have a constant volume system and you need to use enthalpy, not internal energy.

But how are they related to each other...I don't mean the mathematical formulae but the real intuitive relations.

@OsheenSachdev I'm not exactly sure what you mean. I can't think of an example where you analyze one system with both internal energy and enthalpy and try to learn something about their difference. Sometimes its just appropriate to use one, and sometimes the other. There are some cases where you would consider the enthalpic vs entropic contributions to Gibbs energy. Critical solution temperatures are one such application that you might be interested to read about. Can you clarify what you mean by "intuitive relations" thats different than "how one would use them"?


text:
Stronger reducing agent
Why H$_2$Te is a good reducing agent than H$_2$O .
I think it should be opposite as in water , oxygen has only tendency to release electrons .

Are you sure about the statement about oxygen's affinity for electrons?

@airhuff yes it is given in my book

According to this Wikipedia entry:

"[electronegativity] is a chemical property that describes the
tendency of an atom to attract electrons (or electron density) towards
itself.

then according to an electronegativity scale called the Allen electronegativity:

"...neon has the highest electronegativity of all elements, followed
by fluorine, helium, and oxygen."

By any scale, oxygen is among the most electronegative elements. More to the point, electronegativity decreases as you go down a period, so  $\ce{Te}$ is much less electronegative, much more able to give up its electrons, than oxygen.  Thus, $\ce{H2Te}$ makes a better reducing agent than $\ce{H2O}$.


text:
What reactions occur when mixing copper sulfate and sodium hydroxide?
I tried reacting copper sulfate with sodium hydroxide to get copper hydroxide, which should precipitate, according to the following equation:
$$\ce{2NaOH + CuSO4 -> Cu(OH)2(s) + Na2SO4}$$
I looked on Google Images, and the color of copper hydroxide is light-blue, but something interesting happened when I mixed these two solutions: the precipitate formed - and was originally light-blue, as expected - however, it turned into this black-green sludge within a few seconds. I'm thinking a complex ion might have formed between the $\ce{Na2SO4}$, and the $\ce{Cu(OH)2}$. I just saw a video on this, posted by NileRed, but he doesn't go into detail. Does anyone have an idea of what's going on?

Well, many compounds of copper are green. There's for example https://pl.wikipedia.org/wiki/Brochantyt

As noted in your question and in one of the comments, copper forms many different complexes having a variety of colors from red to green to blue to black and probably more.  Of course the situation here is even more "complex" (sorry) as you can end up with mixtures of the different copper complexes, as appears to be the case over the course of your experiment.
The formation of some cupric oxide ($\ce{CuO}$) would account for the appearance of an insoluble black precipitate.  This could happen from the decomposition of some of the $\ce{Cu(OH)2}$.
$\ce{Cu(OH)2}$ itself should be a greenish to pale-blue insoluble precipitate.
It is possible that the original $\ce{Cu(OH)2}$ precipitate appeared more blue than green due to the blue $\ce{CuSO4}$ solution.  As the reaction proceeded, $\ce{CuO}$ and more $\ce{Cu(OH)2}$ precipitated to give the black-green precipitate you observed.

The wiki link shows Cu(OH)2 as a "blue" solid. It looks blusih-green to me. Sort of turquoise color. I think you also have to consider "wet" vs "dry" Cu(OH)2. Cu(SO4)*(H2O)5 is much darker blue than Cu(SO4).

Good point about the hydration @MaxW.  I also tried to give a better description of the turquoise-ish color.  It seems to me to change with the angle of my monitor, so I included the description given in the text;)

Hydroxide ppts are notorious for absorbing other ions. In a strong hydroxide solution I'd expect the ppt to drag extra $\ce{OH^-}$ out of solution. I'd guess that the decomposition of $\ce{Cu(OH)2}$ is base catalyzed. So $$\ce{Cu(OH)2 + OH^- -> CuO(OH)^- + H2O}$$ $$\ce{CuO(OH)^-  -> CuO + OH^-}$$ Out of an ammonia solution the ppt must form slowly enough so as to absorb very little extra $\ce{OH^-}$.

If I remember correctly, anhydrous copper sulfate is white, but that should hardly be an issue here.

Yes, as noted by @airhuff, there is not one but two reactions involving, one is formation of greenish-pale blue copper sulfate, $\ce{CuSO4}$ and other one is formation of black-brown copper(II) oxide $\ce{CuO}$. Also, there are also solvation reactions occurring involving formation of solvated complex but we are going to neglect that.
\begin{align}
\ce{CuSO4 + 2NaOH &-> Cu(OH)2 + Na2SO4}\tag{1}\label{one}\\
\ce{2CuSO4 + 2NaOH &-> [CuO + H2O] + Na2SO4}\tag{2}\label{two}
\end{align}
Some copper(II) hydroxide decomposes to form copper(II) oxide and water,
$$\ce{Cu(OH)2 -> CuO + H2O},$$
and that's how reaction $\eqref{two}$ proceeds.


text:
What is a good undergraduate bioorganic chemistry book?
What is a good undergraduate bioorganic chemistry book? I would like one of these as I'm tired of non-university books on the subject.

Voet, D.; Voet, J. G. Biochemistry, 4th ed.; Wiley: Hoboken, NJ, 2011. From this question

Could you specify what sort of topics you want to be covered?


text:
How would you confirm that a colorless liquid given to you is pure water?
How would you confirm that a colorless liquid given to you is pure water?
A very easy solution is to take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point. Also observe carefully that after some time whole liquid will convert into vapour without leaving any residue.
Now are there any other ways to understand?

I'm afraid there's too much of them. Analytic chemistry as a whole is dedicated to things like that.

BP would only be a crude indication of purity. A measurement of 100 ºC is only good to 1% relative. It is easy to measure ppm levels of chemicals in water. So you'd need to measure BP to 100.0000 ºC. With that sort of precision a correction for the ambient atmospheric pressure would be needed.

Take a drop, dissolve it in $\ce{CD3OD}$, walk to your closest NMR spectrometer and record a spectrum.

Measurement of properties like freezing point and boiling point would give you a quick, qualitative though likely correct, answer.  The simplest to do is the measurement of the liquid's freezing point.
If you can, partially freeze your solution in a kitchen freezer, stir well, then measure the temperature while stirring.  Since the freezing point of water is relatively insensitive to atmospheric pressure, you don't have to worry about correcting for the elevation where you live.  This method is commonly used as a quick way to check the calibration of a thermometer.  Of course a measurement of ${0 ^oC}$ would mean that you likely have water.  If it doesn't freeze, assuming your freezer is cold enough to freeze water, then it's not water.  If this is what you observe, drop a small ice crystal or a grain of salt into the solution to initiate ice formation just to be sure it is not supercooled.  Also, if it doesn't freeze, measure the temperature of the solution to be sure it is at least several degrees below ${0 ^oC}$.
A shortcut, or as a further check, if you have plenty solution to spare you could take a portion and add enough crushed ice that it doesn't all melt upon stirring.  If you get the same value (${0 ^oC}$) as in the first experiment, you can be pretty confident that your solution is water.  If it were some mixture of solutions with a freezing point of ${0 ^oC}$ (unlikely), this test would give you a different measurement than in the first test due to either colligative properties if the amount of ice is small relative to the whole solution or by creating a new, bulk solution with a different melting point than the original solution.
If both of these experiments give you the same result of ${0 ^oC}$, I'll bet my house that it's water.

You could measure various physical properties of the unknown liquid and compare the results to literature values for pure water or to measurements of a known sample of pure water.
However, as a quick and simple test, you could just carefully add pure water with the same temperature to your unknown liquid. If you observe mixing boundaries (schlieren patterns) or even two phases, a change in temperature, or any chemical reaction, your unknown liquid is not pure water.


text:
Are nuclear spin isomers "allotropes"?
In my book there's a question:

Does $\ce{H2}$ show allotropy ? Describe its allotropes and their applications.

Wikipedia doesn't list any allotropes of $\ce{H2}$. However, when searching on the Internet, I found a few resources which call the nuclear spin isomers of dihydrogen (ortho- and para-$\ce{H2}$) "allotropes".
Can nuclear spin isomers of a compound be considered as allotropes (if there is only one element present, as is the case in $\ce{H2}$? If so, then how does the nuclear spin isomer affect their physical form/structure?

Some peer review journal articles and an India high school textbook and various study guides do refer to the spin isomers as allotropes.
There are physical differences such as different heat capacities.  See Orthohydrogen, Parahydrogen and Heavy Hydrogen for comprehensive information.
However, aside from the spin isomers, there are genuine allotropes of hydrogen.  Solid hydrogen has two normal pressure allotropes, a face centered cubic and a hexagonal closest packed allotrope.
Also, at high pressure, there is metallic hydrogen.

So is this a general consensus among chemistry community to consider NSI  to be allotrope? Are physical difference other than structural composition sufficient to define an element as "Allotrope"(The definition of allotrope i've heard is based only on structure and rest of the physical properties as a consequences of it)? As a follow-up question:Is there any example of element in which NSIs have different physical structures/effect on structure?

@Xasel The IUPAC definition of "allotropes" is "Different structural modifications of an element" https://goldbook.iupac.org/A00243.html so I would say, no, it is not a consensus.

@Xasel one article says "In H2, the crystal structure as a
function of temperature and pressure depends upon the
ortho-para (o-p) concentration. At low pressure, single
molecules are in almost free-rotor states. Parahydrogen
molecules are in the spherically symmetric state with rotational
quantum number J=0 and cannot orientationally
order, whereas o-H2 molecules are in the J=1 states
and the molecules can order along crystallographic directions.
Orthohydrogen has an hcp (hexagonal-closepacked)
to fcc (face-centered-cubic) transition at zero
pressure and 2.8 K..."

@Xasel https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.63.2080

Apart from the sources that DavePhD listed, there is also the 1932 Nobel Prize in Physics, awarded to Werner Heisenberg:

"for the creation of quantum mechanics, the application of which has, inter alia, led to the discovery of the allotropic forms of hydrogen."

The "allotropic forms" of hydrogen mentioned here refer to the nuclear spin isomers ortho-$\ce{H2}$ and para-$\ce{H2}$. So, there is some sort of precedent for it; although I'm not sure where it originated. Heisenberg's original paper (in German) can be found at Z. Physik 1927, 43 (3-4), 172–198. Sadly, I don't know German, so I can't verify whether he used the term "allotrope" or anything similar there.
To the best of my knowledge, this usage of "allotrope" is not in common use any more.  Most textbooks just use "nuclear spin isomers".

I wonder if eventual acceptation of NSI as allotropes would open gates for similar acceptance of different molecular states as allotropes. E.g. would be singlet O2 an oxygen allotrope if o-and p-H2 were allotropes ?

Hmmm I totally missed your comment @Poutnik. It's a slippery slope—those are just different electronic states and after that soon we might be accepting different vibrational states, or perhaps even different conformers, as allotropes... In any case, the current trend is in the opposite direction, i.e. once upon a time we called NSI allotropes and nowadays we don't, so I think we're safe.

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


text:
Why does ammonia causes unsymmetric cleavage of diborane, but carbon monoxide causes symmetric cleavage?
Diborane reacts with Lewis bases to form adducts.
I did some research on my own, and found that ammonia breaks it unsymmetrically but $\ce{CO}$ and $\ce{PF3}$ break it symmetrically. I cannot find the reason.

http://chemistry.stackexchange.com/questions/50156/mechanism-for-the-cleavage-of-diborane http://chemistry.stackexchange.com/questions/27826/why-is-diborane-cleaved-symmetrically-by-trimethylamine-but-unsymmetrically-by-a

Clearly not a duplicate, since "sterics" are no longer a good answer: $\ce{NH3}$ is intermediate in volume between $\ce{CO}$ and $\ce{PF3}$.


text:
Reaction between 1,2-dicyanoethylene and 2-phenylbuta-1,3-diene in toluene and heat
I believe this may be a Diels-Alder reaction, but the benzyl group on the edge does not serve a an electron donating group and more as a stabilizing group.
I am at a loss for what reaction would take place if the Diels-Alder reaction doesn't.

The presence of an electron-donating group on the diene, or an electron-withdrawing group on the dienophile, is not an absolute requirement for the Diels-Alder reaction to occur. There are many Diels-Alder reactions that don't have such features.
In any case, the cyano groups on the dienophile are electron-withdrawing groups. So, it's still a pretty favourable scenario for the Diels-Alder to occur.

Since the dienophile is symmetrical there aren't any regioselectivity issues, and the relative stereochemistry of the two cyano groups in the product will be determined by the configuration of the double bond in the dienophile. A trans or (E) double bond will lead to the trans dinitrile (both enantiomers of it), and a cis or (Z) double bond will give rise to the cis dinitrile.
By the way, according to your title, there's no benzyl group on the diene, only a phenyl group. (Not that it makes much of a difference, though.)


text:
How to calculate the percentage yield for a reaction with excess reagent?

A chemist mixed $12~\mathrm{g}$ of phosphorus with $35.5\ \mathrm{g}$ of chlorine gas to synthesize phosphorus(III) chloride (phosphorus trichloride). The yield was $42.4~\mathrm{g}$ of $\ce{PCl3}$. The equation is
  $$\ce{2P + 3Cl2 -> 2PCl3}$$
  Calculate the percentage yield.

I think that the amount of phosphorus the chemist used is in excess. I figured that only 1/3 of the amount of phosphorus the chemist used reacted to produce phosphorus trichloride. But the question didn't mention anything about products in excess that is why I am not so sure of my answer, which is $92.51$.
First I found the amount of substance of phosphorus used which is $12/31$. Then I calculated the amount of substance of chlorine gas used (and it's $0.5$). But $12/31\ \mathrm{mol}$ of $\ce{P}$ needs $1.5 \times (12/31)\ \mathrm{mol}$ of chlorine gas, so $\ce{P}$ is in excess. The amount of substance of $\ce{P}$ that will react will be $$0.5/1.5 = 1/3$$ Now $1/3\ \mathrm{mol}$ of $\ce{P}$ produces $1/3\ \mathrm{mol}$ of $\ce{PCl3}$ and the molecular mass of $\ce{PCl3}$ is $137.5$ so the mass that is supposed to be produced is $137.5 \times 1/3$. Then I divided $42.4$ by $(137.5 \times 1/3)$ and multiplied by $100$.

The yield is typically calculated according to equation $(1)$.
$$\text{yield} = \frac{\text{actual amount of substance obtained}}{\text{theoretical amount of substance to be obtained}}\tag{1}$$
(Whether it is given in percent or as a fractional number is not important, but percentages are the usual value.) So for any reaction, you need to figure out:

the reaction equation
the amount of each reactant substance involved
the amount of desired product theoretically possible by stoichiometric coefficients
the actual amount of desired product.

The first is rather easy, especially since the equation $(2)$ has been given previously.
$$\ce{2 P + 3 Cl2 -> 2 PCl3}\tag{2}$$
Thus, for every mole of phosphorus, we expect one mole of product and for each mole of chlorine gas, we expect two-thirds of a mole of product. The amounts of the reactants are easily calculated using molar masses:
$$\begin{align}M &= \frac mn\tag{3}\\
n (\ce{P}) &= \frac{12~\mathrm{g}}{30.97~\mathrm{g/mol}}\\
&= 0.39~\mathrm{mol}\tag{4}\\[1em]
n(\ce{Cl2}) &= \frac{35.5~\mathrm{g}}{70.9~\mathrm{g/mol}}\\
&= 0.50~\mathrm{mol}\tag{5}\end{align}$$
For each of these two we would need to check the amount of product we would expect using the stoichiometric coefficients.
$$\begin{align}n_\text{theor}(\ce{PCl3}) &= \frac 23 n(\ce{Cl2})\\
&= 0.33~\mathrm{mol}\tag{6}\\[1em]
n_\text{theor}(\ce{PCl3}) &= n(\ce{P})\\
&= 0.39~\mathrm{mol}\tag{7}\end{align}$$
Since $0.33 < 0.39$, chlorine is the limiting reagent and our theoretically possible yield is calculated from the amount of chlorine added to the reaction. Therefore, we now know our denominator which is $0.33~\mathrm{mol}$. What is the numerator? Again, it is easily calculated:
$$\begin{align}n_\text{exp}(\ce{PCl3}) &= \frac{42.4~\mathrm{g}}{137.32~\mathrm{g/mol}}\\
&= 0.31~\mathrm{mol}\end{align}\tag{8}$$
And now all we need to do is calculate the yield:
$$\frac{0.31~\mathrm{mol}}{0.33~\mathrm{mol}} = 0.93 = 93~\%\tag{9}$$
This is consistent with your result.

I'd of done this using a different approach.
Frankly I'd first try a shortcut of sorts. Such chemistry problems usually consume all of one of the reactants.
Knowing that 42.4 g of $\ce{PCl3}$ were produced
g P = $\dfrac{30.97}{137.3} \cdot 42.4 = 9.56$ grams
the rest must be Cl so
g Cl = 42.4 - 9.6 = 32.8 grams
Since neither of these amounts is equal to the mass of the starting reagents I need to look at the moles of the reactants to determine which is the limiting reagent.
moles P = $\dfrac{12}{30.97} = 0.387 $
moles Cl atoms = $\dfrac{35.5}{35.5} = 1 $
But every mole of $\ce{PCl3}$ needs 3 moles of Cl atoms. So 35.5 grams of Cl can make at most 0.333 moles of $\ce{PCl3}$.
So theoretically the yield should have been
$0.333 \cdot 137.3 = 45.8$ grams
% Yield = $\dfrac{42.4}{45.8}\cdot 100\% = 92.6\%$


text:
Change inchikey directly
Preamble:
Now, I am aware that directly editing the inchikey is, generally speaking, a bad idea, but...
Can I just change the 2nd block of an inchikey to "UHFFFAOY" to get the correct main-layer-only version of the inchikey of that molecule?
example where this seems to be the case (from pubchem):
BSAIUMLZVGUGKX-BQYQJAHWSA-N = trans-2-nonenal
BSAIUMLZVGUGKX-FPLPWBNLSA-N = cis-2-nonenal
BSAIUMLZVGUGKX-UHFFFAOYSA-N = 2-nonenal
I have seen lots of other examples (100s if not 1000s) to support this hypothesis (both for cis/trans and for sterochemistry) and never a counter-example.

If there is no stereo-information then the second layer will be empty. The hash of an empty layer will always be the same. So this is pretty safe to do. However, I have no idea why you would want to do that.

@Martin My purpose was finding matches between databases where one might have 2nd block info and the other one doesn't, or for molecules that differ only on stereo and cis/trans. A simple inchikey string manipulation would be faster and less coding than looking up inchi, plugging it into software, and generating the 1st-block-only inchikey.

According to the InCHI FAQ, the 2nd block encodes "encodes stereochemistry and isotopic substitution information". So if you want to generate the corresponding InChIKey for the case where there is no stereo present nor isotopes specified, then it looks like this is possible. If you also want to force the neutral form, you could try setting the final character to 'N' (it may instead be P or O).
Your mileage may vary. To be honest, I wouldn't do this if I were you. I would generate a SMILES without isotopic or stereo information (or remove this programmatically), and then get the InChIKey of that. This would avoid any nasty surprises.

Why would you revert to SMILES when you are working on InChI?

Thanks @Noel. The isotopic substitution info is something I had not considered. Also, all my inchikeys are neutral (just because of the databases I'm working with).


text:
Bolt head corrosion in soup. Toxic? Zinc? Cadmium?
This may not be quite the right forum for this question, but please help me if you can.
I recently repaired the lid of my grandma's crockpot using a bolt I found in a miscellaneous fastener box in the garage. I've been using the crockpot quite a lot since then, and I just noticed that the bolt head is corroding with white gunk. I also noticed a white film floating on the middle of the soup I left cooking all night, obviously from some of the corrosion dripping onto the soup. My grandma and I had both tasted the soup before I noticed the corrosion and film. I've been doing a little research to try and identify the bolt's coating by its appearance, and I am concerned that it might be zinc or cadmium, which are apparently highly toxic. [EDIT: apparently zinc is not highly toxic, so I guess my main concern is: is this cadmium? Also, the bolt may well be over 50 years old, as it's from my grandparents' garage]
I'm hoping someone here who's more familiar with these materials can tell me:

What metal this corrosion looks like
Whether I should be concerned about toxicity considering the quantities present

I'm including a picture of another bolt that matches the one in question, with yellow zinc plated wood screws and some U.S. currency for visual reference.

I will of course be replacing the bolt with something explicitly food-safe, but having already consumed some of the soup, I want to know if we should be high-tailing it to the emergency room, or if in fact it's safe to finish the soup (it's good soup!).

Personal medical questions are off-topic here. But where did you get the information that zinc be highly toxic?

@Jan Mention of zinc toxicity here, Wikipedia article too.

That’s chronic toxicity, i.e. long-term uptake of high levels. Luckily, we are not plants, invertebrates or some fish ;)

You can test for cadmium at home, but the ~US$30 price for the test kit (with 1,5-diphenylcarbazone) is probably more than the soup is worth, and the test might not be sensitive enough to detect sub-acute toxic amounts of Cd.
Considering the highly toxic nature of the metal, you might check with your family physician if you and anyone else sampling the soup should have a blood or urine test.
As has been mentioned, the small amount of zinc that might have been absorbed had the bolt been galvanized (unlikely, anyway) is not significant. The US NIH recommends ~10 mg of Zn in the diet per day as a nutrient.

We're not supposed to discuss medical advice to avoid getting sucked into giving bad medical advice. Having said that, this advice seems over the top. If it we me, I'd switch out the bolt for a stainless steel one and call it a day. I'd then be sure to watch the stainless steel bolt for a while to be sure that it didn't corrode too.


text:
What is more polar?
Is the bond $\ce{Cs-F}$ more polar or $\ce{C-F}$ ? I am not sure because I thought $\ce{Cs-F}$ is classified as an ionic bond, and $\ce{C-F}$ as a polar covalent one. Based on electronegativity values, $\ce{Cs-F}$ has the highest electronegativity difference, so would that be the most polar bond?

Also the ranges are usually .5<x<2 = polar

and > 2 electronegativity difference is ionic

Instead of posting follow-up comments you may wish to [edit] your question.

We don't even have to look at the electronegativity difference. Any ionic bond is more polar than any covalent bond.

$\ce{Cs-F}$ bond is more polar than $\ce{C-F}$ bond.
Electronegativity (rather electropositivity if I may say for $\ce{Cs}$) is $0.69$ . Electronegativity for $\ce{C}$ is $2.55$ and for $\ce{F}$ is $3.98$. These values are according to Pauling scale.
As you have correctly mentioned in your question, the bond for which the difference in electronegativity of bonding atoms is large, that bond is classified as ionic.
(Difference $> 2$ is classified as ionic and if $< 2$ is classified as covalent.)

As a doubt I would like to ask this : Can we call ionic bonds polar covalent bonds ?

No. You can't call ionic bonds as polar covalent bonds. The extent of the effect of the Columb force is different in both cases.

Since you can't call ionic bonds polar covalent, then is Cs-F really the more polar bond?

Yes. Polarity is the existence of a dipole in a molecule. Here in polar covalent bond, the charge on either centers is very less compared to the charge on the centers in the ionic compound.

The electrostatic force here in ionic compound is solely responsible for the charge on the centers. A complete charge transfer takes place between the two centers in a ionic compound. Hence we write it as Cs+ and F-. But in polar covalent bonds, the C-F where C has partial positive charge and F has same amount of partial negative charge. But it's not equal to single unit of electron charge. Is it clear?


text:
The compounds barium sulfate and potassium chloride are the products of a chemical equation. Write a balanced chemical equation for this reaction
I have been teaching my cousin grade 10 science.
Unfortunately, I have never taken chemistry before and am unsure how to do these type of questions.
How does one go about approaching this question?

I would prompt him for everything he knows, get him to write down the part of the equation given in the question.  Take a look at this video https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equations/v/balancing-chemical-equations-introduction and take a look here http://www.kentchemistry.com/Worksheets/Regents/Units/MathofChem/WSBalancing21.pdf for some examples, then just do some more internet searching for "balancing chemical equations".

Thank you for your reply. The problem I am having is not balancing the equations but writing them down as an equation, mainly, getting the subscripts correct.

Since they're ionic compounds you basically just interchange the atoms, while keeping in mind the charges.
Since you know the products are $$\ce{BaSO_4}$$ and $$\ce{KCl}$$
you need to figure out the original composition. If you look at the periodic table and also learning your common ions, you know Barium is 2+ and sulfide is 2-. But the problem is when you interchange them, you have a 2- charge from sulfate with potassium which is 1+, so you must have 2 potassium. Same thing with chloride. Now that you have $$\ce{BaCl_2}$$ and $$\ce{K_2SO_4}$$
you need to balance the equation by putting a 2 in front of potassium chloride.
$$\ce{BaCl_2 + K_2SO_4 <=> BaSO_4 + 2KCl}$$
To sum it up just recognize they are ionic compounds and that you need to balance the equation and charges.

You did not use the compounds given in the question.  You used sulfite rather than sulfate.  The concept is right, the answer is wrong.

Yep, fixed that, my bad.

Yea, such a simple question and we each made a mistake (see comments under my answer).  BTW, you still have barium sulfite in your answer.

You definitely need to look at the links in the comments and search yourself to find the systematic methods for balancing equations. But, just to walk you through this one as an example:
Start him off with
$$\_\_\_\_ + \_\_\_\_\ce{-> BaSO4 + KCl}$$
as the problem states, and ask him to fill in the blanks.  The key is having the same number of each element on each side of the equation.  To start off with, for this problem, the left side should have one $\ce{Ba}$, one $\ce{S}$, four $\ce{O}$'s, one $\ce{K}$ and one $\ce{Cl}$ just like the right side.  You will need to do some searching to learn about changing the amounts of each element in the equation so that you end up with the same number of each element.  That cannot be properly explained in this space, but I will work out this specific very simple (grade 10 level) problem.
This problem is simple because you can just switch around the ions (look up ions if needed) so you start with $\ce{K2SO4}$ and $\ce{BaCl2}$.  The only complication here is that the starting material has twice as many $\ce{K+}$ and $\ce{Cl-}$ ions, so to make things balance you need to have two on the right side. So  he needs to take that information to get:
$$\ce{K2SO4 + BaCl2-> BaSO4 + 2KCl}$$
Again, the problem with my description is that it is just to kind of give you the basic idea by walking through a single example.  There are well defined methods for doing this and that is what your student needs to learn.  There are many good YouTube videos that would be worth watching.  I would follow the links in my comment, find some more sites on your own then try this problem now that you have the answer and some guidance from a different point of view (mine).

Well the reaction is really $$\ce{K2SO4 + BaCl2-> BaSO4 v + 2KCl}$$ The reactants and the KCl product are really the solvated ions not the salts.

@MaxW, absolutely.  Given the context of this question I thought achieving mass balance was the key point to make without explaining phase changes.  Maybe it would have been simple enough to define the appropriate terms though.

LOL - Obvious to us, but not to the OP Ali...


text:
Emission Spectrum of Hydrogen and Helium?
Why do hydrogen and helium have different emission spectra? I thought that emission spectra were the result of electrons 'jumping' from one energy level to the next - since hydrogen and helium both have electrons on n=1, l=0 and only there, why are their emission spectra different?

Hydrogen has a nuclear charge of +1, but helium has a nuclear charge of +2. So although each has one electron the spectra are not the same.

We have solved the Schrödinger equation for hydrogen-like atoms such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, and $\ce{B^4+}$. The energy levels are neatly arranged:
$$E_n = - \frac {E_0} {n^2}$$
where:

$E_n$ is the energy of the $n$-th energy level
$n$ is a positive integer ($1$, $2$, $\dots$)
$E_0 = 13.6~\mathrm{eV}$

This is possible because we only need to consider two particles: the nucleus which has a positive charge and the electron which has a negative charge. We only need to consider the interaction between those two particles. Also, the nucleus is not moving, which simplifies some calculations
However, for higher atoms such as helium, where there are more than one electron, the Schrödinger equation contains two terms for the two nucleus-electron attractions and a term for the electron-electron repulsion. The term for the electron-electron repulsion makes things difficult because both electrons can be moved. That term makes the equation impossible to solve analytically.

For hydrogen, the energy of an energy level is determined solely by its principal quantum number. That means, the $2s$ orbital has the same energy as the $2p$ orbital (ignoring the virtual particle effect).
This is not true in higher atoms such as helium. The energy levels of helium can be seen here:

(The energy levels of hydrogen is on the right of the diagram for comparison.)
Here, the azimuthal quantum number has an effect on the energy of the energy levels, which makes for more possible differences between energy levels, which makes for more lines as well as different on the emission spectrum.

Note that for 'Hydrogenic' atoms such as $\ce{Li^{2+}}$ the energy equation should be modified slightly to $E_n=-Z^2R/n^2$ where Z is the atomic number and R the Rydberg (which you label as $E_0$ and is $R_{\infty}=13.6057 \pu{eV} = 109737 \pu{cm^{-1}}$ ) but varies slightly as it depends on reduced mass (for nucleus of mass M, $R=R_{\infty}(1+m_e/M)$. The $Z^2$ terms is why the energy is so much lower for helium than hydrogen

Electronic energy level for $\ce{H}$ and $\ce{He}$ are different because in $\ce{He}$ nucleus there is two proton and for $\ce{H}$ it has single proton.
As their energy are different they require different amount of energy to excite them, so different spectra comes out with different wavelength.


text:
Equivalent weight of calcium phosphate
In a reaction, calcium phosphate is producing calcium hydrogen phosphate. Then we have to find the equivalent weight of calcium phosphate in this process.
I tried as:
Oxidation number of calcium in calcium phosphate is $\frac43$ and in calcium hydrogen phosphate it is $1.$
So the equivalents of calcium phosphate is $(\frac13)(3)$
Equivalent weight should be $\frac{310}{1}$
Am I correct?

Work on this some more... The oxidation number of calcium is +2 in both salts. The equivalent weight is certainly not 310.

@MaxW so oxidation number of which element is changing

Aditya is right that calcium is $\ce{Ca^{2+}}$ in both salts so calcium isn't changing its oxidation number. In fact there is no redox reaction at all. The reaction is:
$\ce{Ca3(PO4)2 + 2H^+ -> 2CaHPO4 + Ca^{2+}}$
So for one mole of the $\ce{Ca3(PO4)2}$ reacts with 2 moles of $\ce{H+}$ so  1/2 mole of $\ce{Ca3(PO4)2}$ reacts per mole of $\ce{H+}$. Thus the equivalent weight of  $\ce{Ca3(PO4)2}$ is 1/2 of its molecular weight.
eq wt = $\dfrac{1}{2}\cdot 310.2 = 155.1 $
NOTE
Equivalent weight is sort of a peculiar notion and is sometimes not unique. So the use of equivalent weight has fallen out of favor. Look at this redox reaction with the permanganate ion
$\ce{8 H+ + MnO4^- + 5e− -> Mn^{2+} + 4 H2O}$
There are 8 $\ce{H+}$ ions but 5 $\ce{e^-}$, so should the molecular weight of the permanganate ion be divided by 8 or 5 to get the equivalent weight?

Calcium phosphate is $\ce{Ca3(PO4)2}$. The oxidation state of calcium is $+2$ here.
Calcium hydrogen phosphate is $\ce{CaHPO4}$. The oxidation state of calcium is also $+2$ here.
The equivalent weight of a salt is calculated by $\frac {\text{molecular weight of salt}} {\text{charge on the anion or cation}}$.
$$\ce{2 Ca3(PO4)2 -> 6 Ca^2+ + 4 (PO4)^3-}$$
Molar mass of calcium phosphate = $310~\mathrm{g}~\mathrm{mol}^{-1}$.
Equivalent weight = $\displaystyle \frac {2 \times 310} {6 \times 2}$ which gives you around $51~\mathrm{g}$.

But the answer given as 155 g

@user123733 Can you please post the reaction in the question?

There is no reaction given

@MaxW has given the right answer. The equation which he mentioned was what I expected. Equivalent weight is a tricky business. For the same compound, there can be different equivalent weights possible according to the reactions it participates in.


text:
Alkali metal carbonates, their thermal stability and solubility
From Li to Cs, thermal stability of carbonates increases. From Li to Cs, due to larger ion size, hydration enthalpy decreases. So, solubility should decrease from Li to Cs. But, experimentally, order is reverse. I cannot wrap my head around this.

Order of what is reversed?

Found my mistake (probably) thermal stability is related to decomposition, I wanted to break lattice into gaseous ions, whose energy change is given by lattice energy. Both |hydration energy| of constituent ions and lattice energy of M2(CO3) decreases from Li to Cs as in Li>Na>K>Rb>Cs. Entropy change plays a big part too here. I meant order of solubility of M2(CO3) in water.

Yes. Even ignoring entropy (to a first approximation we might assume that the entropy change is constant for all the metals), solubility depends on both lattice energy and hydration enthalpy so you can't just look at one trend. While it's more favourable to solvate Li+, it's also harder to break up Li2CO3 into its constituent gaseous ions. Just so happens that here, the lattice energy term predominates.

A side question, as the hydrated ionic radius is highest and mobility least for Li among the above compounds, would Li2(CO3) have least entropy change of solution?

It's difficult to say. The entropy term also has to be split up into lattice breakage (probably roughly the same for all metals) and solvation effects. In the absence of any data my hypothesis would be that Li+, being the smallest (highest charge density), will lead to the greatest ordering of the solvent molecules around it. Consequently the entropy change would be the least positive. This effect is also reflected in the large hydrated ionic radius that you speak of. (I don't have data and I'll take your word for it.)

Lattice energy is also decreasing down the group. hence thermal stability is decreasing. But since lattice energy is decreasing squarely and hydration energy is decreasing linearly, the effect of decrease in lattice energy is higher than the effect of decrease in hydration energy. So if energy required to break lattice is very less, then the lattice of the bottom elements in alkali group will break easily compared to the lighter ones. Thus if not lattice form, then they prefer the aqueous state and hence they love to stay dissolved and not form lattice. So solubility increases.


text:
What is the group reagent for the zero group (ammonium cation) in inorganic salt analysis?
Some texts suggest sodium hydroxide is the group reagent for the zero group.
However, some say there's no group reagent for zero group

So which one's right?

Yes their dates are out,but actually i was laughed upon by my friends on saying that zero and sixth group have no group reagent.

Indeed, there is no group reagent for group zero and six.

My theory about group reagent is different. I think that group reagent is that reagent that when reacted with all ions present in a  particular group will give the same reaction. Ions present in other group will not give that reaction and that's how ions are differentiated and that is the basis of salt analysis.
I am stressing on the word group and all because all the groups except group zero and six* have more that one ions that will give the same reaction when reacted with the group reagent. Since, group zero has only one cation i.e. ammonium ion, $\ce{NH4+}$, there is no need of calling them group reagent because the group has only one cation. When group has more than one ion, then the reagent should be called group reagent. For instance, the reagent that detect presence of ammonium ion i.e Nessler's reagent is not necessarily called group reagent neither is sodium hydroxide.
Group six is no exception. Though, there are three ions present in group six ($\ce{Na+, K+, Mg^2+}$) mentioned in a detail salt analysis book(Vogel), there is no group reagent for this group as three ions are detected separately and have separate test for their detection. But for the time being, you only need $\ce{Mg^2+}$. :)
*You might think that group 1 has only one ion and still has group reagent. Actually there are three ions $\ce{Pb^2+, Ag+, Hg2^2+}$, but again for the time being, you only need $\ce{Pb^2+}$. :)

Ammonium salts are identified by the release of ammonia upon warming with base.
$$\ce{NH4+ + OH- ->[\Delta] NH3 ^ + H2O}$$
Vogel's Qualitative Inorganic Analysis 5th ed., p 400 writes

6. Test for ammonium ions Boil a little (c. 0.1 g) of the substance with sodium hydroxide solution. The evolution of ammonia, detected by its odour and its action upon red litmus paper and upon filter paper soaked in mercury(I) nitrate solution,* indicates the presence of an ammonium salt.
* Drop-reaction paper, treated with tannic acid and silver nitrate solution, may also be used.

The "Group reagent" is a cocktail of reagents that serves as a prima facie indication of a particular cation, characterized by the formation of a  precipitate during a positive test/result and...well... no precipitation in the event of a negative test/result.
You often come across use of "Group reagents" in a school chemistry lab where, as I mentioned earlier, it's used as a preliminary "probe" (if you will) to narrow down possible cations present in a salt.
As far as I know, most common ammonium salts (sulfates. chlorides, etc) do not precipitate when treated with aqueous NaOH, in fact, they dissolve fairly well, resulting in a clear (sometimes slightly turbid) solution. So that's not exactly a "Group reagent" now, is it?
However, treating ammonium salts with NaOH produces vapors of ammonia, in addition to water and the corresponding sodium salt. Vapors of ammonia react with vapors of hydrochloric acid to form dense, white fumes (which you can actually condense over a chilled surface and collect) of ammonium chloride. This reaction is often used as a preliminary test for the ammonium cation.
So essentially what you do is:
Dissolve your salt (let's assume you already know that it's an ammonium salt) in a little (aqueous) NaOH, and heat the test tube. Once the solution begins to boil, it gives off (faintly white) ammonia vapors. Apart from catching a whiff of the characteristic smell of ammonia, you can also introduce a glass rod (that's been dipped in a concentrated solution of HCl) to the mouth of the test tube. You will then observe dense, white fumes of ammonium chloride rising from the mouth of the test tube.


text:
Transition state optimisation on the surface of periclase
I want to model a reaction catalysed by periclase ($\ce{MgO}$) using DFT.
I have a good guess on the transition state (TS) of the reaction that goes in gas phase/solvent (produced using MOPAC).
The proposal for this reaction goes by the Eley–Rideal mechanism and I have a good guess on how the reactant adsorbs.
Now comes the question: I would like to optimise the TS on the surface using DFT (either ORCA or NWChem), but in tractable time.
I am tempted to use a simple, fixed model for the surface, e.g., a set of fixed atoms in bulk arrangement.
Can this work?
How am I supposed to find a structure with just one imaginary frequency using a partially fixed model (maybe using some clever vibrational projections)?
Even though NEB could be applied, the implementation of it in NWChem does not work with fixed atomic coordinates.
Besides, I am confident on the TS I found without the catalyst, so I would like to use that instead of starting all over again.
Edit: what about using a very small cluster model, say, 8-16 atoms, and let it relax? Do you think it could work (in the sense of giving a reliable TS in terms of geometry and energy)?

It will not be easy because the slab (that's what a cutout of a solid state structure is often called) would need to be optimized in order to supply a decent initial Hessian for the TS search. I doubt that any TS search algorithm deals well  with fixed atoms. If you manage that, I would strongly suggest modelling the surrounding Mg/O by pseudopotentials and point charges. My old group did similar work for polarization properties: DOI:10.1002/cphc.201100521.

@TAR86, thanks for the reference.
So it seems I am with a tough problem at hand.
I edited the question with a new idea, what do you think? Is a cluster model a good idea?

Several comments: mopac ts maybe qualitatively wrong; surface and vacuum ts-s are generally very different, therefore your approach can easily be wrong; orca cannot calculate slabs; slab calculations takes forever, but you cannot cut corners most cases, and so on

With a cluster model, you may get an answer. As Greg points out: whether that has much to do with the reality, I have my doubts, but also lack experience: my old group mostly dealt with gas-phase TS. As a first step, you might check whether your cluster relaxes to a reasonable structure when surrounded by the artificial environment (point charges etc.), because without a relaxed geometry, I don't think you will get anywhere. Next, I would look at a trivial model reaction and test the remaining approach. Given those results, I would assess the feasibility. And don't forget dispersion/vdW.

I would question if there is a fast/easy/cheap solution that does not require experience (or even just a fast one)

The standard scheme here is to take a slab and keep 2-4 atom layers outermost from the reaction center fixed in crystallographic positions while letting everything else to relax naturally. There is a plenty of works done in this fashion. The size of the model used (i.e. how much atomic layers counting from the reaction center) to be used is ideally a matter of trial and asymptotic analysis, but in my experience in published works usually 2-4 layers from reaction center are relaxed and 2-4 more are fixed.

Hi did this ever work? If not let me know. I'd like to discuss this if possible. I have a better approach than NEB if you are interested.

Hi @CodyAldaz, I ended up not calculating the TS, only investigating the adsorption process. I would like to hear about your approach! :)

To study surface reactions I recommend the growing string method (GSM) for surfaces. Here is a nice quote from the paper which developed it (Ref. 1)

GSM’s efficacy was confirmed by comparison with CI-NEB on
  an extensive set of reactions characteristic of modern surface chemistry studies. In these cases, GSM reduces the computational cost (in terms of gradient computations) by about 45% on average over CI-NEB. In addition to high efficiency, GSM has the advantage of operating in single-ended way to enable explorative study of chemical reactions.

Key to the success of GSM is the use of the efficient hybrid delocalized coordinate system which greatly improves optimization and interpolation steps in comparison to NEB which uses Cartesian coordinates.
The exact code to reproduce their work is not publicly available, although you may be able to request a copy. However you should be able to reproduce it and more, with this python version of the code that I have developed.
Also, as others have pointed out, you should consider a higher level of theory like periodic wavefunction DFT (e.g. as implemented in VASP) .
References:

Jafari, M, Zimmerman, P. M.. "Reliable and efficient reaction path and transition state finding for surface reactions with the growing string method", J. Comput. Chem. 2017, 38, 645– 658. https://doi.org/10.1002/jcc.24720

Good answer and very interesting method. I'll take a look at it and accept your answer, thanks!


text:
Obtaining activity coefficients of conjugate acids of some common carboxylic acid molecules
Recently, I have been wondering about justifying the notion on how some common acids (e.g. carboxylic acids) are themselves weaker bases than water to the point that we don't usually consider them to act as base. To begin, we use the formula for $\mathrm{p}K_\mathrm{b}$ and consider for example acetic acid (written in organic chemistry convention, with $\ce{OAc^-}$ being the acetate group $\ce{CH_3COO^-}$ for unfamiliar readers)
$$\mathrm{p}K_\mathrm{b} = -\log \left(\frac{a_{\ce{H2OAc+}}a_{\ce{OH-}}}{a_{\ce{HOAc}}}\right)$$
where everything is measured at room temperature, 1 atm.
As usual, if the solution is dilute enough, we can approximate the activity with concentration.
However, the species $\ce{H2OAc+}$ seemed to be not very popular as NIST have few data on its thermodynamic parameters. While it is in principle possible to perform an experiment to measure $a_{\ce{H2OAc+}}$, if it is indeed very weak as a base, then the measurement result will only be reflecting the acetate ionisation equilibrium and masked any tiny result of $a_{\ce{H2OAc+}}$ if any.

How does one obtain the necessary thermodynamic parameters for a potentially highly unstable protonated species in aqueous medium in order to work out the $\mathrm{p}K_\mathrm{b}$ of the common carboxylic acids?

You can (and should!) use $\ce{...}$ to format chemical formulae. See http://meta.chemistry.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know/3046#3046 I've also introduced the shorthand OAc for acetate, which is ubiquitously used in chemistry.

If no specific activity coefficient data are available, I would use the general formula for ionic compounds of small enough ion cincentations:
$$
\begin{align}
I &= \frac{1}{2}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2} \\
\log{\gamma} &= -0.509 . ( z_+^m . |z_-|^n )^{\frac{1}{m+n}}. \sqrt{I} \quad (*)\\
\end{align}
$$
$c_i$ is the ion concentration.
$z_i$ is the ion charge.
$I$ is the Ionic strength.
$\gamma$ is the activity ciefficient by the Debye-Hückel equation.
(*) - See the link for more info and more complex formulas for the medium $I$.